Nilam Publication module Chemistry Form 4 answer

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CONTENTS KANDUNGAN

1

THE STRUCTURE OF ATOMS STRUKTUR ATOM

2

CHEMICAL FORMULA AND EQUATIONS FORMULA DAN PERSAMAAN KIMIA

22

3

PERIODIC TABLE JADUAL BERKALA

49

4

CHEMICAL BOND IKATAN KIMIA

72

5

ELECTROCHEMISTRY ELEKTROKIMIA

88

6

ACID AND BASES ASID DAN BES

114

7

SALT GARAM

139

8

MANUFACTURED SUBSTANCES IN INDUSTRY BAHAN KIMIA DALAM INDUSTRI

168

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Chemistry Form 4 • MODULE

1

THE STRUCTURE OF ATOMS STRUKTUR ATOM MATTER / JIRIM

• PARTICLE THEORY OF MATTER / TEORI ZARAH JIRIM – To state the particle theory of matter Menyatakan teori zarah jirim

– To differentiate and draw the three types of particles i.e. atom, ion and molecule Membezakan dan melukis tiga jenis zarah jirim iaitu atom, ion dan molekul

– To describe the laboratory activity to investigate the diffusion of particles in gas, a liquid and a solid. (To prove that matter is made up of tiny and discrete particles) Menghuraikan aktiviti makmal untuk mengkaji resapan zarah dalam gas, cecair dan pepejal (Untuk membuktikan bahawa jirim terdiri daripada zarah-zarah yang halus dan diskrit)

• KINETIC THEORY OF MATTER / TEORI KINETIK JIRIM – To state the kinetic theory of matter Menyatakan teori kinetik jirim

– To relate the change of physical states of matters with energy change Menghubungkaitkan perubahan keadaan jirim dengan perubahan tenaga

– To relate the change of energy in the particles with kinetic particle theory of matter Menghubungkaitkan perubahan tenaga dalam zarah dengan perubahan tenaga kinetik zarah

THE STRUCTURE OF ATOMS / STRUKTUR ATOM

• HISTORY OF ATOMIC MODELS DEVELOPMENT / SEJARAH PERKEMBANGAN MODEL ATOM

– To state the contribution of scientists in the atomic structure model such as the scientists who discovered electron, proton, nucleus, neutron and shell Menyatakan sumbangan ahli sains kepada perkembangan model struktur atom dan ahli sains yang menemui elektron, proton, nukleus, neutron dan petala

• SUBATOMIC PARTICLES / ZARAH-ZARAH SUBATOM

– To compare and differentiate subatomic particles i.e. proton, neutron and electron from the aspect of charge, relative mass and location Membanding dan membezakan zarah-zarah subatom iaitu proton, neutron dan elektron dari segi cas, jisim relatif dan kedudukan

– To state the meaning of proton number and nucleon number based on the subatomic particle Menyatakan maksud nombor proton dan nombor nukleon berdasarkan zarah subatom

– To write the symbol of elements with proton number and nucleon number Menulis simbol unsur yang mengandungi nombor proton dan nombor nukleon

• ISOTOPE / ISOTOP

– To state the meaning, examples and the use of isotopes Menyatakan maksud isotop, contoh-contoh isotop dan kegunaan isotop

• ELECTRON ARRANGEMENT / SUSUNAN ELEKTRON

– To know the number of electron shells and number of electrons in the 1st, 2nd and 3rd shell Mengetahui bilangan petala elektron serta bilangan elektron yang diisi dalam petala 1, 2 dan 3

– To write the electron arrangement of atoms based on proton number or number of electrons and state the number of valence electron

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Menulis susunan elektron bagi suatu atom berdasarkan nombor proton atau bilangan elektron dan seterusnya menyatakan bilangan elektron valens

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MODULE • Chemistry Form 4

MATTER / JIRIM

Matter is any substance that has mass and occupies space. Jirim adalah sebarang bahan yang mempunyai jisim dan memenuhi ruang. The Particle Theory of Matter / Teori Zarah Jirim

Matter is made up of tiny and discrete particles. Three types of tiny particles are atoms ,

1

atom

Jisim terdiri daripada zarah yang halus dan diskrit. Tiga jenis zarah tersebut ialah

,

ions

ion

and molecules . molekul .

dan

Matter can be classified as element or compound. / Jirim boleh dikelaskan sebagai unsur atau sebatian. Complete the following: / Lengkapkan yang berikut:

2 3

MATTER / JIRIM ELEMENT / UNSUR satu type of atom. A substance made from only satu Bahan yang terdiri daripada jenis atom sahaja.

COMPOUND / SEBATIAN two or more A substance made from elements which are bonded together. dua Bahan yang terdiri daripada

atau

different

lebih

unsur berbeza yang terikat secara kimia.

Types of particles / Jenis zarah

Atom / Atom The smallest neutral particle of an element (Normally pure metals, noble gases and a few non-metal elements such as carbon and silicon). Zarah neutral yang paling kecil bagi suatu unsur (Biasanya logam tulen, gas adi dan beberapa unsur bukan logam seperti karbon dan silikon).

Example:

Types of particles / Jenis zarah

Molecule / Molekul A neutral particle consists of similar non-metal atoms which are covalently-bonded.

Molecule / Molekul A neutral particle consists of different non-metal atoms which are covalently-bonded.

Zarah neutral terdiri daripada atom-atom bukan logam serupa terikat secara ikatan kovalen.

Zarah neutral terdiri daripada atom-atom bukan logam berlainan terikat secara ikatan kovalen.

Example:

Example:

Contoh:

Contoh:

Oxygen gas, O2

Carbon dioxide gas, CO2

Gas oksigen, O2

Gas karbon dioksida, CO2

Contoh:

Sodium metal, Na

O O

Logam natrium, Na

O O

Na Na Na Na Na Na Na Na Na Na Na Na Na Na Na Na

O O

O

C

O

O

C

O

Ne

O

Example: Contoh:

Na+ Cl – Na+ Cl – Na+

Air, H2O

H H H H

Natrium klorida, NaCl

Water, H2O

Gas hidrogen, H2

H H

Ne

C

Zarah bercas positif atau negatif terbentuk dari logam dan bukan logam terikat secara ikatan ion. Daya tarikan antara dua ion yang berlawanan cas membentuk ikatan ion.

Sodium chloride, NaCl Hydrogen gas, H2

Neon gas, Ne Gas Neon, Ne

O

Ion / Ion Positively or negatively charged particles, which are formed from metal atom and non-metal atom respectively. The force of attraction between the two oppositely charged ions forms an ionic bond.

H H

O O

Cl – Na+ Cl – Na + Cl –

H H

O

H

H

Ne

Na+ Cl – Na+ Cl – Na+

Calcium oxide, CaO Kalsium oksida, CaO

Ca2+ O 2– Ca2+ O 2– Ca2+ O 2– Ca2+ O 2– Ca2+ O 2– Ca2+ O 2– Ca2+ O 2– Ca2+

– Elements can be identified as metal or non-metal by referring to the Periodic Table. Unsur boleh dikenal pasti sebagai logam atau bukan logam dengan merujuk kepada Jadual Berkala Unsur.

– Formation of molecule and ion will be studied in Chapter 4 (Chemical Bond). Publica

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4

Determine the type of particles in the following substances: Tentukan jenis zarah bagi setiap bahan berikut: Substances

Type of particle

Substances

Type of particle

Substances

Type of particle

Molecule

Sulphur dioxide (SO2) Sulfur dioksida (SO2)

Molecule

Tetrachloromethane (CCl4) Tetraklorometana (CCl4)

Molecule

Copper(II) sulphate (CuSO4) Kuprum(II) sulfat (CuSO4 )

Ion

Iron (Fe) Ferum (Fe)

Atom

Zink chloride (ZnCl2) Zink klorida (ZnCl2 )

Ion

Argon (Ar) Argon (Ar)

Atom

Carbon (C) Karbon (C)

Atom

Hydrogen peroxide (H2O2) Hidrogen peroksida (H2O2)

Molecule

Bahan

Jenis zarah

Hydrogen gas (H2) Gas hidrogen (H2)

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Bahan

Jenis zarah

Bahan

Jenis zarah

Diffusion Resapan

(a) The tiny and discrete particles that made up matter are constantly moving. In gases, these particles are very far apart from each other, in liquids, the particles are closer together and in solids, they are arranged closely packed. Jirim terdiri daripada zarah-zarah halus dan diskrit yang sentiasa bergerak. Dalam gas, susunan zarah-zarahnya adalah berjauhan antara satu sama lain, dalam cecair, zarah-zarahnya disusun lebih rapat dan dalam pepejal, zarah-zarahnya disusun dengan sangat padat dan teratur.

(b) Diffusion occurs when particles of a substance move between the particles of another substance. Resapan berlaku apabila zarah-zarah suatu bahan bergerak di antara zarah-zarah bahan lain.

(c) Diffusion occurs in a solid, liquid and gas. Complete the following table: Resapan berlaku dalam pepejal, cecair dan gas. Lengkapkan jadual berikut: Diffusion in a gas Resapan dalam gas

Experiment Eksperimen

A few drops of bromine liquid Beberapa titis cecair bromin

After few minutes Selepas beberapa minit

Diffusion in a liquid Resapan dalam cecair

Water Air

After a few hours Selepas beberapa jam

Potassium manganate(VII) Kalium manganat(VII)

Observation Pemerhatian

Explanation Penerangan

The brown colour of bromine vapour, far Br2 spreads throughout the two jars.

The purple colour of solid potassium manganate(VII), KMnO4 spreads slowly throughout the water.

Diffusion in a solid Resapan dalam pepejal

Gel

Agar-agar

Copper(II) sulphate

After a day Selepas sehari

Kuprum(II) sulfat

The blue colour of copper(II) sulphate, CuSO4 spreads very slowly throughout the gel.

Warna perang wap bromin, Br2 merebak cepat memenuhi kedua-dua dengan balang gas.

Warna ungu pepejal kalium manganat(VII), perlahan KMnO4 merebak dengan di dalam air.

Bromine vapour, Br2 and air are made up of molecules . Wap bromin, Br2 dan udara terdiri molekul daripada . molecules diffuse Bromine quickly between large

Potassium manganate(VII) is Copper(II) sulphate, CuSO4 is made made up of potassium ions and up of copper(II) ions and manganate(VII) ions. The ions sulphate ions . The ions diffuse slowly between close diffuse very slow between space of water particles which is in closely packed space of gel particles liquid form. which is in solid form.

space of air particles which is in gas form.

Kuprum(II) sulfat, CuSO4 terdiri daripada ion ion kuprum(II) dan Ion-ion sulfat. ini meresap dengan perlahan sangat antara ruang padat zarah agar-agar yang berbentuk pepejal. n io

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Molekul bromin meresap pantas besar melalui ruang antara zarahzarah udara yang berbentuk gas.

Kalium manganat(VII) terdiri daripada ion kalium dan ion manganat(VII). Ion-ion perlahan ini meresap rapat antara ruang zarah air yang berbentuk cecair.

Warna biru kuprum(II) sulfat, sangat perlahan CuSO4 merebak di dalam agar-agar.

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(d) Conclusions: Kesimpulan:

gas than in liquid. There is (i) Diffusion occurs faster in gas gas a than a liquid. Particles in a are closer are together.

larger

space in between the particles of

further

apart. The particles in the liquid

gas Resapan berlaku lebih cepat di dalam berbanding di dalam cecair. Terdapat ruang yang gas gas berbanding dengan cecair. Zarah-zarah adalah antara zarah-zarah lebih rapat antara satu sama lain. antara satu sama lain. Zarah-zarah cecair adalah

liquid than in solid. There is (ii) Diffusion occurs faster in a liquid of a than a solid. The particles in the solid are very cecair

Resapan berlaku lebih cepat di dalam cecair antara zarah-zarah dan

padat

larger

lebih besar berjauhan

space in between the particles

close

together. lebih besar

berbanding di dalam pepejal. Terdapat ruang yang

rapat

berbanding dengan pepejal. Zarah-zarah pepejal tersusun sangat

antara satu sama lain.

(iii) Bromine gas, potassium manganate(VII) and copper(II) sulphate are made up of particles that are constantly moving/constant motion .

tiny

and

halus

Gas bromin, kalium manganat(VII) dan kuprum(II) sulfat terdiri daripada zarah-zarah sentiasa bergerak . yang

discrete

diskrit

dan

The Kinetic Theory of Matter / Teori Kinetik Jirim

solid

Matter exists in three different states which are

1

pepejal

Jirim wujud dalam tiga keadaan iaitu

Matter that made up of

2

tiny

Jirim terdiri daripada zarah-zarah

and halus

As the temperature increases, the

3

Apabila suhu meningkat, tenaga

,

discrete

kinetic

kinetik

,

cecair

liquid

and

gas

gas

.

dan

. moving

particles which are always in constantly

dan

diskrit

yang sentiasa

bergerak

.

.

energy of particles increases and the particles move

zarah-zarah akan bertambah dan zarah-zarah akan bergerak dengan

faster

.

lebih cepat

.

Particles in different states of matter have different arrangement, strength of forces between them, movement and energy content.

4

Zarah-zarah

dalam keadaan jirim yang berbeza mempunyai susunan, daya tarikan antara zarah, pergerakan dan kandungan

tenaga yang berbeza.

Complete the following table: / Lengkapkan jadual di bawah:

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State of matter

Solid

Keadaan jirim

Liquid

Pepejal

Gas

Cecair

Gas

Draw the particles arrangement. Each particle (atom/ ion/ molecule) is represented by Lukis susunan zarah. Setiap zarah (atom / ion / molekul) diwakili dengan ‘ ’

Particles arrangement Susunan zarah

The particles are arranged closely packed in orderly

manner.

Zarah-zarah tersusun teratur dan .

Particles movement

m

orderly manner padat

.

Zarah-zarah tersusun tidak teratur tetapi

fixed position.

throughout the liquid.

Zarah bergetar dan berputar pada kedudukan tetap.

terpisah jauh Zarah-zarah antara satu sama lain.

.

Particles can vibrate , rotate and move

from

each other.

padat

Particles can only vibrate and rotate about their

The particles are very widely separated

Zarah bergetar , berputar dan bergerak dalam cecair.

Particles can vibrate , rotate and move

freely.

Zarah bergetar , berputar dan bergerak bebas.

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Pergerakan zarah

The particles are arranged closely packed but not in

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Strong

Attractive forces between the particles Daya tarikan antara zarah

forces between the particles but weaker than

Very strong forces between the particles.

the forces in the solid.

Daya tarikan yang sangat kuat antara zarah-zarah.

kuat Daya tarikan yang antara zarah-zarah tetapi lebih lemah berbanding di

Weak

forces between

the perticles lemah

Daya tarikan yang antara zarah-zarah.

dalam pepejal.

Energy content of the particles Kandungan tenaga zarah

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Energy content is very low . Kandungan tenaga sangat rendah .

Energy content is higher than solid but less than in a gas. Kandungan tenaga lebih tinggi daripada pepejal tetapi lebih rendah daripada gas.

very

Energy content is high. Kandungan tenaga tinggi.

sangat

Changes in the state of matter Perubahan keadaan jirim

(a) Matter undergoes change of state when

heat

energy is absorbed or haba

Jirim mengalami perubahan keadaan apabila tenaga

serap

di

(i) When heat energy is absorbed by the matter (it is heated), the increases and they vibrate faster. diserap Apabila tenaga haba oleh jirim (semasa dipanaskan), tenaga dan zarah tersebut bergerak dengan lebih cepat.

(ii) When matter releases heat energy (it is cooled), the they vibrate less vigorously. dibebaskan Apabila tenaga haba zarah tersebut bergerak kurang cergas.

kinetic

released/lose atau di

kinetic

:

bebaskan

:

energy of the particles

kinetik

zarah

bertambah

energy of the particles decreases and

oleh jirim (semasa disejukkan), tenaga kinetik zarah

berkurang

dan

(b) Inter - conversion of the states of matter: Perubahan keadaan jirim: Solid Pepejal 7

Melting / Peleburan Freezing / Pembekuan

Liquid Cecair

Boiling/Evoporation / Pendidihan/Penyejatan Condensation / Kondensasi

Gas Gas

Determination of melting and freezing points of naphthalene Penentuan takat lebur dan takat beku naftalena

Materials / Bahan: Naphthalene powder, water Apparatus / Radas: Boiling tube, conical flask, beaker, retort stand, thermometer 0 – 100°C, stopwatch, Bunsen burner and wire gauze Procedure / Prosedur: I. Heating of naphthalene / Pemanasan naftalena Set-up of apparatus: / Susunan radas: Thermometer / Termometer Boiling tube / Tabung didih Water / Air Naphthalene / Naftalena

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(a) A boiling tube placed into it. Tabung didih

3 - 5 cm

is filled

height with naphthalene powder and a 3 – 5 cm

diisi dengan serbuk naftalena setinggi

dan

thermometer

termometer

is

diletakkan

di dalamnya.

(b) The boiling tube is immersed in a water bath as shown in the diagram so that the water level in the water bath is higher than naphtalene powder in the boiling tube. Tabung didih dimasukkan ke dalam kukus air seperti di dalam gambar rajah dan pastikan aras air dalam kukus air lebih tinggi daripada aras naftalena dalam tabung didih.

(c) The water is

heated

Air dipanaskan dan naftalena

and the naphthalene is dikacau

stirred

perlahan-lahan dengan

slowly with termometer

thermometer

.

.

60°C , the stopwatch is started. The temperature of (d) When the temperature of naphthalene reaches 90°C naphthalene is recorded at 30 seconds intervals until the temperature of naphthalene reaches . 60°C

Apabila suhu naftalena mencapai 90°C

sehingga suhunya mencapai

II.

, mulakan jam randik. Suhu naftalena dicatat setiap

30 saat

.

Cooling of naphthalene / Penyejukan naftalena

Naphthalene Naftalena

(a) The boiling tube and its content is removed from the water bath and put into a in the diagram.

conical flask kelalang kon

Tabung didih dan kandungannya dikeluarkan daripada kukus air dan dipindahkan ke dalam dalam gambar rajah.

as shown seperti

stirred constantly with thermometer throughout cooling (b) The content in the boiling tube is supercooling process to avoid (the temperature of cooling liquid drops below freezing point, without the appearance of a solid). dikacau Kandungan dalam tabung didih perlahan-lahan dengan termometer sepanjang proses penyejukan untuk penyejukan lampau (Suhu cecair yang disejukkan turun melepasi takat beku tanpa pembentukan mengelakkan pepejal).

(c) The temperature of naphthalene is recorded every 60°C to . Suhu naftalena dicatat setiap

30 saat

30 seconds

sehingga suhunya mencapai

interval until the temperature drops 60°C

.

(d) A graph of temperature against time is plotted for the heating and cooling process respectively.

m

suhu

melawan

masa

dilukis untuk proses pemanasan dan penyejukan.

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The Explanation of the Heating Process of Matter / Penerangan Proses Pemanasan 1

The heating curve of naphthalene: Lengkung pemanasan naftalena: Temperature/°C Suhu/°C

F D B

E

C

A Time/s Masa/s

2

faster

When a solid is heated, the particles absorb heat and move absorbed energy is , the state of matter will change.

as its energy content increases. As the heat

lebih cepat Apabila pepejal dipanaskan, zarah-zarah menyerap haba dan bergerak diserap menyebabkan perubahan keadaan jirim. Tenaga haba

Point Titik

A to B A ke B

State of Matter

Explanation

Keadaan jirim

Penerangan

absorbed

Heat energy is kinetic

Solid

energy to diserap

pepejal

oleh zarah-zarah

lebih cepat

dan zarah bergetar dengan

absorbed

Heat energy

Solid and Liquid

solid

by the particles in the increase and vibrate

naphthalene causing their

faster

. The temperature

increases. Tenaga haba bertambah

B to C B ke C

disebabkan kandungan tenaga bertambah.

overcome

naftalena menyebabkan tenaga meningkat . Suhu semakin

by the particles in the

forces between particles so that the remains constant

temperature

diserap

Tenaga haba yang

liquid solid

C to D C ke D

Liquid

Tenaga haba bertambah

turn to

pepejal oleh zarah-zarah dalam naftalena pepejal cecair berubah menjadi

liquid by the particles in the increase energy to and move oleh zarah-zarah

digunakan

absorbed

naphthalene causing their faster . The temperature

by the particles in the

liquid

naphthalene is

the forces of attraction between particles. The particles begin to move

to form a

gas

Tenaga haba

diserap

. The temperature oleh zarah-zarah dalam

remains constant cecair

absorbed

akan

used

to

freely

.

naftalena digunakan untuk mengatasi bebas gas untuk membentuk . Suhu

gas by the particles in the faster . The temperature energy to incerease and move

Heat energy is

causing their increases .

diserap oleh zarah-zarah gas naftalena menyebabkan tenaga Tenaga haba lebih cepat meningkat dan zarah-zarah bergerak dengan . Suhu semakin

kinetik

kinetic

akan bertambah

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Gas

untuk mengatasi tetap .

. Suhu adalah

kinetik naftalena menyebabkan tenaga lebih cepat meningkat . Suhu semakin .

daya tarikan antara zarah-zarah. Zarah-zarah mula bergerak tetap adalah .

E to F E ke F

to . The

cecair

dan zarah-zarah bergerak dengan

overcome

Liquid and Gas

liquid

. diserap

Heat energy D to E D ke E

used

.

absorbed

increases

akan

.

naphthalene is

daya tarikan antara zarah-zarah supaya

Heat energy kinetic

kinetik

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completely changes to become a liquid is called the melting point . During the melting process, the temperature remains unchanged because heat energy absorbed by the particles used is to overcome the forces between particles so that the solid change to turn into a liquid . The constant temperature at which a

3

solid

pepejal

takat lebur berubah kepada keadaan cecair dipanggil diserap oleh zarah-zarah Semasa proses peleburan, suhu tidak berubah kerana haba yang mengatasi cecair daya tarikan antara zarah supaya pepejal berubah menjadi .

Suhu tetap di mana suatu

. digunakan

liquid

completely changes to become a gas is called the absorbed During the boiling process, the temperature remains unchanged because heat energy The constant temperature at which a

4

used

is

to

overcome

untuk

boiling point

.

by the particles

the forces between particles so that the liquid change to turn into a gas.

Suhu tetap di mana suatu bahan dalam keadaan

cecair

takat didih berubah kepada keadaan gas dipanggil . diserap digunakan oleh zarah-zarah untuk

Semasa proses pendidihan, suhu tidak berubah kerana haba yang mengatasi daya tarikan antara zarah supaya cecair berubah menjadi gas.

The Explanation for the Cooling Process of Matter: / Penerangan Proses Penyejukan Bahan:

The cooling curve of naphthalene:

1

Lengkung penyejukan naftalena: Temperature/°C Suhu/°C

P Q

R S Time/s Masa/s

slower When the liquid is cooled, the particles in the liquid release energy and move decreases. As the energy is released to the surrounding, the state of matter will change.

2

cecair Apabila cecair disejukkan, zarah membebaskan tenaga dan dibebaskan ke persekitaran. berubah semasa tenaga Point Titik

P to Q P ke Q

bergerak

as its energy content

semakin perlahan. Keadaan jirim

State of matter

Explanation

Liquid

Heat is released/given out to the surrounding by the particles in the liquid naphthalene. liquid lose their kinetic energy and move slower. The The particles in the temperature decreases .

Keadaan jirim

Penerangan

dibebaskan ke persekitaran oleh zarah-zarah dalam Haba cecair kinetik kehilangan tenaga dan bergerak

cecair

naftalena. Zarah-zarah dalam semakin perlahan. Suhu semakin menurun

.

The heat released to the surrounding by the particles in liquid naphthalene is balanced heat solid . energy released as the particles attract one another to form a by the Q to R Q ke R

R to S

m

Solid

The temperature

remains constant

.

dibebaskan cecair diimbangi ke persekitaran oleh zarah-zarah dalam naftalena oleh Haba haba terbebas tenaga yang apabila zarah-zarah tertarik antara satu sama lain untuk membentuk pepejal tetap . Suhu adalah .

The particles in the solid naphthalene releases heat and vibrate decreases . Zarah-zarah dalam pepejal naftalena menurun Suhu semakin .

membebaskan

slower . The temperature

tenaga dan bergetar dengan

lebih perlahan

.

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Liquid and Solid

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3

freezing point changes to a solid is called . During the freezing process, the temperature remains unchanged because the heat released to the surrounding is balanced by the solid heat released when the liquid particles rearrange themselves to become a .

The constant temperature at which a

liquid

takat beku berubah kepada keadaan pepejal dipanggil . Semasa proses dibebaskan diimbangi ke persekitaran oleh haba yang terbebas pembekuan, suhu tidak berubah kerana haba yang pepejal . apabila zarah-zarah cecair menyusun semula untuk membentuk

Suhu tetap di mana suatu

cecair

Keadaan Fizik Bahan pada Sebarang Suhu: / Physical State Of A Substance At Any Given Temperature: 1

A substance is in

solid

state if the temperature of the substance is below melting point pepejal

Suatu bahan berada dalam keadaan 2

A substance is in

liquid

state if the temperature of the substance is between melting and boiling points. cecair

Suatu bahan berada dalam keadaan 3

A substance is in

gas

jika suhu bahan tersebut lebih rendah daripada takat leburnya.

jika suhu bahan tersebut berada antara takat lebur dan takat didihnya.

state if the temperature of the substance is above boiling point.

Suatu bahan berada dalam keadaan

gas

jika suhu bahan tersebut lebih tinggi daripada takat didihnya.

EXERCISE / LATIHAN 1

The table below shows substances and their chemical formula. Jadual di bawah menunjukkan bahan dan formula kimia masing-masing. Substance / Bahan

Chemical formula / Formula kimia

Type of particle / Jenis zarah

Silver / Argentum

Ag

Atom

Potassium oxide / Kalium oksida

K2O

Ion

Ammonia / Ammonia

NH3

Molecule

Chlorine / Klorin

Cl2

Molecule

(a) State the type of particles that made up each substance in the table. Nyatakan jenis zarah yang membentuk bahan dalam jadual di atas.

(b) Which of the substances are element? Explain your answer. Yang manakah antara bahan tersebut merupakan suatu unsur? Jelaskan jawapan anda.

Silver and chlorine. Silver and chlorine are made up of one type of atom (c) Which of the substance are compound? Explain your answer. Yang manakah antara bahan tersebut merupakan suatu sebatian? Jelaskan jawapan anda.

Potassium oxide and ammonia. Potassium oxide and ammonia are made up of two different elements 2

The table below shows the melting and boiling points of substance P, Q and R. Jadual di bawah menunjukkan takat lebur dan takat didih bagi bahan P, Q dan R. Melting point / Takat lebur / °C

Boiling point / Takat didih / °C

P

–36

6

Q

–18

70

R

98

230

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Substance / Bahan

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(a) (i)

What is meant by ‘melting point’? Apakah yang dimaksudkan dengan ‘takat lebur’?

The constant temperature at which a solid charges to a liquid at particular pressure

(ii) What is meant by ‘boiling point’? Apakah yang dimaksudkan dengan ‘takat didih’?

The constant temperature at which a liquid changes to a gas at particular pressure (b) Draw the particles arrangement of substances P, Q and R at room condition. Lukis susunan zarah P, Q dan R pada keadaan bilik.

Substance P / Bahan P

(c) (i)

Substance Q / Bahan Q

Substance R / Bahan R

What is the substance that exist in the form of liquid at 0°C. Nyatakan bahan yang wujud dalam keadaan cecair pada suhu 0°C.

P, Q

(ii) Give reason to your answer. Jelaskan jawapan anda.

The temperature 0°C is above the melting point of Q and below the boiling point of Q (d) (i)

Substance Q is heated from room temperature to 100°C. Sketch a graph of temperature against time for the heating of substance Q. Bahan Q dipanaskan dari suhu bilik hingga 100°C. Lakarkan graf suhu melawan masa bagi pemanasan bahan Q terhadap masa untuk pemanasan bahan Q.

Temperature/°C

70

Time/s (ii) What is the state of matter of substance Q at 70°C? Apakah keadaan fizik bahan Q pada 70°C?

Liquid and gas (e) Compare the melting point of substances Q and R. Explain your answer. Bandingkan takat lebur bahan Q dan R. Terangkan jawapan anda.

The melting point of substance R is higher than subtance Q. The attraction force between particles in substance R

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is stronger than Q. More heat is needed to overcome the force between particles in substance R.

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3

The melting point of acetamide can be determined by heating solid acetamide until it melts as shown in the diagram below. The temperature of acetemide is recorded every three minutes when it is left to cool down at room temperature. Takat lebur asetamida boleh ditentukan dengan memanaskan pepejal asetamida sehingga lebur seperti dalam rajah di bawah. Suhu asetamida dicatatkan setiap tiga minit semasa disejukkan pada suhu bilik. Thermometer / Termometer Boiling tube / Tabung didih Water / Air Acetamide / Asetamida

(a) What is the purpose of using water bath in the experiment? Apakah tujuan menggunakan kukus air dalam eksperimen ini?

To ensure even heating of acetemide. Acetamide is easily combustible. (b) State the name of another substance which its melting point can also be determined by using water bath as shown in the above diagram. Namakan satu bahan lain yang mana takat leburnya boleh ditentukan dengan menggunakan kukus air seperti rajah di atas.

Naphthalene (c) Sodium nitrate has a melting point of 310°C. Can the melting point of sodium nitrate be determined by using the water bath as shown in the diagram? Explain your answer. Natrium nitrat mempunyai takat lebur 310°C. Bolehkah takat lebur natrium nitrat ditentukan dengan menggunakan kukus air seperti yang ditunjukkan dalam rajah di atas? Jelaskan jawapan anda.

No, because the melting point of water is 100°C which is less than the melting point of sodium nitrate. (d) Why do we need to stir the acetemide in the boiling tube in above experiment? Mengapakah asetamida dalam tabung didih itu perlu dikacau sepanjang eksperimen?

To make sure the heat is distributed evenly (e) The graph of temperature against time for the cooling of liquid acetamide is shown below. Rajah di bawah menunjukkan graf suhu melawan masa untuk penyejukan cecair asetamida. Temperature / Suhu/ °C

T3 T2

Q

R

T1

(i) What is the freezing point of acetamide?

Time / Masa/s

Apakah takat beku asetamida?

T2°C (ii) The temperature between Q and R is constant. Explain. Suhu antara titik Q dan R adalah tetap. Jelaskan.

The heat lost to the surrounding is balanced by the heat released when the liquid particles rearrange themselves to become solid. (f) Acetemide exists as molecules. State the name of another compound that is made up of molecules. Asetamida wujud sebagai molekul. Namakan sebatian lain yang terdiri daripada molekul.

Water/naphthalene (g) What is the melting point of acetamide? Apakah takat lebur asetamida? n io

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T2°C

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The Atomic Structure / Struktur Atom History of the development of atomic models:

1

Sejarah perkembangan model atom: Scientist Saintis

Atomic Model

Discovery

Model atom

Penemuan

(i)

Matter is made up of particles called

atoms

.

Jirim terdiri daripada zarah-zarah dipanggil

atom

.

created

(ii) Atoms cannot be

Dalton

dicipta

Atom tidak boleh

, destroyed or dimusnah

,

Positively charged sphere Sfera bercas

positif

Thomson Electron charges negative Elektron

(i)

sama

nukleus

bercas positif

(ii)

Nukleus mengandungi proton

, zarah subatom yang pertama.

Proton

yang merupakan pusat bagi atom dan

.

is a part of the nucleus.

Proton

(iii)

yang mengandungi zarah

Discovered the nucleus as the centre of an atom and positively charged . Menjumpai

bergerak di luar

mengandungi Nucleus that contain proton

elektron

positif Atom adalah sfera yang bercas elektron bercas negatif dipanggil .

Electron moves outside the nucleus

Rutherford

.

(ii) Atom is sphere of positive charge which embedded with negatively charged particles called electrons .

(i)

nukleus

.

.

Discovered the electrons , the first subatomic particle. Menjumpai

bercas negatif

Elektron

atau

identical .

(iii) Atoms from the same element are Atom daripada unsur sama adalah

divided

dibahagi

adalah sebahagian daripada nukleus.

Electron move outside the nucleus. Elektron

bergerak di sekeliling nukleus.

(iv) Most of the mass of the atom found in the Nukleus

nucleus .

mempunyai hampir semua jisim atom.

Shell

Neils Bohr

Nucleus that contain proton

(i)

Discovered the existence of electron

Nukleus mengandungi proton

(ii) Electrons move in the

petala

Menjumpai kewujudan Elektron

shells

.

elektron.

shells

around the nucleus.

bergerak di dalam petala mengelilingi

nukleus

.

Electron

Shell

James Chadwick

Nucleus that contain proton and neutron Nukleus mengandungi proton dan neutron

Electron

(i)

Discovered the existence of

proton

.

Nukleus mengandungi zarah-zarah neutral dipanggil proton zarah-zarah bercas positif dipanggil .

neutron

dan

(iii) The mass of a neutron and proton is almost the same. neutron

dan

proton

adalah hampir sama.

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(ii) Nucleus of an atom contains neutral particles called neutron and positively charged particles called

Jisim m

neutron .

neutron

Menjumpai kewujudan

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2

The structure of an atom: / Struktur Atom: Shell / Petala Nucleus that contain proton and neutron Nukleus yang mengandungi proton dan neutron Electron / Elektron

nucleus

(a) An atom has a central Atom mempunyai

nucleus

(b) The

Nukleus

nukleus

shells

and electrons that move in the

around the nucleus.

petala

di tengahnya dan elektron bergerak di dalam

mengelilingi nukleus tersebut.

contains protons and neutrons. mengandungi proton dan neutron.

+1 . Each electron has an electrical charge of –1 . The neutron has no (c) Each proton has charge of charge neutral (it is ). An atom has the same number of protons and electrons, so the overall charge zero of atom is . Atom is of ion will be studied in Chapter 4)

. (If an atom loses or gains electrons it is called an ion – formation

(ianya adalah neutral ). sifar . Atom Setiap atom mempunyai bilangan proton dan elektron yang sama, oleh itu cas keseluruhan bagi atom adalah neutral . (Suatu atom akan membentuk ion apabila ia kehilangan atau menerima elektron – pembentukan ion akan adalah

Setiap proton bercas

+1

neutral

–1

. Setiap elektron bercas

. Neutron tidak mempunyai

cas

dipelajari dalam Tajuk 4.)

(d) The relative mass of a neutron and a proton which are in the nucleus is 1. The mass of an atom is obtained mainly from the number of proton and neutron . proton

Jisim relatif proton dan neutron di dalam nukleus ialah 1. Jisim suatu atom diperoleh daripada jumlah bilangan neutron dan bilangan .

1 (e) The mass of an electron can be ignored as the mass of an electron is about times the size of a proton or 1 840 neutron. Jisim elektron boleh diabaikan kerana ia terlalu kecil iaitu 3

1 daripada jisim proton dan neutron. 1 840

Complete the following table: Lengkapkan jadual di bawah: Subatomic particles

Symbol

Charge

Relative atomic mass Jisim atom relatif

Kedudukan

Electron/Elektron

e

– (negative)

1 = 0 1 840

In the shells

Proton/Proton

p

+ (positive)

1

In the nucleus

Neutron/Neutron

n

neutral

1

In the nucleus

Zarah subatom

4

Simbol

Cas

Position

Atom is the smallest neutral particle of an element. Atom adalah zarah neutral paling kecil dalam suatu unsur.

Complete the following diagram: / Lengkapkan yang berikut: Na

Na

Na

Na

Sodium element natrium

Unsur

natrium

Na

Na

Na

Na

Na

Na

Sodium element Unsur

natrium

Na

Na

Na

Sodium Atom

atom natrium n io

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Unsur

Sodium element

Na

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Proton number of an element (Refer to Periodic table of an element)

5

Nombor proton sesuatu unsur (Rujuk Jadual Berkala Unsur)

(a) Proton number of an

element

atom

is the number of proton in its

.

atom

Nombor proton sesuatu unsur adalah bilangan proton yang terdapat dalam

.

(b) The number of proton of an atom is also equal to the number of electrons in the atom because atom is neutral . Bilangan proton sesuatu atom adalah sama dengan bilangan elektron dalam atom kerana atom adalah

neutral

.

(c) Every element has its own proton number: Setiap unsur mempunyai nombor protonnya tersendiri:

– Proton number of potassium, K is 19. Potasium in the shells.

atom

Atom

Nombor proton untuk kalium, K ialah 19. 19 elektron di dalam petala.

– Proton number of oxygen, O is 8. Oxygen in the shells. Nombor proton untuk oksigen, O ialah 8. 8 elektron di dalam petala.

has 19 protons in the nucleus and 19 electrons 19 proton

kalium mempunyai

atom

8 protons

has

Atom

di dalam nukleus dan

in the nucleus and 8 electrons 8 proton

oksigen mempunyai

di dalam nukleus dan

Nucleon number of an element (Refer to Periodic table of an element)

6

Nombor nukleon sesuatu unsur (Rujuk Jadual Berkala Unsur)

(a) Nucleon number of an element is the total number of protons and neutrons in the nucleus of its Nombor nukleon sesuatu unsur adalah jumlah bilangan proton dan neutron di dalam nukleus sesuatu

atom

atom

.

.

(b) Nucleon number is also known as a mass number. Nombor nukleon juga dikenali sebagai nombor jisim.

(c) Nucleon number = number of proton + number of neutron. Nombor nukleon = bilangan proton + bilangan neutron. Symbol of Element And Standard Representation For An Atom of Element Simbol Unsur dan Perwakilan Piawai bagi Atom Sesuatu Unsur

The symbol of an element is a short way of representing an element. If the symbol has only one letter, it must be a capital letter. If it has two letters, the first is always a capital letter, while the second is always a small letter.

1

Simbol unsur adalah cara mudah untuk mewakilkan unsur. Jika simbol hanya terdiri daripada satu huruf, maka ia mesti ditulis dengan huruf besar. Tetapi jika simbol terdiri daripada dua huruf, maka huruf pertama merupakan huruf besar dan huruf kedua merupakan huruf kecil.

Example: / Contoh: Element Unsur

Element

Symbol

Element

Symbol

O

Nitrogen/Nitrogen

N

Calcium/Kalsium

Ca

Mg

Sodium/Natrium

Na

Copper/Kuprum

Cu

Potassium/Kalium

K

Chlorine/Klorin

Cl

Simbol

Oxygen/Oksigen Magnesium/Magnesium Hydrogen/Hidrogen

Symbol

H

Unsur

Simbol

Unsur

Simbol

The first letter of each element is capitalised to show that it is a new element. This is helpful when writing a chemical formula. For example KCl. There are two elements chemically bonded in KCl because there are two capital letters represent potassium and chlorine. Huruf yang pertama bagi setiap unsur ditulis dengan huruf besar untuk menunjukkan ia adalah unsur yang baru. Ini sangat berguna semasa menulis formula kimia. Contohnya KCl. Terdapat dua unsur yang terikat secara kimia dalam KCl kerana adanya dua huruf besar yang mewakili kalium dan klorin.

Standard representation symbol represents

2

m

one atom

of an element. It can be written as:

sesuatu unsur. Ianya boleh ditulis sebagai:

Nucleon number/Nombor nukleon

A

Proton number/Nombor proton

Z

X

Symbol of an element/Simbol unsur

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Simbol perwakilan piawai mewakili

satu atom

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Example: / Contoh:

27 A1 13 – The element is Aluminium. Unsur itu adalah Aluminium.

27

Nombor nukleon Aluminium adalah

– Aluminium has

. .

, 14 neutrons and

13 proton

Atom Aluminium mempunyai 3

13 13

13 protons

. .

– The proton number of Aluminium is Nombor proton Aluminium adalah

27

– The nucleon number of Aluminium is

13

14 neutron

,

electrons. 13 elektron.

dan

Isotope / Isotop (a) Isotopes are atoms of the same element with same number of protons but different number of neutrons. Isotop ialah atom-atom unsur yang mempunyai bilangan proton yang sama tetapi bilangan neutron yang berbeza.

Or / Atau

Isotopes are atoms of the same element with same

proton

number but different

proton

Isotop ialah atom-atom unsur yang mempunyai nombor berbeza.

nucleon

number.

nukleon

yang sama tetapi nombor

yang

Example: / Contoh: 1 1 H

2 1 H

Nucleon number/Nombor nukleon = 1

Nucleon number/Nombor nukleon = 2

Proton number/Nombor proton = 1

Proton number/Nombor proton = 1

Number of neutron/Bilangan neutron = 0

Number of neutron/Bilangan neutron = 1

– Hydrogen-1 and Hydrogen-2 are isotopes. Hydrogen-1 and Hydrogen-2 atoms have the same proton number or the same number of protons but

different

in nucleon number because of the difference in the number of

Atom Hidrogen-1 dan Hidrogen-2 mempunyai nombor proton atau bilangan neutron kerana perbezaan .

– Isotopes have the same arrangements. Isotop mempunyai sifat

chemical kimia

bilangan proton

properties but different

yang sama tetapi nombor nukleon yang

physical

neutron

.

berbeza

properties because they have the same electron

yang sama kerana mempunyai susunan elektron yang sama tetapi sifat

fizik

yang berbeza.

(b) Examples of the usage of isotopes: Contoh kegunaan isotop:

i.

Medical field Bidang perubatan

–

To detect brain cancer.

–

To detect thrombosis (blockage in blood vessel).

–

Sodium-24 is used to measure the rate of iodine absorption by thyroid gland.

–

Cobalt-60 is used to destroy cancer cells.

–

To kill microorganism in the sterilising process.

Untuk mengesan barah otak. Untuk mengesan trombosis (saluran darah tersumbat). Untuk mengukur kadar penyerapan iodin oleh kelenjar tiroid. Contoh: Natrium-24 Untuk memusnahkan sel barah. Contoh: Kobalt-60 Untuk membunuh mikroorganisma semasa proses pensterilan.

ii.

In the industrial field Bidang industri

–

To detect wearing out in machines.

–

To detect any blockage in water, gas or oil pipes.

Untuk mengesan kehausan enjin.

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Untuk mengesan saluran paip air, gas atau minyak yang tersumbat.

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–

To detect leakage of pipes underground.

–

To detect defects/cracks in the body of an aeroplane.

Untuk mengesan kebocoran paip bawah tanah. Untuk mengesan keretakan atau kecacatan pada badan kapal terbang.

iii.

In the agriculture field Bidang pertanian

–

To detect the rate of absorption of phosphate fertilizer in plants.

–

To sterile insect pests for plants.

Untuk mengesan kadar penyerapan baja fosfat oleh tumbuhan. Untuk memandulkan serangga perosak tumbuhan.

iv.

In the archeology field Bidang arkeologi

–

Carbon-14 can be used to estimate the age of artifacts. Karbon-14 untuk menentukan usia sesuatu artifak.

Electron Arrangement

4

Susunan elektron

(a) The electrons are filled in specific shells. Every shell can be filled only with a certain number of electrons. For the elements with atomic numbers 1-20: Elektron diisi dalam petala tertentu. Setiap petala hanya boleh diisi dengan bilangan elektron tertentu. Bagi unsur-unsur yang mempunyai nombor proton 1–20:

2

– First shell can be filled with a maximum of

electrons. 2

Petala pertama boleh diisi dengan bilangan maksimum

– Second shell can be filled with a maximum of Petala kedua boleh diisi dengan bilangan maksimum Petala ketiga boleh diisi dengan bilangan maksimum

electrons.

8

8

– Third shell can be filled with a maximum of

elektron.

8

elektron. electrons.

8

elektron.

First shell is filled with 2 electrons (duplet) Petala pertama diisi 2 elektron (duplet)

Second shell is filled with 8 electrons (octet) Petala kedua diisi 8 elektron (oktet)

Third shell is filled with 8 electrons (octet) Petala ketiga disi 8 elektron (oktet)

(b) Valence electrons are the electrons in the outermost shell of an atom. Elektron valens: Elektron yang diisi dalam petala paling luar suatu atom.

Complete the following table:

5

Lengkapkan jadual berikut:

(a) Draw the electron arrangement and complete the description for each element: Lukis susunan elektron bagi atom dan penerangan bagi setiap unsur berikut: Standard representation of an element Perwakilan piawai unsur

Electron arrangement of an atom Lukiskan susunan elektron bagi atom

Hydrogen Atom Atom Hidrogen

m

Number of protons/Bilangan proton

1

Number of eletrons/Bilangan elektron

1

Number of neutrons/Bilangan neutron

0


1


1

Electron Arrangement/Susunan elektron

1

H

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1 H 1

Description Penerangan

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Sodium Atom Atom Natrium

23 Na 11

Na

Number of protons/Bilangan proton

11

Number of electrons/Bilangan elektron

11

Number of neutrons/Bilangan neutron

12


11


23

Electron Arrangement/Susunan elektron

2.8.1

(b) Choose the correct statement for the symbol of element X. Pilih pernyataan yang betul bagi simbol unsur X. 23 Na 11 Statement Pernyataan

Element X has 11 proton number. Unsur X mempunyai 11 nombor proton.

The proton number of element X is 11. Nombor proton unsur X ialah 11.

The proton number of atom X is 11. Nombor proton atom X ialah 11.

The number of proton of element X is 11. Bilangan proton unsur X ialah 11.

The number of proton of atom X is 11. Bilangan proton atom X ialah 11.

Nucleon number of element X is 23. Nombor nukleon unsur X ialah 23.

Nucleon number of atom X is 23. Nombor nukleon atom X ialah 23.

Number of nucleon of element X is 23. Bilangan nukleon unsur X ialah 23.

Atom X has 23 nucleon number. Atom X mempunyai 23 nombor nukleon.

Neutron number of atom X is 12. Nombor neutron atom X ialah 12.

Number of neutron of atom X is 12. Bilangan neutron atom X ialah 12.

Number of neutron of element X is 12.

Tanda ( 3 / 7 )

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Bilangan neutron unsur X ialah 12.

Tick ( 3 / 7 )

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EXERCISE / LATIHAN

Complete the following table:

1

Lengkapkan jadual berikut:

Element Unsur

Hydrogen Hidrogen

Helium Helium

Boron Boron

Carbon Karbon

Nitrogen Nitrogen

Neon Neon

Sodium Natrium

Magnesium Magnesium

Calcium Kalsium

Symbol of element Simbol unsur

Number of proton Bilangan proton

Number of electron Bilangan elektron

Number of neutron Bilangan neutron

Proton number Nombor proton

Nucleon number Nombor nukleon

Electron arrangement Susunan elektron atom

Number of valence electron Bilangan elektron valens

1 1 H

1

1

0

1

1

1

1

4 2 He

2

2

2

2

4

2

2

11 5 B

5

5

6

5

11

2.3

3

12 6 C

6

6

6

6

12

2.4

4

14 7 N

7

7

7

7

14

2.5

5

20 Ne 10

10

10

10

10

20

2.8

8

23 Na 11

11

11

12

11

23

2.8.1

1

24 Mg 12

12

12

12

12

24

2.8.2

2

40 Ca 20

20

20

20

20

40

2.8.8.2

2

The diagram below shows the symbol of atoms P, R and S.

2

Rajah di bawah menunjukkan simbol atom P, R dan S.

35 P 17

12 R 6

37 S 17

(a) What is meant by nucleon number / Apakah maksud nombor nukleon? Nucleon number of an element is the total number of protons and neutrons in the nucleus of its atom (b) What is the nucleon number of P / Apakah nombor nukleon atom P? 35 (c) State the number of neutron in atom P / Nyatakan bilangan neutron atom P. 18 (d) State number of proton in atom P / Nyatakan bilangan proton atom P. 17 (e) (i)

What is meant by isotope / Apakah maksud isotop?

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Isotopes are atoms of the same element with same number of proton but different number of neutrons

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(ii) State a pair of isotope in the diagram shown / Nyatakan sepasang isotop dalam rajah yang ditunjukkan. P and S

(iii) Give reason for your answer in (e)(ii) / Berikan sebab bagi jawapan di (e)(ii). Atom P and S have same proton number but different nucleon number//number of neutron

(f) An isotope of R has 8 neutron. Write the symbol for the isotope R. Isotop bagi atom R mempunyai 8 neutron. Tuliskan simbol bagi isotop R.

14 R 6 3

The table below shows the number of proton and neutron of atoms of elements P, Q and R. Jadual di bawah menunjukkan bilangan proton dan neutron bagi atom unsur P, Q dan R. Element Unsur

Number of proton Bilangan proton

Number of neutron Bilangan neutron

P

1

0

Q

1

1

R

6

6

(a) Which of the atoms in the above table are isotope? Explain your answer. Berdasarkan jadual di atas, atom yang manakah merupakan isotop? Terangkan jawapan anda.

P and Q. Atom P and Q have same number of proton but different number of neutron // nucleon number. (b) (i)

Write the standard representation of element Q. Tuliskan perwakilan piawai untuk unsur Q.

2 Q 1

(ii) State three information that can be deduced from your answer in (b)(i). Nyatakan tiga maklumat yang boleh didapati daripada jawapan anda di (b)(i).

The proton number of element Q is 1 // Number of proton of atom Q is 1 Nucleon number of element Q is 2 // Atomic mass of atom Q is 2 Number of neutron of atom Q is 1 Nucleus of atom Q contains 1p and 1n (c) (i)

Draw atomic structure for atom of element R. Lukiskan struktur atom bagi atom unsur R.

6 protons + 6 neutrons

(ii) Describe the atomic structure in (c)(i). Huraikan struktur atom di (c)(i).

– The atom consists of 2 parts: the centre part called nucleus and the outer part called electron shell. – The nucleus consists of 6 protons which are positively charged and 6 neutrons which are neutral. – The electrons are in two shells, the first shell consists of two electrons and the second shell consists of four electrons.

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– Electrons move around nucleus in the shells.

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(d) Element R react with oxygen and to produce liquid Z at room temperature. The graph below shows the sketch of the graph when liquid Z at room temperature, 27°C is cooled to –5°C. Unsur R bertindak balas dengan oksigen dan menghasilkan cecair Z pada suhu bilik. Rajah di bawah menunjukkan lakaran graf apabila cecair Z pada suhu bilik, 27°C disejukkan sehingga –5°C. Temperature /°C Suhu /°C

Time /s

0

t1

Masa /s

t2

−5

(i) What is the state of matter of liquid Z from t1 to t2? Explain why is the temperature remain unchanged from t1 to t2. Apakah keadaan jirim Z daripada t1 hingga t2? Terangkan mengapa suhu tidak berubah daripada t1 hingga t2.

Liquid and solid. Heat lost to the surrounding is balanced by the heat released when the particles at 0 °C (ii) Draw the arrangement of particles of Z at 20°C. Lukiskan susunan zarah-zarah Z pada suhu 20°C.

(iii) Describe the change in the particles movement when Z is cooled from room temperature to –5°C. Nyatakan perubahan dalam pergerakan zarah-zarah apabila cecair Z disejukkan daripada suhu bilik ke –5°C.

The particles move slower Objective Questions / Soalan Objektif The diagram shows the arrangement of particles for a type of matter that undergoes a change in physical state through process X.

1

3

The diagram below shows the heating curve for substance X. Rajah di bawah menunjukkan lengkung pemanasan bahan X.

Temperature / Suhu °C

Rajah di bawah menunjukkan susunan zarah sejenis bahan yang mengalami perubahan keadaan fizik melalui proses X.

U S Q

X

T

R

P

Time (m) Masa (m)

Which region of the graph does boiling process occur?

What is process X?

Bahagian manakah pada graf berlaku proses pendidihan?

Apakah proses X ?

A B

Melting Peleburan

Boiling Pendidihan

C D

Sublimation Pemejalwapan

Antara bahan berikut, yang manakah mengalami pemejalwapan apabila dipanaskan?

A

Sulphur Sulfur

Ammonium chloride Ammonium klorida

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C D

Sodium chloride Natrium klorida

C D

ST TU

Which of the following information is true? Antara pernyataan berikut, yang manakah adalah betul? Change of state Perubahan keadaan

Process Proses

Heat energy Tenaga haba

A

Solid → Liquid

Melting

Released

Pepejal → Cecair

Peleburan

Dibebaskan

B

Liquid → Gas

Evaporation

Released

Cecair → Gas

Penyejatan

Dibebaskan

C

Gas → Solid

Sublimation

Released

Gas → Pepejal

Pemejalwapan

Dibebaskan

D

Gas → Liquid

Condensation Kondensasi

Absorbed

Glucose Glukosa

PQ QR

Gas → Cecair

Diserap

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Which of the following substances can undergo sublimation when heated?

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Freezing Pembekuan

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The diagram below shows the graph of temperature against time when a liquid Y is cooled. Rajah di bawah menunjukkan graf suhu melawan masa apabila cecair Y disejukkan.

Substance Bahan

Melting point/°C Takat lebur/°C

Boiling point/°C Takat didih/°C

S

–182

–162

T

–23

77

U

–97

65

V

41

182

W

132

290

Temperature / Suhu °C t3

P Q

t2

R

Which substance exists as liquid at room temperature? t1

Bahan yang manakah wujud sebagai cecair pada suhu bilik?

S

A

Time (m) Masa (m)

B

Which of the following statements are true about the curve? Antara pernyataan berikut, yang manakah adalah betul tentang lengkung itu?

I

At Q, liquid Y begins to freeze.

II

At PQ, particles in Y absorb heat from the surroundings.

III

Liquid Y freezes completely at S.

IV

The freezing point of Y is t2°C.

8

Pada Q, cecair Y mula membeku.

C

S only S sahaja

D

S and T only S dan T sahaja

Rajah di bawah menunjukkan perwakilan piawai atom kuprum.

64 Cu 29

Cecair Y membeku dengan lengkap pada S.

A B 6

C

I and III only I dan III sahaja

D

I and IV only I dan IV sahaja

Which of the following is correct based on the symbol the diagram?

II and III only

Antara berikut, yang manakah betul berdasarkan rajah di atas?

II dan III sahaja

Rajah di bawah menunjukkan graf suhu melawan masa apabila pepejal Z dipanaskan.

Temperature / Suhu °C 9

0

1

2

3

4

5

6

7

8

9

29

64

29

B

35

29

64

C

64

35

29

D

29

64

35

The diagram below shows the standard representation of beryllium atom.

Apakah bilangan elektron valens bagi atom berillium?

Masa (m)

Antara berikut, yang manakah adalah benar pada minit keempat?

7

A

What is the number of valence electrons of beryllium atom? A B

Which of the following is true during the fourth minute?

D

Number of electron Bilangan elektron

9 Be 4 Time (m)

C

Nucleon number Nombor nukleon

Rajah di bawah menunjukkan perwakilan piawai atom berillium.

80

B


II and IV only II dan IV sahaja

The diagram below shows the graph of temperature against time when solid Z is heated.

A

V and W only V dan W sahaja

The diagram below shows standard representation of an atom copper.

Pada PQ, zarah dalam Y menyerap haba dari persekitaran.

Takat beku bagi Y adalah t2°C.

T and U only T dan U sahaja

All the molecules are in random motion. Semua molekul bergerak secara rawak. All the molecules are closely packed and in random motion. Semua molekul sangat rapat dan bergerak secara rawak. All the molecules are vibrating at fixed positions. Semua molekul bergetar pada kedudukan tetap. Some of the molecules are vibrating at fixed positions but some are in random motion. Sebahagian molekul bergetar pada kedudukan tetap dan sebahagian bergerak secara rawak.

The table shows the melting points and boiling points of substances S, T, U, V and W.

C D

4 7

The table below shows the proton number and the number of neutrons for atoms of elements W, X, Y and Z. Jadual di bawah menunjukkan nombor proton dan bilangan neutron bagi atom unsur W, X, Y dan Z. Element Atom


Number of neutrons Bilangan neutron

W 7 7 X 8 8 Y 8 9 Z 9 10 Which of the following pair of elements is isotope? Antara pasangan berikut, yang manakah adalah isotop?

A B

W and X W dan X

W and Y W dan Y

C D

X and Y X dan Y

Y and Z Y dan Z

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Jadual di bawah menunjukkan takat lebur dan takat didih bahan S, T, U, V dan W.

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2

CHEMICAL FORMULA AND EQUATIONS FORMULA DAN PERSAMAAN KIMIA RELATIF MASS / JISIM RELATIF

• RELATIVE ATOMIC MASS / JISIM ATOM RELATIF (JAR)

– To state the meaning of relative mass and solve numerical problems Menyatakan maksud jisim atom relatif dan menyelesaikan masalah pengiraan

• RELATIVE FORMULA MASS / JISIM FORMULA RELATIF (JFR)

– To state the meaning of RAM, RMM and RFM based on carbon-12 scale Menyatakan maksud JAR, JMR dan JFR berdasarkan skala karbon-12

• RELATIVE MOLECULAR MASS / JISIM MOLEKUL RELATIF (JMR)

– To calculate RAM, RMM and RFM using the chemical formulae of various substances Menghitung JAR, JMR dan JFR menggunakan formula kimia beberapa bahan

MOLE CONCEPT / KONSEP MOL

• MOLE AND THE NUMBER OF PARTICLES / MOL DAN BILANGAN ZARAH

– To solve numerical problems involving mole and the number of atoms/ ions/ molecules Menyelesaikan masalah pengiraan melibatkan mol dan bilangan atom, ion dan molekul

• MOLE AND THE MASS OF SUBSTANCES / MOL DAN JISIM BAHAN

– To solve numerical problems involving mole and the mass of substances, number of particles and volume of gas using mole concept Menyelesaikan masalah pengiraan melibatkan mol, jisim bahan, bilangan zarah dan isipadu gas menggunakan konsep mol

• MOLE AND THE VOLUME OF GAS / MOL DAN ISIPADU GAS

– To solve numerical problems involving mole and the mass of substances, number of particles and volume of gas using mole concept Menyelesaikan masalah pengiraan melibatkan mol, jisim bahan, bilangan zarah dan isipadu gas menggunakan konsep mol

CHEMICAL FORMULA AND EQUATIONS / FORMULA DAN PERSAMAAN KIMIA

• EMPIRICAL FORMULA / FORMULA EMPIRIK

– Stating the purpose and describe the empirical formula laboratory activities to determine the formula empirical Menyatakan maksud formula empirik dan menghuraikan aktiviti makmal untuk menentukan formula empirik

• MOLECULAR FORMULA / FORMULA MOLEKUL

– Solve calculation problems involving empirical formula Menyelesaikan masalah pengiraan melibatkan formula empirik

• CHEMICAL FORMULAE / FORMULA KIMIA

– To write formula of anion and cation and to write chemical formula for ionic compounds Menulis formula kation dan anion dan menulis formula kimia untuk sebatian ion

• CHEMICAL EQUATIONS / PERSAMAAN KIMIA

– Write a balanced chemical equation and solve problems arrangements involving the mole concept

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Menulis persamaan kimia seimbang dan menyelesaikan masalah pengiraan yang melibatkan konsep mol

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RELATIVE ATOMIC MASS (RAM) / JISIM ATOM RELATIF (JAR) 1

A single atom is too small and light and cannot be weighed directly. Satu atom adalah terlalu ringan, kecil dan tidak dapat ditimbang secara langsung.

2

The best way to determine the mass of a single atom is to compare its mass to the mass of another atom of an element that is used as a standard. Cara yang paling sesuai untuk menentukan jisim satu atom ialah dengan membandingkan jisimnya dengan jisim suatu atom unsur lain yang dianggap sebagai piawai.

3

Hydrogen was the first element to be chosen as the standard for comparing mass because the hydrogen atom is the lightest atom with a mass of 1.0 a.m.u (atomic mass unit). Hidrogen adalah unsur pertama dipilih sebagai piawai untuk membandingkan jisim kerana atom hidrogen adalah unsur yang paling ringan dengan jisim 1.0 u.j.a (unit jisim atom).

Example: Contoh:

• The mass of one helium atom is four times larger than one hydrogen atom. Jisim satu atom Helium adalah 4 kali lebih besar daripada satu atom hidrogen.

• RAM for He is 4. JAR untuk He ialah 4. 4

On the hydrogen scale, the relative atomic mass of an element means the mass of one atom of the element compared to the mass of a single hydrogen atom: Pada skala hidrogen, jisim atom relatif suatu unsur ditakrifkan sebagai jisim satu atom unsur berbanding jisim satu atom hidrogen:

Relative atomic mass of an element (RAM) / Jism atom relatif suatu unsur (JAR) =

The average mass of one atom of the element / Jisim purata satu atom unsur Mass of one hydrogen atom / Jisim satu atom hidrogen

• RAM has no unit. JAR tiada unit.

• The new standard used today is the carbon-12 atom. Piawai yang digunakan sekarang adalah berdasarkan atom karbon-12.

1 • RAM based on the carbon-12 scale is the mass of one atom of the element compared with of the mass of an 12 atom of carbon-12: JAR berdasarkan skala atom karbon-12 adalah jisim satu atom unsur berbanding dengan

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• Relative atomic mass of an element (RAM) / Jisim atom relatif suatu unsur (JAR) The average mass on one atom of the element / Jisim purata satu atom unsur = 1 × The mass of an atom of carbon-12 / Jisim satu atom karbon-12 12

1 jisim satu atom karbon-12: 12

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RELATIVE MOLECULAR MASS (RMM) / RELATIVE FORMULA MASS (RFM) JISIM MOLEKUL RELATIF (JMR) / JISIM FORMULA RELATIF (JFR)

RMM / JMR =

1

The average mass on one atom of the element / Jisim purata satu molekul

1 × The mass of an atom of carbon-12 / Jisim satu atom karbon-12 12 RMM is obtained by adding up the RAM of all the atoms that are present in the molecule.

2

JMR diperoleh dengan menambahkan JAR semua atom yang terdapat dalam satu molekul. Molecular substance

Molecular formula

Relative molecular mass

O2

2 × 16 = 32

Water / Air

H2O

2 × 1 + 16 = 18

Carbon dioxide / Karbon dioksida

CO2

12 + 2 × 16 = 44

Ammonia / Ammonia

NH3

14 + 3 × 1 = 17

Bahan molekul

Formula molekul

Oxygen / Oksigen

Jisim molekul relatif

[Relative atomic mass / Jisim atom relatif : O = 16, H = 1, C = 12, N = 14]

For ionic substances, RMM is replaced with Relative Formula Mass (RFM).

3

Untuk sebatian ion, JMR digantikan dengan Jisim Formula Relatif (JFR). Substance

Chemical formula

Relative molecular mass

Sodium chloride / Natrium klorida

NaCl

23 + 35.5 = 58.5

Potassium oxide / Kalium oksida

K2O

2 × 39 + 16 = 94

CuSO4

64 + 32 + 4 × 16 = 160

(NH4)2CO3

2 [14 + 4 × 1] + 12 + 3 × 16 = 96

Aluminium nitrate / Aluminium nitrat

Al(NO3)3

27 + 3 [14 + 3 × 16] = 213

Calcium hydroxide / Kalsium hidroksida

Ca(OH)2

40 + 2 [16 + 1] = 74

Lead(II) hydroxide / Plumbum(II) hidroksida

Pb(OH)2

207 + 2 [16 + 1] = 241

CuSO45H2O

64 + 32 + 4 × 16 + 5 [2 × 1 + 16] = 250

Bahan

Copper(II) sulphate / Kuprum(II) sulfat Ammonium carbonate / Ammonium karbonat

Hydrated copper(II) sulphate / Kuprum(II) sulfat terhidrat

Formula kimia

Jisim formula relatif

[Relative atomic mass / Jisim atom relatif : Na = 23, Cl = 35.5, K = 39, O = 16, Cu = 64, S = 32, N = 14, H = 1, C = 12, Al = 27, Ca = 40, Pb = 207]

(i) The formula of metal oxide of M is M2O3. Its relative formula mass is 152. What is the relative atomic mass of metal M? Oksida logam M mempunyai formula M2O3. Jisim formula relatif ialah 152. Apakah jisim atom relatif logam M?

M = RAM for M 2M + 3 × 16 = 152 M = 52

m

Fosforus membentuk sebatian klorida dengan formula PClx. Jisim molekul relatifnya adalah 208.5. Hitungkan nilai x. [Relative atomic mass / Jisim atom relatif : P = 31, Cl = 35.5]

1 + x × 35.5 = 208.5 3 35.5x = 208.5 – 31 35.5x = 177.5 x = 5

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(ii) Phosphorus forms a chloride with a formula PClx. Its relative molecular mass is 208.5. Calculate the value of x.

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MOLE CONCEPT / KONSEP MOL Mole and the Number of Particles / Bilangan Mol dan Bilangan Zarah 1 2

To describe the amount of atoms, ions or molecules, mole is used. Untuk menyatakan jumlah atom, ion atau molekul, unit mol digunakan. A mole is an amount of substance that contains as many particles as the number of atoms in exactly 12 g of carbon-12. Satu mol ialah jumlah bahan yang mengandungi bilangan zarah seperti mana yang terdapat dalam 12 g atom karbon-12.

3

A mole of a substance is the amount of substance which contains a constant number of particles (atoms, ions, molecules), which is 6.02 × 1023. Satu mol bahan adalah jumlah bahan yang mengandungi bilangan zarah yang tetap (atom, molekul, ion) iaitu 6.02 × 1023.

4 5

The number 6.02 × 1023 is called the Avogadro Constant or Avogadro Number (NA). Nombor 6.02 × 1023 dikenali sebagai Pemalar Avogadro atau Nombor Avogadro (NA ). For compounds that exist as molecules/ions, the number of atoms/ions in that compound must be known. Bagi sebatian yang wujud dalam bentuk molekul/ion, bilangan atom/ion dalam sebatian itu mestilah diketahui.

6 7

The symbol of mole is mol. Simbol untuk mol ialah mol. Complete the following table: Lengkapkan jadual berikut: Substance

Formula

Bahan

Formula

Type of particles

Model / Figure

Number of atom per molecule/ Number of positive and negative ion

Model / Rajah

Jenis zarah

Bilangan atom per molekul/ Bilangan ion positif dan negatif

Chlorine / Klorin

Cl2

Molecule

Cl Cl

Water / Air

H2O

Molecule

H O H

Ammonia / Ammonia

NH3

Molecule

H H N H

Sulphur dioxide / Sulfur dioksida

SO2

Molecule

O S O

MgCl2

Ion

[Cl]– [Mg]2+ [Cl]–

Magnesium chloride / Magnesium klorida

Cl : 2 H : 2 O : 1

N : 1 H : 3

S : 1 O : 2

Mg2+ : 1 Cl– : 2

Al : 2 3+

Aluminium oxide / Aluminium oksida 8

Al2O3

Ion

[O]2– [A1]3+ [O]2– [A1]3+ [O]2–

O2– : 3

Relationship between number of moles and number of particles (atoms/ions/molecules): Hubungan bilangan mol dan bilangan zarah (atom/ion/molekul):

Number of moles Bilangan mol 9

× Avogadro Constant / Pemalar Avogadro ÷ Avogadro Constant / Pemalar Avogadro

Number of particles Bilangan zarah

Complete the following: [Differentiate between “mole” dan “molecule”] Lengkapkan yang berikut: [Bezakan antara “mol” dan “molekul”]

(a) 1 mol of Cl2 [Chlorine gas] 1 mol Cl2 [Gas klorin]

(b) 1 mol of NH3 [Ammonia gas]

molecules of chlorine, Cl2 / molekul klorin, Cl2

2 × 6.02 × 1023 atoms of chlorine, Cl / atom klorin, Cl 6.02 × 1023 4

molecules of ammonia, NH3 / molekul ammonia, NH3 1 mol of nitrogen atom, N / mol atom nitrogen, N mol atoms / mol atom 3 mol of hydrogen atoms, H / mol atom hidrogen, H n io

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1 mol NH3 [Gas ammonia]

6.02 × 1023

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0.25 × 6.02 × 1023 molecules of ammonia, NH / molekul ammonia, NH 3 3 0.25 mol of N atoms / mol atom N, 23 number of N atoms / bilangan atom N = 0.25 × 6.02 × 10 1 mol of atoms

1 mol of NH3 4 [Ammonia gas]

(c)

1 mol NH3 4 [Gas ammonia]

1

mol atom

0.75 mol of H atoms / mol atom H, 23 number of H atoms / bilangan atom H = 0.75 × 6.02 × 10

2 mol of Mg2+ ions / mol ion Mg2+, 23 number of Mg2+ ions / bilangan ion Mg2+ = 2 × 6.02 × 10

(d) 2 mol of MgCl2 [Magnesium chloride] 2 mol MgCl2 [Magnesium klorida]

4 mol of Cl– ions / mol ion Cl–, 23 number of Cl- ions / bilangan ion Cl– = 4 × 6.02 × 10 2 × 6.02 × 1023 molecules of SO / molekul SO 2 2 2 mol of S atoms / mol atom S, 23 number of S atoms / bilangan atom S = 2 × 6.02 × 10 3 × 2 = 6 mol of atoms

(e) 2 mol of SO2 [Sulphur dioxide] 2 mol SO2 [Sulfur dioksida]

3 × 2 = 6 mol atom

4 mol of O atoms / mol atom O, 23 number of O atoms / bilangan atom O = 4 × 6.02 × 10

10 Complete the table below: Lengkapkan jadual berikut: Number of moles

Number of particles

Bilangan mol

0.5 0.5

Bilangan zarah

mole of carbon, C

3.01 × 1023 atoms of carbon

mol atom karbon, C

3.01 × 1023 atom karbon

0.2 moles of hydrogen gas, H2

(i)

0.2 mol gas hidrogen, H2

1 1

(ii)

0.2 × 6.02 × 1023

molecules of hydrogen / molekul hidrogen

2 × 0.2 × 6.02 × 1023 atoms of hydrogen / atom hidrogen

6.02 × 1023 molecules of carbon dioxide contains:

mole of carbon dioxide molecules, CO2

6.02 × 1023 molekul karbon dioksida mengandungi:

mol molekul karbon dioksida, CO2

6.02 × 1023 6.02 × 10

23

atoms of C and atom C dan

2 × 6.02 × 1023

2 × 6.02 × 1023

atoms of O.

atom O.

NUMBER OF MOLES AND MASS OF SUBSTANCE / BILANGAN MOL DAN JISIM BAHAN Molar mass / Jisim molar (a) Molar mass is the mass of one mole of any substance / Jisim molar adalah jisim satu mol sebarang bahan. (b) Molar Mass is the relative atomic mass, relative molecular mass and relative formula mass of a substance in g mol–1.

1

Jisim molar adalah jisim atom relatif, jisim molekul relatif dan jisim formula relatif suatu bahan dalam g mol–1. (c) Molar mass of any substance is numerically equal to its relative mass (Relative atomic mass/ relative formula

mass/relative molecular mass). Jisim molar sebarang bahan mempunyai nilai yang sama dengan jisim relatif (Jisim atom relatif/ jisim formula relatif/ jisim molekul relatif).

Example / Contoh: Molar mass of H2O = 18 g mol–1

2

Jisim molar H2O = 18 g mol–1

× RAM/ /RFM/RMM

Mass of 1 mol of H2O = 18 g

Jisim 1 mol H2O = 18 g

Mass of 2 mol of H2O = 2 mol × 18 g mol = 36 g

Jisim 2 mol H2O = 2 mol ×

–1

18

g mol–1 =

Mass of 2.5 mol of H2O = 45 g

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÷ RAM/ /RFM/RMM

Mass in gram Jisim dalam gram

÷ JAR/JFR/JMR

mol H2O = 45 g

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Number of moles

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Complete the following table: Lengkapkan jadual berikut: Element/ Compound

Unsur/Sebatian

Copper

Chemical formula

RAM/RMM/RFM

Formula kimia

Cu

RAM/JAR = 64

NaOH

RFM/JFR = 40

Kuprum

Sodium hydroxide

Calculate

JAR/JMR/JFR

Natrium hidroksida

Penghitungan –1 (a) Mass of 1 mol of Cu / Jisim 1 mol Cu : 1 mol × 64 g mol = 64 g 2 mol × 64 g mol–1 = 128 g (b) Jisim 2 mol / Jisim 1 mol : 1 mol × 64 g mol–1 = 32 g 1 1 2 (c) Jisim mol / Jisim mol: 2 2 32 g (d) Mass of 3.01 × 1023 Cu atoms / Jisim 3.01 × 1023 atom Cu:

(a) Mass of 3 mol of sodium hydroxide: Jisim 3 mol natrium hidroksida:

120 g

120 g

(b) Number of moles in 20 g sodium hydroxide: 0.5 mol Bilangan mol natrium hidroksida dalam 20 g:

Oxygen gas Gas oksigen

O2

RMM/JMR = 32

(a) Mass of 2.5 mol of oxygen gas: Jisim 2.5 mol gas oksigen:

0.5 mol

2.5 mol × 32 g mol–1 = 80 g

2.5 mol × 32 g mol–1 = 80 g

(b) Number of moles is 1.5 mol oxygen gas: Bilangan molekul dalam 1.5 mol gas oksigen:

1.5 mol × 6.02 × 1023 1 (c) Number of molecules in mol of oxygen gas: 2 1 Bilangan molekul dalam mol gas oksigen: 2 0.5 mol × 6.02 × 1023 (d) Number of atoms in 2 mol of oxygen gas:

Bilangan atom dalam 2 mol gas oksigen:

Sodium chloride

NaCl

RFM/JFR = 58.5

Zn(NO3)2

RFM/JFR = 189

Natrium klorida

Zinc nitrate Zink nitrat

2 × 2 × 6.02 × 1023

Mass of 0.5 mol of NaCl / Jisim bagi 0.5 mol NaCl: 0.5 mol × 58.5 g mol–1 = 29.25 g Number of moles in 37.8 g of zinc nitrate: Bilangan mol dalam 37.8 g zink nitrat:

37.8 g/189 g mol–1 = 0.2 mol [Relative atomic mass / Jisim atom relatif: Cu = 64, Na = 23, O = 16, H = 1, Cl = 35.5, Zn = 65, N = 14]

NUMBER OF MOLES AND VOLUME OF GAS / BILANGAN MOL DAN ISI PADU GAS 1

Molar volume of a gas: Volume occupied by one mole of any gas is 24 dm3 at room conditions and 22.4 dm3 at

standard temperature and pressure (STP). Isi padu molar gas: Isipadu yang dipenuhi oleh satu mol sebarang gas iaitu 24 dm3 pada keadaan bilik dan 22.4 dm3 pada suhu dan tekanan piawai (STP). 2

The molar volume of any gas is 24 dm3 at room conditions and 22.4 dm3 at STP. Isi padu molar sebarang gas adalah 24 dm3 pada keadaan bilik dan 22.4 dm3 pada STP.

3

Generalisation: One mole of any gas always occupies the same volume under the same temperature and pressure: Umumnya: satu mol sebarang jenis gas menempati isi padu yang sama pada suhu dan tekanan yang sama.

Example / Contoh: (i) 1 mol of oxygen gas, 1 mol ammonia gas, 1 mol helium gas dan 1 mol sulphur dioxide gas occupy the same volume of 24 dm3 at room conditions. 1 mol gas oksigen, 1 mol gas ammonia, 1 mol gas helium dan 1 mol gas sulfur dioksida menempati isi padu yang sama iaitu 24 dm3 pada keadaan bilik.

44.8

(ii) 2 mol of carbon dioxide gas occupies 44.8

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2 mol gas karbon dioksida menempati

dm3 pada STP.

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(iii) 16 g of oxygen gas = 0.5 mol of oxygen gas. Therefore, 16 g of oxygen gas occupies a volume of at room conditions [Relative atomic mass: O =16] 0.5 mol gas oksigen. Oleh itu, 16 g gas oksigen menempati isi padu 16 g gas oksigen = [Jisim atom relatif; O = 16]

Number of moles of gas Bilangan mol gas

× 24 dm3 mol–1/ 22.4 dm3 mol–1

12

12

dm3

dm3 pada keadaan bilik.

Volume of gas in dm2 Isi padu gas dalam dm3

÷ 24 dm3 mol–1/ 22.4 dm3 mol–1

Formula for conversion of unit: Formula untuk penukaran unit:

Volume of gas in dm3 Isi padu gas dalam dm3

÷ 24 dm3 mol–1/ 22.4 dm3 mol–1 ÷ (RAM/ /RFM/RMM) g mol–1 Mass in gram (g) Jisim dalam gram (g)

× 24 dm3 mol–1/ 22.4 dm3 mol–1

÷ (JAR/JFR/JMR) g mol–1

Number of moles

÷ (6.02 × 1023)

× (RAM/ /RFM/RMM) g mol–1

Bilangan mol

× (6.02 × 1023)

× (JAR/JFR/JMR) g mol–1

Number of particles Bilangan zarah

EXERCISE / LATIHAN

Relative atomic mass of calcium is 40 based on the carbon-12 scale.

1

Jisim atom relatif kalsium berdasarkan skala karbon-12 ialah 40.

(a) State the meaning of the statement above. Nyatakan maksud penyataan di atas.

Mass of calcium atom is 4 times greater than

1 mass of carbon-12 atom. 12

(b) How many times is one calcium atom heavier than one oxygen atom? [Relative atomic mass: O = 16] Berapa kalikah satu atom kalsium lebih berat daripada satu atom oksigen? [JAR: O = 16]

Relative atomic mass of calcium 40 = = 2.5 times Relative atomic mass of oxygen 16

(c) How many calcium atoms have the same mass as two atoms of bromine? [RAM Br = 80] Berapakah bilangan atom kalsium yang mempunyai jisim yang sama dengan dua atom bromin? [Jisim atom relatif: Br = 80]

Number of calcium atom × 40 = 2 × 80 2 × 80 Number of calcium atom = = 4 40 A sampel of chlorine gas weighs 14.2 g. Calculate / Suatu sampel gas klorin berjisim 14.2 g. Hitungkan: [Relative atomic mass / Jisim atom relatif : Cl = 35.5] (a) Number of moles of chlorine atoms / Bilangan mol atom klorin. 14.2 Number of mol of chlorine atoms, Cl = = 0.4 mol 35.5

2

(b) Number of moles of chlorine molecules (Cl2) / Bilangan mol molekul klorin (Cl2 ). 14.2 Number of mol of chlorine molecule, Cl2 = = 0.2 mol 71 (c) Volume of chlorine gas at room conditions / Isi padu gas klorin pada keadaan bilik. [Molar volume of gas = 24 dm3 mol–1 at room temperature and pressure] [Isi padu molar gas = 24 dm3 mol–1 pada suhu dan tekanan piawai]

m

Volume of chlorine gas = 0.2 mol × 24 dm3 mol–1 = 4.8 dm3

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(a) Calculate the number of atoms in the following substances / Hitungkan bilangan atom yang terdapat dalam bahan berikut: [Relative atomic mass: N = 14; Zn = 65; Avogadro Constant = 6.02 × 1023]

[Jisim atom relatif: N = 14; Zn = 65; Pemalar Avogadro = 6.02 × 1023] (i) 13 g of zinc / 13 g zink

(ii) 5.6 g of nitrogen gas / 5.6 g gas nitrogen 5.6 Number of mol of N atom = = 0.4 mol 14 Number of N atom = 0.4 × 6.02 × 1023 = 2.408 × 1023

13 = 0.2 mol 65 Number of zinc atom = 0.2 × 6.02 × 1023 = 1.204 × 1023 Number of mol of zinc atom =

(b) Calculate the number of molecules in the following substances / Hitungkan bilangan molekul dalam bahan berikut: [Relative atomic mass: N = 14, H = 1, Cl = 35.5, Avogadro Constant = 6.02 × 1023]

[Jisim atom relatif: N = 14, H = 1, Cl = 35.5, Pemalar Avogadro = 6.02 × 1023] (i) 8.5 g of ammonia gas, NH3 / 8.5 g gas ammonia, NH3

(ii) 14.2 g of chlorine gas, Cl2 / 14.2 g gas klorin, Cl2 14.2 × 6.02 × 1023 71 = 1.2 × 1023

4

A gas jar contains 240 cm3 of carbon dioxide gas. Calculate:

8.5 × 6.02 × 1023 17 = 2.408 × 1023

Suatu balang gas berisi 240 cm3 gas karbon dioksida. Hitungkan:

[Relative atomic mass: C = 12, O = 16; Molar volume of gas = 24 dm3 mol–1 at room conditions] [Jisim atom relatif: C = 12, O = 16; Isi padu molar gas: 24 dm3 mol–1 pada keadaan bilik] (a) Number of moles of carbon dioxide gas / Bilangan mol gas karbon dioksida:

Number of moles of CO2 =

240 = 0.01 mol 24 000

(b) Number of molecules of carbon dioxide gas / Bilangan molekul gas karbon dioksida: Number of molecules of CO2 = 0.01 × 6.02 × 1023 = 6.02 × 1021 (c) Mass of carbon dioxide gas / Jisim gas karbon dioksida: Mass of CO2 = 0.01 mol × [12 + 2 × 16] g mol–1 = 0.44 g

5

What is the mass of chlorine molecules (Cl2) that contains twice as many molecules as that found in 3.6 g of water?

Berapakah jisim molekul klorin (Cl2 ) yang mengandungi dua kali ganda bilangan molekul yang terdapat dalam 3.6 g air? [Relative atomic mass / Jisim atom relatif : H = 1, O = 16, Cl = 35.5]

Number of moles of chlorine molecule = 2 × no of mol in H2O 3.6 = 2 × = 0.4 mol 18 Mass of Cl2 = 0.4 × 71= 28.4 g 6

Calculate the mass of carbon that has the same number of atoms as found in 4 g of magnesium. Hitungkan jisim karbon yang mempunyai bilangan atom yang sama seperti yang terdapat dalam 4 g magnesium. [Relative atomic mass / Jisim atom relatif : C = 12, Mg = 24]

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2 g

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Compare the number of molecule in 32 g of sulphur dioxide (SO2) with 7 g of nitrogen gas (N2). Explain your answer.

7

Bandingkan bilangan molekul dalam 32 g sulfur dioksida (SO2 ) dengan 7 g gas nitrogen (N2 ). Terangkan jawapan anda. [Relative atomic mass / Jisim atom relatif : S = 32, O = 16, N = 14]

Number of moles of molecules in 32 g SO2 =

32 = 0.5 mol 64

7 = 0.25 mol 28 Number of molecule in 32 g SO2 is two times more than 7 g N2. Number of mole in sulphur dioxide molecule is two times more than number of mole of nitrogen molecule. Number of moles of molecules in 7 g N2 =

Compare number of atoms in 1.28 g of oxygen to the number of atoms in 1.3 g of zinc. Explain your answer.

8

Bandingkan bilangan atom dalam 1.28 g oksigen dengan bilangan atom dalam 1.3 g zink. Terangkan jawapan anda. [Relative atomic mass / Jisim atom relatif : O = 16, Zn = 65]

1.28 = 0.08 mol 16 1.30 Number of mol of Zn atoms in 1.3 g Zn = = 0.04 mol 65 Number of oxygen atoms in 1.28 g oxygen is 2 times more than number of zinc atoms in 1.3 g zinc. Number of mol of oxygen atom is 2 times more than zinc atom.

Number of mol of O atoms in 1.28 g SO2 =

CHEMICAL FORMULAE AND CHEMICAL EQUATIONS / FORMULA KIMIA DAN PERSAMAAN KIMIA

Symbol of elements – use capital letters for the first alphabet and use small letters if there is a second alphabet.

1

Simbol unsur – gunakan huruf besar untuk huruf pertama dan huruf kecil jika ada huruf kedua.

Example / Contoh: Potassium / Kalium – K, Calcium / Kalsium – Ca, Iron / Ferum – Fe,

Sodium / Natrium – Na Nitrogen / Nitrogen – N Fluorine / Fluorin – F

Chemical Formula – A set of chemical symbols for atoms of elements in whole numbers representing chemical substances. Formula kimia – Satu set simbol kimia bagi atom-atom unsur dengan gandaan nombor bulat yang mewakili bahan kimia. Chemical substance Bahan kimia

Water Air

Ammonia Ammonia

Propane

m

Notes

Formula kimia

H2O NH3 C3H8

Catatan

2 atoms of H combines with 1 atom of O. 2 atom H bergabung dengan 1 atom O.

3 atoms of H combines with 1 atom of N. 3 atom H bergabung dengan 1 atom N.

3 atoms of C combines with 8 atoms of H. 3 atom C bergabung dengan 8 atom H.

2

Information that can be obtained from the chemical formula / Maklumat yang diperoleh daripada formula kimia: (i) All the elements present in the compound / Jenis unsur yang terdapat dalam sebatian, (ii) Number of atoms of each element in the compound / Bilangan atom setiap unsur yang terdapat dalam sebatian, (iii) Calculation of RMM/RFM of the compound / Pengiraan JMR/JFR bagi sebatian.

3

Two types of chemical formula / Dua jenis formula kimia: (i) Empirical formula / Formula empirik, (ii) Molecular formula / Formula molekul.

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EMPIRICAL FORMULA / FORMULA EMPIRIK 1 2

A formula that shows the simplest whole number ratio of atoms of each element in a compound. Formula yang menunjukkan nisbah nombor bulat teringkas bagi bilangan atom setiap unsur yang terdapat dalam sebatian. The formula can be determined by calculating the simplest ratio of moles of atoms of each element in the compound. Formula itu boleh ditentukan dengan menghitung nisbah bilangan mol atom bagi setiap unsur yang terdapat dalam sebatian.

3

Experiments to determine empirical formula of metal oxide / Formula empirik bagi oksida logam diperoleh dengan cara: Empirical formula of magnesium oxide

Empirical formula of copper(II) oxide

Formula empirik magnesium oksida

Set-up of apparatus / Susunan radas:

Formula empirik kuprum(II) oksida

Set-up of apparatus / Susunan radas: Copper(II) oxide Kuprum(II) oksida

Magnesium Magnesium

Hydrogen gas Gas hidrogen

Heat

Heat

Panaskan

Panaskan

Reaction occurs / Tindak balas yang berlaku:

Reaction occurs / Tindak balas yang berlaku:

Magnesium dipanaskan dengan kuat di dalam mangkuk pijar untuk bertindak balas dengan oksigen membentuk magnesium oksida.

Gas hidrogen dilalukan melalui kuprum(II) oksida yang dipanaskan. Hidrogen menurunkan kuprum(II) oksida kepada kuprum dan air.

Balanced equation / Persamaan kimia seimbang:

Balanced equation / Persamaan kimia seimbang:

This method can also be used to determine the empirical formulae of reactive metals such as aluminium oxide and zinc oxide.

This method can also be used to determine the empirical formulae of less reactive metals such as lead(II) oxide and tin(II) oxide.

Kaedah ini boleh juga digunakan untuk menentukan formula empirik oksida logam reaktif seperti aluminium oksida dan zink oksida.

Kaedah ini boleh juga digunakan untuk menentukan formula empirik oksida logam kurang reaktif seperti plumbum(II) oksida and stanum(II) oksida.

Magnesium is burnt in a crucble to react with oxygen to form magnesium oxide.

2Mg + O2 → 2MgO

4

Hydrogen gas is passed through heated copper(II) oxide. Hydrogen reduces copper(II) oxide to form copper and water.

CuO + H2 → Cu + H2O

Experiment to Determine Empirical Formula of Magnesium Oxide Eksperimen untuk Menentukan Formula Empirik Magnesium Oksida

In this experiment, magnesium reacts with oxygen in the air to form white fumes, magnesium oxide: Semasa eksperimen ini, magnesium bertindak balas dengan oksigen dalam udara untuk membentuk asap putih, magnesium oksida:

Magnesium + Oxygen → Magnesium oxide Magnesium + Oksigen → Magnesium oksida

Material / Bahan: Magnesium ribbon, sand paper

Apparatus / Radas: Crucible with lid, tongs, Bunsen burner, tripod stand and balance


Magnesium ribbon

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Heat

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Procedure / Langkah:

crucible

(a) A

Mangkuk pijar

penutup

dengan

pita magnesium

Pita magnesium

coiled

gulung

di

crucible

(d) The

ditimbang.

sand paper

..

loosely and placed in the crucible.

dan diletakkan dalam mangkuk pijar.

magnesium ribbon are weighed again.

together with the lid and

Mangkuk pijar

.

kertas pasir

dibersihkan dengan menggunakan

magnesium ribbon is

(c) The

are weighed.

magnesium ribbon is cleaned with

(b) 10 cm of 10 cm

lid

and its

pita magnesium

bersama dengan penutup dan

ditimbang.

(e) The apparatus is set up as shown in the diagram. Radas disusun seperti dalam gambar rajah.

strongly

(f) The crucible is heated burn

Mangkuk pijar dipanaskan dengan terbakar , mangkuk pijar ditutup dengan

lid

pita magnesium Apabila kuat selama 2 minit lagi.

crucible

, lid and its content are heating

penutup

is removed and the crucible is

dibuka dan mangkuk pijar dipanaskan dengan

weighed again

.

.

ditimbang sekali lagi

cooling

,

suhu bilik

.

repeated

and weighing are

until a

mass is obtained.

pemanasan tetap

penyejukan

,

dan penimbangan

diulang

beberapa kali sehingga jisim

diperoleh.

Observation / Pemerhatian:

Magnesium burns

brightly

Magnesium terbakar dengan

white fumes

to release

terang

membebaskan

and

wasap putih

white solid

is formed. pepejal putih

dan kemudiannya membentuk

.

Inference / Inferens:

Magnesium is a

reactive

metal. reaktif

Magnesium adalah logam yang

Magnesium reacts with

oxygen

Magnesium bertindak balas dengan

.

in the air to form oksigen

magnesium oxide

dalam udara membentuk

.

magnesium oksida

.

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,

, penutup dan kandungannya

constant

lid

lid and its content are allowed to cool down to room temperature .

(k) The process of

m

.

, the

, penutup dan kandungannya dibiarkan sejuk ke

Mangkuk pijar

burning terbakar

berhenti

crucible

Mangkuk pijar

Proses

. Apabila pita magnesium mula

dibuka sekali sekala dengan menggunakan penyepit.

(h) When the magnesium ribbon stops heated strongly for another 2 minutes.

(j) The

tanpa penutup

.

penutup

of the crucible is lifted from time to time using a pair of tongs.

Penutup

(i) The

. When the magnesium starts to

lid

, the crucible is covered with its kuat

(g) The

lid

without its

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Precaution steps / Langkah berjaga-jaga: Step taken / Langkah yang diambil

Purpose / Tujuan

Magnesium ribbon is cleaned with sand paper . Pita magnesium perlu digosok dengan

kertas pasir

To remove the oxide layer on the surface of the magnesium ribbon.

.

Untuk membuang lapisan oksida pada permukaan magnesium oksida.

The crucible lid is lifted from time to time.

To allow

replaced

from the air to react with magnesium .

Untuk membenarkan oksigen masuk dan bertindak balas dengan magnesium .

Penutup mangkuk pijar dibuka sekali sekala.

The crucible lid then

oxygen

To prevent fumes of magnesium oxide from escaping.

quickly.

Untuk mengelakkan wasap magnesium oksida dari terbebas.

Penutup mangkuk pijar kemudian ditutup semula dengan cepat.

The process of heating , cooling and weighing are To ensure magnesium react completely with repeated until a constant mass is obtained. for magnesium oxide .

Result / Keputusan: Description / Penerangan

Mass (g) / Jisim (g)

Mass of crucible + lid

x

Jisim mangkuk pijar + penutup

Mass of crucible + lid + magnesium

y

Jisim mangkuk pijar + penutup + magnesium

Mass of crucible + lid + magnesium oxide

z

Jisim mangkuk pijar + penutup + magnesium oksida

Calculation / Pengiraan: Element / Unsur

Mg

O


y–x

z–y

Number of mole of atoms / Bilangan mol atom

y–x 24

z–y 16

Simplest ratio of moles / Nisbah mol teringkas

p

q

MgpOq

Empirical formula of magnesium oxide is Mg O

Formula empirik magnesium oksida ialah 5

to

lengkap Untuk memastikan semua magnesium telah bertindak balas oksigen dengan untuk membentuk magnesium oksida .

pemanasan , penyejukan penimbang dan Proses jisim tetap diulang beberapa kali sehingga diperoleh.

oxygen

p

q

. .

Experiment to Determine Empirical Formula of Copper(II) Oxide Eksperimen untuk Menentukan Formula Empirik Kuprum(II) Oksida

Copper(II) Oxide + Hidrogen → Copper + Water Kuprum(II) oksida + Hidrogen → Kuprum + Air


Copper(II) oxide Burning of hydrogen gas Hydrogen gas Combustion tube Heat

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Anhydrous calcium chloride, CaCl2

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Observation / Pemerhatian:

The

black

Warna

colour of copper(II) oxide turns

hitam

brown

perang

kuprum(II) oksida menjadi

.

.

Inference / Inferens:

copper metal

Copper(II) oxide reacts with hydrogen to produce the brown Kuprum(II) oksida bertindak balas dengan hidrogen untuk menghasilkan

.

logam kuprum

yang berwarna perang.

Precaution steps / Langkah berjaga-jaga: Step taken / Langkah yang ambil

Purpose / Tujuan

Hydrogen gas is passed through anhydrous calcium chloride.

hydrogen gas.

Gas hidrogen dialirkan melalui kalsium klorida kontang.

mengering

Kalsium klorida kontang menyerap wap air untuk

Dry hydrogen is passed through the combustion tube for 5 to 10 minutes.

To remove all the

air

gas hidrogen.

in the combustion tube. air explodes when lighted).

(The mixture of hydrogen gas and

kering dialirkan melalui tabung pembakaran Gas hidrogen selama 5 hingga 10 minit.

udara dalam tabung pembakaran. Untuk mengeluarkan semua udara (Campuran hidrogen dan menghasilkan letupan apabila dinyalakan)

If the gas burns quietly without ‘pop’ sound , all the has been removed from the combustion tube.

The gas that comes out from the small hole is collected in the test tube. Then, a lighted wooden splinter is placed

dry

Anhydrous calcium chloride absorb water vapour to

bunyi ‘pop’ Jika gas terbakar tanpa daripada tabung pembakaran.

at mouth of the test tube.

Gas yang keluar daripada lubang kecil dikumpul dalam sebuah tabung uji. Kayu uji menyala di letakkan di mulut tabung uji.

The flow of hydrogen gas must be continuous throughout the experiment.

, semua gas telah

air

dikeluarkan

To prevent hot copper from reacting with oxygen to form copper(II) oxide again.

Gas hidrogen dialirkan secara berterusan sepanjang eksperimen.

Untuk mengelakkan kuprum panas daripada bertindak balas dengan oksigen dan membentuk kuprum(II) oksida .

The process of heating , cooling and weighing are To ensure all copper(II) oxide has changed to copper . Untuk memastikan semua kuprum(II) oksida telah bertukar kepada kuprum . repeated until a constant mass is obtained. pemanasan , penyejukan Proses diulang beberapa kali sehingga jisim

dan tetap

penimbang diperoleh.

Result / Keputusan: Description / Penerangan


Mass of combustion tube + porcelain dish

x

Jisim tabung pembakaran + piring tanah liat

Mass of combustion tube + porcelain dish + copper(II) oxide

y

Jisim tabung pembakaran + piring tanah liat + kuprum(II) oksida

Mass of combustion tube + porcelain dish + copper

z

Jisim tabung pembakaran + piring tanah liat + kuprum

Calculation / Pengiraan: Element / Unsur

Cu

O


z–x

y–z


z–x 64

y–z 16


p

q

Empirical formula of copper(II) oxide is m

CupOq

. .

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6

Explain why the set-up of apparatus to determine the empirical formula in both the experiments is different. Terangkan mengapa susunan radas untuk menentukan formula empirik dalam kedua-dua eksperimen itu berbeza.

reactive

(a) Magnesium is

magnesium oxide

metal (above hidrogen in reactivity series). Magnesium

reacts

easily to form

.

reaktif Magnesium adalah logam membentuk magnesium oksida .

teroksida

(terletak di atas hidrogen dalam siri kereaktifan. Magnesium mudah

(b) Copper is below hydrogen in the metal reactivity series. Oxygen in copper(II) oxide can be reduced/removed by

hydrogen gas

Kuprum di bawah gas hidrogen 7

to form copper and water. hidrogen

dalam siri kereaktifan. Kuprum(II) okida boleh

diturunkan/disingkirkan

oleh

untuk membentuk kuprum dan air.

To calculate the empirical formula of a compound, use the following table: Untuk menghitung formula empirik suatu sebatian, jadual di bawah boleh digunakan sebagai panduan:

Calculation steps / Langkah pengiraan:

Element / Unsur

(a) Calculate the mass of each element in the compound. Hitungkan jisim setiap unsur dalam sebatian.

Mass of element (g) / Jisim unsur (g)

(b) Convert the mass of each element to number of mole of atom.

Number of mole of atom / Bilangan mol atom

Tukar jisim setiap unsur kepada bilangan mol atom.

(c) Calculate the simplest ratio of moles of atom of the elements.


Hitungkan nisbah bilangan mol atom teringkas unsur-unsur tersebut.


When 11.95 g of metal X oxide is reduced by hydrogen, 10.35 g of metal X is produced. Calculate the empirical formula of metal X oxide. Apabila 11.95 g oksida logam X diturunkan oleh hidrogen, 10.35 g logam terhasil. Hitungkan formula empirik bagi oksida logam X. [RAM / JAR: X = 207, O = 16] X

O


10.35

1.6


0.05

0.1

Ratio of moles / Nisbah mol

1

2


1

2

Element / Unsur

Empirical formula / Formula empirik: 2

XO2

.

A certain compound contains the following composition / Satu sebatian mengandungi komposisi unsur seperti berikut: Na = 15.23%, Br = 52.98%, O = 31.79 % [Relative atomic mass / Jisim atom relatif: O = 16, Na = 23, Br = 80] (Assume that 100 g of substance is used / Anggap 100 g bahan digunakan) Element / Unsur

Na

Br

O


15.23

52.98

31.79


0.66

0.66

1.99

Ratio of moles / Nisbah mol

1

1

3.01


1

1

3

NaBrO3

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Empirical formula / Formula empirik:

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2.08 g of element X combines with 4.26 g of element Y to form a compound with formula XY3. Calculate the relative atomic mass of element X. [RAM: Y = 35.5]

3

2.08 g unsur X bergabung dengan 4.26 g unsur Y untuk membentuk sebatian dengan formula XY3 . Hitung jisim atom relatif unsur X. [JAR: Y = 35.5] Element / Unsur Mass of element (g) Jisim unsur (g)

Number of mole of atoms Bilangan mol atom

Simplest ratio of moles Nisbah mol teringkas

X

Y

2.08

4.26

2.08 x

4.26 = 0.12 35.5

1

3

x = relative atomic mass of X Mol X = 1

Mol Y 3 2.08 x 1 = 0.12 3 x = 52

2.07 g of element Z reacts with bromine to form 3.67 g of a compound with empirical formula ZBr2. Calculate the relative atomic mass of element Z. [RAM: Br = 80]

4

2.07 g unsur Z bertindak balas dengan bromin membentuk 3.67 g sebatian dengan formula empirik ZBr2. Hitung jisim atom relatif bagi unsur Z. [JAR: Br = 80] Element / Unsur Mass of element (g) Jisim unsur (g)

Number of mole of atoms Bilangan mol atom

Simplest ratio of moles Nisbah mol teringkas

Z

Br

z = relative atomic mass of Z

2.07

1.6

Mol Z

2.07 z

1.6 = 0.02 80

1

2

= 1 Mol Br 2 2.08 z 1 = 2 0.02 z = 207

The statement below is about compound J / Pernyataan berikut adalah mengenai sebatian J.

5

• It is black solid / Merupakan pepejal hitam. • Contains 12.8 g copper and 0.2 mol of oxygen / Mengandungi 12.8 g kuprum dan 0.2 mol oksigen. [Relative atomic mass / Jisim atom relatif : Cu = 64] (a) What is meant by empirical formula / Apakah maksud formula empirik? A formula that shows the simplest whole number ratio of atoms of each element in a compound. (b) (i)

(ii)

Calculate the number of mol of copper atom / Hitung bilangan mol atom kuprum. 12.8 = 0.2 mol 64 What is the empirical formula of compound J / Apakah formula empirik sebatian J ? 0.2 mol Cu : 0.2 mol O. 1 mol Cu : 1 mol O. Empirical formula of Compound J is CuO.

(c) Compound J reacts completely with hydrogen to produce copper and compound Q. Sebatian J bertindak balas lengkap dengan hidrogen menghasilkan kuprum dan sebatian Q.

(i)

State one observation for the reaction / Nyatakan satu pemerhatian daripada tindak balas tersebut. Black solid change to brown

(ii) Name two the substances that can be used to prepare hydrogen gas. Namakan dua bahan yang digunakan untuk menyediakan gas hidrogen.

Zinc/magnesium and hydrochloric acid/nitric acid/sulphuric acid.

(iii) Name compound Q / Nama sebatian Q. Water

(iv) Write a balanced equation for the reaction. Tuliskan persamaan kimia yang seimbang bagi tindak balas tersebut.

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(d) Draw a labelled diagram of the set-up of apparatus for the experiment. Lukiskan gambar rajah berlabel susunan radas bagi tindak balas tersebut.

Gas hidrogen

Compound J

Heat (e) (i)

Why is hydrogen gas passed through the combustion tube after heating has stpopped? Mengapakah gas hidrogen dilalukan melalui tiub pembakaran selepas pemanasan dihentikan?

To avoid copper produced react with oxygen to form copper(II) oxide.

(ii) State how to determine that the reaction between compound J and hydrogen has completed. Nyatakan bagaimana menentukan tindak balas antara sebatian J dengan hidrogen telah lengkap.

By repeating the process of heating, cooling and weighing until constant mass is obtained. (f) (i)

Can the empirical formula of magnesium oxide be determined by the same method? Explain your answer. Bolehkah formula empirik bagi magnesium oksida ditentukan dengan cara yang sama? Jelaskan jawapan anda.

Cannot. Magnesium is more reactive than hydrogen. Hydrogen cannot reduce magnesium oxide to form magnesium.

(ii) Magnesium can reduce copper oxide to copper. Explain why the empirical formula of the copper oxide cannot be determined by heating the mixture of copper oxide and magnesium powder. Magnesium boleh menurunkan kuprum oksida kepada kuprum. Terangkan mengapa formula empirik kuprum oksida tidak boleh ditentukan dengan pemanasan campuran kuprum oksida dengan serbuk magnesium.

Magnesium oxide and copper produced are in solid form, copper cannot be separated from magnesium oxide. The mass of copper cannot be weighed.

MOLECULAR FORMULA / FORMULA MOLEKUL 1

Molecular formula of a compound shows the actual number of atoms of each element that are present in a molecule of the compound. Formula molekul suatu sebatian menunjukkan bilangan sebenar atom bagi setiap unsur yang terdapat dalam satu molekul sebatian.

Molecular Formula = (empirical formula)n, where n is a integer. Formula molekul = (Formula empirik)n, di mana n adalah integer. 2

Example / Contoh: Compound

Molecular formula

Empirical formula

Value of n

Water / Air

H2O

H2O

1

Carbon dioxide / Karbon dioksida

CO2

CO2

1

H2SO4

H2SO4

1

Ethene / Etena

C2H4

CH2

2

Benzene / Benzena

C6H6

CH

6

Glucose / Glukosa

C6H12O6

CH2O

6

Sebatian

Sulphuric acid / Asid sulfurik

Formula molekul

Formula empirik

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The molecular formula and the empirical formula of a compound may be the same if the value of n = 1 but different if the value is n > 1. Formula molekul dan formula empirik suatu sebatian akan sama sekiranya nilai n = 1 tetapi akan berbeza sekiranya nilai n > 1.

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EXERCISE / LATIHAN

The empirical formula of compound X is CH2 and relative molecular mass is 56. Determine the molecular formula of compound X. [Relative atomic mass: H = 1; C = 12]

1

Formula empirik sebatian X adalah CH2 dan JMR adalah 56. Tentukan formula molekul sebatian X. [Jisim atom relatif: H = 1; C = 12]

(12 + 2)n = 56 56 n = = 4 14 Molecular formula = (CH2)4 = C4H8 2.58 g of a hydrocarbon contains 2.16 g of carbon. The relative molecular mass of the hydrocarbon is 86.

2

2.58 g suatu hidrokarbon mengandungi 2.16 g karbon. Jisim molekul relatif bagi hidrokarbon ini ialah 86. [Relative atomic mass / Jisim atom relatif : H = 1; C = 12] (i) Calculate the empirical formula of the hydrocarbon / Hitungkan formula empirik bagi hidrokarbon ini. Element

C

H

Mass of element (g)

2.16

0.42

Number of mole of atoms

0.18

0.42

Ratio of moles

1

2 1 = 7 3 3

Simplest ratio of moles

3

7

Empirical formula = C3H7

(ii) Determine the molecular formula of the hydrocarbon / Tentukan formula molekul hidrokarbon tersebut. (12 × 3 + 7 × 1)n = 86 86 n = = 2 43 Molecular formula = (C3H7)2 = C6H14 The diagram below shows the structural formula for benzene molecule.

3

Rajah di bawah menujukkan formula struktur bagi benzena. H H

C

H

C

C C

C

H

C

H

H

(a) Name the element that make up benzene / Namakan unsur yang membentuk benzena. Carbon and hydrogen (b) What are the molecular formula and empirical formula for benzene? Apakah formula molekul dan formula empirik bagi benzena?

Molecular formula / Formula molekul: C6H6 Empirical formula / Formula empirik: CH

(c) Compare and contrast the molecular formula and empirical formula for benzene. Banding dan bezakan formula molekul dan formula empirik bagi benzena.

• Both empirical formula and molecular formula shows benzene is made up of elements. Kedua-dua fomula molekul dan formula empirik menunjukkan benzena terdiri dari unsur

actual

carbon

and hydrogen

karbon

dan

hidrogen

.

number of atoms and hydrogen atoms in benzene • Molecular formula shows the molecule . Each benzene molecule consists of 6 carbon atoms and 6 hydrogen atoms.

m

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sebenar Formula molekul menunjukkan bilangan bagi atom molekul 6 benzena terdiri daripada benzena. Setiap

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• Empirical formula shows the simplest ratio of number carbon atoms to hydrogen atoms, the simplest carbon hydrogen 1 : 1 ratio of number of atoms to atoms in benzene is . Formula empirik benzena menunjukkan

nisbah paling ringkas

Nisbah paling ringkas bilangan atom

karbon

kepada

hidrogen

karbon bilangan atoms kepada atom hidrogen 1 : 1 adalah .

.

PERCENTAGE COMPOSITION BY MASS OF AN ELEMENT IN A COMPOUND PERATUS KOMPOSISI UNSUR MENGIKUT JISIM DALAM SEBATIAN

Total RAM of the element in the compound × 100% 1

% composition by mass of an element = % komposisi unsur mengikut jisim

2

Jumlah JAR unsur dalam suatu sebatian × 100%

RMM/RFM of compound/JMR/JFR sebatian

Example / Contoh: Calculate the percentage composition by mass of nitrogen in the following compounds: Hitungkan peratusan nitrogen mengikut jisim dalam sebatian berikut: [Relative atomic mass / Jisim atom relatif : N = 14, H = 1, O = 16, S = 32, K = 39]

(i) (NH4)2SO4 2 × 14 × 100% 132 = 21.2%

%N =

(ii) KNO3

14 × 100% 101 = 13.9%

%N =

CHEMICAL FORMULA FOR IONIC COMPOUNDS / FORMULA KIMIA BAGI SEBATIAN ION 1

Chemical formula of an ionic compound comprising of the ions Xm+ and Yn– is by exchanging the charges on each ion. The formula obtained will be XmYn.

Formula kimia sebatian ion yang mengandungi ion X m+ dan Y n– boleh diperoleh melalui pertukaran bilangan cas setiap ion. Formula yang diperoleh ialah XnYm. 2

Example / Contoh: (i) Sodium oxide / Natrium oksida Ion / Ion

Na+

O2–

+1

–2

Exchange of charges / Pertukaran bilangan cas

2

1

Smallest ratio / Nisbah teringkas

2

1

2 Na+

O2–

Charges / Bilangan cas

Number of combining ions / Bilangan ion yang bergabung Formula / Formula

(ii) Copper(II) nitrate / Kuprum(II) nitrat Cu2+ +2

1 2 (Ratio / Nisbah) ⇒ Cu(NO3)2

(iii) Zinc oxide / Zink oksida Zn2+ +2

O2– –2

2

2

1 ⇒ ZnO

1 (Ratio / Nisbah) n io

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NO3– –1

Na2O

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CuO Copper(II) oxide

CaO Calcium oxide

Ag2O Silver oxide

Na2O Sodium oxide

K2O Potassium oxide

Ion aluminium

Al 3+ Aluminium ion

Ion plumbum(II)

Pb2+ Lead(II) ion

Ion zink

Zn2+ Zinc ion

PbCO3 Lead(II) carbonate

ZnCO3 Zinc carbonate

MgCO3 Magnesium carbonate

CuCO3 Copper(II) carbonate

PbSO4 Lead(II) sulphate

ZnSO4 Zinc sulphate

MgSO4 Magnesium sulphate

CuSO4 Copper(II) sulphate

CaSO4 Calcium sulphate

(NH4)2SO4 Ammonium sulphate

(NH4)2CO3 Ammonium carbonate

H2SO4 Sulphuric acid

Na2SO4 Sodium sulphate

Ag2SO4 Silver sulphate

CaCO3 Calcium carbonate

Ion klorida

Cl–, Chloride ion Ion bromida

Br–, Bromide ion

CaBr2 Calcium bromide

NH4Br Ammonium bromide

AgBr Silver bromide

HBr Hydrobromic acid

NaBr Sodium bromide

AlCl3 Aluminium chloride

PbCl2 Lead(II) chloride

ZnCl2 Zinc chloride

MgCl2 Magnesium chloride

PbI2 Lead(II) iodide

ZnI2 Zinc iodide

Ion nitrat

NO3–, Nitrate ion

Ca(NO3 )2 Calcium nitrate

NH4NO3 Ammonium nitrate

AgNO3 Silver nitrate

HNO3 Nitric acid

NaNO3 Sodium nitrate

Pb(OH)2 Lead(II) hydroxide

Zn(OH)2 Zinc hydroxide

Al(NO3)3 Aluminium nirate

Pb(NO3 )2 Lead(II) nitrate

Zn(NO3 )2 Zinc nitrate

Mg(NO3 )2 Magnesium nitrate

Cu(OH)2 Cu(NO3 )2 Copper(II) hydroxide Copper(II) nitrate

Ca(OH)2 Calcium hydroxide

AgOH Silver hydroxide

NaOH Sodium hydroxide

KOH KNO3 Potassium hydroxide Potassium nitrate

Ion hidroksida

OH–, Hydroxide ion

Mg(OH)2 MgI2 Magnesium Magnesium iodide hydroxide

CuI2 Copper(II) iodide

CaI2 Calcium iodide

NH4I Ammonium iodide

AgI Silver iodide

HI Hydroiodic acid

NaI Sodium iodide

KI Potassium iodide

Ion iodida

I–, Iodide ion

Al(OH)3 AlBr3 AlI3 Aluminium Aluminium bromide Aluminium iodide hydroxide

PbBr2 Lead(II) bromide

ZnBr2 Zinc bromide

MgBr2 Magnesium bromide

CuCl2 CuBr2 Copper(II) chloride Copper(II) bromide

CaCl2 Calcium chloride

NH4Cl Ammonium chloride

AgCl Silver chloride

HCl Hydrocloric acid

NaCl Sodium chloride

K2SO4 KCl KBr Potassium sulphate Potassium chloride Potassium bromide

Ion sulfat

SO42–, Sulphate ion

Ag2CO3 Silver carbonate

H2CO3 Carbonic acid

Na2CO3 Sodium carbonate

K2CO3 Potassium carbonate

Ion karbonat

CO32–, Carbonat ion

Al2(SO4 )3 Al2O3 Al2(CO3 )3 Aluminium Aluminium oxide Aluminium carbonate sulphate

PbO Lead(II) oxide

ZnO Zinc oxide

MgO Mg2+ Magnesium ion Magnesium Ion magnesium oxide

Ion kuprum(II)

Cu2+ Copper(II) ion

Ion kalsium

Ca2+ Calcium ion

Ion ammonium

NH4 + Ammonium ion

Ion argentum

Ag+ Silver ion

Ion hidrogen

H+ Hydrogen ion

Ion natrium

Na+ Sodium ion

Ion kalium

K+ Potassium ion

Ion oksida

O2–, Oxide ion

Aktiviti 1: TULIS FORMULA KIMIA DAN NAMA BAGI BAHAN KIMIA BERIKUT

ACTIVITY 1: WRITE THE CHEMICAL FORMULAE AND NAMES OF THE FOLLOWING COMMON COMPOUNDS


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Ion aluminium

Aluminium ion

Ion plumbum(II)

Lead(II) ion

Ion zink

Zinc ion

Ion magnesium

Magnesium ion

Ion kuprum(II)

Copper(II) ion

Ion kalsium

Calcium ion

Ion ammonium

Ammonium ion

Ion argentum

Silver ion

Ion hidrogen

Hydrogen ion

Ion natrium

Sodium ion

Ion kalium

Potassium ion

Al2O3

PbO

ZnO

MgO

CuO

CaO

Ag2O

Na2O

K2O

Ion oksida

Oxide ion

Al2(CO3)3

PbCO3

ZnCO3

MgCO3

CuCO3

CaCO3

(NH4 )2CO3

Ag2CO3

H2CO3

Na2CO3

K2CO3

Ion karbonat

Carbonat ion

Al2(SO4 )3

PbSO4

ZnSO4

MgSO4

CuSO4

CaSO4

(NH4 )2SO4

Ag2SO4

H2SO4

Na2SO4

K2SO4

Ion sulfat

Sulphate ion

AlCl3

PbCl2

ZnCl2

MgCl2

CuCl2

CaCl2

NH4Cl

AgCl

HCl

NaCl

KCl

Ion klorida

Chloride ion

AlBr3

PbBr2

ZnBr2

MgBr2

CuBr2

CaBr2

NH4 Br

AgBr

HBr

NaBr

KBr

Ion bromida

Bromide ion

AlI3

PbI2

ZnI2

MgI2

CuI2

CaI2

NH4 I

AgI

HI

NaI

KI

Ion iodida

Iodide ion

Al(OH)3

Pb(OH)2

Zn(OH)2

Mg(OH)2

Cu(OH)2

Ca(OH)2

AgOH

NaOH

KOH

Ion hidroksida

Hydroxide ion

AKTIVITI 2: TANPA MERUJUK KEPADA JADUAL AKTIVITI 1, TULISKAN FORMULA KIMIA BAGI SEBATIAN BERIKUT

Al(NO3 )3

Pb(NO3 )2

Zn(NO3 )2

Mg(NO3 )2

Cu(NO3 )2

Ca(NO3 )2

NH4 NO3

AgNO3

HNO3

NaNO3

KNO3

Ion nitrat

Nitrate ion

ACTIVITY 2: WITHOUT REFERRING TO THE TABLE IN ACTIVITY 1, WRITE THE CHEMICAL FORMULAE OF THE FOLLOWING COMPOUNDS


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ACTIVITY 3: WRITE THE CHEMICAL FORMULAE AND TYPE OF PARTICLES FOR THE FOLLOWING ELEMENT/COMPOUND AKTIVITI 3: TULIS FORMULA KIMIA DAN JENIS ZARAH UNTUK UNSUR/SEBATIAN BERIKUT

Compound / Element Sebatian/Unsur

Sodium sulphate Natrium sulfat

Ammonium carbonate Ammonium karbonat

Magnesium nitrate Magnesium nitrat

Hyrochloric acid Asid hidroklorik

Potassium oxide Kalium oksida

Magnesium oxide Magnesium oksida

Lead(II) carbonate Plumbum(II) karbonat

Iron(III) sulphate Ferum(III) sulfat

Magnesium chloride Magnesium klorida

Zinc sulphate Zink sulfat

Silver nitrate Argentum nitrat

Ammonium sulphate Ammonium sulfat

Zinc oxide Zink oksida

Nitric acid Asid nitrik

Ammonia gas Gas ammonia

Magnesium Magnesium

Zinc Zink

Copper(II) sulphate Kuprum(II) sulfat

Iodine Iodin

Chlorine m

Type of particles

Na2SO4

Ion

(NH4 )2CO3

Ion

Mg(NO3 )2

Ion

HCl

Ion

K2O

Ion

MgO

Ion

PbCO3

Ion

Fe2(SO4)3

Ion

MgCl2

Ion

ZnSO4

Ion

AgNO3

Ion

(NH4 )2SO4

Ion

ZnO

Ion

HNO3

Ion

NH3

Molecule

Mg

Atom

Zn

Atom

CuSO4

Ion

I2

Molecule

Cl2

Molecule

Formula

Jenis zarah

Compound / Element Sebatian/Unsur

Zinc carbonate Zink karbonat

Ammonium carbonate Ammonium karbonat

Silver chloride Argentum klorida

Sulphuric acid Asid sulfurik

Copper(II) nitrate Kuprum(II) nitrat


Carbon dioxide gas Gas karbon dioksida

Oxygen gas Gas oksigen

Aluminium sulphate Aluminium sulfat

Lead(II) chloride Plumbun(II) klorida

Potassium iodide Kalium iodida

Copper(II) carbonate Kuprum(II) karbonat

Potasium carbonate Kalium karbonat

Sodium hydroxide Natrium hidroksida

Aqueous ammonia Ammonia akueus

Ammonium chloride Ammonium klorida

Nitrogen dioxide gas Gas nitrogen dioksida


Silver Argentum

Bromine Bromin

Formula

Type of particles

ZnCO3

Ion

(NH4 )2CO3

Ion

AgCl

Ion

H2SO4

Ion

Cu(NO3 )2

Ion

H2

Molecule

CO2

Molecule

O2

Molecule

Al2(SO4 )3

Ion

PbCl2

Ion

KI

Ion

CuCO3

Ion

K2CO3

Ion

NaOH

Ion

NH3(aq)

Ion and molecule

NH4Cl

Ion

NO2

Molecule

NaCl

Ion

Ag

Atom

Br2

Molecule

Formula

Jenis zarah

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CHEMICAL EQUATIONS / PERSAMAAN KIMIA 1

Two types of equation / Dua jenis persamaan: • Equation in words / Persamaan perkataan – using names of reactants and products / menggunakan nama bahan tindak balas dan hasil tindak balas; • Equation using symbols / Persamaan menggunakan simbol – reactants and products are represented by chemical formulae and have certain meanings menggunakan formula kimia untuk mewakili bahan tindak balas dan hasil tindak balas serta menggunakan pelbagai jenis simbol yang membawa makna tertentu. Symbol / Simbol +

2

Meaning / Maksud

Symbol / Simbol

Meaning / Maksud

Separating 2 reactants / products

(g)

Gaseous state

Mengasingkan 2 bahan / hasil

(g)

Keadaan gas

Produces

(aq)

Aqueous state

Menghasilkan

(ak)

Keadaan akueus

Reversible reaction

Gas released

Tindak balas berbalik

Gas terbebas

(s)

Solid state

Precipitation

(p)

Keadaan pepejal

Bahan termendap

(l)

Liquid state

(ce)

Keadaan cecair

∆

Heating / Heat energy is given Pemanasan / Haba dibekalkan

Information obtained from chemical equation using symbols / Maklumat yang diperoleh daripada persamaan kimia bersimbol: (a) Qualitative aspect / Aspek kualitatif : type of reactants and products involved in the chemical reaction and the state of each reactant and product. jenis bahan / hasil tindak balas yang terlibat dalam tindak balas dan keadaan fizikal bagi setiap bahan / hasil tindak balas. (b) Quantitative aspect / Aspek kuantitatif : number of moles of reactants and products involved in the chemical reaction

that is the coeffficients involved in a balanced equation of the formulae of reactants and products. bilangan mol yang bertindak balas dan hasil tindak balas yang terbentuk iaitu pekali bagi setiap formula bahan dan hasil tindak balas.

Example / Contoh:

Zn (s) + 2HCl (aq) ZnCl2 (aq) + H2 (g) Zn (p) + 2HCl (ak)

ZnCl2 (ak) + H2 (g)

1 mol 2 mol 1 mol 1 mol Interpretation / Tafsiran: 1 mol of zinc reacts with 2 mol of hydrochloric acid to produce 1 mol of zinc chloride and 1 mol of hydrogen. 1 mol zink bertindak balas dengan 2 mol asid hidroklorik menghasilkan 1 mol zink klorida dan 1 mol hidrogen. 3

Writing balanced chemical equations / Menulis persamaan kimia seimbang:

Step 1 / Langkah 1 : Write the correct chemical formulae for each reactant and product. Tulis formula kimia bagi setiap bahan dan hasil tindak balas.

Step 2 / Langkah 2 : Detemine the number of atoms for each element / Tentukan bilangan atom setiap unsur. Step 3 / Langkah 3 : Balance the number of atoms for each element by adjusting the coefficients in front of the chemical formulae.

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Imbangkan bilangan atom setiap jenis unsur dengan menambahkan pekali di hadapan setiap formula kimia.

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EXERCISE / LATIHAN Write a balanced chemical equation for each of the following reactions: Tulis persamaan kimia seimbang bagi setiap tindak balas yang berikut:

Zinc carbonate Zinc oxide + Carbon dioxide / Zink karbonat

1

ZnCO3

Zink oksida + Karbon dioksida

ZnO + CO2

Sulphuric acid + Sodium hydroxide Sodium sulphate + Water / Asid sulfurik + Natrium hidroksida

2

H2SO4 + 2NaOH

Natrium sulfat + Air

Na2SO4 + 2H2O

Silver nitrate + Sodium chloride Silver chloride + Sodium nitrate

3

Argentum nitrat + Natrium klorida Argentum klorida + Natrium nitrat AgNO3 + NaCl AgCl + NaNO3

Copper(II) oxide + Hydrochloric acid Copper(II) chloride + Water

4

Kuprum(II) oksida + Asid hidroklorik Kuprum(II) klorida + Air CuO + 2HCl CuCl2 + H2O

Magnesium + Oxygen Magnesium oxide / Magnesium + Oksigen

5

2Mg + O2

2MgO

Sodium + Water Sodium hydroxide + Hydrogen / Natrium + Air

6

Magnesium oksida

2Na + 2H2O

Natrium hidroksida + Hidrogen

2NaOH + H2

Potassium oxide + Water Potassium hydroxide / Kalium oksida + Air

7

K 2O + H 2O

2KOH

Zinc oxide + Nitric acid Zinc nitrate + Water / Zink oksida + Asid nitrik

8

Kalium hidroksida

ZnO + 2HNO3

Zink nitrat + Air

Zn(NO3 )2 + H2O

Lead(II) nitrate Lead(II) oxide + Nitrogen dioxide + Oxygen

9

Plumbum(II) nitrat Plumbum (II) oksida + Nitrogen dioksida + Oksigen 2Pb(NO3 )2 2PbO + 4NO2 + O2 10 Aluminium nitrate Aluminium oxide + Nitrogen dioxide + Oxygen Aluminium nitrat Aluminium oksida + Nitrogen dioksida + Oksigen 4Al(NO3 )3 2Al2O3 + 12NO2 + 3O2

NUMERICAL PROBLEMS INVOLVING CHEMICAL EQUATIONS / PENGHITUNGAN BERKAITAN PERSAMAAN KIMIA Calculation steps / Langkah perhitungan: S1 / L1 : Write a balanced equation / Tulis persamaan kimia seimbang. S2 / L2 : Write the information from the question above the equation / Tulis maklumat daripada soalan di atas persamaan. S3 / L3 : Write the information from the chemical equation below the equation (information about the number of moles of

reactants/products). Tulis maklumat daripada persamaan kimia di bawah persamaan (Maklumat perhubungan bilangan mol bahan/hasil tindak balas terlibat).

S4 / L4 : Change the information in S2 into moles by using the method shown in the chart below. Tukarkan maklumat L2 kepada mol menggunakan carta di bawah. S5 / L5 : Use the relationship between number of moles of substance involved in S3 to find the answer. Gunakan perhubungan bilangan mol bahan terlibat dalam L3 untuk mencari jawapan. S6 / L6 : Change the information to the unit required using the chart below. Tukar maklumat kepada unit yang dikehendaki dengan menggunakan carta di bawah.

Mass (g)

m

× (RAM/FRM/RMM) g mol–1

No. of moles (n) Bilangan mol (n)

× 24 dm3 mol–1 / 22.4 dm3 mol–1 ÷ 24 dm3 mol–1 / 22.4 dm3 mol–1

Volume of gas (dm3) Isipadu gas (dm3)

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Jisim (g)

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The equation shows the reaction between zinc and hydrochloric acid. Persamaan menunjukkan tindak balas antara zink dengan asid hidroklorik.

Zn + 2HCl

ZnCl2 + H2

Calculate the mass of zinc required to react with excess hydrochloric acid to produce 6 dm3 of hydrogen gas at room conditions. [Relative atomic mass: Zn = 65, Cl = 35.5, 1 mole of gas occupies 24 dm3 at room conditions] Hitungkan jisim zink yang perlu ditindakbalaskan dengan asid hidroklorik berlebihan untuk menghasilkan 6 dm3 gas hidrogen pada keadaan bilik. [Jisim atom relatif: Zn = 65, Cl = 35.5, 1 mol gas menempati 24 dm3 pada suhu bilik]

Mol of H2 =

6 dm3 = 0.25 mol 24 dm3 mol–1

From the equation, 1 mol of H2 : 1 mol of Zn 0.25 mol of H2 : 0.25 mol of Zn Mass of Zn = 0.25 × 65 = 16.2 g

2

The equation shows the reaction between potassium and oxygen. Persamaan berikut menunjukkan tindak balas antara kalium dengan oksigen.

4K + O2

2K2O

Calculate the mass of potassium required to produce 23.5 g of potassium oxide. [Relative atomic mass: K = 39, O = 16] Hitungkan jisim kalium yang diperlukan untuk menghasilkan 23.5 g kalium oksida. [Jisim atom relatif: K = 39, O = 16]

Mol of K2O =

23.5 23.5 = = 0.25 mol (2 × 39 + 16) 94

From the equation, 2 mol of K2O : 4 mol of K 0.25 mol of K2O : 0.5 mol of K Mass of K = 0.5 mol × 39 g mol–1 = 19.5 g

3

The equation shows the decomposition of hydrogen peroxide. Persamaan menunjukkan penguraian hidrogen peroksida.

H2O2

H2O + O2

Balance the equation above. Calculate the number of moles of H2O2 that decomposes if 11.2 dm3 oxygen gas is collected at STP. [Relative Atomic mass: H = 1, O = 16, molar volume of gas = 22.4 dm3 mol–1 at STP] Seimbangkan persamaan di atas. Hitung bilangan mol H2O2 yang telah terurai sekiranya 11.2 dm3 gas oksigen dikumpulkan pada STP. [Jisim atom relatif: H = 1, O = 16, isi padu molar gas = 22.4 dm3 mol–1 pada STP]

Mol of O2 =

11.2 dm3 = 0.5 mol 22.4 dm3 mol–1

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From the equation, 1 mol of O2 : 2 mol of H2O2 0.5 mol of O2 : 1.0 mol of H2O2

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8.0 g of copper(II) oxide powder is added to excess dilute nitric acid and heated. Calculate the mass of copper(II) nitrate produced. [Relative atomic mass: N = 14, O = 16, Cu = 64]

4

8.0 g serbuk kuprum(II) oksida dicampurkan kepada asid nitrik cair yang berlebihan dan dihangatkan. Hitungkan jisim kuprum(II) nitrat yang terhasil. [Jisim atom relatif: N = 14, O = 16, Cu = 64]

CuO + 2HNO3 Cu(NO3 )2 + H2O Mol of CuO =

8 g = 0.1 mol (64 + 16)g mol–1

From the equation, 1 mol of CuO : 1 mol of Cu(NO3)2 0.1 of CuO : 0.1 mol of Cu(NO3)2 Mass of Cu(NO3)2 = 0.1 mol × 188 g mol–1 = 18.8 g

1.3 g of zinc reacts with excess dilute sulphuric acid. The products are zinc sulphate and hydrogen. Calculate the volume of hidrogen gas released at STP. [Relative atomic mass: Zn = 65, 1 mol of gas occupies 22.4 dm3 mol–1at STP]

5

1.3 g zink bertindak balas dengan asid sulfurik cair yang berlebihan. Hasil tindak balas adalah zink sulfat dan hidrogen. Hitungkan isi padu hidrogen yang terbebas pada STP. [Jisim atom relatif: Zn = 65, isipadu molar gas = 22.4 dm3 mol–1 pada STP]

Answer/Jawapan:

448 cm3

0.46 g of sodium burns completely in chlorine gas at room conditions to produce sodium chloride. Calculate the volume of chlorine gas that has reacted. [Relative atomic mass: Na = 23, Molar volume of gas = 24 dm3 mol–1 at room conditions]

6

0.46 g natrium terbakar lengkap dalam gas klorin pada keadaan bilik menghasilkan natrium klorida. Hitungkan isi padu klorin yang diperlukan untuk bertindak balas lengkap. [Jisim atom relatif: Na = 23, isi padu molar gas = 24 dm3 mol–1 pada keadaan bilik]

Answer/Jawapan:

0.24 dm3

The equation shows the combustion of propane gas.

7

Persamaan menunjukkan pembakaran gas propana.

C3H8 + 5O2

3CO2 + 4H2O

720 cm3 of propane gas (C3H8) at room conditions burns in excess oxygen. Calculate the mass of carbon dioxide formed. [Relative atomic mass: C = 12, O = 16, Molar volume of gas = 22.4 dm3 mol–1 at room conditions] 720 cm3 gas propana (C3H8) pada keadaan bilik terbakar dalam oksigen berlebihan. Hitungkan jisim karbon dioksida yang terbentuk. [Jisim atom relatif: C = 12, O = 16, isi padu molar gas = 24 dm3 mol–1 pada keadaan bilik]

3.96 g

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Objective Questions / Soalan Objektif 1

2

The mass of one atom of element Y is two times more than an atom of oxygen. What is the relative atomic mass of element Y? [Relative atomic mass: O = 16] Jisim satu atom unsur Y adalah dua kali lebih daripada satu atom oksigen. Apakah jisim atom relatif bagi unsur Y? [Jisim atom relatif: O = 16] A 12 B 24 C 32 D 36

5

Jadual berikut menunujukkan jisim atom relatif bagi neon, karbon, oksigen dan kalsium. Element/Unsur Relative atomic mass/Jisim atom relatif

The chemical formula for butane is C4H10. Which of the following statements are true about butane? [Relative atomic mass: H = 1, C =12 and O =16, Avogadro Constant = 6 × 1023 mol–1]

Formula empirik butana ialah CH2.

III 1 mol of butane contains a total of 8.4 × 1024 atoms. 6

A I and II only

II, III and IV only

II dan III sahaja

D I, II, III and IV I, II, III dan IV

7

A bottle contains 3.01 × 1023 of gas particles. What is the number of moles of the gas in the bottle?

4

C 3.0 mol D 6.0 mol

16 g oksigen mengandungi 6.02 × 1023 molekul oksigen

Mass of one oxygen atom is 16 times bigger than one carbon atom

A bulb is filled with 1 800 cm3 of argon gas at room conditions. What is the number of argon atom in the bulb? [Molar volume of gas = 24 dm3 mol–1 at room conditions, Avogadro constant = 6.02 × 1023 mol–1]

8

Antara gas berikut, yang manakah mengandungi 0.4 mol atom pada suhu dan tekanan bilik? [Isi padu molar gas = 24 dm3 mol–1 pada suhu dan tekanan bilik]

C D

6.02 × 1023 3.01 × 1023

5 g of element X reacted with 8 g of element Y to form a compound with the formula XY2. What is the relative atomic mass of element X? [Relative atomic mass: Y = 80] 5 g unsur X bertindak balas dengan 8 g unsur Y membentuk sebatian dengan formula XY2. Apakah jisim atom relatif unsur X? [Jisim atom relatif: Y = 80]

A 25 B 40

C 50 D 100

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C 4.8 dm3 CO2 D 4.8 dm3 NH3

C 8.03 × 1022 D 8.03 × 1021

What is the number of hydrogen atom in 0.1 mol of water? [Avogadro constant: 6.02 × 1023 mol–1] A 6.02 × 1022 B 60.2 × 1023

Which of the following gases contains 0.4 mol of atoms at room temperature and pressure? [Molar volume of gas = 24 dm3 mol–1 at room temperature and pressure]

A 4.8 dm3 Ne B 4.8 dm3 O2

40

Berapakah bilangan atom oksigen dalam 0.1 mol air? [Pemalar Avogadro = 6.02 × 1023 mol–1]

Sebuah botol mengandungi 3.01 × 1023 zarah gas. Berapakah bilangan mol zarah gas dalam botol itu?

A 0.5 mol B 1.0 mol

Calcium / Kalsium

A 4.515 × 1022 B 4.515 × 1023

II, III dan IV sahaja

3

16

Sebuah belon diisi dengan 1 800 cm3 gas argon pada keadaan bilik. Berapakah bilangan atom argon dalam belon itu? [Isipadu molar gas = 24 dm3 mol–1 pada keadaan bilik, Pemalar Avogadro = 6.02 × 1023 mol–1]

I dan II sahaja

C

Oxygen / Oksigen

Jisim satu atom oksigen adalah 16 kali lebih besar daripada satu atom karbon

Satu molekul butana mempunyai jisim 84 kali lebih daripada jisim satu atom hidrogen.

II and III only

12

D

Setiap molekul butana terdiri dari 4 atom karbon dan 10 atom hidrogen.

B

Carbon / Karbon

molecule

II Each butane molecule is made up of 4 carbon atoms and 10 hydrogen atoms.

IV One butane molecule has a mass of 84 times higher than the mass of 1 hydrogen atom.

20

Antara pernyataan berikut, yang manakah adalah benar? [Pemalar Avogadro = 6.02 × 1023 mol–1] A Mass of one calcium atom is 40 g Jisim satu atom kalsium ialah 40 g B Mass of 1 mol of neon is 20 g Jisim 1 mol neon ialah 20 g C 16 g of oxygen contains 6.02 × 1023 oxygen

The empirical formula for butane is CH2.

Jumlah bilangan atom dalam 1 mol butana adalah 8.4 × 1024.

Neon / Neon

Which of the following statements is true? [Avogadro constant = 6.0 × 1023 mol–1]

Formula kimia bagi butana ialah C4H10. Antara pernyataan berikut, yang manakah adalah benar tentang butana? [Jisim atom relatif: H = 1, C = 12 dan O = 16, Pemalar Avogadro = 6 × 1023 mol–1]

I

The table below shows the relative atomic mass of neon, carbon, oxygen and calcium.

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The diagram below shows the set-up of apparatus to determine the empirical formula of an oxide metal X. Rajah di bawah menunjukkan susunan radas bagi menentukan formula empirik oksida logam X.

What is the number of oxygen molecules is produced when 7.4 g magnesium nitrate decomposed when heated. [Relative formula mass of Mg(NO3)2 = 148; Avogadro constant = 6.02 × 1023 mol–1]

Logam X

Heat Panaskan

Which of the following is metal X? Antara berikut, yang manakah mungkin bagi logam X?

C

Zink

B

Tin

Stanum

Lead Plumbum

D Copper

Persamaan di bawah menunjukkan penguraian nitrat apabila dipanaskan.

2Mg(NO3)2 2MgO + 4NO2 + O2

Metal X

A Zinc

11 The equation shows a decomposition of magnesium nitrate when heated.

Kuprum

10 The following equation shows the decomposition reaction of lead(II) nitrate when heated at room temperature and pressure. Persamaan tindak balas di bawah menunjukkan penguraian plumbum(II) nitrat apabila dipanaskan pada suhu dan tekanan bilik.

Berapakah bilangan molekul oksigen apabila 7.4 g magnesium nitrat terurai apabila dipanaskan? [Jisim formula relatif Mg(NO3)2 = 148; Pemalar Avogadro = 6.02 × 1023 mol–1]

A B C D

12 The equation below shows the chemical equation of the combustion of ethanol in excess oxygen. Persamaan di bawah menunjukkan persamaan kimia pembakaran etanol dalam oksigen berlebihan.

2Pb(NO3)2 2PbO + 4NO2 + O2 Which of the following are true when 0.1 mol of lead(II) nitrate is decomposed? [Relative atomic mass: N = 14, O = 16, Pb = 207 and 1 mol gas occupies the volume of 24 dm3 at room temperature and pressure] Antara berikut, yang manakah adalah benar apabila 0.1 mol plumbum(II) nitrat terurai? [Jisim atom relatif: N = 14, O = 16, Pb = 207 dan 1 mol gas menempati isipadu 24 dm3 pada suhu dan tekanan bilik]

I

66.2 g of lead(II) oxide is formed 66.2 g plumbum(II) oksida terbentuk

II 22.3 g of lead(II) oxide is formed

22.3 g plumbum(II) oksida terbentuk

III 2.4 dm3 of oxygen gases is given off

2.4 dm3 gas oksigen dibebaskan

IV 4 800 cm3 of nitrogen dioxide given off

4 800 cm3 nitrogen dioksida dibebaskan

A I and III only I dan III sahaja

B

I and IV only

C

II and III only

I dan IV sahaja II dan III sahaja

D II and IV only

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2C2H5OH + 6O2 4CO2 + 6H2O What is the volume of carbon dioxide gas released when 9.20 g ethanol burnt completely? [Relative atomic mass of H = 1, C = 12, O = 16, 1 mol of gas occupies 24 dm3 at room condition] Apakah isi padu gas karbon dioksida dibebaskan apabila 9.20 g etanol terbakar lengkap? [Jisim atom relatif: H = 1, C = 12, O = 16, 1 mol gas menempati 24 dm3 pada keadaan bilik]

A B C D

4.8 cm3 9.6 cm3 96.0 cm3 9 600 cm3

13 What is the percentage by mass of nitrogen content in urea, CO(NH2)2? [Relative atomic mass: C = 12, N = 14, H = 1 and O = 16] Apakah peratus kandungan nitrogen mengikut jisim dalam urea, CO(NH2)2? [Jisim atom relatif: C = 12, N = 14, H = 1 dan O = 16]

A B C D

23.3% 31.8% 46.7% 63.6%

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1.505 × 1022 3.010 × 1022 1.505 × 1023 3.010 × 1023

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3

PERIODIC TABLE JADUAL BERKALA

HISTORICAL DEVELOPMENT / SEJARAH PERKEMBANGAN – To identify the contribution of scientists in the arrangement of elements in the Periodic Table. Mengetahui sumbangan ahli sains untuk penyusunan unsur dalam Jadual Berkala. – To get ideas on the arrangement of elements in the Periodic Table based on their proton numbers. Mendapat idea penyusunan unsur dalam Jadual Berkala berdasarkan nombor proton.

ARRANGEMENT OF ELEMENT IN THE PERIODIC TABLE / PENYUSUNAN UNSUR DALAM JADUAL BERKALA • GROUP / KUMPULAN – To write the electron arrangement for atoms of elements with proton numbers 1 to 20. Menulis susunan elektron bagi atom unsur dengan proton 1 hingga 20. • PERIOD / KALA – To determine the group and period based on the electron arrangement of atoms or otherwise. Menentukan kumpulan dan kala berdasarkan susunan elektron atom dan sebaliknya.

PROPERTIES OF ELEMENTS IN THE PERIODIC TABLE / SIFAT-SIFAT UNSUR DALAM JADUAL BERKALA • GROUP 18 / KUMPULAN 18 – To explain the existence of noble gases as monoatom and their uses. Menerangkan kewujudan gas adi secara monoatom serta kegunaannya. • GROUP 1 / KUMPULAN 1 – To explain physical properties, similar chemical properties (with water, oxygen and chlorine) and the different reactivities. Menerangkan sifat fizik, sifat kimia yang sama (dengan air, oksigen dan dengan klorin) serta kereaktifan yang berbeza. • GROUP 17 / KUMPULAN 17 – To explain physical properties, similar chemical properties (with water, sodium hydroxide and iron) and the different reactivities. Menerangkan sifat fizik, sifat kimia yang sama (dengan air, natrium hidroksida dan ferum) serta kereaktifan yang berbeza. • PERIOD 3 / KALA 3 – To explain changes in atomic size, electronegativity, metallic properties as well as oxide properties across period 3 from left to right. Menerangkan perubahan saiz atom, keelektronegatifan, sifat kelogaman serta sifat oksida merentasi Kala 3 dari kiri ke kanan.

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• TRANSITION ELEMENTS / UNSUR PERALIHAN – To state metallic properties of transition metals and their special characteristics. Menyatakan sifat kelogaman unsur peralihan serta ciri-ciri istimewa unsur peralihan.

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ADVANTAGES OF CLASSIFYING THE ELEMENTS IN THE PERIODIC TABLE KEBAIKAN PENGELASAN UNSUR DALAM JADUAL BERKALA

1

Elements are arranged systematically in the Periodic Table in an increasing order of proton number which enables: Unsur disusun secara sistematik dalam Jadual Berkala mengikut tertib pertambahan nombor proton yang membolehkan:

(a) Chemists to study, understand and remember the chemical and physical properties of all the elements and compounds in an orderly manner, Ahli kimia mempelajari, memahami dan mengingat sifat kimia dan sifat fizik semua unsur dan sebatian secara teratur.

(b) Properties of elements and their compounds to be predicted based on the position of elements in the Periodic Table, Sifat unsur dan sebatiannya diramal berdasarkan kedudukan unsur dalam Jadual Berkala.

(c) Relationship between elements from different groups to be known. Perhubungan unsur dari kumpulan yang berlainan diketahui.

CONTRIBUTION OF SCIENTIST TO THE HISTORICAL DEVELOPMENT OF THE PERIODIC TABLE SUMBANGAN AHLI SAINS DALAM SEJARAH PERKEMBANGAN JADUAL BERKALA

Scientists / Saintis Antoine Lavoisier

Discoveries / Penemuan – Substances were classified into 4 groups with similar chemical properties. Bahan dikelaskan kepada empat kumpulan dengan sifat kimia sama.

J.W Dobereiner

– Substances were arranged in 3 groups / Bahan disusun dalam tiga kumpulan. – Groups with similar chemical properties were called Triads. Kumpulan dengan sifat kimia sama dinamakan triad.

– Triad system was confined to some elements only / Sistem triad terhad kepada beberapa unsur sahaja. John Newlands

– Elements were arranged in ascending atomic mass / Unsur disusun mengikut pertambahan jisim atom. – Law of Octaves because similar chemical properties were repeated at every eighth element. Hukum Oktaf kerana sifat sama berulang pada setiap unsur kelapan.

– This system was inaccurate because there were some elements with wrong mass numbers. Sistem ini tidak tepat kerana ada unsur dengan nombor jisim salah.

Lothar Meyer

Mass of 1 mol (g) / Jisim 1 mol (g) Density (g cm–3) / Ketumpatan (g cm–3) – Plotted graph for the atomic volume against atomic mass / Memplotkan graf isi padu atom melawan jisim atom. – Found that elements with similar chemical properties were positioned at equivalent places along the curve. – The atomic volume / Isipadu atom =

Mendapati unsur dengan sifat kimia sama menduduki tempat setara dalam lengkungan.

Mendeleev

– Elements were arranged in ascending order of increasing atomic mass. Unsur disusun mengikut pertambahan jisim atom.

– Elements with similar chemical properties were in the same group. Unsur dengan sifat kimia sama berada dalam kumpulan sama.

– Empty spaces were allocated for elements yet to be discovered. Ruang kosong disediakan untuk unsur yang belum ditemui.

– Contributor to the formation of the modern Periodic Table. Penyumbang kepada pambentukan Jadual Berkala Moden.

Henry Moseley

– Classified concepts of proton number and elements in ascending order of increasing proton number. Mengelaskan unsur berdasarkan konsep nombor proton dan menyusun unsur-unsur mengikut turutan nombor proton menaik.

– Contributor to the formation of the modern Periodic Table.

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THE ARRANGEMENT OF ELEMENTS IN THE MODERN PERIODIC TABLE SUSUNAN UNSUR DALAM JADUAL BERKALA MODEN

1

Write the electron arrangement for each atom of element in the Periodic Table below. Tuliskan susunan elektron untuk setiap atom unsur dalam Jadual Berkala di bawah. Nucleon number / Nombor nukleon Proton number / Nombor proton

A Z

X

Symbol of an element / Simbol unsur

GROUP / KUMPULAN 1 1

P E R I O D / K A L A

2 3

1

3

Li

11

8 4

19

K

2.8.8.1

11

Be

2.1 Na

13

2 5

2.2 24 12

2.8.1 39

4

2

1

23

3

4

H*

1

7

2

18

LOGAM PERALIHAN

40 20

27 13

2.8.2

3

4

5

6

7

8

9

12

B

6

2.3

TRANSITION METALS

Mg

14 14

C

7

2.4 28

Al

16 16

N

8

2.5 31

Si

14

2.8.3

10 11 12

15

15

2.8.4

32 16

2.8.5

2

17 19

O

9

2.6

P

He

20

F

Ne

10

2.7 35

S

17

2.8.6

40

Cl

80 35

2.8.8.2

Elements in the Periodic Table are arranged horizontally in increasing order of

proton number

.

Unsur-unsur dalam Jadual Berkala disusun secara mendatar mengikut tertib pertambahan

nombor proton

.

Ar

18

2.8.7

Ca

2.8

2.8.8

Br

Two main components of the Periodic Table / Dua komponen utama Jadual Berkala: (a) Group / Kumpulan (b) Period / Kala GROUP / KUMPULAN

1

The vertical column of elements in the Periodic Table arranged according to the number of valance electron in the outermost shell of atoms is called groups. Lajur

menegak

petala terluar

2

dalam Jadual Berkala yang disusun mengikut bilangan

elektron valens

yang terdapat pada

bagi atom dipanggil kumpulan.

There are 18 vertical columns, called Group 1, Group 2, and Group 3 until Group 18. Terdapat 18 lajur disusun secara menegak disebut Kumpulan 1, Kumpulan 2, Kumpulan 3 hingga Kumpulan18. Number of valence electrons

1

2

3

4

5

6

7

Group

1

2

13

14

15

16

17

Bilangan elektron valens Kumpulan

8 (except Helium) 8 (kecuali Helium)

18

For atoms of elements with 3 to 8 valence electrons, the group number is: 10 + number of valence electrons.

Bagi atom unsur dengan 3 hingga 8 elektron valens, nombor kumpulan ialah: 10 + bilangan elektron valens.

Specific name of groups / Nama-nama khas kumpulan: (a) Group 1: Alkali metals # / Kumpulan 1: Logam alkali # (b) Group 2: Alkali-earth metals / Kumpulan 2: Logam alkali bumi (c) Group 3 to 12: Transition elements # / Kumpulan 3 to 12: Unsur peralihan # (d) Group 17: Halogens # / Kumpulan 17: Halogen # n io

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(e) Group 18: Noble gases # / Kumpulan 18: Gas adi # #The important groups that will be studied with respect to chemical and physical properties. # Kumpulan penting yang akan dipelajari dari segi sifat fizik dan sifat kimia.

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Types of substances according to the groups / Jenis bahan mengikut kumpulan: (a) Elements of group 1, 2 and 13 – atoms of each element have 1, 2 and 3 valence electrons respectively are metals. Unsur Kumpulan 1, 2 dan 13 – atom setiap unsur mempunyai 1, 2 dan 3 elektron valens adalah logam.

(b) The elements of group 3 to 12 – transition elements are metals. Unsur Kumpulan 3 hingga 12 – unsur peralihan yang merupakan logam.

(c) The elements of Group 14, 15, 16, 17 and 18 – atoms of each element have 4, 5, 6, 7 and 8 valence electrons respectively are non-metals. Unsur Kumpulan 14, 15, 16, 17 dan 18 – atom setiap unsur mempunyai 4, 5, 6, 7 dan 8 elektron valens adalah bukan logam.

PERIOD / KALA The horizontal row of elements in the Periodic Table, consists of the same number of electrons in an atom called period.

1

Baris unsur secara atom

dalam

2

mendatar

dalam Jadual Berkala, mempunyai bilangan

petala

berisi

shell

occupied with

elektron

yang sama di

disebut sebagai kala.

There are seven horizontal rows of elements known as Period 1, 2, ....., 7 [Refer to the Periodic Table] Terdapat tujuh baris unsur secara mendatar disebut Kala 1, 2, ....., 7 [Rujuk Jadual Berkala]

(a) Period 1 has 2 elements / Kala 1 mengandungi 2 unsur (b) Period 2 and 3 have 8 elements # / Kala 2 dan 3 mengandungi 8 unsur # (c) Period 4 and 5 have 18 elements / Kala 4 dan 5 mengandungi 18 unsur (d) Period 6 has 32 elements / Kala 6 mengandungi 32 unsur (e) Period 7 has 23 elements / Kala 7 mengandungi 23 unsur

Short periods, # Period 3 will be studied in detail with respect to physical and chemical properties / Kala pendek, # Kala 3 akan dipelajari dengan terperinci dari segi sifat fizik dan sifat kimia

Long periods / Kala panjang


Complete the table below / Lengkapkan jadual berikut. Element Unsur


Electron arrangement Susunan elektron

Number of valence electrons

Kumpulan

Group

Number of shell

Period

Bilangan petala

Kala

m

H

1

1

1

1

1

1

He

2

2

2

18

1

1

Li

3

2.1

1

1

2

2

Be

4

2.2

2

2

2

2

B

5

2.3

3

13

2

2

C

6

2.4

4

14

2

2

N

7

2.5

5

15

2

2

O

8

2.6

6

16

2

2

F

9

2.7

7

17

2

2

Ne

10

2.8

8

18

2

2

Na

11

2.8.1

1

1

3

3

Mg

12

2.8.2

2

2

3

3

Al

13

2.8.3

3

13

3

3

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2

The diagram below shows the chemical symbols which represent elements X, Y and Z. Rajah di bawah menunjukkan simbol kimia yang mewakili unsur X, Y dan Z. 23 11

X

12 6

Y

39 19

Z

(a) Explain how to determine the position of element X in the Periodic Table. Terangkan bagaimana menentukan kedudukan unsur X dalam Jadual Berkala.

The proton number of element X is 11 and the number of proton in electrons in atom X is 11 . The electron arrangement of atom X is 1

because three shells

atom 2.8.1

atom

X has one valence electron . Element X is in period occupied with electrons .

X is

11

. The number of

. Element X is located in Group 3 because atom X has

atom X adalah 11 . Bilangan elektron dalam atom Nombor proton unsur X adalah 11 dan bilangan proton dalam 11 . Susunan elektron bagi atom 2.8.1 . Unsur X terletak dalam kumpulan 1 X adalah kerana X adalah atom satu elektron valens 3 atom tiga X mempunyai . Unsur X berada dalam kala kerana X mempunyai berisi elektron dengan . petala

(b) (i)

State the position of element Y in the Periodic Table. / Nyatakan kedudukan unsur Y dalam Jadual Berkala. Element Y is located in Group 14 and Period 2.

(ii)

Explain how to determine the position of element Y in the Periodic Table. Terangkan bagaimana anda menentukan kedudukan unsur Y dalam Jadual Berkala.

– The proton number of element Y is 6 and the number of proton in atom Y is 6. – The electron arrangement of atom Y is 2.4. – Element Y is located in Group 14 because atom Y has 4 valance electron. – Element Y is in Period 2 because atom Y has 2 shells occupied/filled with electrons. (c) Which of the above elements show the same chemical properties? Explain your answer. Antara unsur di atas, yang manakah mempunyai sifat kimia yang sama? Terangkan jawapan anda.

– Element X and element Z. – Electron arrangement of atom X is 2.8.1 and electron arrangement of atom Z is 2.8.8.1. Atoms X and Z have the same number of valence electron.

GROUP 18 (NOBLE GASES) / KUMPULAN 18 (GAS ADI) 1

Consist of Helium (He), Neon (Ne), Argon (Ar), Krypton (Kr), Xenon (Xe) and Radon (Rn). Terdiri dari Helium (He), Neon (Ne), Argon (Ar), Kripton (Kr), Xenon (Xe) dan Radon (Rn). Elements / Unsur

Electron arrangement / Susunan elektron

Helium / Helium Neon / Neon Argon / Argon Krypton / Kripton

2

2 2.8 2.8.8 2.8.18.8

Noble gases are chemically inert because the outermost shell of the atom has achieved duplet electron arrangement for helium and octet electron arrangement for others. Unsur Kumpulan 18 adalah lengai secara kimia kerana petala terluar atomnya telah mencapai susunan elektron duplet untuk helium dan susunan elektron oktet untuk yang lain.

3

Noble gases do not react with other elements (the atom does not lose, gain or share electrons). Unsur Kumpulan ini tidak bergabung dengan unsur lain (atomnya tidak akan menderma, menerima, atau berkongsi elektron).

4

These gases exist as single uncombined atoms and are said to be monatomic gases.

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Gas ini wujud sebagai atom tunggal iaitu sebagai gas monoatom.

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5

Going down Group 18 / Menuruni Kumpulan 18: (a) The atomic size is increasing because the number of Saiz atom bertambah kerana bilangan

petala

shells

increases.

bertambah.

(b) The melting point and boiling points are very low because atoms of noble gases atoms are attracted by weak Van der Waals forces, less energy is required to overcome these forces. However, the melting and boiling points increase going down the group because atomic size increases, causing the Van der Waal forces to more increase and energy is required to overcome these forces. Takat lebur dan takat didih sangat rendah kerana atom-atom gas adi ditarik oleh daya Van der Waals yang lemah , sedikit tenaga diperlukan untuk mengatasi daya tersebut. Walau bagaimanapun, takat lebur dan takat didih bertambah menuruni kumpulan kerana pertambahan saiz atom menyebabkan daya tarikan Van der Waals semakin bertambah, semakin banyak tenaga diperlukan untuk mengatasinya.

(c) The density is low and increases gradually because the mass increases greatly compared to the volume going down the group. Ketumpatan rendah dan semakin meningkat kerana jisim bertambah dengan banyak berbanding dengan isi padu menuruni kumpulan.

6

All noble gases are insoluble in water and cannot conduct electricity in all conditions. Semua gas adi tidak larut dalam air dan tidak dapat mengkonduksikan elektrik dalam semua keadaan.

7

Complete the uses of noble gases in the table below / Lengkapkan jadual kegunaan gas adi. Noble gases / Gas adi

Uses / Kegunaan

Helium / Helium

To fill weather balloons and airship.

Neon / Neon

To fill neon light (for advertisement board).

Argon / Argon

To fill electrical bulb.

Krypton / Kripton

To fill photographic flash lamp.

Radon / Radon

To treat cancer.

GROUP 1 (ALKALI METALS) / KUMPULAN 1 (LOGAM ALKALI) 1

Consist of Lithium (Li), Sodium (Na), Potassium (K), Rubidium (Rb), Cesium (Cs) and Francium (Fr). Terdiri dari Litium (Li), Natrium (Na), Kalium (K), Rubidium (Rb), Sesium (Cs) dan Fransium (Fr). Elements

Symbol

Proton number

Electron arrangement

Number of shells

Lithium / Litium

Li

3

2.1

2

Sodium / Natrium

Na

11

2.8.1

3

Potassium / Kalium

K

19

2.8.8.1

4

Unsur

2

Simbol

Nombor proton

Susunan elektron

Bilangan petala

Physical properties / Sifat fizik: (a) Grey solid with shiny surface / Pepejal kelabu dengan permukaan berkilat. (b) Softer and the density is lower compared to other metals. Lebih lembut dan ketumpatan yang lebih rendah berbanding dengan logam lain.

(c) Lower melting and boiling points compared to other metals. Takat lebur dan takat didih lebih rendah berbanding dengan logam lain.

3

Changes in physical properties going down the group / Perubahan sifat fizik menuruni kumpulan: (a) Atomic size increases because the number of shells increases / Saiz atom bertambah kerana bilangan petala bertambah. (b) Density increases because mass increases faster than the increase in radius. Ketumpatan bertambah kerana pertambahan jisim lebih cepat dari pertambahan jejari

(c) Melting and boiling points decrease because when the atomic size increases, the metal bonds get weaker. Takat didih dan takat lebur berkurang kerana apabila saiz atom bertambah, ikatan logam semakin lemah.

4

Chemical properties of Group 1 elements / Sifat kimia unsur Kumpulan 1: atoms 1 (a) All of elements in Group 1 have valence electron and achieve a stable duplet/octet electron arrangement by releasing

m

Semua melepaskan

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electron to form

+1

charged ions:

1

unsur mempunyai elektron valens dan mencapai susunan elektron oktet/duplet yang stabil dengan satu +1 elektron valens membentuk ion bercas .

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Example / Contoh: (i) Atom releases one electron to achieve stable duplet electron arrangement: Atom litium melepaskan satu elektron untuk mencapai susunan elektron duplet yang stabil:

Li

+e

Electron arrangement / Susunan elektron : 2.1 Number of protons = 3, total charge: +3

Electron arrangement / Susunan elektron : 2 Number of protons = 3, total charge: +3

Bilangan proton = 3,


jumlah cas: +3

jumlah cas: +3

Number of electrons = 3, total charge: –3


Bilangan elektron = 3,


jumlah cas: –3

neutral

Lithium atom is

neutral

Atom litium adalah

(ii)

Li+

Positively

.

charges lithium ion, Li+ is formed. positif

Ion litium bercas

.

jumlah cas: –2 , Li+ terbentuk.

Sodium atom releases one electron to achieve stable octet electron arrangement: Atom natrium melepaskan satu elektron untuk mencapai susunan elektron oktet yang stabil:

Na

Na+

Electron arrangement / Susunan elektron : 2.8.1 Number of protons = 11, total charge: +11 Bilangan proton = 11,


jumlah cas: +11


Number of electrons = 11, total charge: –11 Bilangan elektron = 11,

Atom natrium adalah

neutral

jumlah cas:

+11

–10

Number of electrons = 10, total charge:

jumlah cas: –11

neutral

Sodium atom is

+e


.

Positively

.

jumlah cas:

–10

charges sodium ion, Na+ is formed.

Ion natrium bercas

positif

, Na+ terbentuk.

atoms (b) All elements in Group 1 have similar chemical properties because all in Group 1 have one valence electron releasing electron and achieve the stable duplet/octet arrangement by its valence electron to form a

positively

charged ions.

atom unsur Kumpulan 1 mempunyai bilangan Semua unsur Kumpulan 1 mempunyai sifat kimia yang sama kerana semua elektron yang stabil dengan melepaskan satu elektron valensnya elektron valens yang sama iaitu satu dan mencapai susunan untuk membentuk ion bercas

5

positif

.

The reactivity of alkali metals increases going down the Group 1: Kereaktifan unsur logam alkali bertambah menuruni Kumpulan 1:

– Atoms of Group 1 metals achieve a stable duplet/octet electron arrangement one by releasing valence electron to form +1 charged ion.

Menuruni Kumpulan 1, bilangan

petala

elektron valens pada petala terluar semakin

bertambah, saiz atom bertambah dan jauh dari nukleus.

– The strength of attraction from the proton in the nucleus to the valence weaker . elecron gets Kekuatan tarikan nukleus kepada elektron valens semakin

– The valence electron is loosely held and it is be released.

Li

Na

K

.

easier

for the electron to

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Elektron valens ditarik dengan lemah dan ia makin

lemah

down Group 1

increases, the atomic size further increases and the valence electron in the outer most shell gets away from the nucleus.

increases

shells

menurun Kumpulan 1

– Going down Group 1, the number of

Reactivity

Kereaktifan logam Kumpulan 1 bergantung pada kesenangan atom melepaskan elektron, semakin senang elektron dilepaskan, kereaktifan logam semakin bertambah .

bertambah

– The reactivity of Group 1 metals depends on the tendency for atoms to lose electrons; the easier it loses an electron, the reactivity of the metal increases .

Kereaktifan

Atom logam Kumpulan 1 mencapai susunan elektron gas adi yang stabil dengan satu melepaskan elektron valens membentuk ion bercas +1.

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Chemical reactions of Group 1 elements / Tindak balas kimia unsur Kumpulan 1: (a) Metal Group 1 reacts with water to produce alkali and hydrogen gas. Logam Kumpulan 1 bertindak balas dengan air menghasilkan alkali dan gas hidrogen.

2X + 2H2O

2X + 2H2O

2XOH + H2, X is the metal of Group 1

2XOH + H2 , X adalah logam Kumpulan 1 Lithium / Litium Water / Air

Procedure / Kaedah: (i) Cut a small piece of lithium using a knife and forceps. Potong sepotong litium menggunakan pisau dan forsep.

(ii)

Dry the oil on the surface of the lithium with filter paper. Keringkan minyak pada permukaan litium menggunakan kertas turas.

(iii) Place the lithium slowly onto the water surface in a water trough. Letakkan litium dengan perlahan di atas permukaan air di dalam bekas.

(iv) When the reactions stop, test the solution produced with red litmus paper. Apabila tindak balas berhenti, uji larutan yang terhasil dengan kertas litmus merah.

(v) Record the observation / Catatkan semua pemerhatian. (vi) Repeat steps (i) – (v) using sodium and potassium to replace lithium one by one. Ulang langkah (i) – (v) dengan menggunakan natrium dan kalium menggantikan litium satu demi satu.

Observation / Pemerhatian: Element

Observation

Li

Lithium moves slowly on the water red surface and produces flame. The colourless solution formed turns red litmus to blue .

Unsur

Inference

Pemerhatian

Inferens

perlahan

quickly

Sodium moves

surface and produces

on the water yellow flame. The

colourless solution formed turns red litmus to blue . Natrium bergerak cepat dengan nyalaan kuning di atas permukaan air. Larutan tidak berwarna menukarkan kertas litmus merah kepada biru .

K

Potassium moves

very quickly

on the

yellow

flame. water surface and produce The colourless solution formed turns red litmus blue . to sangat cepat

m

lithium hydroxide: Litium adalah logam yang paling kurang reaktif bertindak balas dengan air membentuk larutan beralkali , litium hidroksida.

Balanced equation / Persamaan kimia seimbang: 2Li + 2H2O 2LiOH + H2 Sodium is reactive metal reacts with water to produce alkaline solution, sodium hydroxide. reaktif bertindak Natrium adalah logam yang balas dengan air membentuk larutan beralkali , natrium hidroksida.

Balanced equation / Persamaan kimia seimbang: 2Na + 2H2O 2NaOH + H2 the most reactive

metal alkaline reacts with water to produce solution, potassium hydroxide. Potassium is

Kalium adalah logam yang paling reaktif bertindak balas dengan air membentuk larutan beralkali , kalium hidroksida.

Balanced equation / Persamaan kimia seimbang: 2K + 2H2O 2KOH + H2

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dengan nyalaan Kalium bergerak kuning di atas permukaan air. Larutan tidak berwarna menukarkan kertas litmus merah kepada biru .

Lithium is the least reactive metal reacts with water to produce alkaline solution,

Reactivity increases down Group 1

Na

Kereaktifan

Kereaktifan bertambah menuruni Kumpulan 1

dengan nyalaan Litium bergerak merah di atas permukaan air. Larutan tidak berwarna menukarkan kertas litmus merah kepada biru .

Reactivity

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(b) Metal Group 1 reacts with oxygen or air to form metal oxide. The metal oxide dissolves in water to produce alkaline solution. Logam Kumpulan 1 bertindak balas dengan oksigen membentuk oksida logam. Oksida logam larut dalam air menghasilkan larutan berakali.

X2O + H2O X2O + H2O

4X + O2 2X2O 2XOH, X is a metal element of Group 1 (Li, Na and K) 2XOH, X adalah logam unsur Kumpulan 1 (Li, Na dan K) Combustion spoon / Sudu pembakaran Gas jar / Balang gas Chlorine gas / Gas klorin Burning lithium / Litium menyala

Procedure / Kaedah: (i) Cut a small piece of lithium using a knife and forceps / Potong secebis kecil litium menggunakan pisau dan forsep. (ii) Dry the oil on the surface of the lithium with filter paper. Keringkan minyak pada permukaan litium dengan kertas turas.

(iii) Place the lithium in a combustion spoon and heat lithium until it start to burn. Letakkan litium pada sudu pembakaraan dan panaskan litium dengan kuat hingga ia menyala.

(iv) Put the burning lithium into a gas jar of oxygen / Letakkan litium yang menyala dalam balang gas berisi oksigen. (v) When the reaction stop, add water to dissolve the compound formed. Apabila tindak balas berhenti, tambahkan air untuk melarutkan sebatian yang terbentuk.

(vi) Add a few drops of universal to the solution formed. Tambahkan beberapa titis penunjuk universal kepada larutan yang terbentuk.

(vii) Record the observation / Catatkan pemerhatian. (viii) Repeat steps (i) – (vii) using sodium and potassium to replace lithium one by one. Ulang langkah (i) – (vii) menggunakan natrium dan kalium untuk menggantikan litium satu demi satu.


Observation

Unsur

Li

Inference

Pemerhatian

Litium adalah paling kurang reaktif terhadap oksigen.

form

colourless

– Lithium reacts with oxygen to produce lithium oxide .

soluble in water to solution.

Litium bertindak balas dengan oksigen membentuk litium oksida .

Pepejal putih larut dalam air membentuk tidak berwarna . larutan

– The solution turns green indicator to purple .

Balanced equation / Persamaan kimia seimbang: 4Li + O2 2Li2O

universal

Larutan itu menukarkan warna penunjuk hijau kepada ungu universal dari

.

– Lithium reacts with water to form alkaline solution, lithium hydroxide. Litium oksida bertindak balas dengan air membentuk larutan beralkali, litium hidroksida


– Lithium is the least reactive metal towards oxygen.

Litium terbakar perlahan dengan nyalaan merah menghasilkan pepejal putih .

white solid

Kereaktifan


– Lithium burns slowly with a red flame to produce white solid .

– The

Reactivity

Inferen

.

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Balanced equation / Persamaan kimia seimbang: Li2O + H2O 2LiOH

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Na

– Sodium burns brightly with a yellow flame to produce white solid . terang Natrium terbakar kuning menghasilkan

– The form

white solid

dengan nyalaan pepejal putih .

soluble in water to

colourless

solution.



universal

Larutan itu menukarkan warna penunjuk universal dari hijau kepada ungu

.

– Sodium is

reactive

Natrium adalah logam

metal towards oxygen. reaktif

terhadap oksigen.

– Sodium reacts with oxygen to produce sodium oxide . Natrium bertindak balas dengan oksigen membentuk natrium oksida .

Balanced equation / Persamaan kimia seimbang: 4Na + O2 2Na2O – Sodium reacts with water to form alkaline solution, sodium hydroxide. Natrium bertindak balas dengan air membentuk larutan beralkali , natrium hidroksida.

Balanced equation / Persamaan kimia seimbang: Na2O + H2O 2NaOH K

– Potassium burns very brightly with a purple flame to produce

– Potassium is the most reactive towards oxygen.

white solid . Kalium terbakar sangat terang dengan nyalaan ungu menghasilkan pepejal putih .

– The form

white solid

soluble in water to

colourless

solution.



universal

Larutan itu menukarkan warna penunjuk universal dari hijau kepada ungu

.

Kalium adalah logam oksigen.

paling reaktif

metal terhadap

– Potassium reacts with oxygen to produce potassium oxide . Kalium bertindak balas dengan oksigen membentuk kalium oksida .

Balanced equation / Persamaan kimia seimbang: 4K + O2 2K2O – Potassium reacts with water to form alkaline solution, potassium hydroxide. Kalium oksida bertindak balas dengan air membentuk larutan beralkali , kalium hidroksida.

Balanced equation / Persamaan kimia seimbang: K2O + H2O 2KOH

(c) Metal Group 1 reacts with with chlorine to produce metal chloride. Logam Kumpulan 1 bertindak balas dengan klorin menghasilkan logam klorida.

2X + Cl2

2X + Cl2 + 2H2O

X is a metal element of Group 1 (Li, Na and K) 2XCl , X adalah logam unsur Kumpulan 1 (Li, Na dan K)

Combustion spoon / Sudu pembakaran Gas jar / Balang gas Chlorine gas / Gas klorin Burning of metal Group 1

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Observation

Unsur

Li

Inference

Pemerhatian

slowly

Lithium burns flame to produce

red

with a

white

Reactivity

Inferen

solid.

Litium terbakar perlahan dengan nyalaan merah putih menghasilkan pepejal

least reactive

– Lithium is chlorine. .

Litium adalah klorin.

Kereaktifan

metal towards

paling kurang reaktif

terhadap

– Lithium reacts with chlorin to produce lithium chloride . Litium bertindak balas litium klorida .

dengan klorin membentuk

Sodium burns brightly with a yellow flame to produce white solid. terang

– Sodium is

reactive

metal towards chlorine. reaktif

Natrium adalah logam

terhadap klorin.

reacts

with chlorine to produce – Sodium sodium chloride .

dengan nyalaan Natrium terbakar kuning menghasilkan pepejal putih .

Natrium bertindak balas dengan klorin membentuk natrium klorida .

Balanced equation / Persamaan kimia seimbang: 2Na + Cl2 2NaCl K

very brightly with Potassium burns a purple flame to produce white

– Potassium is the most reactive metal towards chlorine.

dengan nyalaan Kalium terbakar ungu menghasilkan pepejal putih .

paling reaktif

Kalium adalah logam klorin.

solid. sangat terang


Na


Balanced equation / Persamaan kimia seimbang: 2Li + Cl2 2LiCl

terhadap

– Potassium reacts with chlorine to produce sodium chloride . Kalium

bertindak balas

kalium klorida

dengan klorin membentuk

.

Balanced equation / Persamaan kimia seimbang: 2K + Cl2 2KCl

GROUP 17 (HALOGENS) / KUMPULAN 17 (HALOGEN) 1

Consist of Fluorine (F2), Chlorine (Cl2), Bromine (Br2), Iodine (I2) and Astatine (At2).

Terdiri dari Fluorin (F2 ), Klorin (Cl2 ), Bromin (Br2 ), Iodin (I2 ) dan Astatin (At2 ). Elements

Symbol

Proton number

Fluorine / Fluorin

F2

Chlorine / Klorin

Cl2

Bromine / Bromin

Unsur

Iodine / Iodin

2


Number of shells

19

2.7

2

17

2.8.7

3

Br2

35

2.8.18.7

4

I2

53

2.8.18.18.7

5

Simbol

Nombor proton

Susunan elektron

Bilangan petala

Physical properties: Halogens cannot conduct heat and electricity in all state.

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Sifat fizik: Halogen tidak boleh mengkonduksi elektrik dan haba dalam semua keadaan.

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Changes in physical properties going down the group / Perubahan sifat fizik menuruni kumpulan: (a) The melting and boiling points are low because the molecules are attracted by weak Van der Waals forces, and small amount of energy is required to overcome these forces. However the melting and boiling points increase going down the group. Takat didih dan takat lebur adalah rendah kerana molekul ditarik oleh tarikan Van der Waals yang lemah, sedikit tenaga diperlukan untuk mengatasi daya itu. Walau bagaimanapun, takat lebur dan takat didih meningkat menuruni kumpulan. Explanation / Penerangan:

–

The atomic size increases molecules get larger. bertambah

Saiz atom

–

shell

going down the Group 17 because of increasing in number of

menuruni kumpulan kerana dengan pertambahan bilangan

petala

, the size

, saiz molekul semakin besar.

The inter molecular forces of attraction (Van der Waals forces) between molecules become stronger. Daya tarikan antara molekul (daya Van der Waals) antara molekul semakin kuat.

–

More heat is needed to overcome the stronger forces between molecules during melting or boiling. Lebih banyak tenaga diperlukan untuk mengatasi daya antara molekul yang lebih kuat semasa peleburan atau pendidihan.

(b) Physical properties change from gas (fluorine and chlorine) to liquid (bromine) and to solid (iodine) at room temperature due to increase in the strength of inter molecular forces from fluorine to iodine. Keadaan fizik berubah dari gas (flourin dan klorin) kepada cecair (bromin) dan kepada pepejal (iodin) pada suhu bilik kerana pertambahan kekuatan tarikan antara molekul dari flourin kepada iodin.

(c) The density is low and increases going down the group. Ketumpatan adalah rendah dan semakin meningkat apabila menuruni kumpulan.

darker (d) The colour of the elements becomes going down the group: fluorine (light yellow), chlorine (greenish yellow), bromine (brown) and iodine (purplish black). Warna unsur semakin iodin (ungu kehitaman).

4

gelap

menuruni kumpulan iaitu flourin (kuning muda), klorin (kuning kehijauan), bromin (perang) dan

Chemical properties of Group 17 elements / Sifat kimia unsur Kumpulan 17: atoms seven (a) All of elements in Group 17 have valence electrons and achieve a stable octet electron one negatively arrangement by accepting electron to form charged ions. atom

Semua menerima

satu

unsur Kumpulan 17 mempunyai elektron membentuk ion bercas

Example / Contoh: (i) Fluorine atom Atom

elektron valens, mencapai susunan elektron oktet yang stabil dengan negatif .

receives one electron to achieve stable duplet electron arrangement:

flourin menerima satu elektron untuk mencapai susunan elektron oktet yang stabil:

F

+e

F–




jumlah cas: +9


Number of electrons = 9,

total charge: –9



jumlah cas: –9


neutral

Fluorine atom is Atom flourin adalah

(ii)

tujuh

Negatively

.

neutral

.

jumlah cas: +9 jumlah cas: –10

charged fluoride ion, F– is formed.

Ion flourida, F– bercas

negatif

terbentuk.

Chlorine atom receives one electron to achieve stable octet electron arrangement: Atom klorin menerima satu elektron untuk mencapai susunan elektron oktet yang stabil:

Cl

+e

Cl–

Electron arrangement / Susunan elektron : 2.8.7 Number of protons = 17, total charge: +17

Electron arrangement / Susunan elektron : 2.8.8 Number of protons = 17, total charge: +17



jumlah cas: +17

jumlah cas: +17





Chlorine atom is Atom klorin adalah

jumlah cas: –17

neutral neutral

. .

Negatively

jumlah cas: –18

charged chloride ion, Cl– is formed.

Ion klorida, Cl– bercas

negatif

terbentuk.

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(b) All elements in Group 17 have similar chemical properties because atoms in Group 17 have seven valence electron and achieve the stable octet electron arrangement by receiving one electron to form a negatively charged ion.

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atom tujuh unsur Kumpulan 17 mempunyai Semua unsur Kumpulan 17 mempunyai sifat kimia yang sama kerana menerima elektron valens sama dalam atom, mencapai susunan elektron oktet yang stabil dengan satu elektron membentuk ion negatif . bercas

5

Reactivity of halogens decreases going down the group / Kereaktifan halogen berkurang menuruni kumpulan: – All the atoms of Group 17 have seven valence electrons and achieve a stable octet electron arrangement by accepting one electron to form negatively charged ion.

– Going down Group 17, the number of Apabila menuruni Kumpulan 17, bilangan

petala

atom menerima elektron.

increases, atomic bertambah,

saiz

size

increases.

atom bertambah.

jauh

dari nukleus.

– The strength of attraction from the proton in the nucleus to attract one electron into the outermost occupied shell becomes weaker . Kekuatan tarikan daripada proton dalam nukleus untuk menarik satu elektron ke dalam petala luar semakin lemah .

– The strength of a halogen atom to attract electron astatine (electronegativity decreases). Kekuatan atom halogen untuk menarik elektron (keelektronegatifan berkurang).

6

decreases

berkurang

Cl

down Group 17

Petala luar semakin

menurun Kumpulan 17

– Outer shell becomes further from the nucleus.

F

decreases

shells

berkurang

kecenderungan

Kereaktifan unsur Kumpulan 17 bergantung pada

of the atom to receive

Reactivity

tendency

– The reactivity of a halogen atom depends on the electron.

Kereaktifan

Semua atom unsur Kumpulan 17 mempunyai tujuh elektron valens dan mencapai susunan elektron oktet yang stabil dengan menerima satu elektron membentuk ion bercas negatif .

Br

from fluorine to

dari fluorin ke astatin

Elements in group 17 exist as diatomic molecules. Two atoms of element sharing one pair of valence electrons to achieve stable octet electron arrangement. Unsur Kumpulan 17 wujud sebagai molekul dwiatom. Dua atom unsur berkongsi sepasang elektron valens untuk mencapai susunan elektron oktet yang stabil.

Example: Two fluorine atoms share one pair of electron to form one fluorine molecule: Contoh: Dua atom fluorin berkongsi sepasang elektron untuk membentuk molekul fluorin:

F

Share / Kongsi kongsi kongsi

Fluorine atom / Atom fluorin

F

F

F

Fluorine Molekul moleculeflorin / Molekul fluorin

Fluorine atom / Atom fluorin

Chlorine, bromine and iodine exists as diatomic molecules. (Cl2, Br2 and I2)

Klorin, bromin dan iodin wujud sebagai molekul dwiatom (Cl2 , Br2 dan I2 )

7

Chemical properties reactions of Group 17 elements / Tindak balas kimia unsur Kumpulan 17: (a) Halogen reacts with water with different reactivity / Halogen bertindak balas dengan air dengan kereaktifan berbeza: X2 + H2O

HX + HOX, X is halogen. (Cl2, Br2 and I2) / X2 + H2O

Chlorine gas / Gas Klorin

HX + HOX, X adalah halogen. (Cl2 , Br2 dan I2 )

Bromine water / Air Bromin

Chlorine

Iodine crystals / Hablur Iodin

Bromine

Gas klorin

Bromin

Water

water

Water air Air

Procedure / Kaedah: – Chlorine gas is passed through water in a test tube. Gas klorin dilalukan melalui air dalam tabung uji.

– The solution produced tested with blue litmus paper. Larutan yang terhasil diuji dengan kertas litmus biru.

Water / Air Procedure / Kaedah: – A few drops of bromine water are added to water in a test tube. Beberapa titis air bromin ditambah kepada air dalam tabung uji.

– The test tube is shaken. Tabung uji digoncang.

– The solution produced tested with blue litmus paper.

Iodine cystals Hablur iodin

Procedure / Kaedah: – Some iodine crystals are added to water in a test tube. Sedikit hablur iodin ditambah kepada air dalam tabung uji.

– The test tube is shaken. Tabung uji digoncang.

– The solution produced tested with blue litmus paper. Larutan yang terhasil diuji dengan kertas litmus biru. n io

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Larutan yang terhasil diuji dengan kertas litmus biru.

Air

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Observation / Pemerhatian: – Chlorine dissolves rapidly in water to form light yellow solution: Klorin larut dengan cepat dalam air menghasilkan larutan berwarna kuning muda:

Cl2 + H2O

Observation / Pemerhatian: – Bromine dissolves slowly in water to form brown solution:

Bromin larut dengan perlahan dalam air menghasilkan larutan berwarna perang:

HCl + HOCl

– The solution changes red litmus paper to

Br2 + H2O

blue

qucikly decolourises it.

slowly decolourises it.

kepada merah dan dengan cepat.

kepada merah dan dengan perlahan.

Larutan menukarkan kertas litmus biru

blue and

Larutan menukarkan kertas litmus biru

melunturkannya

sedikit dalam Iodin larut dengan air menghasilkan larutan berwarna perang:

HBr + HOBr


and

Observation / Pemerhatian: – Iodine dissolves slightly in water to form brown solution:

melunturkannya

I2 + H2O

HI + HOI


blue

. The litmus paper is not decolourises . Larutan menukarkan kertas litmus biru kepada merah . Kertas litmus tidak dilunturkan .

Inference / Inferens: – Chlorine, bromine and iodine react water to form acidic solution. Klorin, bromin dan iodin bertindak balas dengan air membentuk larutan berasid.

– Solubility decreases from chlorine to iodine / Keterlarutan berkurang dari klorin kepada iodin.

(b) Halogens react with sodium hydroxide solution / Halogen bertindak balas dengan larutan natrium hidroksida: X2 + 2NaOH

X2 + 2NaOH

NaX + NaOX + H2O, X2 is halogen. (Cl2, Br2 and I2)

NaX + NaOX + H2O, X2 adalah halogen. (Cl2 , Br2 dan I2 )

Complete the following / Lengkapkan yang berikut: (i)

Cl2 + 2NaOH

(ii)

Br2 + 2NaOH

(iii) I2 + 2NaOH

NaCl + NaOCl + H2O

Reactivity decreases

NaBr + NaOBr + H2O

Kereaktifan berkurang

NaI + NaOI + H2O

(c) Halogens react with hot iron to form brown solid, iron(III) halide. Halogen bertindak balas dengan besi panas membentuk pepejal perang, ferum(III) halida. Iron wool / Wul besi Iodine Iodin

Chlorine or Bromine Klorin atau Bromin

Iron wool

Heat

Wul besi

Haba

NaOH to absorb chlorine/bromine NaOH untuk menyerap klorin/bromin

2Fe + 3X2 2Fe + 3X2

2FeX3, X2 represents any halogen. (Cl2, Br2 or I2)

2FeX3, X2 mewakili sebarang halogen. (Cl2 , Br2 atau I2 )

Halogen

Observation

Halogen

Chlorine Klorin

Bromin

Iron wool burns when cooled.

very brightly

Iodin

dan membentuk pepejal perang

dan membentuk pepejal

2Fe + 3Cl2

2FeCl3

2Fe + 3Br2

2FeBr3

with a dull glow and forms a

perlahan

dan membentuk pepejal perang

2Fe + 3I2

2FeI3

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and forms a brown solid

sangat terang

Iron wools glows slowly brown solid when cooled. Wul besi berbara dengan apabila sejuk.

m

terang

Persamaan kimia

Iron wool burns brightly and forms a brown solid when cooled. Wul besi berbara dengan perang apabila sejuk.

Iodine

Chemical equation

Pemerhatian

Wul besi terbakar dengan apabila sejuk.

Bromine

Heat / Haba

d. Bh


12/9/2011 5:57:55 PM


similar

Experiment (a), (b) dan (c) show that all halogens have decreases going down the group:

chemical properties but their reactivity sama

Eksperimen (a), (b) dan (c) menunjukkan semua halogen menunjukkan sifat kimia yang berkurang apabila menuruni kumpulan. Reactivity

decreases

tetapi kereaktifannya

berkurang

/ Kereaktifan

F2, Cl2, Br2 and I2 / F2 , Cl2 , Br2 dan I2

PERIOD / KALA 1 2 3

Horizontal rows in the periodic table / Baris mendatar dalam Jadual Berkala. There are seven periods known as Period 1, 2, 3, 4, 5, 6, 7 / Terdapat 7 kala ditulis sebagai Kala 1, 2, 3, 4, 5, 6, 7. The number of period of an element represents the number of shells occupy with electrons in each atom of element. Nombor kala suatu unsur mewakili bilangan petala yang diisi oleh elektron di dalam setiap atom unsur. Elements

Proton number

Unsur

4

Nombor proton

Number of shells

Period

Susunan elektron

Bilangan petala

Kala

Li

3

2.1

2

2

Na

11

2.8.1

3

3

K

19

2.8.8.1

4

4

Period 3 elements (complete the following table): / Unsur Kala 3 (lengkapkan jadual berikut): Elements / Unsur

Na

Mg

Al

Si

P

S

Cl

Ar

11

12

13

14

15

16

17

18

2.8.1

2.8.2

2.8.3

2.8.4

2.8.5

2.8.6

2.8.7

2.8.8

3

3

3

3

3

3

3

3

+11

+12

+13

+14

+15

+16

+17

+18

0.191

0.160

0.130

0.118

0.110

0.102

0.099

0.095



Number of shells Bilangan petala

Positive charge in the nucleus Bilangan cas positif dalam nukleus

Radius (nm) Jejari (nm)

5


Physical changes across the Period 3 (from left to right) / Perubahan fizik merentasi Kala 3(dari kiri ke kanan): (a) Change in atomic radius across Period 3 / Perubahan jejari atom merentasi Kala 3: The atomic radius of the atoms decreases from sodium to chlorine berkurang dari natrium kepada klorin

Jejari atom

Na

Mg

Al

Si

S

P

Cl

16 p

Atom / Atom

Na

Mg

Al

Si

P

S

Cl

Number of proton / Bilangan proton

11 p

12 p

13 p

14 p

15 p

16 p

17 p

Positive charge / Cas positif

+11

+12

+13

+14

+15

+16

+17


2.8.1

2.8.2

2.8.3

2.8.4

2.8.5

2.8.6

2.8.7

–

All the atoms of elements have 3

shells

occupied with electrons

.

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Semua atom unsur mempunyai

3

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–

The proton number

increases

by one unit from sodium to chlorine.

Nombor proton bertambah satu unit dari natrium kepada klorin.

– –

Increase in proton number causes the number of

positive

Pertambahan nombor proton menyebabkan bilangan cas

positif

increase

charge in the nucleus to

.

bertambah .

pada nukleus

increases .

The strength of attraction from the proton in the nucleus to the electrons in the shells bertambah . across period 3 / Jejari atom unsur berkurang

Daya tarikan proton dalam nukleus terhadap elektron dalam petala

– The size of atom decreases (b) Change in electronegativity / Perubahan keelektronegatifan: –

The atomic radius decreases sodium to chlorine. berkurang

Jejari atom klorin.

–

elektron

berkurang

Keelektonegatifan

bertambah

dari natrium kepada

dari natrium kepada klorin.

increases

Tendency of atoms to attract electron to the outermost shells The electronegativity

ke arah nukleusnya.

from sodium to chlorine.

bertambah

Kekuatan nukleus menarik elektron kepada petala paling luar

–

towards its nucleus.

due to the increasing nuclei attraction on the electrons in the shells from

kerana daya tarikan nukleus terhadap elektron dalam petala

decreases

The size of atom Saiz atom

–

electron

Electronegativity: The strength of an atom in a molecule to attract Kelektronegatifan: Kekuatan suatu atom dalam molekul menarik

–

merentasi Kala 3.

increases

bertambah

from sodium to chlorine.

dari natrium kepada klorin.

across Period 3 from sodium to chlorine.

merentasi Kala 3 dari natrium kepada klorin.

(c) Physical state / Keadaan fizik: (i) The physical state of elements in a period changes from solid to gas from left to right. Keadaan fizik unsur-unsur dalam suatu kala berubah dari pepejal kepada gas dari kiri ke kanan.

(ii)

Metals on the left are solid while non-metals on the right are usually gases. Logam di sebelah kiri adalah pepejal dan bukan logam di sebelah kanan kebanyakannya adalah gas.

(d) Changes in metallic properties and electrical conductivity / Perubahan sifat kelogaman dan kekonduksian elektrik: Na

Element / Unsur

Al

Si

P

S

Cl

Ar

Metallic properties

Semi metal

Non-metal

Sifat kelogaman

Logam

Separa logam

Bukan logam

Electrical conductivity

Good conductors of electric.

Weak conductor of electric but it increases with the presence of boron or phosphorous.

Cannot conduct electricity

Konduktor elektrik yang baik.

Konduktor elektrik yang lemah tetapi bertambah dengan kehadiran boron atau fosforus. Uses: semi-conductor / Kegunaan: semi konduktor

Kekonduksian elektrik

6

Mg

Metal

Tidak boleh mengkonduksi elektrik

Changes in properties of oxide of elements Period 3 / Sifat oksida unsur Kala 3: Na

Mg

Basic oxide / Oksida bes Basic oxide + Water Oksida bes + Air

Alkali

Alkali

Oksida bes + Asid

Si

Amphoteric oxide + Acid Oksida amfoterik + Asid Oksida amfoterik + Alkali

Salt + Water

Garam + Air

Example / Contoh: MgO + 2HCl MgCl2 + H2O

Salt + Water

Garam + Air

Amphoteric oxide + Alkali

Example / Contoh: Na2O + H2O 2NaOH Basic oxide + Acid

Al Amphoteric oxide / Oksida amfoterik

P

Acidic oxide + Water Oksida asid + Air

Salt + Water

Garam +Air

Example / Contoh: Al2O3 + 6HNO3 2Al(NO3)3 +3H2O Al2O3 + 2NaOH 2NaAlO2 + H2O

S

Cl

Acidic oxide / Oksida asid Acid

Asid

Example / Contoh: SO2 + H2O H2SO3 Acidic oxide + Alkali Oksida asid + Alkali

Salt + Water

Garam + Air

Example / Contoh: SiO2 + 2NaOH Na2SiO3 + H2O

(a) Elements in Period 3 can be classified as metals and non-metals based on basic and acidic properties of their oxides / Unsur Kala 3 boleh dikelaskan sebagai logam dan bukan logam berdasarkan sifat kebesan dan keasidan oksidanya. acid salt (i) Basic oxide is metal oxide that can react with to form and water .

m

asid

membentuk

garam

dan

air

.

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Oksida bes adalah oksida logam yang boleh bertindak balas dengan

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(ii)

Acidic oxide is non-metal oxide that can react with

alkali

to form

Oksida asid adalah oksida bukan logam yang boleh bertindak balas dengan air .

membentuk

to form

and

alkali

Oksida amfoterik adalah oksida yang boleh bertindak balas dengan air . dan

asid

dan

water .

and

alkali

acid

(iii) Amphoteric oxide is oxide that can react with both

salt

alkali

garam

salt

dan

and

water . garam

untuk membentuk

(b) Complete the following table / Lengkapkan jadual berikut: (i) Reaction with water / Tindak balas dengan air: Oxide

Solubility in water

Oksida

Sodium oxide, Na2O

pH

Type of oxide

pH larutan

Jenis oksida

14

Basic oxide

9

Basic oxide

–

–

–

–

3

Acidic oxide

3

Acidic oxide

White solid dissolve in water

Natrium oksida, Na2O

Pepejal putih larut dalam air

Magnesium oksida, MgO

White solid slightly dissolve in water

Magnesium oksida, MgO

Pepejal putih larut separa dalam air

Aluminium oxide, Al2O3

Insoluble

Aluminium oksida, Al2O3

Tidak larut

Silicon oxide, SiO2

Insoluble

Silikon oksida, SiO2

Tidak larut

Phosphorous oxide, P4O10


Fosforus oksida, P4O10


Sulphur dioxide, SO2


Sulfur dioksida, SO2

(ii)

Keterlarutan dalam air


Reaction between the oxide of elements Period 3 with nitric acid and sodium hydroxide solution: Tindak balas antara oksida unsur Kala 3 dengan asid nitrik dan larutan natrium hidroksida: Observation / Pemerhatian Oxide Oksida

Reaction with dilute nitric acid

Reaction with sodium hydroxide solution

The white solid dissolve to form colourless solution.

No change. The white solid does not dissolve.

Tindak balas dengan asid nitrik cair

Magnesium oxide, MgO Magnesium oksida, MgO

Pepejal putih larut membentuk larutan tanpa warna.

Aluminium oxide, Al2O3 Aluminium oksida, Al2O3

Silicon oxide, SiO2

Tiada perubahan. Pepejal putih tidak larut.



Jenis oksida

Tindak balas dengan natrium hidroksida


The white solid dissolve to form colourless solution. Pepejal putih larut membentuk larutan tanpa warna.

The white solid dissolve to form colourless solution. Pepejal putih larut membentuk larutan tanpa warna.

Basic oxide

Amphoteric oxide

Acidic oxide

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Silikon oksida, SiO2


Type of oxide

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7

Steps to compare and explain the change in atomic size/ radius/ electronegativity across Period 3, reactivity down Group 1 and Group 17: Langkah-langkah untuk membanding dan menerangkan perubahan saiz atom/ jejari/ keelekronegatifan merentasi Kala 3, kereaktifan menuruni Kumpulan 1 dan Kumpulan 17:

16 p

Na

K

(a) To Compare Atomic Size/ Radius and Electronegativity Across Period 3: Membanding Jejari/ Saiz Atom dan Keelektronegatifan Merentasi Kala 3: (i) Compare number of shells in each atom. Bandingkan bilangan petala dalam setiap atom. (ii) Compare number of proton in the nucleus. Bandingkan bilangan proton dalam nukleus. (iii) Compare the strength of attraction from the nucleus to the electrons in the shells . Bandingkan kekuatan tarikan dari proton dalam nukleus kepada elektron dalam petala . (iv) Compare the atomic size/ Compare the electronegativity. Bandingkan saiz atom/ Bandingkan keelektronegatifan. (b) To Compare Reactivity Down Group 1 and Group 17: Membanding Kereaktifan Menuruni Kumpulan 1 dan Kumpulan 17: (i)

Compare number of shells in each atom. Bandingkan bilangan petala dalam setiap atom.

(ii)

Compare the strength of proton in the nucleus to attract valence electron (Group 1)// to attract electron to the outermost shells (Group 17). Bandingkan kekuatan proton dalam nukleus menarik elektron valens (Kumpulan 1) // menarik elektron ke petala paling luar (Kumpulan 17).

(iii) Compare tendency of the atom to release electron (Group 1)// receive electron (Group 17). Bandingkan kecenderungan atom untuk melepaskan elektron (Kumpulan 1) // menerima elektron (Kumpulan 17).

Reactivity decreases down Group 17/Kereaktifan berkurang menurun Kumpulan 17

Reactivity increases down Group 1/Kereaktifan bertambah menurun Kumpulan 1

Li

Atomic radius of the atoms decreases across Period 3 from sodium to chlorine. Jejari atom berkurang merentasi Kala 3 dari natrium kepada klorin.

F

Cl

Br

TRANSITION ELEMENT / UNSUR PERALIHAN 1

Situated between Group 2 and 13. The examples of transition element are Sc, Ti, V, Cr, Mn, Fe, Co, Ni, Cu and Zn. Terletak antara Kumpulan 2 dan 13. Contoh unsur peralihan adalah Sc, Ti, V, Cr, Mn, Fe, Co, Ni, Cu dan Zn.

2

Show metal properties: Shiny, conducts heat and electricity, malleable, high tensile strength, high melting point and density. Mempunyai sifat logam: Permukaan berkilat, konduktor haba dan elektrik, bersifat mulur, boleh ditempa, kekuatan tegangan yang tinggi, takat lebur dan ketumpatan tinggi. Special characteristics / Ciri istimewa: (a) Form coloured compound / Membentuk sebatian berwarna. Example / Contoh:

3

Iron(III) chloride is brown, iron(II) chloride is green and copper(II) sulphate is blue. Ferum(III) klorida adalah perang, ferum(II) klorida adalah hijau dan kuprum(II) sulfat adalah biru. (b) Form different oxidation numbers / Membentuk nombor pengoksidaan berbeza. (c) Form complex ions: MnO4–, Cr2O72–, CrO42–, etc / Membentuk ion kompleks: MnO4–, Cr2O72–, CrO42–, dan sebagainya. (d) Useful as a catalyst in industries / Berguna sebagai mangkin dalam industri. Example / Contoh: Iron: Haber Process in the manufacture of ammonia / Ferum: Proses Haber dalam penghasilan ammonia.

N2 + 3H2

Fe

2NH3

Vanadium(V) Oxide: Contact Process in the manufacture of sulphuric acid. Vanadium(V) Oksida: Proses Sentuh dalam penghasilan asid sulfurik.

m

Platinum: Ostwald Process in the manufacture of nitric acid / Platinum: Proses Ostwald dalam penghasilan asid nitrik.

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2SO2 + O2

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The diagram below shows the electron arrangement for atoms P and Q. Rajah di bawah menunjukkan susunan elektron bagi atom P dan Q.

P

Q

(a) Elements P and Q are placed in the same group in Periodic Table. State the group. Unsur P dan Q terletak dalam kumpulan yang sama dalam Jadual Berkala. Nyatakan kumpulan itu.

Group 1 (b) How is elements P and Q kept in the laboratory? Give reason for your answer. Bagaimanakah unsur P dan Q disimpan di dalam makmal? Berikan sebab bagi jawapan anda.

In paraffin oil. To prevent them from reacting with oxygen or water vapour in the atmosphere. (c) (i)

Write chemical equation for the reaction between elements P with water. Tuliskan persamaan kimia untuk tindak balas antara unsur P dengan air.

2P + 2H2O

(ii)

2POH + H2

What is the expected change of colour when a few drops of phenolphthalein are added into the aqueous solution of the product? Explain your answer. Apakah perubahan warna yang dijangkakan apabila beberapa titik fenolftalein dititiskan ke dalam larutan akueus yang terhasil? Terangkan jawapan anda.

Colourless to purple/ pink. The solution formed is alkaline. (iii) Between element P and element Q, which is more reactive in the reaction with water? Antara unsur P dan Q, yang manakah lebih reaktif apabila bertindak balas dengan air?

Element Q is more reactive than P. (iv) Explain your answer in (c)(iii) / Terangkan jawapan anda dalam (c)(iii). The size of atom Q is larger than atom P. The valence electron of atom Q is further away from the nucleus compared to atom P. The attraction forces between proton in the nucleus to the valence electron of atom Q is weaker than atom P. Atom Q is easier to release the valence electron compared to atom P. (d) Name one element that has the same chemical properties as P and Q. Namakan satu elemen yang mempunyai ciri-ciri kimia yang sama dengan P dan Q.

Potassium 2

The diagram below shows the information regarding elements W and X which are from the same group in the Periodic Table. Rajah di bawah menunjukkan maklumat mengenai unsur W dan X yang terletak di kumpulan yang sama dalam Jadual Berkala. 19 9

(a) (i) (ii)

W

35 17

X

Write the electron arrangement of atom of elements W and X / Tuliskan susunan elektron bagi atom unsur W dan X. 2.7 2.8.7 Atom W / Atom W : Atom X / Atom X : State the position of elements W and X in the Periodic Table. Nyatakan kedudukan unsur W dan X dalam Jadual Berkala.

Element W / Unsur W : Group 17, Period 2 Element X / Unsur X : Group 17, Period 3

(iii) Do elements W and X show the same chemical property? Explain your answer. Adakah unsur W dan X menunjukkan sifat kimia yang serupa? Terangkan jawapan anda.

Elements W and X have the same chemical property. Atoms W and X have the same number of valence electrons.

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(b) State the type of particles of substances W and X / Nyatakan jenis zarah yang terdapat pada W dan X. Molecule

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(c) (i)

Compare the boiling point of elements W and X. Explain the difference. Bandingkan takat didih unsur W dan X. Terangkan perbezaan itu.

The boiling point of element X is higher than element W. The size of molecule X2 is bigger than molecule W2 . The forces of attraction between molecules X2 is stronger than molecule W2. More heat energy is needed to overcome the stronger forces between molecules. (d) (i)

Element X can react with sodium element to form a compound. Write the chemical equation for the reaction. Unsur X boleh bertindak balas dengan unsur natrium untuk membentuk sebatian. Tulis persamaan kimia untuk tindak balas tersebut.

X2 + 2Na (ii)

2NaX

How does the reactivity of element W and element X differ? Explain your answer. Bagaimanakah kereaktifan unsur W dan X berbeza? Terangkan jawapan anda.

Element W is more reactive than element X. The size of atom W is smaller than atom X. The outermost occupied shell of atom W is nearer to the nucleus compare to atom X. The strength of the nucleus of atom W to attract electron to the outermost shell is stronger than atom X. 3

The table below shows the number of neutron and relative atomic mass of eight elements represented as P, Q, R, S, T, U and W. Jadual di bawah menunjukkan bilangan neutron dan jisim atom relatif bagi lapan unsur yang diwakili oleh huruf P, Q, R, S, T, U, V dan W. Atom / Unsur Number of neutron Bilangan neutron dalam atom

Relative atomic mass Jisim atom relatif

Number of proton Bilangan proton dalam atom

Electron arrangement Susunan elektron dalam atom

P

Q

R

S

T

U

V

W

12

12

14

14

16

16

18

22

23

24

27

28

31

32

35

40

11

12

13

14

15

16

17

18

2.8.1

2.8.2

2.8.3

2.8.4

2.8.5

2.8.6

2.8.7

2.8.8

(a) Complete the above table by writing the number of proton and electron arrangement for the atom of each element. Lengkapkan jadual di atas dengan menulis bilangan proton dan susunan elektron bagi atom setiap unsur.

(b) (i)

State the period of element P – W in the Periodic Table. Explain your answer. Nyatakan kala manakah unsur P – W terletak dalam Jadual Berkala? Terangkan jawapan anda.

Period 3 because P – W atoms have three shells occupied with electrons. (ii)

What is the proton number of another element that is in the same group as P? Nyatakan bilangan proton bagi unsur lain yang sama kumpulan dengan P.

3/19 (c) Write the standard representation for element Q / Tuliskan simbol perwakilan piawai untuk unsur Q. 24 Q 12 (d) Which element exist as / Unsur yang manakah wujud sebagai W monoatomic gas / gas monoatom? (e) (i)

diatomic gas / gas dwiatom?

T/ U/ V

Which element can react vigorously with water to produce hydrogen gas? Unsur yang manakah bertindak balas cergas dengan air untuk menghasilkan gas hidrogen?

P

m

Write the balanced equation for the reaction in (e)(i) / Tuliskan persamaan seimbang untuk tindak balas (e)(i). 2P + 2H2O 2POH + H2

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(f)

State the arrangement of elements T, U and V in the order of increasing atomic radius. Explain your answer. Nyatakan susunan unsur T, U dan V dalam tertib pertambahan jejari atom. Terangkan jawapan anda.

V, U and T. Atoms of T, U, and V have three shells occupied with electrons. The proton number // positive charges in the nucleus increases from T to V. The forces of attraction between proton in the nucleus and the electrons in the shells increase from T to V. The shells filled with electrons are pulled nearer to the nucleus from T to V. 4

The diagram below shows part of the Periodic Table of Elements. X, Y, A, B, D, E, F and G do not represent the actual symbols. Rajah di bawah menunjukkan sebahagian daripada Jadual Berkala Unsur. X, Y, A, B, D, E, F dan G tidak mewakili simbol yang sebenar. X Y A F

(a) (i) (ii)

B

D

E

G

State the position of element B in the Periodic Table / Nyatakan kedudukan unsur B dalam Jadual Berkala. Period 3, Group 13 Explain your answer in (a)(i) / Terangkan jawapan anda dalam (a)(i). Electron arrangement atom B is 2.8.3. Atom B has three valence electrons, element B is in Group 13. Atom B has three shells occupied with electrons, element B is in Period 3.

(b) (i) (ii)

Which element is monatomic gas / Unsur yang manakah adalah gas monoatom? Element Y Explain your answer in (b)(i) / Terangkan jawapan anda dalam (b)(i). Atom Y has achieved octet electron arrangement // has electron arrangement 2.8.

(c) Choose an element that / Pilih unsur yang: (i)

exists in the form of molecule / wujud dalam bentuk molekul

(ii)

forms acidic oxide / membentuk oksida asid

(iii)

has atoms that have no neutron / atom yang tiada neutron

(iv)

is an alkali metal / logam alkali

(v)

forms amphoteric oxide / membentuk oksida amfoterik

B

(vi)

has a proton number of 15 / mempunyai nombor proton 15

C

(vii) is most electropositive / paling elektropositif (viii) forms basic oxide / membentuk oksida bes (ix)

forms coloured compound / membentuk sebatian berwarna

X/D/E D/E X A/F

F A/F G

(d) Arrange Y, A, B, D and E according to the order of increasing atomic size. Susun Y, A, B, D dan E mengikut tertib pertambahan saiz atom.

Y, E, D, B, A (e) (i) (ii)

Write the electron arrangement for an atom of element / Tulis susunan elektron bagi atom unsur: 2.8.5 2.8.7 D: E: Compare electronegativity of elements D and E / Bandingkan keelektronegatifan unsur D dan E. Element E is more electronegative than element D.

(iii) Explain your answer in (e)(ii) / Terangkan jawapan anda dalam (e)(ii). Atoms E and D have the same number of shells occupied with electrons. The number of proton in the nucleus of atom E is more than atom D. The strength of proton in nucleus to attract electrons to the outermost shells n io

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Proton number of element P is 8. What is the position of this element in the Periodic Table of Elements?

Period / Kala

16

2

A

2

B

16

3

C

18

2

D

18

3

Potassium reacts with element Q from Group 17 in Periodic Table. Which of the following chemical equations is correct? Kalium bertindak balas dengan unsur Q dalam Kumpulan 17 dalam Jadual Berkala Unsur. Antara persamaan kimia berikut, yang manakah betul?

A K + Q KQ B K+ + Q – KQ 3

C 2K + Q2 2KQ D K + Q2 KQ2

Y oxide

Z oxide

A

Amphoteric

Acidic

Basic

Amfoterik

Asid

Bes

B

Amphoteric

Basic

Acidic

Amfoterik

Bes

Asid

C

Acidic

Amphoteric

Basic

Asid

Amfoterik

Bes

D

Acidic

Acidic

Basic

Asid

Asid

Bes

Oksida X

Nombor proton unsur P adalah 8. Apakah kedudukan unsur P dalam Jadual Berkala Unsur?

Group/Kumpulan

X oxide

6

Jadual di bawah menunjukkan sifat oksida unsur X, Y dan Z yang berada dalam Kala 3 Jadual Berkala Unsur.

X Y Z

Antara pernyataan berikut, yang manakah benar?

I II

A B C D 5

Calcium / Kalsium III Potassium / Kalium Sulphur / Sulfur IV Nitrogen / Nitrogen I and II only / I dan II sahaja I and III only / I dan III sahaja II and IV only / II dan IV sahaja III and IV only / III dan IV sahaja

The diagram below shows the standard representation for elements X, Y and Z.

– Oxide of Z reacts with sodium hydroxide solution. Oksida Z bertindak balas dengan larutan natrium hidroksida.

– Oxide of Z reacts with nitric acid.

A All the elements can conduct electricity.

Antara berikut, yang manakah dapat membentuk oksida asid?

Oksida Y bertindak balas dengan larutan natrium hidroksida.

Oksida Y tidak bertindak balas dengan asid nitrik.

Z

Which of the following elements can form acidic oxide?

– Oxide of Y reacts with sodium hydroxide solution. – Oxide of Y does not react with nitric acid

Which of the following statements is true?

4

Oksida X bertindak balas dengan asid nitrik.

natrium hidroksida.

Y

D

– Oxide of X reacts with nitric acid.

– Oxide of X does not react with sodium hydroxide solution./Oksida X tidak bertindak balas dengan larutan

Rajah di bawah menunjukkan kedudukan unsur X, Y dan Z dalam Jadual Berkala Unsur.

C

Property of the oxide formed Sifat oksida yang terbentuk

Element Unsur

The diagram below shows the position of elements X, Y and Z in the Periodic Table.

B

Oksida Z

The table below shows the properties of the oxide of elements X, Y and Z which are located in Period 3 of the Periodic Table.

X

Semua unsur boleh mengkonduksi elektrik. All the elements exist as gas at room temperature. Semua unsur wujud dalam bentuk gas pada suhu bilik. The boiling points of the elements increase from X Y Z. Takat didih unsur bertambah dari X → Y → Z. The density of the elements decreases going down from X Y Z. Ketumpatan unsur berkurang dari X → Y → Z.

Oksida Y

Oksida Z bertindak balas dengan asid nitrik.

What is the correct arrangement of elements X, Y and Z from left to right in Period 3 of the Periodic Table? Apakah susunan yang betul bagi unsur X, Y dan Z dari kiri ke kanan Kala 3 Jadual Berkala Unsur?

A Z, X, Y B X, Z, Y 7

C X, Y, Z D Y, Z, X

The following statements describe the characteristic of an element: Pernyataan berikut menerangkan sifat suatu unsur.

– Used as a catalyst / Digunakan sebagai mangkin. – Forms coloured ions or compound. Membentuk ion atau sebatian berwarna.

– Shows different oxidation number in its compound. Menunjukkan numbor pengoksidaan yang berbeza.

Which of the following is the position of the element in the Periodic Table of Element? Antara berikut, yang manakah adalah kedudukan unsur tersebut dalam Jadual Berkala Unsur?

Rajah di bawah menunjukkan simbol unsur X, Y dan Z. 27 13

X

32 16

Y

23 11

Z

What type of oxides are formed by X, Y and Z? m

A B

C

D

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8

The table below shows the proton number of elements in Period 3 of the Periodic Table of Elements. Jadual di bawah menunjukkan nombor proton unsur dalam Kala 3 Jadual Berkala Unsur.

10 The table below shows the proton numbers of elements X and Y.

Jadual di bawah menunjukkan nombor proton unsur X dan Y.

Elements / Unsur

Proton number / Nombor proton

Jejari (nm)

X

11

11

0.191

Y

19

Mg

12

0.160

Which statements are true about elements X and Y?

Al

13

0.130

Antara pernyataan berikut, yang manakah benar tentang unsur X dan Y?

Si

14

0.118

I

P

15

0.110

S

16

0.102

Cl

17

0.099

Ar

18

0.095

Elements

Proton number

Radius (nm)

Na

Unsur

Nombor proton

II III IV

Why does the atomic radius of the atoms decrease from sodium to argon in the period? Mengapakah saiz atom berkurang dari natrium ke argon dalam kala?

A The number of valence electrons increases. Bilangan elektron valens bertambah. The electronegativity of the elements increases. Keelektronegatifan unsur bertambah. The properties of the elements change from metallic to non-metallic. Sifat unsur berubah dari logam kepada bukan logam. The strength of attraction of the nucleus to the electrons in the shells increases. Kekuatan tarikan nukleus kepada elektron dalam petala bertambah.

B C D

9

A B C D

Atoms X and Y have one valence electron. Atom X dan Y mempunyai satu elektron valens. Elements X is more reactive than element Y. Unsur X lebih reaktif daripada unsur Y. Atom X has a bigger atomic size than atom Y. Saiz atom X lebih besar daripada saiz atom Y. Elements X and Y are in the same group in the Periodic Table. Unsur X dan Y berada dalam kumpulan sama dalam Jadual Berkala. I and III only / I dan III sahaja I and IV only / I dan IV sahaja II and III only / II dan III sahaja II and IV only / II dan IV sahaja

The table below shows proton number for elements P, Q and R. Jadual di bawah menunjukkan nombor proton bagi unsur P, Q dan R.

Elements / Unsur


P

11

Q

17

R

19

Which of the following statements about these elements are true? Antara pernyataan berikut, yang manakah benar tentang unsur-unsur tersebut?

I II III IV

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A B C D

P and R has the lowest number of valence electrons. P dan R mempunyai bilangan elektron valens paling rendah. P and R have similar chemical properties. P dan R mempunyai sifat kimia yang sama. Size of atom R is bigger than size of atom Q. Saiz atom R lebih besar daripada saiz atom Q. Element R is more electronegative than element Q. Unsur R lebih elektronegatif daripada unsur Q. I, II and III / I, II dan III I, II dan IV / I, II dan IV I, III dan IV / I, III dan IV II, III dan IV / II, III dan IV

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CHEMICAL BOND

4

IKATAN KIMIA

TYPE OF CHEMICAL BOND / JENIS IKATAN KIMIA • IONIC BOND / IKATAN ION – To predict the formulae of ionic compounds based on the electron arrangement. Meramal formula sebatian ion berdasarkan susunan elektron. – To describe the formation of ionic bond / Menghuraikan pembentukan ikatan ion. – To draw the diagram of the formation of ionic bond / Melukis rajah pembentukan ikatan ion. • COVALENT BONDS / IKATAN KOVALEN – To predict the formulae of molecules of elements or covalent compounds as well as the types of covalent bond. Meramal formula molekul unsur atau molekul sebatian kovalen serta jenis ikatan kovalen. – To describe the formation of covalent bonds / Menghuraikan pembentukan ikatan kovalen. – To draw the diagram of the formation of covalent bonds / Melukis rajah pembentukan ikatan kovalen.

PROPERTIES OF IONIC AND COVALENT COMPOUNDS / SIFAT SEBATIAN ION DAN KOVALEN • IONIC COMPOUNDS / SEBATIAN ION – To state and explain the properties from the aspect of melting point and electrical conductivity in solid and molten state. Menyatakan dan menerangkan sifat dari segi takat lebur, kekonduksian elektrik dalam keadaan pepejal dan leburan. • COVALENT COMPOUNDS / SEBATIAN KOVALEN – To state the solubility in water and organic solvents / Menyatakan keterlarutan dalam air dan pelarut organik. – To differentiate between ionic and covalent compounds / Membezakan sebatian ion dengan sebatian kovalen.

CHEMICAL BONDS BETWEEN ATOMS / IKATAN KIMIA ANTARA ATOM

Chemical bonds are formed when two or more atoms of elements bond together. Atoms form chemical bonds to achieve a stable duplet or octet electron arrangement. There are two types of chemical bond, that is Ionic Bond and Covalent Bond.

1

Ikatan kimia dibentuk apabila dua atau lebih atom-atom unsur berpadu. Atom-atom membentuk ikatan kimia untuk mencapai susunan elektron yang stabil iaitu susunan elektron duplet atau oktet. Terdapat dua jenis ikatan kimia iaitu Ikatan Ion dan Ikatan Kovalen.

IONIC BOND / IKATAN ION

Ionic bond is formed between atoms of metal elements that release electrons to atoms of non-metal elements.

1

Ikatan ion terbentuk antara atom unsur logam yang melepaskan elektron kepada atom unsur bukan logam yang menerima elektron.

Atom of an element is neutral because the number of protons is equal to the number of electrons.

2

Atom suatu unsur adalah neutral kerana bilangan proton adalah sama dan dengan bilangan elektron.

Atoms of elements that release the electrons form positive ions and atoms that receive the electrons form negative ions to achieve a stable octet or duplet electron arrangement:

3

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Atom unsur yang melepaskan elektron membentuk ion positif dan atom yang menerima elektron membentuk ion negatif untuk mencapai susunan elektron oktet atau duplet yang stabil.

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Complete the following table / Lengkapkan jadual di bawah: Changes

Na

Na+ + e

Ca


2.8.1

2.8

2.8.2

Total of positive charges (From number of proton)

+11

+11

Total of negative charges (From number of proton)

–11

Perubahan

Ca2+ + 2e

O + 2e

O2–

Cl + e

Cl–

2.8

2.6

2.8

2.8.7

2.8.8

+12

+12

+8

+8

+17

+17

–10

–12

–10

–8

–10

–17

–18

0

+1

0

+2

0

–2

0

–1

Sodium atom

Sodium ion

Calcium atom

Oxygen atom

Oxide ion

Chlorine atom

Susunan elektron

Jumlah cas positf (Dari bilangan proton)

Jumlah cas negaitf (Dari bilangan proton)

Total changes Jumlah cas

Type of particles Jenis zarah

Atom natrium

3

Atom kalsium

Calcium ion

Atom oksigen

Atom klorin

Chlorine ion

The positive ions and negative ions are attracted to one another with strong electrostatic forces. The electrostatic force between the positive and negative ions forms ionic bond. Ion positif dan ion negatif tertarik antara satu sama lain dengan daya elekrostatik yang kuat. Daya elektrostatik antara ion positif dan ion negatif membentuk ikatan ion.

4

Ionic bond is usually formed between atoms from Groups 1, 2 and 13 (metal) with atoms from Groups 15, 16 and 17 (non-metal). Ikatan ion biasanya dibentuk antara atom-atom daripada Kumpulan 1, 2 dan 13 (logam) dengan atom-atom dari Kumpulan 15, 16 dan 17 (bukan logam).

5

The maximum number of electrons transferred in the formation of ionic bond is usually three: Bilangan maksimum elektron yang berpindah dalam pembentukan ikatan ion biasanya tiga.

(a) Atoms of elements in Groups 1, 2 and 13 release 1, 2 and 3 electrons respectively to form positively charged ions (+1, +2 and +3). Atom unsur Kumpulan 1, 2 dan 13 masing masing melepaskan 1, 2 dan 3 elektron membentuk ion bercas positif (+1, +2 dan +3).

(b) Atoms of elements in Groups 15, 16 and 17 receive 3, 2 and 1 electrons respectively to form negatively charged ions (–3, –2 and –1) Atom unsur Kumpulan 15, 16 dan 17 masing-masing menerima 3, 2 dan 1 elektron membentuk ion bercas negatif (–3, –2 dan –1). 6

Examples / Contoh-contoh: (i) Sodium chloride / Natrium klorida Predict the formula / Ramal formula: Element

Proton number


Na

11

2.8.1

Cl

17

2.8.7

Unsur

Nombor proton

Susunan elektron

Na Cl + e

Na+ + e Cl–

Na+

Cl–

1

1

NaCl

Draw the electron arrangement of the compound formed. Lukiskan susunan elektron bagi setiap sebatian yang terbentuk.

Transfer Pindah Na Na

C1 Cl

Na Na

C1 Cl

Sodium atom, Na Atom natrium, Na

Chlorine atom,Cl Cl Atom klorin,

Sodium ion, Na Ion natrium, Na +

Chloride ion,Cl Cl– Ion klorida,

–

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Explanation / Penerangan:

2.8.1

(a) Electron arrangement of sodium atom is stable Therefore sodium atom is not

one

. Sodium atom has one . Sodium atom releases

octet electron arrangement to form sodium ion , Na+ with electron arrangement

valence electron.

electron to achieve a stable 2.8 .

2.8.1 . Atom natrium mempunyai satu elektron valens. Dengan itu atom Susunan elektron atom natrium ialah stabil . Atom natrium melepaskan satu elektron ini untuk mencapai susunan elektron oktet yang natrium tidak stabil membentuk ion

natrium , Na+ dengan susunan elektron

2.8

.

2.8.7

seven (b) Electron arrangement of chlorine atom is . Chlorine atom has valence electrons. one Chlorine atom receives electron to achieve stable octet electron arrangement to form chlorine

2.8.8

ion, Cl– with an octet arrangement of electron 2.8.7

Susunan elektron bagi atom klorin ialah

. tujuh

. Atom klorin mempunyai elektron valens. Atom klorin satu elektron membentuk ion klorida , Cl– dengan

mencapai susunan elektron oktet yang stabil dengan menerima 2.8.8 . susunan elektron

(c)

Sodium ions , Na+ and chloride ions , Cl– ions are attracted with strong bond formed is called ionic bond. Ion natrium , Na+ dan ikatan ion.

electrostatic

force. The

ion klorida , Cl– ditarik dengan daya elektrostastik yang kuat. Ikatan yang terbentuk dinamakan

(ii) Magnesium oxide / Magnesium oksida Predict the formula / Ramal formula: Element

Proton number


Mg

12

2.8.2

O

8

2.6

Unsur

Nombor proton

Susunan elektron

Mg Mg+ + 2e O + 2e O2–

Draw the electron arrangement of the compound formed.

Mg2+

O2–

2

2

1

1

MgO

Lukiskan susunan elektron bagi setiap sebatian yang terbentuk. 2+

Mg Mg

Pindah Transfer

Magnesium atom, Mg Atom magnesium, Mg

O O

2−

O O

Mg Mg

Oxygen atom, OO Atom oksigen,

Magnesium ion, Mg Mg 2+ Ion magnesium, 2+

2– 2− Oxide ion, OO Ion oksida,


2.8.2 (a) The electron arrangement of magnesium atom is . Magnesium atom has stable electrons. Therefore magnesium atom is not . Magnesium atom releases electrons to achieve a stable octet electron arrangement to form 2.8 arrangement .

magnesium ion , Mg

2+

two

valence

2

valence

with electron

2.8.2 . Atom magnesium mempunyai dua elektron di petala terluar. Maka atom Susunan elektron atom magnesium stabil dua . Atom magnesium melepaskan elektron valens untuk mencapai susunan elektron oktet magnesium tidak yang stabil membentuk

ion magnesium , Mg2+ dengan susunan elektron

2.8

.

2.6 (b) The electron arrangement of oxygen atom is . Oxygen atom is also unstable. Oxygen atom receives two electrons to achieve a stable octet electron arrangement to form oxide ion , O2– with electron arrangement

2.8

Susunan elektron atom oksigen ialah

. 2.6

. Atom oksigen juga tidak stabil, atom oksigen

m

ion oksida

menerima

, O dengan susunan elektron

dua elektron 2.8

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2–

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(c)

Magnesium ion , Mg2+ and formed is called ionic bond. Ion magnesium , Mg2+ dan dinamakan ikatan ion.

oxide ion , O2– are attracted by strong

ion oksida

electrostatic

force. The bond

, O2– ditarik dengan daya elektrostatik yang kuat. Ikatan yang terbentuk

(iii) Magnesium chloride /Magnesium klorida Predict the formula / Ramal formula: Element

Proton number


Mg

12

2.8.2

Cl

17

2.8.7

Unsur

Nombor proton

Mg Cl + e

Susunan elektron

Mg2+ + 2e Cl–

Mg2+

Cl–

1

2

MgCl2

Draw the electron arrangement of the compound formed. Lukiskan susunan elektron bagi setiap sebatian yang terbentuk. 2+

Transfer Pindah

Transfer Pindah Mg

C1

Chlorine atom, Atom klorin, Cl Cl

C1

Magnesium atom,Mg Mg Atom magnesium,

Mg

C1

Chlorine atom, Atom klorin, Cl Cl

C1

Chlorine ion,ClCl– Ion Magnesium ion,Mg Mg2+2+ Ion Chlorine ion, Ion klorida, magnesium, klorida, ClCl–


2.8.2

(a) The electron arrangement of magnesium atom is

2 . Magnesium atom has stable . Magnesium atom releases

in the outer shell. Therefore, magnesium atom is not

valence electrons to achieve a stable octet electron arrangement to form 2.8 electron arrangement .

electrons 2

magnesium ion , Mg2+ with

2.8.2 . Atom magnesium mempunyai 2 elektron di petala terluar. Maka atom Susunan elektron atom magnesium stabil 2 . Atom magnesium melepaskan elektron valens untuk mencapai susunan elektron oktet magnesium tidak ion magnesium , Mg2+ dengan susunan elektron

yang stabil membentuk

2.8

.

2.8.7 (b) The electron arrangement of chlorine atom is . Chlorine atom is also unstable. Chlorine atom receives one electron to achieve a stable octet electron arrangement to form chloride ion , Cl– with electron arrangement

2.8.8

Susunan elektron atom klorin ialah

. 2.8.7

ion klorida , Cl dengan susunan elektron

mencapai susunan elektron oktet yang stabil membentuk

(c) As such, Oleh itu,

one satu

Daya

elektrostatik

–

magnesium atom releases atom magnesium melepaskan

(d) Strong electrostatic ionic bond.

menerima

. Atom klorin juga tidak stabil. Atom klorin

2

force is formed between

yang kuat terbentuk antara

2

2

electrons to

elektron kepada

2

2.8.8

.

chlorine atoms. atom klorin.

magnesium ion , Mg

ion magnesium , Mg2+ dan

satu elektron untuk

2+

and

chloride ion , Cl– to form

ion klorida , Cl– membentuk ikatan ionik.

COVALENT BONDS / IKATAN KOVALEN 1

This bond is formed when two or more similar or different atoms share valence electrons between them, so that each atom achieves the octet or duplet electron arrangement that is a stable electron arrangement for noble gases. Ikatan ini terbentuk apabila dua atau lebih atom yang sama atau berlainan berkongsi elektron valens antara satu sama lain supaya setiap atom mencapai susunan elektron oktet atau duplet iaitu susunan elektron gas adi yang stabil.

2

Normally, this bond is formed when similar or different non-metal atoms bond together. [Atoms from Groups 14, 15, 16 and 17] n io

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Ikatan ini biasanya terbentuk apabila atom-atom bukan logam berpadu. [Atom-atom dari Kumpulan 14, 15, 16 dan 17]

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When atoms of non-metals share their valence electrons from their outermost shells to achieve stable duplet or octet sharing atoms electron arrangement, covalent bonds are formed. The product of the of electrons between

3

is called

molecule .

Apabila atom-atom bukan logam berkongsi elektron pada petala terluar untuk mencapai susunan elektron duplet atau oktet yang stabil, ikatan kovalen terbentuk. Hasil daripada perkongsian elektron antara atom-atom ini membentuk molekul .

The molecules are neutral as there is no electron transfer involved. During the formation of covalent bonding , each atom contributes same number of electrons for sharing. The number of electrons shared can be one pair, two pairs or three pairs.

4

neutral kerana tidak melibatkan pemindahan elektron. Semasa pembentukan ikatan kovalen , setiap atom akan Molekul adalah menyumbang bilangan elektron yang sama untuk dikongsi. Bilangan elektron yang dikongsi boleh jadi sepasang, dua pasang atau tiga pasang.

The forces that exist between molecules are Van der Waals forces that are weak. These forces become stronger when the molecule size increases.

5

Daya yang wujud antara molekul adalah

daya Van der Waals

yang lemah. Daya ini semakin kuat apabila saiz molekul bertambah.

Examples / Contoh: (i) Hydrogen molecule / Molekul hidrogen: (a) Hydrogen atom has one electron in the first shell, with an electron arrangement of 1 needs one electron to achieve a stable duplet electron arrangement.

6

Atom hidrogen mempunyai satu elektron pada petala pertama dengan susunan elektron 1 memerlukan satu elektron untuk mencapai susunan elektron duplet yang stabil.

(b)

Two hydrogen atoms share a pair of electrons to form a hydrogen molecule. Dua atom hidrogen berkongsi sepasang elektron membentuk satu molekul hidrogen.

(c)

Both hydrogen atoms achieve a stable duplet arrangement of electron. Kedua-dua atom hidrogen mencapai susunan elektron duplet yang stabil.

Draw the electron arrangement of the compound formed / Lukiskan susunan elektron bagi sebatian yang terbentuk. Share

Kongsi Kongsi Share

H H

H

H H

H

The number of electron pairs shared is

one

pair. Single covalent bond is formed.

Bilangan pasangan elektron dikongsi adalah

satu

pasang. Ikatan kovalen tunggal terbentuk.

(ii) Oxygen molecule / Molekul oksigen: (a)

Oxygen atom with an electron arrangement 2.6 needs two electrons to achieve a stable arrangement. Atom oksigen dengan susunan elektron 2.6 memerlukan dua elektron untuk mencapai susunan elektron

(b)

Two oxygen atoms share

octet oktet

electron yang stabil.

two

pairs of electrons to achieve a stable octet arrangement of electron, form an oxygen molecule. Each oxygen atom achieves stable octet electron arrangement. dua

pasang elektron untuk mencapai susunan elektron oktet yang stabil, membentuk satu oktet yang stabil. molekul oksigen. Setiap atom oksigen mencapai susunan elektron Dua atom oksigen berkongsi

Draw the electron arrangement of the compound formed / Lukiskan susunan elektron bagi sebatian yang terbentuk.

O O

Kongsi Share

m


O O

O

Oxygen molecule, O22 Molekul oksigen, O


2

pairs. Double covalent bond is formed.


2

pasang. Ikatan kovalen ganda dua terbentuk.

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(iii) Nitrogen molecule / Molekul nitrogen: (a)

3

Atom nitrogen dengan susunan elektron 2.5 memerlukan stabil.

(b)

3

Nitrogen atom with an electron arrangement 2.5 needs arrangement.

electrons to achieve stable

elektron untuk mencapai susunan elektron

octet oktet

yang

3

pairs of electrons to achieve a stable octet arrangement, form a nitrogen molecule. Each nitrogen atom achieves stable octet electron arrangement. Two nitrogen atoms share

3

oktet pasang elektron untuk mencapai susunan elektron oktet yang stabil. satu molekul nitrogen. Setiap atom nitrogen mencapai susunan elektron Dua atom nitrogen berkongsi

yang stabil membentuk

Draw the electron arrangement of the compound formed / Lukiskan susunan elektron bagi sebatian yang terbentuk. Share Kongsi

N Nitrogen atom, N N Atom nitrogen,

N

N N

Nitrogen atom, NN Atom nitrogen,

N N

Nitrogen Molekulmolecule, nitrogen,NN2 2


3

pairs. Triple covalent bond is formed.


3

pasang. Ikatan kovalen ganda tiga terbentuk.

(iv) Hydrogen chloride molecule /Molekul hidrogen klorida Predict the formula / Ramal formula: Element

Proton number


H

H

1

1

Cl

Cl

17

2.8.7

Unsur

Nombor proton

needs

1 electron

perlu

1 elektron

needs 1 electron 1 elektron

perlu

Cross the number of electrons each atom needs: HCl Silangkan bilangan elektron yang diperlukan oleh setiap atom: HCl


H

Share Kongsi

Hydrogen atom, H H Atom hidrogen,

H H

Cl C1

Chlorine atom, Atom klorin, ClCl

Cl C1

Hydrogen chloride molecule, HCl Molekul hidrogen klorida, HCl


(a)

Hydrogen atom with an electron arrangement duplet electron arrangement. Atom

hidrogen dengan susunan elektron

1

1

memerlukan

needs satu

one

electron to achieve a stable


duplet yang stabil.

(b)

Chlorine atom with an electron arrangement 2.8.7 needs electron arrangement. Atom

klorin dengan susunan elektron 2.8.7 memerlukan yang stabil.

(c)

One chloride Satu

one chlorine atom share molecule with the formula atom klorin berkongsi

HCl

electron to achieve stable


pair of electrons with HCl . pasang elektron dengan

one satu

octet oktet

hydrogen atom to form hydrogen atom hidrogen membentuk

molekul

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hidrogen klorida dengan formula

satu

satu

one

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(d)

One chlorine atom contributes electron for sharing. Satu

satu

atom klorin menyumbang dikongsi bersama.

One

(e)

Satu

(f)

Chlorine duplet Atom

one

chlorine atom forms

one

atom klorin membentuk

satu

atom

hydrogen atom contributes one

atom hidrogen menyumbang satu elektron untuk

single covalent bond with

one

hydrogen atom.

ikatan kovalen tunggal dengan

satu

atom hidrogen.

electron arrangement and hydrogen

atom

achieves stable

electron arrangement. oktet

klorin mencapai susunan elektron

duplet

satu

elektron dan

octet

achieves stable

one

electron and

atom

yang stabil dan

hidrogen mencapai susunan elektron

yang stabil.

(v) Water molecule /Molekul air Predict the formula / Ramal formula: Element

Proton number


H

H

1

1

O

O

8

2.6

Unsur

Nombor proton

needs

1 electron

perlu

1 elektron

needs 2 electrons 2 elektron

perlu

Cross the number of electrons each atom needs: H2O

Silangkan bilangan elektron yang diperlukan oleh setiap atom: H2O


H

Kongsi Share

Hydrogen atom, H Atom hidrogen,

Kongsi Share

O O


H H

H H

Hydrogen atom, H Atom hidrogen,

O O

H H

Water molecule, O Molekul air, HHO 2 2


(a)

Hydrogen atom with an electron arrangement electron arrangement. Atom

1

hidrogen dengan susunan elektron

1

memerlukan

needs electron to achieve a stable satu

duplet


duplet yang stabil.

(b)

(c)

Oxygen octet

atom

with an electron arrangement

two

needs

electrons to achieve stable

electron arrangement.

Atom

oksigen dengan susunan elektron

oktet

yang stabil.

One

oxygen atom share molecule with the formula

2.6

two

Satu

atom oksigen berkongsi air dengan formula H2O .

(d)

2.6

memerlukan

dua

pairs of electrons with

H2O

.

dua

pasang elektron dengan

One

two

oxygen atom contributes electron for sharing to form single


two dua

hydrogen atoms form water atom hidrogen membentuk

molekul

electrons and each of the two hydrogen atoms contributes one

covalent bond.

Satu

dua

atom oksigen menyumbang elektron dan setiap satu daripada dua atom hidrogen menyumbang satu dikongsi bersama membentuk ikatan kovalen tunggal. elektron untuk

(e)

One Satu

(f)

Oxygen duplet

m

atom oksigen membentuk

atom

achieves stable

single covalent bonds with

dua

ikatan kovalen tunggal dengan

octet

two dua

hydrogen atoms. atom hidrogen.

electron arrangement and hydrogen

atom

achieves

electron arrangement. oksigen mencapai susunan elektron

oktet

yang stabil dan

atom

hidrogen mencapai susunan elektron

duplet yang stabil.

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(vi) The molecule formed between carbon and chlorine /Molekul yang terbentuk antara karbon dan klorin Predict the formula / Ramal formula: Element

Proton number


C

C

6

2.4

Cl

Cl

17

2.8.7

Unsur

Nombor proton

needs

Cross the number of electrons each atom needs: CCl4

4 electrons

perlu

4 elektron

Silangkan bilangan elektron yang diperlukan oleh setiap atom: CCl4

1 electron

needs

1 elektron

perlu


Cl

Cl

Cl

Cl


(a)

(b)

(c)

Carbon octet

atom

karbon dengan susunan elektron

oktet

yang stabil.

atom

2.4

empat

memerlukan


klorin dengan susunan elektron

oktet

yang stabil.

2.8.7

2.8.7

memerlukan

satu

four carbon atom share pairs of electrons with CCl4 . tetrachloromethane molecule with the formula atom karbon berkongsi empat tetraklorometana berformula CCl4 .

four


one

needs

One

One

electrons to achieve a stable

electron to achieve a stable

electron arrangement..

Atom

Satu

(d)

four

needs

electron arrangement.

Atom

Chlorine octet

2.4


pasang elektron dengan


four empat

four carbon atom contributes electrons and each of the electron for sharing to form single covalent bond.

chlorine atoms to form atom klorin membentuk

four

chlorine atoms contributes

Satu

satu atom karbon menyumbang empat elektron dan setiap daripada empat satu elektron untuk dikongsi bersama membentuk ikatan kovalen tunggal . menyumbang

(e)

One Satu

(f)

carbon atom forms atom karbon membentuk

Carbon and chlorine

atoms

empat

single covalent bonds with ikatan kovalen tunggal dengan

achieve stable

octet

karbon dan atom klorin mencapai susunan elektron

four empat

atom klorin

chlorine atoms. atom klorin.

electron arrangement. oktet

yang stabil.

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Atom

four

molekul

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7

Comparing the Formation of Ionic and Covalent Bonds / Perbandingan Pembentukan Ikatan Ion dan Kovalen Ionic Bond / Ikatan Ion

Covalent Bond / Ikatan Kovalen

Type of element involved

Between metals (Groups 1, 2 and 13) and non-metals (Groups 15, 16 and 17).

Electron

Electron is released by metal atoms and received by non-metal atoms (electron transfer).

Jenis unsur terlibat

Elektron

Jenis zarah yang dihasilkan

How to predict the formulae

Antara bukan logam 14, 15, 16 dan 17).

logam (Kumpulan 1, 2 dan 13) dengan Antara bukan logam (Kumpulan 15, 16 dan 17).

Elektron dilepaskan oleh atom logam dan atom bukan logam (elektron berpindah).

Type of particle produced

Between non-metal 14, 15, 16 and 17).

diterima

oleh

Metal atom forms positive ion. Non-metal atom forms negative ion. positif . Atom logam membentuk ion negatif Atom bukan logam membentuk ion

different

dengan

non-metals

(Groups

bukan logam

(Kumpulan

of electrons are shared non-metal atoms.

Pasangan elektron dikongsi sama atau berlainan.

by the same or

oleh atom-atom bukan logam

Neutral molecule . Molekul

yang neutral.

.

Determine the coefficient of the charge of the ions and criss cross. Tentukan pekali cas pada ion dan silangkan.

Determine the number of electrons is needed to achieve stable duplet or octet electron arrangement and criss cross. Tentukan bilangan elektron yang diperlukan untuk mencapai susunan elektron duplet atau oktet yang stabil dan silangkan.

Bagaimana meramal formula

Example of electron arrangement in the particles

Pairs

and

+

A

2–

E

+

A

Contoh susunan elektron dalam zarah

Strong electrostatic forces between ions Daya elektrostatik yang kuat antara ion

Strong covalent bond between atoms in the molecules

Example of ionic and covalent compounds

m

# Covalent bond is the shared pairs of electrons between atoms in a molecule. # Ikatan kovalen terhasil daripada perkongsian pasangan elektron antara atom-atom dalam molekul.

Lead(II) bromide, PbBr2

Naphthalene, C8H10

Sodium chloride, NaCl

Acetamide, CH3CONH2

Copper(II) sulphate, CuSO4

Hexane, C6H14

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Contoh sebatian ion dan kovalen

Ikatan kovalen yang kuat antara atom dalam molekul

# Ionic bond is the strong electrostatic force of attraction between positively charged ion and negatively charged ion. # Ikatan ion terhasil daripada daya tarikan elektrostatik yang kuat antara ion bercas positif dan ion bercas negatif.

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PHYSICAL PROPERTIES OF IONIC AND COVALENT COMPOUND SIFAT FIZIK SEBATIAN ION DAN KOVALEN Ionic compound / Sebatian ion Example

Covalent compound / Sebatian kovalen

Sodium chloride, NaCl / Natrium klorida, NaCl

Carbon dioxide, CO2 / Karbon dioksida, CO2

Contoh

Weak Van der Waals forces between molecules

Daya Van der Waals yang lemah antara molekul

Strong electrostatic forces between positive and negative ions Daya elektrostatik yang kuat antara ion

Strong covalent bond between atoms in the molecules Ikatan kovalen yang kuat antara atom dalam molekul

Type of forces between particles

Strong electrostatic force between ions.

Melting and boiling points

–

Weak Van der walls forces (intermolecular force) between molecule.

Daya elekrostatik yang kuat antara ion.

Daya Van der Waals yang lemah antara molekul.

Jenis daya antara zarah

Takat lebur dan takat didih

High

melting and boiling points because positive ions and negative ions are attracted by strong electrostatic force . Takat lebur dan takat didih tinggi kerana ion positif dan ion negatif ditarik oleh daya tarikan elektrostatik yang kuat.

–

Large amount of energy is needed to overcome it. Banyak tenaga haba diperlukan untuk

– Low melting and boiling points because of the weak “Van der Waals” force between molecules. Takat lebur/takat didih rendah kerana daya "Van der Waals" yang lemah antara molekul.

–

Small

amount of energy is needed to overcome it. Sedikit

mengatasinya .

tenaga haba diperlukan untuk

mengatasinya .

– Giant molecules such as silicon dioxide have very high melting and boiling points. Molekul raksaksa seperti silikon dioksida mempunyai takat didih dan lebur yang amat tinggi.

Electrical conductivity Kekonduksian elektrik

– Cannot conduct electricity when in

solid

but is able to conduct electricity when in or aqueous form.

form molten

Tidak boleh mengkonduksi elektrik dalam keadaan pepejal tetapi boleh mengkonduksi elektrik dalam keadaan leburan atau akueus .

– In solid form, the ions are not Dalam bentuk pepejal, ion-ion tidak bergerak .

free bebas

to

move . untuk

–

Cannot

conduct electricity in all state.

Tidak boleh keadaan.

mengkonduksi elektrik dalam semua

– Covalent compound is made up of neutral molecules . Sebatian kovalen terdiri daripada

molekul

yang neutral.

– No free moving ions in molten or aqueous state. Tidak ada ion bebas bergerak dalam keadaan leburan atau akueus.

– In molten or aqueous state, the ions are free to move to be attracted to the anode or cathode. bebas

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Dalam keadaan leburan atau akueus, ion-ion bergerak untuk ditarik ke anod atau katod.

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Ionic compound / Sebatian ion Solubility

Keterlarutan

– Most are soluble organic solvent*. Kebanyakannya larut dalam pelarut organik*

Covalent compound / Sebatian kovalen

in water and insoluble in

Insoluble in water but soluble in organic solvents* (example: ether, alcohol, benzene, tetrachloromethane and propanone). This is because covalent molecules and organic solvents are both held together by weak Van der Waals forces.

–

dalam air tetapi tidak larut

– This is because the polarisation of water molecule. Water molecules have partially positive end (the hydrogen end) and partially negative end (the oxygen end).

Tidak larut dalam air tetapi larut dalam pelarut organik* (contoh: eter, alkohol, benzena, tetraklorometana dan propanon). Ini kerana molekul kovalen dan pelarut organik ditarik oleh daya tarikan Van der Waals yang lemah. * Organic solvents are covalent compounds that exist as liquid at room temperature. * Pelarut organik adalah sebatian kovalen yang wujud dalam bentuk cecair pada suhu bilik.

Ini kerana air adalah molekul yang berkutub. Molekul air mempunyai bahagian bercas separa positif (bahagian hidrogen) dan bahagian bercas separa negatif (bahagian oksigen).

EXERCISE / LATIHAN

The Table below shows the proton number of elements D, E, F, G, J and L.

1

Jadual di bawah menunjukkan nombor proton bagi unsur D, E, F, G, J dan L. Element / Unsur

D

E

F

G

J

L


1

6

17

11

18

8

(a) Which element in the table are metal and non-metal / Unsur yang manakah merupakan logam dan bukan logam? (i) Metal / Logam : G (ii) Non-metals / Bukan logam : D, E, F, J, L (b) State an element that exists as monoatomic gas. Explain your answer. Nyatakan unsur yang wujud sebagai gas monoatom. Terangkan jawapan anda.

Element J, Atom J has 8 electrons in the outermost shell, the atom has achieved stable octet electron arrangement. (c) Write the formula for the ion formed from an atom of element L. Tuliskan formula ion yang terbentuk daripada atom unsur L.

L2– (d) Element E reacts with element L to form a compound / Unsur E bertindak balas dengan unsur L untuk membentuk sebatian. (i) State the type of bond present in this compound / Nyatakan jenis ikatan yang wujud dalam sebatian ini. Covalent bond (ii)

Write the formula of the compound formed / Tuliskan formula bagi sebatian yang terbentuk. EL2

(iii) Explain how a compound is formed between element E and element L based on their electron arrangement. Jelaskan dari segi susunan elektron bagaimana unsur E dan unsur L bergabung membentuk sebatian.

– E atom with electron arrangement 2.4 needs four electrons to achieve stable octet electron arrangement. – L atom with an electron arrangement 2.6 needs two electrons to achieve octet electron arrangement. – One E atom share four pairs of electrons with two L atoms to form a molecule with the formula EL2. – One E atom contributes four electrons and each of the two L atoms contributes two electrons for sharing to form double covalent bond. – One E atom forms two double covalent bond with two L atoms. – E atom and L atom achieve stable octet electron arrangement that is 2.8. (e) (i)

Draw the electron arrangement of the compound formed / Lukiskan susunan elektron bagi sebatian yang terbentuk.

m

E

L

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(ii)

State one physical property of the compound / Nyatakan satu sifat fizik sebatian tersebut. Low melting/boiling point // does not dissolve in water // dissolves in organic solvents // does not conduct electricity in aqueous solution or molten state.

(f)

When element G is burnt in L gas, G burns rapidly and brightly with a yellow flame and produces white fumes. Apabila unsur G dibakar dalam gas L, G terbakar cergas dengan nyalaan kuning terang dan menghasilkan wasap putih.

(i)

Write the equation for the reaction between element G and gas L. Tuliskan persamaan kimia bagi tindak balas antara unsur G dan gas L.

4G + L2 (ii)

2G2L

.

Explain how a compound is formed between elements G and L based on their electron arrangement. Jelaskan dari segi susunan elektron bagaimana unsur G dan L bergabung membentuk sebatian.

– The electron arrangement of G atom is 2.8.1. G atom is not stable. G atom releases one valence electron to form G+ ion and achieve stable octet electron arrangement 2.8. – The electron arrangement of L atom is 2.6. L atom is also unstable. L atom receives 2 electrons to form L2– ion and achieves a stable octet electron arrangement 2.8.8. – Therefore two G atoms release two electrons to one L atom, a strong electrostatic force is formed between G+ and L2– ions. (iii) Draw the electron arrangement of the compound formed / Lukiskan susunan elektron bagi sebatian yang terbentuk. +

+

2–

G

L

G

(g) Compare the boiling point of the compounds formed in 1(d) and 1(e). Explain your answer. Bandingkan takat didih sebatian yang terbentuk di 1(d) dan 1(e). Jelaskan jawapan anda.

– The boiling point of compound G2L is high and EL2 is low. – The boiling point of compound G2L is high because positive ions and negative ions are attracted by strong electrostatic force. Large amount of energy is needed to overcome it. – The boiling point of EL2 is low because the molecules are attracted by weak Van der Waals forces, small amount of energy is needed to overcome it. 2

The diagram below shows the electron arrangement of compound A. Compound A is formed from the reaction between element X and element Y. Rajah di bawah menunjukkan susunan elektron bagi sebatian A. Sebatian A terbentuk dari tindak balas antara unsur X dan unsur Y.

+ X

(a) (i) (ii)

– Y

Write the electron arrangement for atom of elements X and Y / Tuliskan susunan elektron bagi atom unsur X dan Y. X: 2.8.1 Y: 2.8.7 Compare the size of atoms of elements X and Y. Explain your answer. Bandingkan saiz atom unsur X dan unsur Y. Jelaskan jawapan anda.

– Atom Y is smaller than atom X. n io

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– Atom X and atom Y have the same number of shells occupied with electrons.

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– The number of proton in the nucleus of atom Y is more than X. – The strength nuclei attraction to the electrons in the shells of atom Y is stronger than X. (b) How are X ion and Y ion formed from their respective atoms? Bagaimana ion X dan ion Y terbentuk daripada atom masing-masing?

X ion / Ion X : Atom X releases one electron Y ion / Ion Y : Atom Y receives one electron (c) (i) (ii)

Write the formula for compound A / Tuliskan formula sebatian A. XY Name type of bond in compound A / Namakan jenis ikatan dalam sebatian A. Ionic compound

(iii) Write the chemical equation for the reaction between element X and element Y to form compound A. Tuliskan persamaan kimia untuk tindak balas antara unsur X dan unsur Y untuk membentuk sebatian A.

2X + Y2

2XY

.

(d) Y can react with carbon to form a compound. Draw the electron arrangement for the compound formed. [Given that proton number for carbon is 6] Y bertindak balas dengan karbon untuk membentuk suatu sebatian. Lukiskan susunan elektron bagi sebatian yang terbentuk. [Diberi nombor proton karbon ialah 6]

The table below shows the nucleon number, the number of neutrons and number of electrons in particles X, Y, Z, Q, R, T and U.

3

Jadual di bawah menunjukkan nombor nukleon, bilangan neutron dan bilangan elektron bagi zarah X, Y, Z, Q, R, T dan U. Particles / Zarah

X

Y

Z

Q

R

T

U

Nucleon number / Nombor nukleon

20

24

23

16

12

27

35

Number of proton / Bilangan proton

10

12

11

8

6

13

17

Number of neutron / Bilangan neutron

10

12

12

8

6

14

18

Number of electron / Bilangan elektron

10

10

11

10

6

10

17

(a) What is meant by nucleon number / Apakah maksud nombor nukleon? The total number of proton and neutron in the nucleus of an atom. (b) Complete the number of proton of the particles in the table above. Lengkapkan bilangan proton bagi zarah dalam jadual di atas.

(c) State a particle which is / Nyatakan zarah yang merupakan (i)

an atom of a non-metal / atom bukan logam

(ii)

an atom of a metal / atom logam

Y/T

(iv) a negative ion / ion negatif

Q

(v)

T

a positive ion with charge 3+ / ion positif dengan cas 3+

X

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(iii) a positive ion / ion positif

(vi) an atom of a noble gas / atom gas adi m

X/R

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(d) Particle Y combines with particle Q to form a compound / Zarah Y bergabung dengan zarah Q untuk membentuk sebatian. (i) State the type of compound formed / Nyatakan jenis sebatian yang terbentuk. Ionic compound (ii)

Write chemical formula for the compound formed / Tuliskan formula kimia bagi sebatian yang terbentuk. YQ

(iii) Draw the electron arrangement of the compound formed / Lukiskan susunan elektron bagi sebatian yang terbentuk. 2+

2–

Q

Y

(e) Particle R combines with particle U to form a compound. Zarah R bergabung dengan zarah U untuk menghasilkan suatu sebatian.

(i)

State the type of compound formed / Nyatakan jenis sebatian yang terbentuk. Covalent compound

(ii)

(f)

Write a chemical formula for the compound formed / Tuliskan formula kimia bagi sebatian yang terbentuk. RU4

Compare the electrical conductivity of the compounds formed in 3(d) and 3(e). Explain your answer. Bandingkan kekonduksian elektrik bagi sebatian yang terbentuk di 3(d) dan di 3(e). Jelaskan jawapan anda.

– Compound in YQ cannot conduct electricity in solid state but can conduct electricity in molten or aqueous solution. Compound RU4 cannot conduct electricity in molten and aqueous states. – In solid form the ions in compound YQ are not free to move but in molten and aqueous state, the ions are free to move to be attracted to the anode and cathode. – Compound RU4 only consists of neutral molecules, there are no free moving ions in molten or aqueous state. 4

The table below shows the melting point and electrical conductivity of substances W, X, Y and Z. Jadual di bawah menunjukkan takat lebur dan kekonduksian elektrik bagi bahan W, X , Y dan Z. Substance Bahan

Electrical conductivity / Kekonduksian elektrik

Melting point (°C) Takat Lebur (°C)

V

–7

W

80

X

808

Y

1 080

Solid / Pepejal

Molten / Leburan



Tidak mengkonduksi elektrik







Conduct electricity


Mengkonduksi elektrik

Conduct electricity

Conduct electricity



(a) Which of the substance is copper? Give reason for your answer. Antara bahan di atas, yang manakah kuprum? Beri sebab bagi jawapan anda.

Y. It can conduct electricity in solid and molten state. (b) (i)

State the type of particles in substances V and W / Nyatakan jenis zarah dalam bahan V dan W. Molecule

(ii)

Explain why substances V and W cannot conduct electricity in solid and molten state. Jelaskan mengapa bahan V dan W tidak boleh mengkonduksi elektrik dalam keadaan pepejal dan leburan.

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Substances V and W are made up of neutral molecules. No free moving ions in molten state.

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(c) The boiling point of substance V is 59°C. What is the physical state of substance V at room temperature? Takat didih bahan V adalah 59°C. Apakah keadaan fizikal bahan V pada suhu bilik?

Liquid (d) Draw the arrangement of particle V at room temperature / Lukiskan susunan zarah V pada suhu bilik.

(e) Explain why the melting and boiling points of substances V and W are low? Jelaskan mengapa takat lebur dan takat didih bahan V dan W rendah?

– Van der Waals / intermolecular forces between molecules are weak. – Small amount of heat energy is required to overcome it. (f)

(i) (ii)

State the type of particle in substance X / Nyatakan jenis zarah dalam sebatian X. Ion . Explain why substance X cannot conduct electricity in solid but can conduct electricity in molten state. Jelaskan mengapa bahan X tidak boleh mengkonduksi elektrik dalam keadaan pepejal tetapi boleh mengkonduksi elektrik dalam keadaan leburan.

Ions are not freely moving // ions are in a fixed position in solid state. Ion can move freely in molten state.


Which substance is an ionic compound? Antara bahan berikut, yang manakah adalah sebatian ion?

A B C D 2

4

Methane, CH4 / Metana, CH4 Carbon dioxide, CO2 / Karbon dioksida, CO2 Propanol, C3H7OH / Propanol, C3H7OH Copper(II) oxide, CuO / Kuprum(II) oksida, CuO

12

Apakah susunan elektron bagi ion yang terbentuk dari atom T?

A B C D

3

Volatile / Mudah meruap Has a low melting point Mempunyai takat lebur rendah

Insoluble in water / Tidak larut dalam air Conducts electricity in the molten state Mengalirkan arus elektrik dalam keadaan leburan

The diagram below shows the electron arrangement of a compound formed between atoms X and Y.

5

Jadual di bawah menunjukkan susunan elektron atom P, Q, R dan S.

Y X

Y

Y

Which of the following statements is true about the compound? Antara pernyataan berikut, yang manakah adalah benar tentang sebatian itu?

A B C

m

Atom / Atom


P

2.4

Q

2.8.1

R

2.8.2

S

2.8.7

Which pair of atoms forms a compound by transferring of electrons? Antara pasangan berikut, yang manakah membentuk sebatian secara perpindahan elektron?

A B C D

P and S / P dan S P and R / P dan R Q and S / Q dan S Q and R / Q dan R

Sebatian itu mempunyai takat lebur yang tinggi.

The compound conducts electricity. Sebatian itu boleh mengkonduksi elektrik.

The compound is formed by sharing of electrons. Sebatian terbentuk secara perkongsian elektron.

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D

It is an ionic compound / Ia adalah sebatian ion. The compound has high melting point.

2.8 2.8.2 2.8.8 2.8.8.8

The table below shows the electron arrangements of atoms P, Q, R and S.

Rajah di bawah menunjukkan susunan elektron dalam sebatian yang terbentuk antara atom X and atom Y.

Y

T

What is the electron arrangement of ion formed by an atom of T?

Which of the following is a property of zinc chloride?

C D

Rajah menunjukkan simbol unsur T. 24

Antara berikut, yang manakah adalah sifat zink klorida?

A B

The diagram shows symbol of an element T.

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6

The table below shows the proton number of four elements P, Q, R and S. Jadual di bawah menunjukkan nombor proton unsur P, Q, R dan S.

9

The diagram below shows the electron arrangement for an ion of element Q. Rajah di bawah menunjukkan susunan elektron ion unsur Q.

Element / Unsur

P

Q

R

S


6

8

17

20

2–

Which of the following pairs will form a compound with high melting and boiling points?

Q

Antara pasangan berikut, yang manakah membentuk sebatian dengan takat lebur dan takat didih yang tinggi?

A B 7

C D

P and Q / P dan Q Q and S / Q dan S

P and R / P dan R Q and R / Q dan R

The table below shows the proton number of elements X and Y.

What are the number of protons and electrons in an atom of element Q? Apakah bilangan proton dan elektron dalam atom unsur Q?

Number of protons

Number of electrons

A

20

20

Jadual berikut menunjukkan nombor proton unsur X dan Y.

Element / Unsur

X

Y


6

8

Bilangan elektron

What type of bond and the chemical formula of the compound formed between atoms X and Y?

B

20

18

C

16

16

Apakah jenis ikatan dan formula kimia bagi sebatian yang terbentuk antara atom X dan Y ?

D

18

18

Type of bond

Chemical formula

A

Ion / Ion

YX2

B

Ion / Ion

XY2

Element / Unsur

P

Q

R

C

Covalent / Kovalen

XY2


10

11

12

D

Covalent / Kovalen

YX2

Jenis ikatan

8

Bilangan proton

Formula kimia

The diagram below shows the electron arrangement of ion X +. Rajah di bawah menunjukkan susunan elektron ion X +.

X

10 The table below shows the proton number of elements P, Q and R. Jadual di bawah menunjukkan nombor proton unsur P, Q dan R.

Which of the following particles contain 10 electrons? Antara berikut, yang manakah adalah zarah yang mengandungi 10 elektron?

I II III IV A B C

Which of the following is the position of element X in the Periodic Table?

D

Q P Q+ R2+ I, II and III only I, II dan III sahaja

I, II and IV only I, II dan IV sahaja

I, III and IV only I, III dan IV sahaja

II, III and IV only II, III dan IV sahaja

Antara berikut, yang manakah adalah kedudukan unsur X dalam Jadual Berkala?

Group / Kumpulan

Period / Kala

A

1

3

B

18

3

1

4

D

18

4

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5

ELECTROCHEMISTRY ELEKTROKIMIA

ELECTROLYSIS / ELEKTROLISIS • CONDUCTOR AND ELECTROLYTE / KONDUKTOR DAN ELEKTROLIT – To differentiate between electrolyte and conductor with regard to electrical conductivity and any chemical changes that may occur. Membezakan elektrolit dan konduktor dari segi kebolehan mengkonduksikan elektrik dan sebarang perubahan kimia yang berlaku.

– To list examples of substances which are classified as electrolytes and conductors. Menyenaraikan contoh-contoh bahan yang dikelaskan sebagai elektrolit dan konduktor.

• ELECTROLYSIS CELL / SEL ELEKTROLISIS – To draw and label the electrolytic cell / Melukis dan melabelkan sel elektrolisis. – To identify anode and cathode in the electrolytic cell diagram / Mengenali anod dan katod dalam rajah sel elektrolisis. • IONIC THEORY / TEORI ION – To relate the existence of free moving ions in an electrolyte with the electron flow in an external circuit. Mengaitkan kewujudan ion-ion yang bebas bergerak dalam elektrolit dengan proses pengaliran elektron dalam litar luar.

– To explain the electrolysis process / Menerangkan proses elektrolisis. – To conclude that electrolysis process involve changes from electrical to chemical energy. Membuat kesimpulan proses elektrolisis sebagai perubahan tenaga elektrik kepada tenaga kimia.

• FORMATION OF FREE MOVING IONS / PEMBENTUKAN ION BEBAS BERGERAK – To differentiate molten and aqueous electrolytes / Membezakan elektrolit lebur dan akueus. – To write the ionisation equation of molten and aqueous electrolytes. Menulis persamaan pengionan untuk elektrolit lebur dan akueus.

• REACTION AT ELECTRODE / TINDAK BALAS DI ELEKTROD – To write the discharge equation at the anode, where the anion releases electron. Focus on ions that are normally selected for discharge, such as chloride, hydroxide and bromide ions. Menulis persamaan di anod yang melibatkan anion melepaskan elektron. Fokus adalah kepada ion-ion yang biasa terpilih untuk nyahcas seperti ion klorida, ion hidroksida dan ion bromida.

– To write the discharge equation at the cathode, where the cation receives electron. Focus on ions that are normally selected for discharge, such as hydrogen, copper(II) and silver ions. Menulis persamaan di katod yang melibatkan kation menerima elektron. Fokus adalah kepada ion yang biasa terpilih untuk nyahcas seperti ion hidrogen, ion kuprum(II) dan ion argentum.

• FACTORS THAT AFFECT REACTIONS AT THE ELECTRODES FAKTOR-FAKTOR YANG MEMPENGARUHI TINDAK BALAS DI ELEKTROD (i) The position of ions in the electrochemical series – for dilute solutions and inert electrodes. Kedudukan ion dalam siri elektrokimia – bagi larutan cair dan elektrod lengai

(ii) The concentration – for concentrated solutions and inert electrodes / Kepekatan – bagi larutan pekat dan elektrod lengai. (iii) The types of electrode – for diluted solutions and reactive electrodes / Jenis elektrod – bagi larutan cair dan elektrod tak lengai. • ELECTROLYSIS IN INDUSTRY / KEGUNAAN ELEKTROLISIS DALAM INDUSTRI – Electrolysis in electroplating, purifying and extracting metals / Elektrolisis dalam penyaduran, penulenan dan pengekstrakan logam.

VOLTAIC CELL / SEL KIMIA • ELECTROCHEMICAL SERIES / SIRI ELEKTROKIMIA – To define and memorise the sequence of metal including hydrogen in the Electrochemical Series. Menakrif dan menghafal siri logam termasuk hidrogen dalam Siri Elektrokimia.

• APPLICATION OF ELECTROCHEMICAL SERIES IN DISPLACEMENT OF METALS APLIKASI SIRI ELEKTROKIMIA DALAM PENYESARAN LOGAM – To predict the displacement of metal reactions based on the positions of metals in the Electrochemical Series. Meramal tindak balas penyesaran logam berdasarkan kedudukan logam dalam Siri Elektrokimia.

– To write the equation of displacement reaction and to state the observations. Menulis persamaan tindak balas penyesaran dan menyatakan pemerhatian.

– To describe the metal displacement experiment to construct the Electrochemical Series. Menghuraikan eksperimen penyesaran logam bagi membina Siri Elektrokimia.

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• APPLICATION OF ELECTROCHEMICAL SERIES IN VOLTAIC CELL APLIKASI SIRI ELEKTROKIMIA DALAM SEL KIMIA – To determine the negative and positive terminals of a voltaic cell / Menentukan terminal negatif dan positif suatu sel kimia. – To predict the voltage of voltaic cell / Meramal voltan sel kimia. – To determine the direction of electron flow / Menentukan arah pengaliran elektron.

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ELECTROLYSIS / ELEKTROLISIS 1

Three types of substances that can be classified based on electrical conductivity. Bahan boleh dibahagikan kepada tiga jenis berdasarkan kekonduksian elektrik. Type of substance

Definition

Conductor Konduktor

Example

Definisi

Jenis bahan

Contoh

Element that can conduct electricity Copper, lead, tin, silver and carbon Kuprum, plumbum, stanum, argentum dan karbon at solid or molten state without any chemical changes , normally metals and carbon. Unsur yang boleh mengkonduksi arus elektrik dalam keadaan pepejal atau leburan tanpa perubahan kimia , biasanya logam dan karbon.

Electrolyte Elektrolit

Compounds that can conduct electricity in *molten state or *aqueous solution and undergo chemical changes . Sebatian yang boleh mengkonduksikan arus elektrik dalam keadaan *lebur atau *akueus serta mengalami perubahan kimia . * Molten state: a solid that is heated until it melts. * Lebur: pepejal yang dipanaskan sehingga cair. * Aqueous solution: a solid that is dissolved in water. * Akueus: pepejal yang larut di dalam air.

– Aqueous solution of ionic compound such as copper(II) sulphate solution and sodium chloride solution. Larutan akueus bagi sebatian ion contohnya larutan kuprum(II) sulfat dan larutan natrium klorida.

– Aqueous solution of *acid or alkali such as hydrochloric acid (HCl) and ammonia solution (NH3). Larutan akueus *asid atau alkali contohnya asid hidroklorik (HCl) dan larutan ammonia (NH3 ).

– Molten ionic compounds such as molten lead(II) bromide, molten sodium chloride and molten aluminium oxide. ion contohnya leburan plumbum(II) Leburan sebatian bromida, leburan natrium klorida dan leburan aluminium oksida.

* HCl and NH3 are covalent compounds, exist in form of molecule without water but ionised in water. (Explanation is in the next topic i.e acid and base) * HCl dan NH3 adalah sebatian kovalen, yang terdiri daripada molekul dalam keadaan tanpa air tetapi ianya terion dalam air (akan dijelaskan dalam tajuk seterusnya iaitu dalam asid dan bes)

2

Non- electrolyte

Compounds that cannot conduct electricity Molten covalent compound such as naphthalene, molten in molten and aqueous solution. sulphur and liquid bromine.

Bukan elektrolit

Sebatian kimia yang tidak boleh mengkonduksikan elektrik dalam keadaan lebur dan akueus.

process

Electrolysis is a current passes

3

contohnya naftalena, sulfur lebur dan cecair

whereby an electrolyte is decomposed to its constituent elements when electric penguraian elektrolit kepada unsur juzuknya apabila

Energy change in electrolysis process is electric energy to

chemical energy

Perubahan tenaga dalam proses elekrolisis adalah dari tenaga elektrik kepada 4

kovalen

through it. proses

Elektrolisis adalah

Leburan sebatian bromin.

arus elektrik

dialirkan melaluinya.

.

tenaga kimia

.

Conductor which is dipped into electrolyte which carries electric current in and out of electrolyte is called an

electrode

.

Electrode

is normally made up of

inert

substance such as carbon.

Konduktor yang dicelup dalam elektrolit yang mengalirkan arus elektrik ke dalam dan keluar daripada elektrolit dipanggil Elektrod lengai biasanya terdiri daripada bahan seperti karbon. 5

An electrolytic cell is a set-up of apparatus that contains two electrodes which are dipped in an battery and produce a chemical reaction when connected to a (source of electricity). elektrolit

.

electrolyte

dan menghasilkan

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elektrod yang dicelup ke dalam Sel elektrolisis adalah susunan radas yang terdiri daripada dua bateri . (sumber arus elektrik). tindak balas kimia apabila disambungkan kepada

elektrod

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Example of electrolytic cell / Contoh sel elektrolisis:

(i)

(ii)

(iii) Electrolyte

A

Electrodes Elektrod

Elektrolit

Electrode

Electrode

Electrolyte

Elektrod

Elektrod

Elektrolit

Electrodes

Electrolyte Elektrolit

Heat

Panaskan

Electrolysis of aqueous electrolyte (No gas released)

Electrolysis of molten electrolyte Elektrolisis elektrolit lebur

Elektrod

A

Electrolysis of aqueous electrolyte (Gas is released)

Elektrolisis elektrolit dalam bentuk akueus (Tiada gas dibebaskan)

Elektrolisis elektrolit dalam bentuk larutan (Gas dibebaskan)

Electric current from the battery flows into the electrolyte through the electrode. There are two types of electrode in the electrolytic cell:

6

Arus elektrik dari bateri mengalir ke dalam elektrolit melalui elektrod. Terdapat dua jenis elektrod dalam sel elektrolisis:

(a) Anode: An electrode that is connected to the Anod: Elektrod yang disambung kepada

positive terminal

terminal positif

bateri dalam sel elektrolisis.

negative terminal

(b) Cathode: An electrode that is connected to the Katod: Elektrod yang disambung kepada

of the battery.

terminal negatif

of the battery.

bateri dalam sel elektrolisis.

An electrolyte consists of free moving ions because it is in a molten or aqueous state. Each ion moves to the opposite charge electrode. There are two types of ions in electrolyte:

7

Dalam keadaan lebur atau akueus, elektrolit terdiri daripada ion-ion yang bergerak bebas. Setiap ion bergerak kepada elektrod yang bertentangan cas. Terdapat dua jenis ion dalam elektrolit:

(a) Anions: Negative ions which are attracted and move to the positively charged electrode, Anion: Ion

(b) Cations: Kation: Ion

negatif

Positive

akan tertarik dan bergerak ke arah elektrod

anod

yang bercas

positif

ions which are attracted and move to the negatively charged electrode,

positif

akan tertarik dan bergerak ke arah elektrod

katod

yang bercas

negatif

anode

.

cathode

.

. .

Electrolysis occurs at the electrode when electric current flows in the electrolytic cell. The stages in electrolysis process are:

8

Proses elektrolisis berlaku di elektrod apabila arus elektrik mengalir melalui sel elektrolisis. Peringkat dalam proses elektrolisis adalah seperti berikut:

(a) Anions (negative ions) are attracted and move to the

anode

of anode and become neutral atoms or molecule. The anions are Anion (ion negatif) akan tertarik dan bergerak ke arah menjadi atom/molekul. Anion dinyahcaskan pada anod.

(b) Electrons flow from the Elektron mengalir dari

anode anod

to the ke

katod

anod

cathode

. The anions release electrons to the surface discharged at the anode.

. Anion melepaskan elektron pada permukaan anod dan

through the connecting wire in the external circuit .

melalui wayar penyambung dalam

(c) Cations (positive ions) are attracted and move to the

litar luar

.

cathode

. The cations receive electrons at the surface of cathode and become neutral atoms or molecules. The cations are discharged at the cathode. katod Kation (ion positif) akan tertarik dan bergerak ke arah menjadi atom/molekul. Kation dinyahcaskan pada katod.

. Kation menerima elektron pada permukaan katod dan

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– Electrons flow through the external circuit / Elektron mengalir melalui litar luar. – Chemical changes occur at the anode and cathode / Perubahan kimia berlaku di anod dan katod.

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FORMATION OF FREE MOVING IONS IN THE ELECTROLYTE PEMBENTUKAN ION BERGERAK BEBAS DALAM ELEKTROLIT 1

Ionisation equation is an equation to determine the ions present in molten or aqueous electrolyte. Persamaan pengionan adalah persamaan yang menunjukkan ion yang hadir dalam elektrolit sama ada dalam keadaan leburan atau akueus.

(a) Example of ionisation of molten electrolyte (a compound that is heated until it melts) Contoh pengionan elektrolit dalam keadaan leburan (sebatian yang dipanaskan sehingga lebur)

(i)

Molten sodium chloride / Natrium klorida lebur:

NaCl (s)

Na+(l) + Cl–(l)

(ii) Molten lead (II) bromide / Plumbum (II) bromida lebur: PbBr2 (s)

Pb2+(l) + 2Br –(l)

(iii) Molten sodium oxide / Natrium oksida lebur:

Na2O (s)

2Na+(l) + O2–(l)

(iv) Molten aluminium oxide / Aluminium oksida lebur:

Al2O3 (s)

2Al3+(l) + 3O2–(l)

(b) Example of the ionisation on an aqueous electrolyte (a compound that is dissolved in water): Contoh pengionan elektrolit dalam keadaan akueus (sebatian yang dilarutkan dalam air): Na+(aq) + Cl+(aq) (i) Sodium chloride solution / Larutan natrium klorida: NaCl(aq / ak )

2

H+(aq) + OH–(aq)

(ii) Copper(II) sulphate solution / Larutan kuprum(II) sulfat: CuSO4(aq / ak )

Cu2+ + SO42–

H2O

H2O

(iii) Sulphuric acid / Asid sulfurik:

H2SO4(aq / ak )

H2O

H+ + OH–

2H+ + SO42– H+ + OH–

Ionisation of molten electrolyte produces cation and anion of the compound only. However the ionisation of an aqueous electrolyte produces cation and anion from the ionisation of the compound and water. Pengionan elektrolit dalam keadaan lebur hanya menghasilkan kation dan anion dari sebatian itu sahaja. Pengionan elektrolit dalam keadaan akueus menghasilkan kation dan anion daripada sebatian dan air.

Example / Contoh: molten

(i) Ionisation of Pengionan

leburan

sodium chloride produces Na+ and Cl– only. natrium klorida menghasilkan Na+ dan Cl– sahaja.

aqueous

(ii) Ionisation of

sodium chloride produces Na+, H+, Cl– and OH–.

akueus

Pengionan larutan

natrium klorida menghasilkan Na+, H+, Cl– dan OH–.

REACTIONS AT THE ELECTRODES / TINDAK BALAS DI ELEKTROD 1

The process of cation gaining electron at the cathode or anion losing electrons at the anode is called discharged : nyahcas

Proses apabila kation menerima elektron di katod atau anion melepaskan elektron di anod dipanggil

(a) A cation is Kation

discharged

dinyahcaskan dinyahcaskan

(c) When ions are Apabila ion 2

apabila

discharged

(b) An anion is Anion

when it

discharged

dinyahcaskan

menerima

when it

apabila

receives

electrons at the cathode.

elektron di katod.

releases

melepaskan

electrons at the anode.

elektron di anod.

, they become neutral

atom

atom

atau

, ianya akan menjadi

:

or molecule . molekul

The ionic equation that occurs at the anode and cathode to produce neutral ‘half equation’. atom

atau

atom

or molecule is called

molekul

neutral dipanggil ‘persamaan

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Persamaan ion yang berlaku di anod dan di katod untuk menghasilkan setengah’.

yang neutral.

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Common half equation at the anode (anion/metal atom releases electrons):

3

Persamaan setengah yang biasa di anod (anion/atom logam melepaskan elektron): Half equation

Explanation

Persamaan setengah

4OH–

Penerangan

2H2O + O2 + 4e

2Cl–

Cl2 + 2e

2Br–

Cu

Cu2+ + 2e

Ag

Ag+ + e

Empat ion hidroksida

melepaskan

Two chloride ions

release

melepaskan

Dua ion klorida

Br2 + 2e

release

Four hydroxide ions molecule .

Two bromide ions Dua ion bromida

Copper atom

releases

releases

Atom argentum

melepaskan

oksigen.

.

klorin.

molekul

dua elektron membentuk satu

molecule

.

bromin.

copper(II) ion

ion kuprum(II)

dua elektron membentuk

Silver atom

molekul

two electrons to form one bromine

two electrons to form

melepaskan

molecule

two electrons to form one chlorine

release

molekul

empat elektron membentuk dua molekul air dan satu


melepaskan

Atom kuprum

four electrons to form two water molecules and one oxygen

.

.

one electron to form

silver ion

satu elektron membentuk

ion argentum .

.

Common half equation at the cathode (cation receives electrons):

4

Persamaan setengah yang biasa di katod (kation menerima elektron): Half equation

Explanation

Persamaan setengah

2H+ + 2e

Ag+ + e

Cu2+ + 2e

H2

Ag

Cu

Penerangan

Two hydrogen ions Dua ion hidrogen

Silver ion

receive

Ion argentum

menerima

Copper(II) ion Ion kuprum(II)

receive

menerima

molekul


receives

satu elektron membentuk satu

atom

two electrons to form one copper dua elektron membentuk satu

atom

.

hidrogen.

atom

one electron to form one silver

menerima

molecule

two electrons to form one hydrogen

.

argentum.

atom

.

kuprum.

Write the equation of discharge of ion

5

Tuliskan persamaan setengah untuk nyahcas ion yang berikut:

(i) Lead(II) ion to lead atom

:

Ag++ e Ag

:

2I– I2 + 2e

Ion argentum kepada atom argentum

(iii) Iodide ion to iodine molecule

Pb2+ + 2e Pb

Ion plumbum(II) kepada atom plumbum

(ii) Silver ion to silver atom /

:

Ion iodida kepada molekul iodin

EXERCISE / LATIHAN

Using lead(II) bromide as an example, explain the electrolysis of molten lead(II) bromide. In your explanation, draw a labeled diagram for the set up of apparatus and show the movement of particles by using arrows that occur in lead(II) bromide and the direction of electron flow in the external circuit.

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Dengan menggunakan plumbum(II) bromida sebagai contoh, jelaskan elektrolisis leburan plumbum(II) bromida. Dalam penerangan anda, lukiskan satu rajah susunan radas berlabel dan tunjukkan dengan anak panah pergerakan zarah yang berlaku dalam plumbum(II) bromida serta arah aliran elektron dalam litar luar.

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Set-up of apparatus / Rajah susunan radas:

Carbon electrodes

Lead(II) bromide

Heat


– The ions present are lead(II) ions/ Pb2+ and bromide ions/ Br –. – Bromide ion/ Br – move to the anode. – Bromide ion/ Br – releases one electron to form bromine atom at the anode. – Two bromine atoms combine to form bromine molecule. – 2Br – Br2 + 2e – Lead(II) ions/Pb2+ move to the cathode. – Lead(II) ions/Pb2+ receive two electrons to form lead atom at the cathode. – Pb2+ + 2e Pb

FACTOR THAT AFFECT THE ELECTROLYSIS OF AN AQUEOUS SOLUTION FAKTOR YANG MEMPENGARUHI ELEKTROLISIS LARUTAN AKUEUS

When more than one type of ion are attracted towards the electrodes during electrolysis, only one type of ion is selected to be discharged at each electrode. Selective discharge only occurs in aqueous solution because it usually has more than one type of ion attracted to the anode or cathode.

1

Apabila lebih dari satu jenis ion bergerak ke elektrod semasa elektrolisis, hanya satu jenis ion sahaja yang akan dipilih untuk dinyahcas pada setiap elektrod. Pemilihan nyahcas ion hanya berlaku di dalam larutan akueus sahaja kerana ia biasanya mempunyai lebih dari satu jenis ion yang tertarik ke anod atau katod.

The selection of ion for discharge depends on three factors / Pemilihan ion untuk nyahcas bergantung pada tiga faktor: (a) The position of ions in the electrochemical series (normally in dilute solution and inert electrode).

2

Kedudukan ion dalam siri elektrokimia (biasanya dalam larutan cair dan elektrod lengai).

(b) The concentration of electrolyte (normally in concentrated solution and inert electrode). Kepekatan elektrolit (biasanya dalam larutan pekat dan elektrod lengai).

(c) The types of electrode (when reactive metal electrode is used). Jenis elektrod (apabila elektrod logam reaktif digunakan). The position of ions in the Electrochemical Series / Kedudukan ion dalam Siri Elektrokimia:

3

(a) When electrolysis is conducted on dilute solution and inert electrodes, the lower position of cation in the Electrochemical Series, or anions in the lower position of the anion discharge series will be selected to be discharged. Apabila elektrolisis dijalankan ke atas larutan cair dan elektrod lengai, kation yang lebih rendah kedudukan dalam Siri Elektrokimia atau anion yang lebih rendah kedudukan dalam siri discas anion akan dinyahcas.

Cation: K+, Na+, Ca2+, Mg2+, Al3+, Zn2+, Fe2+, Sn2+, Pb2+, H+, Cu2+, Ag+, and Au+ Kation: K+, Na+, Ca2+, Mg2+, Al3+, Zn2+, Fe2+, Sn2+, Pb2+, H+, Cu2+, Ag+, dan Au+

Increasing ease of discharge of ion from left to right

Ion semakin mudah dinyahcas dari kiri ke kanan – 2– – Anion: F , SO4 , NO3 , Cl–, Br –, I–, and OH–

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Anion: F–, SO42–, NO3–, Cl–, Br –, I–, dan OH–

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(b) Choose the ion to be discharged from the following pairs of ions. State the electrode where it occurs and write the half equation for the discharge of ion: Pilih ion yang akan dinyahcas dari pasangan ion berikut, nyatakan di elektrod mana ia berlaku dan tulis persamaan setengah untuk nyahcas ion:

(i)

Ion hidroksida & ion sulfat

(ii) Hydroxide & nitrate ions Ion hidroksida & ion nitrat

(v) Hydrogen & silver ions Ion hidrogen & ion argentum

4OH

–

: Half equation: : Persamaan setengah:

4OH

–

Cu + 2e Cu Cu2+ + 2e

: Persamaan setengah:

Ag + e Ag


Ag + e +

katod

di

Ag

di

katod

.

.

cathode

at the

.

.

cathode

at the

H2

+

: Half equation:

katod

di

.

.

cathode

at the

2H + 2e H2


anod

di

.

.

anode

at the

Cu

+

2H+ + 2e

anod

di

2H2O + O2 + 4e

2+

anode

at the

2H2O + O2 + 4e

4OH– 2H2O + O2 + 4e

(iv) Hydrogen & potassium ions : Half equation: Ion hidrogen & ion kalium


(iii) Hydrogen & copper(II) ions : Half equation: Ion hidrogen & ion kuprum(II)

4OH– 2H2O + O2 + 4e

Hydroxide & sulphate ions : Half equation:

.

.

(c) Complete the following table for the electrolysis of 0.1 mol dm–3 sodium nitrate solution using carbon electrode. Lengkapkan jadual berikut bagi elektrolisis larutan natrium nitrat 0.1 mol dm–3 menggunakan elektrod karbon. Set-up of apparatus

Susunan radas

Carbon electrodes

Sodium nitrate

NaNO3 H2O

Equation of electrolyte ionisation Persamaan pengionan elektrolit

Electrode / Elektrod

Na+ + NO3– H+ + OH–

Anode / Anod

Ions that are attracted to the anode and cathode NO3–, OH–

Cathode / Katod Na+, H+

Ion yang ditarik ke anod dan katod

Half equation Persamaan setengah

Name of the products

2H+ + 2e

H2

Hydrogen

Pemerhatian

Gas bubbles are released.


Confirmatory test (method and observations)

– Insert a glowing wooden splinter into – When a lighted wooden splinter is test tube. placed near the mouth of the test tube. – Glowing wooden splinter is lighted up. – A ‘pop’ sound is produced.

Observations

Ujian pengesahan (kaedah dan pemerhatian)

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2H2O + O2 + 4e

Oxygen

Nama hasil

m

4OH–

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(d) Complete the following table for the electrolysis of 0.1 mol dm–3 sulphuric acid using carbon electrodes. Lengkapkan jadual berikut bagi elektrolisis asid sulfurik 0.1 mol dm–3 menggunakan elektrod karbon. Set-up of apparatus Susunan radas

Carbon electrodes

Sulphuric acid

H2SO4

Equation of electrolyte ionisation

H2O

Persamaan pengionan elektrolit

2H+ + SO42– H+ + OH–


Anode / Anod

Cathode / Katod

Ions that are attracted to the anode and cathode

SO42–, OH–

H+


Half equation

4OH–

Persamaan setengah

Name of the products Nama hasil

Observations Pemerhatian

Confirmatory test (method and observations) Ujian pengesahan (kaedah dan pemerhatian)

2H2O + O2 + 4e

2H+ + 2e

H2

Oxygen

Hydrogen



– Insert a glowing wooden splinter into – When a lighted wooden splinter is test tube. placed near the mouth of the test tube. – Glowing wooden splinter is lighted up. – A ‘pop’ sound is produced.

(e) Complete the following table for the electrolysis of 0.1 mol dm–3 copper(II) sulphate solution using carbon electrodes. Lengkapkan jadual berikut bagi elektrolisis larutan kuprum(II) sulfat 0.1 mol dm–3 menggunakan elektrod karbon. Set-up of apparatus Susunan radas

Carbon electrodes

Copper(II) sulphate

H2SO4


H2O


2H+ + SO42– H+ + OH–

Anode / Anod


Cathode / Katod

Ions that are attracted to the anode and SO42–, OH– cathode

Cu2+, H+

Half equation

Cu2+ + 2e

Ion yang ditarik ke anod dan katod Persamaan setengah




4OH–

2H2O + O2 + 4e

Cu

Oxygen

Copper


Brown solid deposited

– Insert a glowing wooden splinter into test tube. – Glowing wooden splinter is lighted up.

–


4

Concentration of electrolyte / Kepekatan elektrolit: (a) When electrolysis is carried out using inert electrodes and concentrated solutions, ions that are more concentrated will be discharged but this is only true for halide ions, which are Cl–, Br – and I–. Apabila elektrolisis dijalankan menggunakan elektrod lengai dan larutan pekat, ion yang lebih pekat akan dinyahcas tetapi ia benar untuk ion-ion halida sahaja iaitu Cl–, Br– dan I–. n io

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Nyatakan ion yang terpilih untuk dinyahcaskan di anod dan di katod bagi larutan pekat di bawah.

Publicat

(b) State the selected ions to be discharged at the anode and cathode for the following concentrated solutions.

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(i)

Concentrated hydrochloric acid solution, using carbon electrodes Larutan asid hidroklorik pekat menggunakan elektrod karbon

Cl– Anode / Anod: Cathode / Katod: (ii) Concentrated potassium iodide solution, using carbon electrodes

l– Anode / Anod: Cathode / Katod: (iii) Concentrated sodium chloride solution, using carbon electrodes

H+

Larutan kalium iodida pekat menggunakan elektrod karbon

K+

Larutan natrium klorida pekat menggunakan elektrod karbon

Anode / Anod:

Cl–

H+

Cathode / Katod:

(c) Complete the following table for the electrolysis of 0.001 mol dm–3 hydrochloric acid and 2.0 mol dm–3 hydrochloric acid, using carbon electrodes. Lengkapkan jadual berikut bagi elektrolisis asid hidroklorik 0.001 mol dm–3 dan asid hidroklorik 2.0 mol dm–3 menggunakan elektrod karbon. Set-up of apparatus Susunan radas

Carbon electrodes

Hydrochloric acid

HCl


H2O


Half equation at the cathode Persamaan setengah di katod

Observation at cathode Pemerhatian di katod

Confirmatory test at cathode (method and observations) Ujian pengesahan (kaedah dan pemerhatian)

Name the products at the cathode Nama hasil di katod

Ions that are attracted to the anode Ion bergerak ke anod

Half equation at the anode Persamaan setengah di anod

H+

H+

2H+ + 2e

H2


2H+ + 2e

H2


– Insert a burning wooden splinter into – Insert a burning wooden splinter into the test tube. the test tube. – A ‘Pop’ sound is produced. – A ‘pop’ sound is produced. Hydrogen gas

Hydrogen gas

Cl– , OH–

Cl– , OH–

4OH–

2H2O + O2 + 4e

2Cl–

Cl2 + 2e


Confirmatory test at anode (method and observations)

– Insert a glowing wooden splinter into – A damp blue litmus paper placed near the test tube. the mouth of the test tube. – Glowing wooden splinter is lighted up. – The gas changed the damp blue litmus paper to red and then bleached it.

Name the product at the anode Nama hasil di anod

The concentration of hydrochloric acid after a while and explanation Kepekatan elektrolit selepas beberapa ketika dan terangkan

Oxygen gas

Greenish yellow gas is released.

Chlorine gas

Concentration of hydrochloric acid Concentration of hydrochloric acid increases . Hydrogen gas is released at decreases . Hydrogen gas released at the cathode and oxygen gas is released the cathode and chlorine gas released at at the anode. Water decomposed to the anode. Concentration of chloride oxygen gas and hydrogen gas. ions decreases. bertambah . Kepekatan asid hidroklorik Gas hidrogen dibebaskan di katod dan gas oksigen dibebaskan di anod. Air terurai kepada gas oksigen dan gas hidrogen .

berkurang Kepekatan asid hidroklorik Gas hidrogen dibebaskan di katod dan gas klorin dibebaskan di anod. Kepekatan ion klorida berkurang.

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HCl 2.0 mol dm–3

Observations at anode / Pemerhatian


m

2.0 mol dm-3 of HCl

HCl 0.001 mol dm–3

Elektrolit

Ion bergerak ke katod

H+ + OH–

0.001 mol dm-3 of HCl

Electrolyte Ions that are attracted to the cathode

H+ + Cl–

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(d) Complete the following table for the electrolysis of 2.0 mol dm–3 sodium iodide solution using carbon electrodes. Lengkapkan jadual berikut bagi elektrolisis larutan natrium iodida 2.0 mol dm–3 menggunakan elektrod karbon. Set-up of apparatus Susunan radas

Carbon electrodes

Sodium iodide

NaI H2O


Na+ + I– H+ + OH–


Anode / Anod

Cathode / Katod

Ions that are attracted to the anode and cathode

I–, OH–

Na+, H+


Half equation

2I–

Persamaan setengah




2H+ + 2e

H2

Iodine

Hydrogen

Brown solution is formed.


– A few drops of starch solution added. – Starch solution turns to dark blue.

– When a lighted wooden splinter is placed near the mouth of the test tube. – A ‘pop’ sound is produced.


5

I2 + 2e

Types of electrode Jenis elektrod: (a) There are two types of electrode Terdapat dua jenis elektrod: (i) Inert electrode – An electrode that acts as a conductor only and does not undergo any chemical changes.

Normally they are made of carbon or platinum.

Elektrod lengai – Elektrod yang bertindak sebagai pengalir arus sahaja dan tidak mengalami perubahan kimia. Biasanya diperbuat daripada karbon atau platinum. (ii) Reactive electrode – An electrode that not only acts as a conductor but also undergoes chemical changes. During the electrolysis, the metal atom at the anode releases electron to form metal ion, metal anode becomes thinner while the less electropositive cation will be selected at the cathode which consist of metal

electrodes such as copper, silver and nickel.

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Elektrod reaktif – Elektrod yang bertindak bukan sahaja sebagai pengalir arus tetapi juga mengalami perubahan kimia. Semasa proses elektrolisis berlaku, atom logam pada anod melepaskan elektron menjadi ion logam, anod logam menjadi nipis manakala ion yang kurang elektropositif akan menyahcas di katod yang terdiri daripada logam seperti kuprum, argentum dan nikel.

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(b) Complete the following table for the electrolysis of 1 mol dm–3 copper(II) sulphate solution with carbon electrode and copper electrode. Lengkapkan jadual berikut bagi elektrolisis larutan kuprum(II) sulfat 1 mol dm–3 menggunakan elektrod karbon dan elektrod kuprum. Set-up of apparatus Susunan radas

Copper(II) sulphate

Copper electrodes

Carbon electrodes

Copper(II) sulphate

CuSO4 (aq / ak ) Cu2+ + SO42– H2O H+ + OH–


Type of electrode

Carbon electrode

Jenis elektrod

The ions that move to the cathode Ion bergerak ke katod

Half equation at the cathode Persamaan setengah di katod

Name the product at the cathode Nama hasil di katod

Observation at cathode Pemerhatian di katod

The ions that move to the anode Ion bergerak ke anod

Half equation at the anode Persamaan setengah di anod

Name the product at anode Nama hasil di anod

Observations at the anode Pemerhatian di anod

Confirmatory test (method and observations) Ujian pengesahan (kaedah dan pemerhatian )

The concentration of copper(II) solution after a while and explanation

m

Cu2+, H+

Elektrod kuprum

Cu2+, H+

Cu2+ + 2e

Cu

Cu2+ + 2e

Cu

Copper

Copper



SO42–, OH–

SO42–, OH–

4OH–

Cu

2H2O + O2 + 4e

Cu2+ + 2e

Oxygen gas

Copper(II) ion

– Gas bubbles are released. – Intensity of blue colour decreases.

– Copper electrode becomes thinner. – Intensity of blue colour remains unchanged.

– Insert a glowing wooden splinter into the test tube. – Glowing wooden splinter is lighted up.

– Concentration of copper(II) sulphate solution decreases. – Copper(II) ions discharge as copper atoms and deposited the cathode.

–

– Concentration of copper(II) sulphate solution remains unchanged. – The number of copper atoms form copper(II) ions at the anode is equal to the number of copper(II) ions form copper atoms at the cathode.

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Kepekatan elektrolit selepas beberapa ketika dan terangkan

Copper electrode

Elektrod karbon

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Complete the table below / Lengkapkan jadual di bawah: Electrolyte Elektrolit

Electrode Factor that affects Elektrod electrolysis

Ions present Ion yang hadir

Faktor yang mempengaruhi elektrolisis

Dilute sulphuric acid

Carbon

H+, SO42–, OH–

Concentration of electrolyte

H+, Cl–, OH–

Karbon

Position of ion in the electrochemical series

Ag+, NO3–, H+, OH–

Silver

Type of electrode

Asid sulfurik cair

Concentrated hydrochloric acid

Persamaan setengah di anod dan pemerhatian


Karbon

Carbon Karbon

Carbon

Larutan argentum nitrat

Silver nitrate solution

4OH–

2H2O + O2 + 4e

Half equation at the cathode and observation Persamaan setengah di katod dan pemerhatian

2H+ + 2e

H2



2Cl–

2H+ + 2e

Cl2 + 2e

H2

Greenish yellow gas is released.


4OH–


Ag+ + e Ag Grey shiny solid deposited.

Ag+, NO3–, H+, OH–

Ag Ag+ + e Anode becomes thinner.

Ag+ + e Ag Grey shiny solid deposited.


K+, I–, H+, OH–

4OH–

2H+ + 2e

Concentration of electrolyte

K+, I–, H+, OH–


K+, SO42–, H+, OH–

Asid hidroklorik pekat


Half equation at the anode and observation

Argentum

2H2O + O2 + 4e

Larutan argentum nitrat

Carbon Dilute Karbon potassium iodide solution

2H2O + O2 + 4e

H2



2I–

2H+ + 2e

Larutan kalium iodida cair

Carbon Concentrated Karbon potassium iodide solution

I2 + 2e

H2

Brown solution formed.


4OH–

2H+ + 2e

Larutan kalium iodida pekat

Dilute potassium sulphate solution

Carbon Karbon

2H2O + O2 +4e


H2


Larutan kalium sulfat cair

2

Electrolysis is carried out on a dilute potassium chloride solution using carbon electrodes. Explain how this electrolysis occurs. Use a labelled diagram to explain your answer. Proses elektrolisis dijalankan ke atas larutan kalium klorida cair menggunakan elektrod karbon. Jelaskan bagaimana proses elektrolisis ini berlaku. Gunakan gambar rajah berlabel untuk menerangkan jawapan anda. Set-up of apparatus / Susunan radas:

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Carbon electrode

Dilute potassium chloride solution Carbon electrode

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K+, H+, Cl–

– Potassium chloride solution consist of

K , H , Cl +

Larutan kalium klorida mengandungi ion

Cl–

–

ion and Cl –

Ion

OH–

– –

Ion OH

OH

dan

dalam siri elektrokimia.

oxygen

2H2O + O2 + 4e

is lower than

K+ ion

water

and oksigen

dipilih untuk dinyahcaskan dengan melepaskan elektron membentuk molekul

– Half equation / Persamaan setengah: K+ ion and H+ ion move to the cathode / –

–

yang bergerak bebas.

is selectively discharged by releasing electrons to form 4OH–

–

ions that move freely.

–

ion in the electrochemical series. Cl –

terletak di bawah ion

–

H+ ion

OH–

and

bergerak ke anod.

Cl–

ion is lower than

OH– ion

–

ions move to the anode.

OH –

dan ion

OH –

Ion

OH–

+

air

dan

.

.

Ion K +

ion H +

dan

bergerak ke katod.

in the electrochemical series.

Ion H +

terletak di bawah

H+ ion

is selectively discharged by receiving electrons to form hydrogen molecules.

Ion H

dipilih untuk dinyahcaskan dengan menerima elektron membentuk molekul

+

molecule.

ion K +

dalam siri elektrokimia

2H + + 2e

– Half equation / Persamaan setengah:

H2

hidrogen

.

.

Describe an experiment to determine the product of electrolysis copper(II) sulphate solution with carbon electrode.

3

Your answer should include the observation, confirmatory test for the product at the anode and half equation at the electrode. Huraikan satu eksperimen untuk menentukan hasil elektrolisis larutan kuprum(II) sulfat menggunakan elektrod karbon. Dalam jawapan anda perlu disertakan pemerhatian, ujian pengesahan untuk hasil yang terbentuk di anod dan persamaan setengah bagi tindak balas yang berlaku di elektrod. Answer / Jawapan:

Apparatus / Radas : Battery / power supply, carbon electrodes, wire, electrolytic cell, test tube, Ammeter [from a

labelled diagram]

Material / Bahan

–3 : 1 mol dm copper(II) sulphate solution

Carbon electrodes

Copper(II) sulphate solution

Procedure / Langkah: –3 (a) Pour 1 mol dm copper(II) sulphate

Masukkan

larutan

kuprum(II) sulfat

solution 1 mol dm–3

in the electrolytic cell until it is ke dalam sel elektrolitik sehingga

(b) The apparatus is set up as shown in the diagram. Fill the anode invert the test tube on the . Radas disusunkan seperti dalam gambar rajah. Isi anod . uji itu pada

tabung uji

test tube dengan

half full

with copper(II) sulphate

larutan

.

separuh penuh

.

solution

and

kuprum(II) sulfat dan terbalikkan tabung

(c) Turn on the switch / Hidupkan suis. (d) Collect the gas produced at the anode (e) Gas produced at the

m

anod

anod / Kumpulkan gas yang terhasil di glowing wooden splinter is tested with a . diuji dengan

kayu uji berbara

.

.

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Gas yang terhasil di

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Observation and half equation / Pemerhatian dan persamaan setengah: Electrodes

Observation

Elektrod

4

Confirmatory test

Pemerhatian

Half equation

Ujian pengesahan

Cathode


Anode

Gas bubbles are released

Persamaan setengah

Cu2+ + 2e Cu

–

– Insert the glowing wooden splinter into the test tube. – The glowing wooden splinter is lighted up.

4OH– 2H2O + O2 + 4e

Copper(II) sulphate solution is electrolysed using copper electrodes. Larutan kuprum(II) sulfat dielektrolisis dengan menggunakan elektrod kuprum.

(a) Write the formula of all the anions present in the solution / Tuliskan formula semua anion yang terdapat dalam larutan itu. SO42–, OH– (b) Write the half equation for the reaction at the / Tuliskan persamaan setengah untuk tindak balas di 2+ (i) anode / anod : Cu Cu + 2e 2+ (ii) cathode / katod : Cu + 2e Cu (c) (i) From your observations, what happen to the intensity of the blue colour of the copper(II) sulphate solution during electrolysis?

Daripada pemerhatian anda, nyatakan apakah yang berlaku ke atas keamatan warna biru larutan kuprum(II) sulfat semasa proses elektrolisis?

The intensity of the blue colour of copper(II) sulphate remains unchanged.

(ii) Explain your answer / Jelaskan jawapan anda. The number of copper(II) ions become copper atoms at the cathode is equal to the number of copper atoms become copper(II) ions at the anode.

(d) If the experiment is repeated with the copper electrodes being replaced by carbon electrodes, state the name of the products formed at the Jika eksperimen diulangi dengan menggantikan elektrod kuprum dengan elektrod karbon, namakan hasil yang terbentuk di

5

(i)

anode / anod:

Oxygen

(ii) cathode / katod:

Copper

The diagram below shows the set-up of apparatus of an electrolytic cell. Rajah di bawah menunjukkan susunan radas bagi sel elektrolisis.

Carbon electrode P

Carbon electrode Q

Elektrod karbon P

Elektrod karbon Q

Copper(II) nitrate solution Larutan kuprum(II) nitrat

(a) Write the formula of all ions present in copper(II) nitrate solution. Tuliskan formula semua ion yang hadir dalam larutan kuprum(II) nitrat.

Cu2+, NO3–, H+ and OH– . (b) Write half equation for the reaction at / Tuliskan persamaan setengah di: 2+ electrode P / elektrod P : Cu + 2e Cu – electrode Q / elektrod Q : 4OH 2H2O + O2 + 4e (c) (i) What is the colour of copper(II) nitrate / Apakah warna larutan kuprum(II) nitrat?

Blue

(ii) What happens to the intensity of the colour of copper(II) nitrate solution? Explain your answer. Apakah yang berlaku kepada keamatan warna larutan kuprum(II) nitrat? Jelaskan jawapan anda.

The intensity of the blue colour of copper(II) nitrate decreases. The concentration of Cu2+ decreases because

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copper(II) ions receive electrons to form copper atom at the cathode.

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ELECTROLYSIS IN INDUSTRY / ELEKTROLISIS DALAM INDUSTRI

Three uses of electrolysis in industries are / Tiga kegunaan elektrolisis dalam industri ialah:

1

Application Aplikasi

(a) Electroplating Penyaduran logam

Example

Electrolyte

Contoh

Silver electroplating

Elektrolit


Penyaduran perak

(b) Purification of metal Penulenan logam

(c) Metal extraction Pengekstrakan logam

Purification of copper

Molten aluminium oxide

Pengekstrakan aluminium

Cathode / Half equation

Anod / Persamaan setengah

Katod / Persamaan setengah

Anode / Anod: Silver metal

Cathode / Katod: Metal to be electroplated

Half equation / Persamaan setengah: Ag Ag+ + e

Half equation / Persamaan setengah: Ag+ + e Ag

Copper(II) Anode / Anod: sulphate solution Impure copper

Penulenan kuprum

Extraction of aluminium

Anode / Half equation

Cathode / Katod: Pure copper

Half equation / Persamaan setengah: Cu Cu2+ + 2e

Half equation / Persamaan setengah: Cu2+ + 2e Cu

Anode / Anod: Carbon

Cathode / Katod: Carbon

Half equation / Persamaan setengah: 2O2– O2 + 4e

Half equation / Persamaan setengah: Al3+ + 3e Al

The following diagram shows the aluminium extraction process.

2

Rajah di bawah menunjukkan proses pengekstrakan aluminium. Substance Z / Bahan Z

Substance Y Bahan Y

Substance X + cryolite Substance W

Bahan X + kriolit

Bahan W

(a) State the name of the following substances / Nyatakan nama bahan-bahan berikut: W : Liquid aluminium X : Molten aluminium oxide Y : Carbon Z : Carbon (b) Which substance acts as anode and cathode / Bahan yang manakah bertindak sebagai anod dan katod? Anode / Anod : Z Cathode / Katod : Y

(c) State the name of the product at anode and cathode / Namakan hasil yang diperoleh di anod dan katod. Anode / Anod : Oxygen Cathode / Katod : Aluminium (d) Write the ionic equation for the reactions at / Tuliskan persamaan ion bagi tindak balas yang berlaku di 2– 3+ anode / anod : 2O O2 + 4e cathode / katod : Al + 3e Al (e) Why is cryolite added to X / Mengapakan kriolit ditambah ke dalam X ? To lower down the melting point of aluminium oxide (from 2 045°C to 900°C ). The diagram below shows the set-up of apparatus used in the purification of copper.

3

Rajah di bawah menunjukkan susunan radas yang digunakan untuk proses penulenan kuprum.

Electrode X

Electrode Y

Elektrod X

Elektrod Y

Electrode Z

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(a) State the name of the substance used as / Nyatakan nama bahan yang dijadikan sebagai: electrode X / elektrod X : Impure copper

: Pure copper electrolyte Z / elektrolit Z : Copper(II) sulphate solution electrode Y / elektrod Y

(b) Write the half equation that occur at the / Tuliskan persamaan setengah yang berlaku di Cu Cu2+ + 2e electrode X / elektrod X : electrode Y / elektrod Y :

Cu2+ + 2e

Cu

(c) What are the observations at the / Apakah pemerhatian di electrode X / elektrod X : Electrode becomes thinner 4

electrode Y / elektrod Y : Brown solid deposited

To purify metal an impure metal / Untuk menulenkan logam tak tulen: (a) The impure metal is used as the anode / Logam tak tulen (b) The

pure metal

dijadikan sebagai anod.

Logam tulen

dijadikan sebagai katod. is used as the cathode / salt solution (c) The electrolyte used is an containing the ions of the purifying metal. Elektrolit adalah 5

larutan garam

yang mengandungi ion logam yang hendak ditulenkan.

A student intends to electroplate an iron spoon with copper. Describe a laboratory experiment to electroplate the iron ring. Your answer should involve the following: Seorang pelajar bercadang untuk menyadurkan sebatang sudu besi dengan kuprum. Huraikan satu eksperimen di dalam makmal untuk menyadur sebatang sudu besi. Jawapan anda perlu mengandungi: – A labelled diagram showing the set-up of apparatus / Rajah berlabel menunjukkan susunan radas. – Procedure / Kaedah. – Half equation for the reactions at both electrodes / Persamaan setengah untuk tindak balas di kedua-dua elektrod. – Observation at both electrodes / Pemerhatian di kedua-dua elektrod.

Answer / Jawapan:

Copper Iron spoon

Copper(II) nitrate solution

Procedure / Kaedah:

(a) Copper plate and iron spoon are cleaned with Kepingan kuprum dan sudu besi dibersihkan dengan

(b)

Copper(II) nitrate solution is poured into a Larutan kuprum(II) nitrat

(c)

sand paper

.

kertas pasir

.

beaker

dituangkan ke dalam bikar sehingga

until

half full

.

separuh penuh .

Iron spoon

is then connected to the negative terminal of battery while the copper plate is connected to the positive terminal of the battery// Iron spoon is made as cathode while copper plate is made as anode. Sudu besi bateri//

disambungkan kepada terminal negatif bateri dan Sudu besi dijadikan katod dan kepingan kuprum

kepingan kuprum

disambungkan kepada terminal positif

dijadikan anod.

(d) The iron spoon and the copper plate are dipped in the copper(II) nitrate solution as shown in the diagram. Sudu besi dan plat kuprum

(e) The circuit is

completed

dicelup

larutan kuprum(II) nitrat ke dalam / Litar dilengkapkan .

seperti ditunjukkan dalam rajah.

Cu2+ + 2e Cu . (f) Half equation at the cathode / Persamaan setengah di katod : (g) Observation of the cathode: Brown solid is deposited / Pemerhatian di katod: pepejal perang

terenap.

Cu Cu2+ + 2e

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(h) Half equation at the anode / Persamaan setengah di anod : . Copper plate becomes thinner (i) Observation of the anode / Pemerhatian di anod :

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To electroplate an object with metal / Untuk menyadur sesuatu objek dengan logam: (a) The metal object to be electroplated is made to be cathode / Objek yang hendak disadur dijadikan anod anode .. (b) The electroplating metal is made to be / Logam penyadur dijadikan

6

katod

..

(c) The electrolyte used is an aqueous salt solution containing the ions of the electroplating metal. Elektrolit

yang digunakan adalah larutan akueus garam yang mengandungi ion logam penyadur.

ELECTROCHEMICAL SERIES / SIRI ELEKTROKIMIA

Electrochemical Series is an arrangement of positive ion.

1

Siri Elektrokimia ialah susunan

logam

metals

according to their tendency to release electrons to form a

mengikut kecenderungan melepaskan elektron membentuk ion bercas

positif

.

The position of metal atoms in Electrochemical Series / Kedudukan atom logam dalam Siri Elektrokimia: K, Na, Ca, Mg, Al, Zn , Fe, Sn ,Pb, Cu, Ag

2

Tendency of metal atom to release/donate electrons increases (electropositivity increases) Kecenderungan untuk atom logam melepaskan/menderma elektron bertambah (keelektropositifan bertambah)

The position of metal ions (cation) in the Electrochemical Series / Kedudukan ion logam (kation) dalam Siri Elektrokimia: K+, Na+, Ca2+, Mg2+, Al3+, Zn2+, Fe2+, Sn2+, Pb2+, *H+, Cu2+

3

Tendency of metal ion (cation) to receive/gain electrons increases Kecenderungan untuk ion logam (kation) untuk menerima elektron bertambah *H+ is also in the series of ion because it is present in aqueous solution of any electrolyte (salt solution/acid/alkali) * H+ juga terdapat dalam siri ion kerana kehadiran ion H+ dalam elektrolit larutan akueus (larutan garam/asid/alkali)

METAL DISPLACEMENT REACTION / TINDAK BALAS PENYESARAN LOGAM

The metal which is situated at a higher position (higher tendency to release electron) in the Electrochemical Series is able to displace metals below it from its salt solution .

1

Logam yang berada di kedudukan atas (kecenderungan melepaskan elektron yang tinggi) dalam Siri Elektrokimia dapat menyesarkan logam yang di bawahnya daripada larutan garam logam tersebut.

Example / Contoh:

2

Experiment / Eksperimen Silver nitrate solution Larutan argentum nitrat

Observation / Pemerhatian – Copper strip becomes thinner . Kepingan kuprum menipis .

– A grey deposited.

solid

Pepejal kelabu terenap.

– The colourless solution turns blue. Copper Kuprum

Larutan tidak berwarna bertukar menjadi biru.

Remark / Catatan Inference / Inferens: – The grey solid is Pepejal

kelabu

adalah

– The blue solution is Larutan biru adalah

silver

.

argentum

.

copper(II) nitrate

.

kuprum(II) nitrat .

Explanation / Penerangan: Silver ion receives electrons to form – Ion

argentum

silver

menerima elektron membentuk atom

– Copper atom releases electrons to form Atom kuprum melepaskan elektron membentuk

silver

– Copper has displaced Kuprum telah menyesarkan

Cu + 2AgNO3

atom.

argentum .

copper(II) ion ion kuprum(II)

. .

from silver nitrate solution.

argentum

dari larutan argentum nitrat.

Cu(NO3)2 + 2Ag .

– Copper is more electropositive than silver// Copper is above silver in the Electrochemical Series of metal.

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lebih elektropositif daripada argentum //Kuprum Kuprum adalah di atas terletak argentum dalam Siri Elektrokimia logam.

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– Magnesium strip becomes thinner .


Kepingan magnesium menipis .

Larutan kuprum (II) sulfat

– The brown deposited.

Inference / Inferens: – The brown solid is copper . perang

Pepejal

solid

Pepejal perang terenap.

– The blue solution turn colourless. Larutan biru bertukar menjadi tidak berwarna.

Magnesium

adalah

kuprum

– The colourless solution is

.

magnesium sulphate magnesium sulfat

Larutan tidak berwarna adalah

. .

Explanation / Penerangan: – Copper(II) ion receives electrons to form copper atom. Ion

kuprum(II)

menerima elektron membentuk atom kuprum.

magnesium ion

– Magnesium atom releases electrons to form

ion magnesium

Atom magnesium melepaskan elektron membentuk

.

.

– Magnesium has displaced copper from copper(II) sulphate solution.

Magnesium

Magnesium telah menyesarkan

kuprum

dari larutan kuprum(II) sulfat.

MgSO4 + Cu

Mg + Cu SO4

.

more

electropositive than copper// Magnesium – Magnesium is above copper in the Electrochemical Series of metal. is Magnesium adalah di atas terletak

No observable changes.

Zinc nitrate solution

Tiada perubahan yang dapat diperhatikan.

Larutan zink sulfat

lebih

elektropositif daripada kuprum// magnesium kuprum dalam Siri Elektrokimia logam.

Inference / Inferens: – No reaction occur. Tiada tindak balas berlaku.

Explanation / Penerangan: – Copper cannot displace Kuprum

tidak boleh

zinc

menyesarkan

from zinc sulphate solution. zink

daripada larutan zink sulfat.

– Copper is more electropositive than zinc// Copper is zinc in the Electrochemical Series of metal.

below

Kuprum adalah kurang elektropositif daripada zink // kuprum terletak di bawah zink dalam Siri Elektrokimia logam.

Copper / Kuprum

VOLTAIC CELL (CHEMICAL CELL) / SEL RINGKAS (SEL KIMIA) 1

A cell that produces electrical energy when chemical reactions occur in it. Sel yang menghasilkan tenaga elektrik apabila berlaku tindak balas kimia di dalamnya.

2

electrical energy

Energy change in voltaic cell is chemical energy to

Perubahan tenaga dalam sel ringkas ialah dari tenaga kimia kepada 3

Produced when two

different

Terhasil apabila dua logam 4

5

.

metals are dipped in an electrolyte and are connected by an

berlainan

elektrolit

dicelup dalam

The voltage of chemical cell depends on the

distance

the further the distance between them, the

higher

Voltan sel kimia bergantung pada tinggi Elektrokimia, semakin

.

tenaga elektrik

jarak

dan disambung dengan

litar luar

external circuit

.

.

between the two metals in the Electrochemical Series, where is the voltage.

antara dua logam dalam Siri Elektrokimia di mana semakin jauh dua logam dalam Siri

voltannya.

A more electropositive metal becomes the positive terminal:

negative

negatif

sel. Logam yang kurang elektropositif akan menjadi terminal

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Logam yang lebih elektropositif akan menjadi terminal positif sel:

terminal of the cell. A less electropositive metal becomes the

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Electrical current produced is detected by the galvanometer Electrical energy) (Chemical energy Arus elektrik terhasil dikesan oleh galvanometer (Tenaga kimia Tenaga elektrik)

Negative terminal / Terminal negatif • More electropositive metal.

G

:

_

Positive terminal / Terminal positif • Less electropositive metal.

+

Logam lebih elektropositif.

:

Logam kurang elektropositif.

• Metal atom will release electrons that will flow through the external circuit. Metal atom becomes metal ion (becomes thinner).

• The electrons that flow from the external circuit are received by the positive ion in the electrolyte through this terminal.

Atom logam akan melepaskan elektron yang akan mengalir di litar luar. Atom logam menjadi ion logam (semakin nipis).

Elektron yang akan mengalir dari litar luar diterima oleh ion positif dalam elektrolit melalui terminal ini.

Example of simple voltaic cell / Contoh voltan sel ringkas:

6

V

Magnesium

Copper

Magnesium

Kuprum

Copper(II) sulphate solution Larutan kuprum(II) sulfat

(a) Magnesium electrode is a negative terminal because magnesium is more electropositive than Elektrod magnesium adalah terminal

– Magnesium atom Atom magnesium

negatif

releases

melepaskan

kuprum :

elektron untuk membentuk ion magnesium, Mg2+.

– Magnesium electrode becomes

Mg Mg2+ + 2e

thinner

copper

kuprum

Elektron mengalir melalui litar luar ke elektrod

positive

.

/ Elektrod magnesium menjadi

– Electron flows through external circuit to the

Elektrod kuprum adalah terminal

daripada

:

electrons to form magnesium ion, Mg2+.

– Half equation / Persamaan setengah :

(b) Copper electrode is a

elektropositif

kerana magnesium lebih

copper

.

electrode.

.

electropositive

terminal because copper is less positif

nipis

kerana kuprum kurang

elektropositif

than

daripada

magnesium

:

magnesium

:

– Electrons from magnesium flow through external circuit to copper electrode. Elektron dari magnesium mengalir melalui litar luar ke elektrod kuprum.

–

Copper(II) ion in the electrolyte Ion

kuprum(II)

dalam elektrolit

receives

menerima

electron to form copper atom.

elektron untuk membentuk atom kuprum.

Cu + 2e Cu – Half equation / Persamaan setengah : . Brown solid is deposited on the surface of copper electrode. – +

Pepejal perang

terenap di permukaan elektrod kuprum.

(c) The concentration of copper(II) sulphate decreases because copper(II) ions discharged to copper atom at the positive terminal. The intensity of blue colour of copper(II) sulphate decreases.

Kepekatan larutan kuprum(II) sulfat berkurang warna biru larutan kuprum(II) sulfat berkurang.

kerana ion kuprum(II) dinyahcaskan kepada atom kuprum.

Keamatan

(d) If the magnesium metal is replaced with a zinc metal, the voltage reading decreases because zinc is nearer to copper in the electrochemical series.

m

berkurang

kerana zink lebih dekat dengan kuprum

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Daniell cell / Sel Daniell (a) It is an example of voltaic cell which consists of zinc electrode dipped in zinc sulphate solution, copper electrode dipped in copper(II) sulphate solution and connected by a salt bridge or porous pot. Merupakan satu contoh sel kimia yang terdiri daripada elektrod zink yang dicelup ke dalam larutan zink sulfat, elektrod kuprum dicelupkan ke dalam larutan kuprum(II) sulfat dan dihubungkan dengan titian garam atau pasu berliang. Zn / ZnSO4 // CuSO4 / Cu

(b) The function of porous pot or salt bridge is to allow the flow of ions through it so that the electric circuit is completed. Fungsi pasu berliang atau titian garam adalah untuk membenarkan ion-ion mengalir melaluinya dan melengkapkan litar.

(c) The diagram below shows the set-up of apparatus of Daniell cell. Rajah di bawah menunjukkan susunan radas bagi sel Daniell.


Copper

Copper Copper(II) sulphate solution

Zinc sulphate

negative

Elektrod zink adalah terminal

– Zinc atom Atom zink

releases

melepaskan

terminal because zinc is more negatif

electropositive elektropositif

kerana zink adalah lebih

Pasu berliang

than

copper

daripada

kuprum

: :

electron to form zinc ion, Zn . 2+

elektron untuk membentuk ion zink, Zn2+.

Zn Zn2+ + 2e

– Half equation / Persamaan setengah : – Zinc electrode becomes

thinner

Elektrod kuprum adalah terminal

copper

kuprum

Elektron mengalir melalui litar luar ke elektrod

positive

. nipis

/ Elektrod zink menjadi

– Electrons flow through external circuit to the

(e) Copper electrode is a

Porous pot

Larutan kuprum(II) sulfat

Zink sulfat

(d) Zinc electrode is a

Zink sulfat


Zink


Zinc / Zink Zinc sulphate

Kuprum

Zinc

Kuprum

electrode.

.

electropositive

terminal because copper is less positif

..

elektropositif

kerana kuprum kurang

than

daripada

zinc

:

zink

:

– Electrons from zinc electrode flow through external circuit to copper electrode. Elektron dari zink mengalir melalui litar luar ke elektrod kuprum.

–

Copper(II) ion in the electrolyte Ion

kuprum(II)

dalam elektrolit

receives

menerima

– Half equation / Persamaan setengah : –

electron to form copper atom.

elektron untuk membentuk atom kuprum.

Cu + 2e Cu 2+

.

Brown solid

is deposited on the surface of copper electrode.

Pepejal perang

terenap di permukaan elektrod kuprum.

(f) The concentration of copper(II) sulphate decreases because copper(II) ions are discharged to copper atoms. The intensity of blue colour of copper(II) sulphate decreases. Kepekatan larutan kuprum(II) sulfat berkurang warna biru kuprum(II) sulfat berkurang.

kerana ion kuprum(II) telah dinyahcaskan kepada atom kuprum.

Keamatan

(g) If zinc metal is replaced with a magnesium metal, the voltage reading increases because magnesium is further from copper in the Electrochemical Series.

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Jika logam zink digantikan dengan logam magnesium, bacaan voltan bertambah kerana jarak antara magnesium dengan kuprum jauh daripada jarak antara zink dengan kuprum dalam Siri Elektrokimia. lebih

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Four main uses of the Electrochemical Series / Kegunaan utama Siri Elektrokimia: (a) To predict the terminal of chemical cell / Untuk meramalkan terminal sel kimia – The more electropositive metal is the negative terminal of the cell.

8

Logam yang lebih elektropositif ialah terminal negatif sel.

– The less electropositive metal is the positive terminal of the cell. Logam yang kurang elektropositif ialah terminal positif sel.

(b) To predict the voltage of chemical cell / Untuk meramalkan voltan sel kimia – The further the distance between two metals in the Electrochemical Series, the higher is the voltage of the chemical cell. Semakin jauh jarak antara dua logam dalam Siri Elektrokimia, semakin tinggi bacaan voltan sel kimia.

(c) To predict the metal displacement reactions / Untuk meramalkan tindak balas penyesaran logam – The more electropositive metal can displace a less electropositive metal from its salt solution. Logam yang lebih elektropositif dapat menyesarkan logam yang kurang elektropositif daripada larutan garamnya.

(d) To predict the selected ion to be discharged at the electrode in an electrolysis Untuk meramalkan pemilihan ion untuk dinyahcas di elektrod dalam proses elektrolisis

EXERCISE / LATIHAN

The table below shows the results of an experiment to construct the Electrochemical Series through the ability of metals to displace other metals from their salt solution.

1

Jadual di bawah menunjukkan keputusan eksperimen untuk membina Siri Elektrokimia berdasarkan keupayaan suatu logam untuk menyesarkan logam lain dari larutan garamnya. Experiment I / Eksperimen I

Experiment II / Eksperimen II

P nitrate solution

R nitrate solution

Larutan P nitrat

Larutan R nitrat

Zinc / Zink

Zinc / Zink

Metal P is displaced, blue colour solution turn colourless. Logam P disesarkan, larutan biru bertukar menjadi tanpa warna.

No reaction. Tiada tindak balas.

(a) Based on the results in the table, arrange metal P, zinc and R in descending order of electropositivity. Berdasarkan keputusan dalam jadual, susunkan logam P, zink dan R dalam tertib menurun keelektropositifan.

R, Zn, P (b) Based on the observation in Experiment I / Berdasarkan pemerhatian dalam Eksperimen I, (i) state the name the suitable metal P / namakan logam yang sesuai bagi P. Copper .

(ii) zinc can displace metal P from P nitrate solution. Explain. zink boleh menyesarkan logam P daripada larutan P nitrat. Terangkan.

Zinc is more electropositive than P. .

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(iii) write the chemical equation for the reaction / tuliskan persamaan kimia untuk tindak balas. Zn + Cu(NO3 )2 Zn(NO3 )2 + Cu

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2

The diagram below shows the set-up of the apparatus to arrange metals W, X, Y and Z based on the potential difference of the metals. Rajah di bawah menunjukkan susunan radas bagi eksperimen untuk menentukan kedudukan logam W, X, Y dan Z berdasarkan beza upaya logam.

V Metal electrode

Metal electrode

Elektrod logam

Elektrod logam

Electrolyte / Elektrolit

The table below shows the results of the experiment. Jadual di bawah menunjukkan keputusan eksperimen. Pair of metals Pasangan logam

Potential difference (V)

Negative terminal

0.50

X

0.30

Y

1.10

Z

Beza keupayaan (V)

W and X W dan X

X and Y X dan Y

W and Z W dan Z

Terminal negatif

(a) Arrange metals W, X, Y and Z in descending order of the electropositivity of metal. Susunkan logam W, X, Y dan Z dalam tertib menurun keelektropositifan logam.

Z, Y, X, W . (b) (i)

Metals X and Z are used as electrodes in the diagram. State which metal acts as positive terminal. Logam X dan Z digunakan sebagai terminal dalam rajah. Nyatakan logam yang manakah akan bertindak sebagai terminal positif.

Metal X

(ii) Give reason for your answer in (b)(i) / Berikan sebab untuk jawapan anda di (b)(i). Metal X is less electropositive than metal Z. .

(c) Predict the voltage of the cell in (b)(i) / Ramalkan nilai voltan dalam sel di (b)(i). 0.6 V 3

The diagram below shows the set-up of apparatus for two types of cell. Rajah di bawah menunjukkan susunan radas untuk dua jenis sel.

Zinc

Copper

Copper

Zink

Kuprum

Kuprum


Cell Y / Sel Y

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Complete the following table to compare cell X and cell Y : Lengkapkan jadual berikut untuk membandingkan sel X dan sel Y : Description

Cell X

Cell Y

Electrolytic cell

Chemical cell

Perkara

Sel X

Type of cell Jenis sel

The energy change

Electrical energy

Perubahan tenaga

Ion presence in the electrolyte

Sel Y

Chemical energy

Chemical energy

Cu2+, H+, SO42–, OH–

Ion hadir dalam elektrolit

Electrical energy

Cu2+, H+, SO42–, OH–

Electrode

Anode / Anod: Copper

Negative terminal / Terminal negatif : Zinc

Elektrod

Cathode / Katod: Copper

Positive terminal / Terminal positif : Copper

Half equation

Anode / Anod: Cu

Negative terminal / Terminal negatif : Zn

Persamaan setengah

Cu2+ + 2e

2+ Cathode / Katod: Cu + 2e

Cu

Zn2+ + 2e

2+ Positive terminal / Terminal positif : Cu + 2e

Cu

Observation

Anode / Anod:

Negative terminal / Terminal negatif :

Pemerhatian

Copper electrode becomes thinner

Zinc electrode becomes thinner

Cathode / Katod: Brown solid deposited

Positive terminal / Terminal positif :

Electrolyte / Elektrolit:

Electrolyte / Elektrolit:

Intensity blue colour of copper(II) sulphate

Intensity blue colour of copper(II) sulphate decreases


remains unchanged

The diagram below shows the set-up of apparatus for an experiment.

4

Rajah di bawah menunjukkan susunan radas bagi suatu eksperimen.

V

–

+

Zinc / Zink

Anode

Cathode Copper

Copper

Kuprum

Kuprum

Zinc sulphate solution


Larutan zink sulfat Pasu berliang


Cell A / Set A

(a)


Porous pot


Cell B / Set B

In the above diagram, label Dalam gambar rajah di atas, label

(i) (ii)

the positive terminal and negative terminal Cell A, terminal positif dan terminal negatif bagi Sel A,

anode and cathode in Cell B. anod dan katod bagi Sel B.

(b) What is the energy change in Cell A and Cell B?

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Cell A / Sel A : Chemical energy to electrical energy

Cell B / Sel B : Electrical energy to chemical energy

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Apakah perubahan tenaga dalam Sel A dan Sel B?

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(c) What is the function of the porous pot in cell A? Apakah fungsi pasu berliang dalam Sel A?

To allow the movement of ions through it. (d) Referring to Cell A. Merujuk kepada Sel A.

(i)

What is the observation at zinc electrode?

Apakah pemerhatian di elektrod zink?

Zinc electrode becomes thinner.

(ii) Write the half equation for the reaction at zinc electrode. Tuliskan persamaan setengah untuk tindak balas di elektrod zink.

Zn

Zn2+ + 2e

(iii) What is the observation at copper electrode / Apakah pemerhatian di elektrod kuprum? Brown solid deposited.

(iv) Write the half equation for the reaction at copper electrode. Tuliskan persamaan setengah untuk tindak balas di elektrod kuprum.

Cu2+ + 2e

Cu .

(v) After 30 minutes, what is the colour change of the copper(II) sulphate solution? Explain why. Selepas 30 minit, apakah perubahan warna larutan kuprum(II) sulfat? Jelaskan mengapa.

– The intensity of blue colour decreases. – Copper(II) ions are discharged to form copper atoms. – Concentration of copper(II) ions in copper(II) sulphate decreases. (e) Referring to Cell B. Merujuk kepada Sel B.

(i)

What is the observation at the anode? Apakah pemerhatian di anod?

Copper electrode becomes thinner.

(ii) Write the half equation for the reaction at the anode. Tuliskan persamaan setengah untuk tindak balas di anod.

Cu

Cu2+ + 2e

(iii) What is the observation at the cathode? Apakah pemerhatian di katod?

Brown solid deposited.

(iv) Write the half equation for the reaction at copper electrode. Tuliskan persamaan setengah untuk tindak balas di katod.

Cu2+ + 2e

Cu

(f) The intensity of blue colour of copper(II) sulphate solution in the Cell B remains unchanged during the experiment. Explain why. Keamatan warna biru larutan kuprum(II) sulfat dalam Sel B tidak berubah semasa eksperimen. Jelaskan mengapa.

– The concentration of copper(II) sulphate remain unchanged. – The rate of copper(II) ions discharged to copper atom at the cathode equals to the rate of copper atoms form

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copper(II) ions at the anode.

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Which of the following is an electrolyte? Antara berikut, yang manakah adalah elektrolit?

C

Copper electrode becomes thicker


Elektrod kuprum semakin tebal

Elektrod kuprum semakin nipis

D



A Glacial ethanoic acid Asid etanoik glasial

B

Molten naphthalene

C

Aqueous solution of zinc chloride

Naftalena lebur


Larutan akueus zink klorida

4

D Hydrogen chloride in methylbenzene Hidrogen klorida dalam metilbenzena

2

Gelembung gas dibebaskan

The diagram below shows the set-up of apparatus of an electrolysis process. Rajah di bawah menunjukkan sususnan radas untuk proses elektrolisis.

The diagram below shows the set-up of apparatus used to electrolyse substance X.

Electrolyte

Rajah di bawah menunjukkan susunan radas untuk elektrolisis bahan X.

Elektrolit

Carbon electrode

P

Elektrod karbon

Q

Carbon electrode Elektrod karbon

Carbon electrodes Elektrod karbon

Which of the following electrolytes produce oxygen gas at electrode Q?

Substance X

Antara elektrolit berikut, yang manakah membebaskan gas oksigen pada elektrod Q?

Bahan X

I

Heat

Asid hidroklorik 1.0 mol dm–3

Panaskan

3

1.0 mol dm–3 hydrochloric acid

II

1.0 mol dm–3 sulphuric acid Asid sulfurik 1.0 mol dm–3

Which of the following compounds can light up the bulb when used as substance X?

III 1.0 mol dm–3 potassium iodide solution

Antara berikut, yang manakah boleh menyalakan mentol apabila digunakan sebagai bahan X? A Copper(II) nitrate / Kuprum(II) nitrat B Lead(II) iodide / Plumbum(II) iodida C Zinc carbonate / Zink karbonat D Sodium carbonate / Natrium karbonat

IV 1.0 mol dm–3 nitric acid

The diagram below shows the set-up of apparatus for electrolysis of copper(II) sulphate solution. Rajah di bawah menunjukkan susunan radas untuk elektrolisis larutan kuprum(II) sulfat.

Larutan kalium iodida 1.0 mol dm–3

A B C D 5

Asid nitrik 1.0 mol dm–3 I and II only / I dan II sahaja II and III only / II dan III sahaja II and IV only / II dan IV sahaja II, III and IV only / II, III dan IV sahaja

The table below shows the observation of electrolysis of a substance Q using carbon electrode. Jadual di bawah menunjukkan pemerhatian bagi elektrolisis bahan Q menggunakan elektrod karbon.

Electrode

Observation

Elektrod

Copper electrode X

Copper electrode Y Elektrod kuprum Y

Elektrod kuprum X


Anode

A greenish-yellow gas released

Anod

Gas kuning kehijauan terbebas

Cathode

A colorless gas which burns with a ‘pop’ sound is released

Katod

Gas tanpa warna terbakar dengan bunyi ‘pop’ dibebaskan


What can be observed at the electrodes X and Y after 30 minutes? Apakah yang dapat diperhatikan pada elektrod X dan Y selepas 30 minit?

A

B

X

Y







Pemerhatian

What is substance Q? Apakah bahan Q?

A B C D

1.0 mol dm–3 of hydrochloric acid. Asid hidroklorik 1.0 mol dm–3.

1.0 mol dm–3 of potassium nitrate solution. Larutan natrium nitrat 1.0 mol dm–3.

1.0 mol dm–3 of copper(II) chloride solution. Larutan kuprum(II) klorida 1.0 mol dm–3.

1.0 mol dm–3 of magnesium bromide solution. Larutan magnesium bromida 1.0 mol dm–3.


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The diagram below shows the set-up of apparatus of a chemical cell that shows the direction of electron flow from zinc to metal Q.

8

The table below shows the information about three voltaic cells. Jadual di bawah menunjukkan maklumat tentang tiga sel kimia.

Rajah di bawah menunjukkan susunan radas sel kimia yang menunjukkan arah pengaliran elektron ke logam Q.

Pair of metals Pasangan logam

Q

Zinc Zink

Larutan natrium klorida cair

A B

Apakah logam Q?

B C D

Copper

9

Iron

Beza upaya (V)

Terminal positif

W, Z

Z

3.1

X, Y

Y

0.3

W, X

X

1.8

Apakah beza upaya sel kimia apabila logam Y dipasangkan dengan logam Z?

What is metal Q? Kuprum

Potential difference (V)

What is the potential difference of the voltaic cell when metal Y is paired with metal Z?

Dilute sodium chloride solution

A

Positive terminal

C D

1.0 V 1.3 V

2.1 V 2.8 V

The diagram below shows the set-up of apparatus in a chemical cell and electrolytic cell. Rajah di bawah menunjukkan susunan radas bagi sel kimia dan sel elekrolisis.

Besi

Aluminium Aluminium

Magnesium Magnesium

7

The diagram below shows the set-up of apparatus used to purify impure copper by using electrolysis method. Rajah di bawah menunjukkan susunan radas yang digunakan untuk menulenkan kuprum tak tulen dengan menggunakan kaedah elektrolisis.

P

Zinc

Elektrolit Z

A B C

Antara berikut, yang manakah adalah kedudukan yang betul untuk kuprum tulen dan kuprum tak tulen?

D

A

Elektrod Y

Impure copper

Pure copper

Kuprum tak tulen

Kuprum tulen

Electrolyte Z Elektrolit Z


B

Pure copper

Impure copper

Kuprum tulen

Kuprum tak tulen

Kuprum

Zinc sulphate solution

Which of the following shows the correct position of pure copper and impure copper?

Electrode Y

Copper


Antara berikut, yang manakah merupakan pemerhatian pada elektrod R?

Electrolyte Z

Elektrod X

S

Which of the following is the observation at electrode R?

Y

Electrode X

R

Copper Kuprum

Larutan zink sulfat

X

Q

Zink

Pure copper

Sulphuric acid

Kuprum tak tulen

Kuprum tulen

Asid sulfurik

D

Pure copper

Impure copper

Sulphuric acid

Kuprum tulen

Kuprum tak tulen

Asid sulfurik

A colourless gas is released Gas tanpa warna terbebas

A brown solid is deposited Pepejal perang terenap

Jadual di bawah menunjukkan keputusan eksperimen untuk mengkaji penyesaran logam daripada larutan garamnya menggunakan logam lain.

Metal Logam P


Impure copper

Electrode R becomes thicker Elektrod R semakin tebal

10 The table below shows the results of an experiment to study the displacement of metal from its solution using other metals.


C

Electrode R becomes thinner Elektrod R semakin nipis

Q

Nitrate of Q Nitrat bagi Q

Nitrate of S Nitrat bagi S

–

– reaction occur / tindak balas berlaku – no reaction / tiada tindak balas Which of the following is the arrangement of metals P, Q and R in ascending order of the tendency of the metals to form ions? Antara berikut, yang manakah adalah susunan logam P, Q dan R dalam susunan menaik kecenderungan logam membentuk ion?

P, S, Q Q, S, P

C D

S, P, Q S, Q, P n io

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ACID AND BASES ASID DAN BES

ACID AND BASES / ACID DAN BES • ACID / ASID – To state the meaning of acid, give examples and write chemical equations and observations for the reaction of acids: Menyatakan maksud asid, memberi contoh dan menulis persamaan tindak balas kimia dan pemerhatian bagi tindak balas asid: (i) with carbonates / dengan karbonat (ii) with metals / dengan logam (iii) with bases / dengan bes • BASICITY OF AN ACID / KEBESAN ASID – To state the meaning of basicity of an acid and to write equations for the ionisation of monoprotic and diprotic acids. Menyatakan maksud kebesan asid dan menulis persamaan pengionan asid monoprotik dan diprotik. – To relate the basicity of acid/alkali with pH values / Mengaitkan kebesan asid /alkali dengan nilai pH. • BASE / ALKALI / BES / ALKALI – To state the meaning of base and to correlate base with alkali / Menyatakan maksud bes dan mengaitkan bes dengan alkali. – To write chemical equations involving alkalis with acids and ammonium salts. Menulis persamaan tindak balas kimia alkali dengan asid dan dengan garam ammonium.

ROLE OF WATER IN ACIDS AND ALKALI / PERANAN AIR DALAM ASID DAN ALKALI – To explain why the acid and alkali properties are shown in the presence of water. Menerangkan mengapa sifat asid dan alkali ditunjukkan dengan kehadiran air.

– To explain why the acid and alkali properties do not show in the absence of water or in non-water solvent. Menerangkan mengapa sifat asid dan alkali tidak ditunjukkan tanpa kehadiran air atau dalam pelarut bukan air.

pH SCALE / SKALA pH – To state the meaning of pH / Menyatakan maksud pH. – To relate the pH value with the concentration of H+ ion for the acids and OH– ions for alkalis. Mengaitkan nilai pH dengan kepekatan ion H+ bagi asid dan ion OH– bagi alkali.

• STRONG / WEAK ACID AND STRONG / WEAK ALKALI / ASID KUAT / LEMAH DAN ALKALI KUAT / LEMAH – To list examples and equations for the ionisation of strong / weak acid and strong / weak alkali. Menyenaraikan contoh dan menulis persamaan pengionan bagi asid kuat / lemah dan alkali kuat / lemah. – To relate the pH value with the strength of acid / alkali / Mengaitkan nilai pH dengan kekuatan asid / alkali.

ACID AND ALKALI CONCENTRATION / KEPEKATAN ASID DAN ALKALI – To state the meaning of concentration in g dm–3 and mol dm–3 / Menyatakan maksud kepekatan dalam unit g dm–3 dan mol dm–3. – To state the meaning of standard solution and to describe the preparation of standard solution. Menyatakan maksud larutan piawai dan menghuraikan eksperimen penyediaan larutan piawai. MV . – To solve various problems with calculations related to the preparation of standard solution using n = 1 000 Menyelesaikan pelbagai masalah pengiraan berkaitan penyediaan larutan piawai menggunakan formula n = MV . 1 000

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• NEUTRALISATION OF ACID AND ALKALI / TINDAK BALAS PENEUTRALAN ASID DAN ALKALI – To describe the titration of acid with alkali and to calculate acid / alkali concentrations if a standard solution are given. Menghuraikan titratan asid dengan alkali dan menghitung kepekatan asid / alkali jika satu larutan piawai diberikan. – To describe the type of indicators used and the colour changes at the end-point. Menyatakan jenis penunjuk yang digunakan dan perubahan warna penunjuk pada takat akhir. – To solve numerical problems involving neutralisation / Menyelesaikan masalah pengiraan berkaitan peneutralan.

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ACID / ASID 1 2

Acid is a chemical substance which ionises in water to produce hydrogen ion. Asid ialah bahan kimia yang mengion dalam air menghasilkan ion hidrogen.

Acid tastes sour and turns moist blue litmus to red. Asid mempunyai rasa yang masam dan menukar kertas litmus biru lembap menjadi merah.

3

Example of acid is hydrochloric acid / Contoh asid ialah asid hidroklorik : (a) Hydrogen chloride gas is a *covalent compound exist in the form of molecule. Gas hidrogen klorida ialah *sebatian kovalen wujud dalam bentuk molekul.

(b) As hydrogen chloride dissolves in water, hydrogen chloride molecule ionises to hydrogen ion and chloride ion in aqueous solution. This aqueous solution is called hydrochloric acid. Apabila hidrogen klorida melarut dalam air, molekul hidrogen klorida mengion kepada ion hidrogen dan ion klorida dalam larutan akueus. Larutan akueus itu dipanggil asid hidroklorik.

HCl (aq / ak ) Hydrochloric acid / Asid hidroklorik

H+ (aq / ak ) Hydrogen ion / Ion hidrogen

+

Cl– (aq / ak ) Chloride ion / Ion klorida

(c) An aqueous hydrogen ion, H+(aq) is actually the hydrogen ion combined with water molecule to form hydroxonium ion, H3O+. However this ion can be written as H+. Ion hidrogen akueus, H+(ak) ialah ion hidrogen yang bergabung dengan molekul air membentuk ion hidroksonium, H3O+. Walau bagaimanapun, ion ini boleh ditulis sebagai H+. HCl (g) + H2O(l/ce) H3O+ (aq/ak ) + Cl– (aq/ak )

Hydrogen chloride

Ion hydroxonium Ion klorida

Hidrogen klorida

Ion hidroksonium

H3O+ Ion hydroxonium 4

H+(aq/ak ) Ion hidrogen

Ion hidroksonium

The ionisation of hydrochloric acid is represented as:

Ion klorida

Pengionan asid hidroklorik diwakili oleh: HCl (aq/ak) H+ (aq/ak) + Cl– (aq/ak)

+ H2O

Ion hidrogen

Basicity of an acid is the number of ionisable of hydrogen atom per molecule of an acid molecule in an aqueous solution / Kebesan asid ialah bilangan atom hidrogen yang boleh mengion bagi setiap molekul asid dalam larutan akueus.

– Monoprotic: One acid molecule ionises to Monoprotik: Satu molekul asid mengion kepada

one

hydrogen ion.

satu

ion hidrogen.

– Diprotic: One acid molecule ionises to

two

hydrogen ion.

Diprotik: Satu molekul asid mengion kepada

dua

ion hidrogen.

– Triprotic: One acid molecule ionises to

three

Triprotik: Satu molekul asid mengion kepada

tiga

hydrogen ion. ion hidrogen.

Hydrochloric is monoprotic acid because one molecule of hydrochloric acid ionises to Asid hidroklorik ialah sejenis asid monoprotik kerana satu molekul asid hidroklorik mengion kepada 5

one satu

hydrogen ion. ion hidrogen.

Examples of acid and their basicity / Contoh-contoh asid dan kebesannya: Ionisation of acid Pengionan asid

HNO3 (aq/ak ) Nitric acid

– + NO3 (aq) Hydrogen ion Nitrate ion

Ion hidrogen

H2SO4 (aq/ak ) Sulphuric acid

2– + SO4 (aq) Hydrogen ion Sulphate ion

Asid sulfurik

H3PO4 (aq/ak ) Phosphoric acid *CH3COOH (aq/ak ) Ethanoic acid Asid etanoik

3H+(aq) Hydrogen ion

Asid fosforik

Ion hidrogen

Kebesan asid

One

Monoprotic

Two

Diprotic

Three

Triprotic

One

Monoprotic

Ion nitrat

2H+(aq)

Ion hidrogen

Basicity of acid

Bilangan ion hidrogen dihasilkan bagi setiap molekul asid

H+(aq)

Asid nitrik

Number of hydrogens ion produce per molecule of acid

Ion sulfat

3– + PO4 (aq) Phosphate ion

Ion fosfat

CH3COO–(aq) + H+(aq) Ethanoate ion Hydrogen ion Ion etanoat

Ion hidrogen n io

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*Not all hydrogen atoms in ethanoic acid are ionisable / *Bukan semua ion hidrogen dalam asid etanoik boleh mengion

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BASES / BES Bases is a chemical substance that reacts with acid to produce salt and water only. For example, Bes ialah sejenis bahan kimia yang bertindak balas dengan asid menghasilkan garam dan air sahaja. Contohnya,

1

(a) Copper(II) oxide (a base) reacts with sulphuric acid to produce copper(II) sulphate (a salt) and water. Kuprum(II) oksida (bes) bertindak balas dengan asid sulfurik menghasilkan kuprum(II) sulfat (garam) dan air.

CuO + H2SO4

CuSO4

+

H2O

(b) Zinc hydroxide (a base) reacts with hydrochloric acid to produce zinc chloride (a salt) and water. Zink hidroksida (bes) bertindak balas dengan asid hidroklorik menghasilkan zink klorida (garam) dan air.

ZnCl2 H2O Zn(OH)2 + 2HCl + Most bases are metal oxide or metal hydroxide which are ionic compound. Example of bases are magnesium oxide, zinc oxide, sodium hydroxide and potassium hydroxide.

2

Kebanyakan bes ialah oksida logam atau hidroksida logam yang merupakan sebatian ion. Contoh-contoh bes ialah magnesium oksida, zink oksida, natrium hidroksida dan kalium hidroksida.

The bases that can dissolve in water (soluble bases) are known as alkali.

3

Bes yang boleh melarut dalam air (bes larut) dikenali sebagai alkali.

Sodium hydroxide and potassium hydroxide are soluble in water and they are called alkali whereas magnesium oxide and zinc oxide are called bases as they are insoluble in water.

4

Natrium hidroksida dan kalium hidroksida larut dalam air dan dipanggil sebagai alkali manakala magnesium oksida dan zink oksida dipanggil sebagai bes kerana tidak terlarut dalam air. Alkali is a base that is soluble in water and ionises to hydroxide ion. For example, Alkali ialah bes yang larut dalam air dan mengion kepada ion hidroksida. Contohnya,

5

(a) Sodium hydroxide dissolves in water and ionises to hydroxide ion. Natrium hidroksida terlarut dalam air dan mengion kepada ion hidroksida.

NaOH (aq/ak ) Na+ (aq/ak ) + OH– (aq/ak ) (b) Ammonia solution is obtained by dissolving ammonia molecule in water, ionisation occur to produce a hydroxide ion, OH–. Larutan ammonia diperoleh dengan melarutkan molekul ammonia dalam air, pengionan berlaku menghasilkan ion hidroksida, OH–. – NH3 (g) + H2O (l/ce ) NH+ 4 (aq/ak ) + OH (aq/ak )

(c) Other examples of alkalis are barium hydroxide and calcium hydroxide. Contoh alkali lain adalah barium hidroksida dan kalsium hidroksida.

Alkali tastes bitter, slippery and turns moist red litmus to blue.

6

Alkali mempunyai rasa yang pahit, licin dan menukarkan kertas litmus merah lembap kepada biru.

EXERCISE / LATIHAN

Complete the following table / Lengkapkan jadual berikut : Soluble base (alkali) / Bes terlarut (alkali) Name / Nama Sodium oxide Natrium oksida

Potassium oxide Kalium oksida

Ammonia Ammonia


Potassium hydroxide Kalium hidroksida

Barium hydroxide Barium hidroksida

Formula / Formula

Ionisation equation / Persamaan pengionan

Na2O

Na2O(s) + H2O 2NaOH(aq) NaOH(aq) Na+ (aq) + OH– (aq)

K2O

K2O(s) + H2O 2KOH(aq) KOH(aq) K+ (aq) + OH– (aq)

NH3

NH3(g)+ H2O

NaOH

NaOH(aq)

KOH

KOH(aq)

Ba(OH)2

Ba(OH)2(aq)

NH4+(aq) + OH–(aq) Na+ (aq) + OH– (aq) K+ (aq) + OH– (aq) Ba2+(aq) + 2OH– (aq)

Insoluble base / Bes tak terlarut Name / Nama

Formula / Formula

Copper(II) oxide Kuprum(II) oksida

Copper(II) hydroxide Kuprum(II) hidroksida

Zinc hydroxide Zink hidroksida

Aluminium oxide Aluminium oksida

Lead(II) hydroxide Plumbum(II) hidroksida

Magnesium hydroxide Magnesium hidroksida

CuO Cu(OH)2 Zn(OH)2 Al2O3 Pb(OH)2 Mg(OH)2

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Bases that can dissolve in water (soluble bases) are known as alkali / Bes yang larut dalam air (bes larut) dipanggil alkali m

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CHEMICAL PROPERTIES OF ACID / SIFAT-SIFAT KIMIA ASID 1

Acid react with metal, base / alkali and metal carbonate / Asid bertindak balas dengan logam, bes/alkali dan karbonat logam: Chemical properties Sifat-sifat kimia

1 Acid + Metal Asid + Logam

Salt + Hydrogen

Garam + Hidrogen

Example of experiment Contoh eksperimen

Zinc + Hydrochloric acid Zink + Asid hidroklorik

Observation Pemerhatian

Remark Catatan

– The grey solid Chemical equation: Persamaan kimia: dissolves. Pepejal kelabu Mg + 2HCl MgCl2 + H2

* Acid react with the metals that are terlarut. Lighted wooden more electropositive than hydrogen – Gas bubbles splinter in electrochemical series, acids do not Inference / Inferens : are released. Kayu uji menyala react with copper and silver (type of – Magnesium reacts with When a Hydrochloric acid reaction is displacement, the metals hydrochloric acid. burning Asid hidroklorik that are placed above hydrogen in wooden Magnesium bertindak balas Electrochemical Series can displace Magnesium powder dengan asid hidroklorik. splinter is hydrogen from acid) Serbuk magnesium placed at the – Hydrogen gas is * Asid bertindak balas dengan logam-logam mouth of the yang lebih elektropositif daripada hidrogen (a) About 5 cm3 of dilute released. test tube, dalam Siri Elektrokimia, asid tidak bertindak hydrochloric acid is poured Gas hidrogen terbebas. balas dengan kuprum dan argentum (jenis ‘pop sound’ is into a test tube. tindak balas ialah penyesaran, logam-logam produced. Sebanyak 5 cm3 asid hidroklorik di atas hidrogen dalam Siri Elektrokimia boleh menyesarkan hidrogen daripada asid)

* Application of the reaction: * Aplikasi tindak balas:

– Preparation of soluble salt (Topic Salt) Penyediaan garam terlarut (Tajuk Garam)

– Preparation of hydrogen gas in determination of the empirical formula of copper(II) oxide (Topic Chemical Formula and Equation) Penyediaan gas hidrogen dalam menentukan formula empirik kuprum(II) oksida (Tajuk Formula dan Persamaan Kimia)

2 Acid + Metal carbonate Water + Carbon dioxide Asid + Karbonat logam Karbon dioksida

Salt +

Garam + Air +

cair dimasukkan ke dalam tabung uji.

(b) One spatula of magnesium powder is added to the acid. Satu spatula serbuk magnesium ditambah kepada asid.

(c) A burning wooden splinter is placed at the mouth of the test tube. Kayu uji menyala diletakkan pada mulut tabung uji.

(d) The observations are recorded. Semua pemerhatian direkodkan.

Calcium carbonate + Nitric acid Kalsium karbonat + Asid nitrik Hydrochloric acid Asid hidroklorik

Lime water Air kapur

*Application of the reaction: *Aplikasi tindak balas:


– Confirmatory test for anion carbonate ion in qualitative analysis of salt (Topic Salt) Ujian pengesahan bagi ion karbonat dalam analisis kualitatif garam (Tajuk Garam)

Gelembung gas dibebaskan. Apabila kayu uji menyala didekatkan pada mulut tabung uji, bunyi ‘pop’ dihasilkan.

Calcium carbonate / Kalsium karbonat

(a) About 5 cm3 of dilute hydrochloric acid is poured into a test tube. Sebanyak 5 cm3 asid hidroklorik cair dimasukkan ke dalam tabung uji.

(b) One spatula of calcium carbonate powder is added into the test tube. Satu spatula serbuk kalsium karbonat dimasukkan ke dalam asid.

– The white solid dissolves. Pepejal putih terlarut.

Chemical equation: Persamaan kimia:

CaCO3 + 2HCl

CaCl2 + H2O + CO2 – Gas bubbles are released. Inference / Inferens : When the – Calcium carbonate gas passed reacts with nitric acid. through lime Kalsium karbonat water, the lime bertindak balas dengan water turns asid hidroklorik. chalky. Gelembung gas – Carbon dioxide gas terbebas. Apabila is released. gas tersebut dilalukan melalui air kapur, air kapur menjadi keruh.

Gas karbon dioksida terbebas.

(c) The gas released is passed through lime water as shown in the diagram. Gas yang dibebaskan dilalukan melalui air kapur seperti ditunjukkan dalam rajah.

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Semua pemerhatian direkodkan.

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3 Acid + Base / Alkali Asid + Bes / Alkali

Salt + Water

Garam + Air

*Acid neutralises base/alkali

Copper(II) oxide + Sulphuric acid Kuprum(II) oksida + Asid sulfurik Sulphuric acid / Asid sulfurik

– The black solid dissolves.

Persamaan kimia:

CuO + H2SO4

Pepejal hitam terlarut.

* Asid meneutralkan bes/alkali

*Application of the reaction: *Aplikasi tindak balas:


Chemical equation:

Copper(II) oxide / Kuprum(II) oksida

CuSO4 + H2O

– The colourless solution turns Inference / Inferens : blue. – Copper(II) oxide reacts Larutan tanpa with sulphuric acid. warna bertukar

(a) Dilute hydrochloric acid is poured into a beaker until half full.

menjadi biru.

Asid hidroklorik cair dimasukkan dalam bikar hingga separuh penuh.

(b) The acid is warmed gently. Asid dihangatkan.

(c) One spatula of copper(II) oxide powders added to the acid.

Kuprum(II) oksida bertindak balas dengan asid sulfurik.

– The blue solution is copper(II) sulphate . Larutan biru tersebut ialah kuprum(II) sulfat .

Satu spatula serbuk kuprum(II) oksida ditambahkan kepda asid tersebut.

(d) The mixture is stirred with a glass rod. Campuran dikacau dengan rod kaca.

(e) The observations are recorded. Semua pemerhatian direkodkan.

Write the chemical formulae for the following compounds / Tuliskan formula kimia bagi sebatian berikut:

2

Compound / Sebatian Hydrochloric acid Asid hidroklorik



Ethanoic acid Asid etanoik


Potassium hydroxide Kalium hidroksida

Calcium hydroxide Kalsium hidroksida

Sodium carbonate Natrium karbonat

Magnesium hydroxide Magnesium hidroksida

Ammonium sulphate Ammonium sulfat

Hydroxide ion Ion hidroksida

Sodium sulphate Natrium sulfat

Carbon dioxide Karbon dioksida

Copper(II) carbonate Kuprum(II) karbonat

Water

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HCl HNO3 H2 SO4 CH3COOH NaOH KOH Ca(OH)2 Na2CO3 Mg(OH)2 (NH4 )2SO4 OH– Na2 SO4 CO2 CuCO3 H2O

Compound / Sebatian Magnesium oxide Magnesium oksida

Calcium oxide Kalsium oksida

Copper(II) oxide Kuprum(II) oksida

Lead(II) oxide Plumbum(II) oksida

Sodium nitrate Natrium nitrat

Potassium sulphate Kalium sulfat

Barium hydroxide Barium hidroksida


Magnesium Magnesium

Zinc Zink

Sodium Natrium

Calcium carbonate Kalsium karbonat


Sodium oxide Natrium oksida


Chemical formulae / Formula kimia MgO CaO CuO PbO NaNO3 K2 SO4 Ba(OH)2 NaCl Mg Zn Na CaCO3 H2 Na2O Mg(NO3 )2

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Air

Chemical formulae / Formula kimia

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Ionic equation / Persamaan ion :

Ionic equation shows particles that change during chemical reaction. Persamaan ion menunjukkan zarah yang berubah semasa tindak balas kimia.

Example / Contoh : (i) Reaction between sulphuric acid and sodium hydroxide solution: Tindak balas antara asid sulfurik dengan larutan natrium hidroksida: Write balanced equation / Tulis persamaan seimbang :

H2SO4 + 2NaOH

Na2SO4 + 2H2O

Write the formula of all the particles in the reactants and products: Tulis formula bagi semua zarah dalam bahan tindak balas dan hasil tindak balas:

2H+ + SO42– + 2Na+ + 2OH–

Remove all the particles in the reactants and products which remain unchanged: Keluarkan semua zarah dalam bahan dan hasil tindak balas yang tidak berubah:

2H+ + SO42– + 2Na+ + 2OH–


2H+ + 2OH–

2Na+ + SO42– + 2H2O

2Na+ + SO42– + 2H2O

2H2O ⇒ H+ + OH–

H2O

(ii) Reaction between zinc oxide and hydrochloric acid / Tindak balas antara zink dengan asid hidroklorik : Write balanced equation / Tulis persamaan seimbang : 2HCl + Zn ZnCl2 + H2

4

Write the formula of all the particles in the reactants and products: Tulis formula bagi semua zarah dalam bahan tindak balas dan hasil tindak balas:

2H+ + 2Cl– + Zn

Remove all the particles in the reactants and products which remain unchanged: Keluarkan semua zarah dalam bahan dan hasil tindak balas yang tidak berubah:

2H+ + 2Cl– + Zn


2H+ + Zn

Zn2+ + 2Cl– + H2

Zn2+ + 2Cl– + H2

Zn2+ + H2

Write the chemical equations and ionic equation for the following reactions: Tulis persamaan kimia dan persamaan ion untuk tindak balas berikut: Reactant / Bahan tindak balas Hydrochloric acid and #magnesium oxide Asid hidroklorik dan #magnesium oksida

Hydrochloric acid and sodium hydroxide Asid hidroklorik dan natrium hidroksida

Hydrochloric acid and magnesium Asid hidroklorik dan magnesium

Hydrochloric acid and #calcium carbonate Asid hidroklorik dan #kalsium karbonat

Sulphuric acid and zinc Asid sulfurik dan zink

Sulphuric acid and #zinc oxide Asid sulfurik dan #zink oksida

Sulphuric acid and #zinc carbonate Asid sulfurik dan #zink karbonat

Nitric acid and #copper(II) oxide Asid nitrik dan #kuprum(II) oksida

Nitric acid and sodium hydroxide Asid nitrik dan natrium hidroksida

Chemical equations / Persamaan kimia MgO + 2HCl

MgCl2 + H2O

HCl + NaOH

NaCl + H2O

2HCl + Mg 2HCl + CaCO3 H2SO4 + Zn H2SO4 + ZnO H2SO4 +ZnCO3 2HNO3 + CuO HNO3 + NaOH

MgCl2 + H2 CaCl2 + CO2 + H2O ZnSO4 + H2 ZnSO4 +H2O ZnSO4 + CO2 + H2O Cu(NO3)2 + H2O NaNO3 + H2O

Ionic equation / Persamaan ion 2H+ + MgO

Mg2+ + H2O

H+ + OH– 2H+ + Mg 2H+ + CaCO3

Mg2+ + H2 Ca2+ + CO2 + H2O

2H+ + Zn

Zn2+ + H2

2H+ + ZnO 2H+ + ZnCO3

H2O

Zn2+ + H2O Zn2+ + CO2 + H2O

2H+ + CuO H+ + OH–

Cu2+ + H2O H2O

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# Ions in magnesium oxide, calcium carbonate, zinc oxide, zinc carbonate and copper(II) oxide cannot be separated because the compounds are insoluble in water and the ions do not ionise. # Ion dalam magnesium oksida, kasium karbonat, zink oksida, zink karbonat dan kuprum(II) oksida tidak boleh diasingkan kerana sebatian tersebut tidak larut dalam air dan ion-ionnya tidak mengion.

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CHEMICAL PROPERTIES OF ALKALIS / SIFAT KIMIA ALKALI

Chemical properties

Write the balance chemical equation for the reaction

Sifat-sifat kimia

1 Alkali + Acid Alkali + Asid

Salt + Water

Penyediaan garam terlarut (Tajuk Garam)

2 Alkali + Ammonium salt Alkali + Garam Ammonium

Salt + Water + Ammonia gas

*Gas ammonia dibebaskan apabila alkali dipanaskan dengan garam ammonium. Gas ammonia mempunyai bau yang sengit dan menukar kertas litmus merah lembap kepada biru. *Application of the reaction / Aplikasi tindak balas :

H2SO4 + 2KOH

K2SO4 + 2H2O

(b) Barium hydroxide and hydrochloric acid: Barium hidroksida dan asid hidroklorik:

2HCl + Ba(OH)2

BaCl2 + H2O

(c) Ammonium chloride and potassium hydroxide: Ammonium klorida dan kalium hidroksida:

Garam + Air + Gas ammonia

*Ammonia gas is released when alkali is heated with ammonium salt. Ammonia gas has pungent smell and turn moist red litmus paper to blue.

Kalium hidroksida dan asid sulfurik :

Garam + Air

*Alkali neutralises acid / Alkali meneutralkan asid. *Application of the reaction / Aplikasi tindak balas : – Preparation of soluble salt (Topic Salt)

Tuliskan persamaan kimia seimbang bagi tindak balas

(a) Potassium hydroxide and sulphuric acid

KOH + NH4Cl

KCl + H2O + NH3

(d) Ammonium sulphate and sodium hydroxide: Ammonium sulfat dan natrium hidroksida:

2NaOH + (NH4)2SO4

Na2SO4 + 2H2O + 2NH3

– Confirmatory test for cations ammonium in qualitative analysis of salt (Topic Salt)

Ujian pengesahan kation ammonium dalam analisis kualitatif garam (Tajuk Garam)

3 Alkali + Metal ion Alkali + Ion logam

Insoluble metal hydroxide

Logam hidroksida tak larut

*Most of the metal hydroxides are insoluble. *Kebanyakan logam hidroksida tak terlarut.

(e) 2OH–(aq/ak) + Mg2+(aq/ak)

Magnesium hydroxide (white precipitate) Magnesium hidroksida (mendakan putih)

*Hydroxides of transition element metals are coloured. *Hidroksida bagi logam peralihan adalah berwarna. *Application of the reaction / Aplikasi tindak balas :

– Confirmatory test for cations in qualitative analysis of salt (Topic Salt)

Mg(OH)2(p)

(f) 2OH–(aq/ak) + Cu2+(aq/ak)

Ujian pengesahan bagi kation dalam analisis kualitatif garam (Tajuk Garam)

Cu(OH)2(p) Copper(II) hydroxide (blue precipitate) Kuprum(II) hidroksida (mendakan biru)

ROLE OF WATER AND THE PROPERTIES OF ACID / PERANAN AIR DAN SIFAT ASID

An acid shows its acidic properties when it is dissolved in water.

1

Asid menunjukkan sifat keasidannya apabila terlarut dalam air.

Acid molecules ionise in aqueous solution to form hydrogen ions. The presence of hydrogen ions is needed for the acid to show its acidic properties.

2

Molekul asid mengion dalam larutan akueus membentuk ion hidrogen. Kehadiran ion hidrogen diperlukan oleh asid untuk menunjukkan sifat keasidannya.

Acid will remain in the form of molecules in two conditions / Asid akan kekal dalam bentuk molekul dengan dua keadaan: (a) Without the presence of water for example dry hydrogen chloride gas and *glacial ethanoic acid.

3

Tanpa kehadiran air seperti gas hidrogen klorida kering dan *asid etanoik glasial

(b) Acid is dissolved in *organic solvent for example solution of hydrogen chloride in methylbenzene and ethanoic acid in propanone.

m

Asid dilarutkan dalam *pelarut organik seperti larutan hidrogen klorida dalam metilbenzena dan asid etanoik dalam propanon. * Glacial ethanoic acid is pure ethanoic acid / Asid etanoik glasial ialah asid etanoik tulen. * Organic solvent is covalent compound that exist as liquid at room temperature such as propanone, methylbenzene and trichloromethane. * Pelarut organik ialah sebatian kovalen yang wujud dalam bentuk cecair pada suhu bilik seperti propanon, metilbenzena dan triklorometana.

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Example / Contoh : Glacial ethanoic acid

Solution of hydrogen chloride in methylbenzene

Asid etanoik glasial

Larutan hidrogen klorida dalam metilbenzena

HCl

CH3COOH CH3COOH

HCl

CH3COOH

HCl

CH3COOH CH3COOH

HCl

HCl

CH3COOH CH3COOH

Larutan hidrogen klorida dalam air (asid hidroklorik)

HCl

H+

H+ Cl-

HCl

• Glacial ethanoic acid molecules do not • Hydrogen chloride molecules in ionise . methylbenzene do not ionise . • Glacial ethanoic exist as molecule only, no hydrogen ions present. Etanoik glasial hanya terdiri daripada molekul CH COOH sahaja, tiada ion 3 hidrogen hadir.

ClH+

Methylbenzene / Metilbenzena

Molekul asid etanoik glasial tidak mengion .

Cl-

HCl HCl

HCl

Solution of hydrogen chloride in water (hydrochloric acid)

Molekul hidrogen klorida dalam metilbenzena tidak mengion .

ClH+

Cl-

H+

Water / Air

• Hydrogen chloride molecule in water ionises : Molekul hidrogen klorida dalam air mengion :

H+ (aq/ak) + Cl– (aq/ak) • Hydrogen chloride exist as molecule HCl (aq/ak) only, there are no hydrogen ions • Hydrogen ions and chloride ions present. present. molekul Hidrogen klorida wujud sebagai Ion

sahaja, tiada ion hidrogen hadir.

hidrogen dan

ion

klorida hadir.

• Hydrogen chloride in water • Glacial ethanoic acid and hydrogen chloride in methylbenzene does not show acidic (hydrochloric acid) shows acidic properties: properties: Asid etanoik glasial dan hidrogen klorida dalam metilbenzena tidak menunjukkan sifat asid: (i) They

do not

Sebatian tersebut

tidak

do not

turn

Sebatian tersebut

tidak

(ii) They

Hidrogen klorida dalam air (asid hidroklorik) menunjukkan sifat asid:

react with metal, base or metal carbonate. bertindak balas dengan logam, bes dan karbonat logam.

blue

litmus paper to

menukarkan warna kertas litmus

red biru

. kepada

(i) Hydrochloric acid react with metal, base or metal carbonate.

merah .

• There are no free moving ions, hydrogen chloride in methylbenzene and glacial ethanoic acid cannot conduct electricity (non-electrolyte). Tidak wujud ion bebas bergerak , asid etanoik glasial dan hidrogen klorida dalam metilbenzena tidak dapat mengkonduksikan elektrik (bukan elektrolit).

Asid hidroklorik bertindak balas dengan logam, bes dan karbonat logam.

(ii) Hydrogen ions turn blue red . litmus paper to

Ion hidrogen menukarkan warna kertas biru kepada merah . litmus

• There are free moving ions, can conduct hydrochloric acid electricity (electrolyte). Terdapat ion yang bebas bergerak , asid dapat hidroklorik mengkonduksikan elektrik (elektrolit).

ROLE OF WATER AND THE PROPERTIES OF ALKALI / PERANAN AIR DAN SIFAT ALKALI 1

In the presence of water, an alkali dissolves and ionises to produce hydroxide ions. For example potassium hydroxide solution and ammonia solution. Dengan kehadiran air, alkali melarut dan mengion menghasilkan ion hidroksida. Contohnya larutan kalium hidroksida dan larutan ammonia:

2

KOH(aq/ak ) K+(aq/ak ) + OH–(aq/ak ) NH3(g) + H2O(l/ce) NH4+(aq/ak ) + OH–(aq/ak )

Without water or in organic solvents, no hydroxide ions are produced, so the alkaline properties are not shown.

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Tanpa air atau dalam pelarut organik, tiada ion hidroksida yang dihasilkan, maka sifat-sifat alkali tidak ditunjukkan.

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EXERCISE / LATIHAN

The diagram below shows the apparatus set-up to investigate the role of water and other solvent in showing the properties of acid and the observations made from the investigation.

1

Rajah di bawah menunjukkan susunan radas untuk mengkaji peranan air atau pelarut lain dalam menunjukkan sifat asid serta pemerhatian yang dibuat. Experiment / Eksperimen

I

II

Set-up of apparatus Susunan radas

Asid hidroklorik dalam air

Hydrochloric acid in tetrachloromethane

Magnesium ribbon

Magnesium ribbon

Hydrochloric acid in water

Asid hidroklorik dalam tetraklorometana

Pita magnesium

Observation

• Bubbles of gas are released

Pemerhatian


Pita magnesium

• No bubble of gas Tiada gelembung gas

• Magnesium ribbon dissolves Pita magnesium larut

(a) What is meant by acid / Apakah yang dimaksudkan dengan asid ? Acid is a chemical substance which ionises in water to produce hydrogen ion. (b) (i)

Name the bubble of gas released in Experiment I / Namakan gas yang terbebas dalam Eksperimen I. Hydrogen gas

(ii) Write the chemical equation for the formation of the bubbles in Experiment I. Tulis persamaan kimia untuk pembentukan gelembung gas dalam Eksperimen I.

Mg + 2HCl MgCl2 + H2

(iii) Write the ionic equation for the chemical equation in (b)(ii). Tulis persamaan ion untuk persamaan kimia dalam (b)(ii).

Mg + 2H+ Mg2+ + H2 (c) Compare observation in Experiment I and Experiment II. Explain your answer. Bandingkan pemerhatian dalam Eksperimen I dan Eksperimen II. Terangkan jawapan anda.

– Hydrochloric acid in water in Experiment I Asid hidroklorik dalam air dalam Eksperimen I

reacts

with magnesium.

bertindak balas

dengan magnesium.

– Hydrochloric acid in tetrachloromethane in Experiment I do not react with magnesium. Asid hidroklorik dalam tetraklorometana dalam Eksperimen II

– Hydrochloric acid in water – H+ ions react with

ionises

magnesium atom

tidak bertindak balas

to H+ / Asid hidroklorik dalam air HCl H+ + Cl–

dengan magnesium. mengion kepada ion H+:

to produce hydrogen molecule:

Ion H+ bertindak balas dengan atom magnesium untuk menghasilkan molekul hidrogen:

Mg + 2H+ Mg2+ + H2

– Hydrochloric acid in tetrachloromethane remains in the form of molecule . No hydrogen ion present.

m

molekul

. Tiada ion

hidrogen

hadir.

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2

The diagram below shows the set-up of apparatus to prepare two solutions of ammonia in solvent X and solvent Y. A piece of red litmus paper is dropped into each beaker. Gambar rajah di bawah menunjukkan susunan radas bagi menyediakan dua larutan ammonia dalam pelarut X dan pelarut Y. Sekeping kertas litmus merah dimasukkan ke dalam setiap bikar. Ammonia

Ammonia

Ammonia

Ammonia

Solvent X

Solvent Y

Pelarut X

Pelarut Y

Beaker A / Bikar A

Beaker B / Bikar B

The table below shows the observation on the red litmus paper in solvent X and solvent Y. Jadual di bawah menunjukkan pemerhatian ke atas kertas litmus merah dalam pelarut X dan pelarut Y. Solution / Larutan

Observation / Pemerhatian

Ammonia in solvent X in beaker A

The red litmus paper turns blue.

Ammonia dalam pelarut X dalam bikar A

Kertas litmus merah bertukar menjadi biru.

Ammonia in solvent Y in beaker B

No visible change in the colour of red litmus paper.

Ammonia dalam pelarut Y dalam bikar B

Tiada perubahan yang nyata pada warna kertas litmus merah.

(a) Name possible substances that can be solvent X and solvent Y. Namakan bahan-bahan yang mungkin bagi pelarut X dan pelarut Y.

Solvent X / Pelarut X : Water Solvent Y / Pelarut Y : Propanone / methylbenzene / trichloromethane

(b) Explain the difference in the observation on the beakers A and B. Terangkan perbezaan antara pemerhatian dalam bikar A dengan bikar B.

– Ammonia gas in beaker A is dissolved in water, ammonia molecules hydroxide ions: Gas ammonia dalam bikar A ion hidroksida :

larut

dalam air, molekul ammonia

NH3 (g) + H2O (l/ce)

ionise

mengion

to ammonium ion and

kepada ion

ammonium

dan

NH 4+ (ak) + OH– (ak)

– The presence of hydroxide ions change the red litmus paper to blue. Kehadiran ion-ion

hidroksida

menukar kertas litmus merah kepada biru.

– Ammonia gas in beaker B is dissolved in molecules do not ionise . Gas ammonia dalam bikar B mengion .

larut

propanone / methylbenzene / trichloromethane propanon / metilbenzena / triklorometana

dalam

, ammonia

, molekul ammonia tidak

– No hydroxide ions present, the red litmus paper remains unchanged. Tiada ion

(c) (i)

hidroksida

, warna merah kertas litmus tidak berubah.

Between solution in beakers A and B, which one is an electrolyte and non-electrolyte? Explain your answer. Antara larutan dalam bikar A dangan bikar B, yang manakah elektrolit dan bukan elektrolit? Terangkan jawapan anda.

an electrolyte – Solution in beaker A is , it contains ionisation of ammonia molecules in water.

free moving ions

elektrolit , ia mengandungi ion-ion yang Larutan dalam bikar A ialah pengionan molekul ammonia dalam air.

– Solution in beaker B is a non-electrolyte , ammonia molecules propanone / methylbenzene / trichloromethane . , molekul ammonia

bebas bergerak

do not ionise

tidak mengion

daripada

in

dalam

.

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bukan elektrolit Larutan dalam bikar B propanon / metilbenzena / triklorometana

from the

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(ii) Draw a labelled diagram to show the set-up of apparatus used to show the electrical conductivity of an electrolyte. Lukiskan gambar rajah berlabel yang menunjukkan susunan radas yang digunakan untuk menunjukkan kekonduksian arus elektrik bagi sesuatu elektrolit.



Elektrolit Electrolyte

THE pH SCALE / SKALA pH

The pH is a scale of numbers to measure the degree of acidity and alkalinity of an aqueous solution based on the concentration of hydrogen ions, H+ or hydroxide ions, OH–.

1

Skala pH ialah skala bernombor untuk mengukur darjah keasidan dan kealkalian suatu larutan akueus berdasarkan kepekatan ion hidrogen, H+ atau ion hidroksida, OH–.

The pH scale has the range of number from 0 to 14 / Skala pH bernombor dari 0 hingga 14 :

2

pH

0

1

2

3

4

5

6

pH < 7: • Acidic solution / Larutan berasid. • The lower the pH value, the higher is the concentration of hydrogen ion, H+.

7

pH = 7 Neutral Neutral

8

9

10

11

12

13

14

pH > 7: • Alkaline solution / Larutan beralkali. • The higher the pH value, the higher is the concentration of hydroxide ion, OH–.

Semakin rendah nilai pH, semakin tinggi kepekatan ion hidrogen, H+.

Semakin tinggi nilai pH, semakin tinggi kepekatan ion hidroksida, OH–.

The pH of an aqueous solution can be measured by / Nilai pH bagi sesuatu larutan akueus boleh diukur dengan menggunakan: (a) pH meter / Meter pH (b) Acid-base indicator / Penunjuk asid-bes Complete the following table / Lengkapkan jadual berikut :

3

Colour / Warna

Indicator Penunjuk

Acid / Asid

Neutral / Neutral

Alkali / Alkali

Litmus solution / Larutan litmus

Red

Purple

Blue

Methyl orange / Metil jingga

Red

Orange

Yellow

Colourless

Colourless

Pink

Red

Green

Purple

Phenolphthalein / Fenolftalein Universal indicator / Penunjuk universal

THE STRENGTH OF ACID AND ALKALI / KEKUATAN ASID DAN ALKALI

The strength of acid and alkali depend on the degree of ionisation or dissociation of the acid and alkali in water.

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Kekuatan asid dan alkali bergantung pada darjah pengionan asid dan alkali dalam air. (a) A strong acid is an acid that ionises completely in water to produce high concentration of hydrogen ion, H+. Asid kuat ialah asid yang mengion sepenuhnya dalam air menghasilkan kepekatan ion hidrogen, H+ yang tinggi. (b) A weak acid is an acid that partially ionises in water to produce low concentration of hydrogen ion, H+. Asid lemah ialah asid yang mengion separa dalam air menghasilkan kepekatan ion hidrogen, H+ yang rendah. (c) A strong alkali is an alkali that ionises completely in water to produce high concentration of hydroxide ion, OH–. Alkali kuat ialah alkali yang mengion sepenuhnya dalam air menghasilkan kepekatan ion hidroksida, OH– yang tinggi. (d) A weak alkali is an alkali that partially ionises completely in water to produce low concentration of hydroxide ion, OH–. Alkali lemah ialah alkali yang mengion separa dalam air menghasilkan kepekatan ion hidroksida, OH– yang rendah.

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Example of different strength of acid and alkali / Contoh asid dan alkali dengan kekuatan yang berbeza. Acid / Alkali Asid / Alkali

Strong acid Asid kuat

Example

Ionisation equation

Contoh

Persamaan ion

Hydrochloric HCl (aq/ak ) acid, HCl H+ (aq/ak ) + Cl– (aq/ak ) Asid hidroklorik, HCl

Nitric acid, HNO3 Asid nitrik, HNO3

Sulphuric acid, H2SO4 Asid sulfurik, H2SO4

Weak acid Asid lemah

Ethanoic acid, CH3COOH Asid etanoik, CH3COOH

Carbonic acid, H2CO3 Asid karbonik, H2CO3

Strong alkali Alkali kuat

Sodium hydroxide, NaOH

HNO3(aq/ak ) H+ (aq/ak ) + NO3– (aq/ak )

H2SO4 (aq/ak ) 2H+ (aq/ak ) + SO42– (aq/ak )

Alkali lemah

Ammonia solution, NH3(aq) Larutan ammonia, NH3(ak)

All hydrogen chloride molecules that H+ and Cl– + – dissolve in water ionises completely into H dan Cl hydrogen ions and

chloride

ions.

All nitric acid ionises completely in water into hydrogen ions and nitrate ions.

H+ and NO3–

All sulphuric acid ionises completely into hydrogen ions and sulphate ions.

H+ and SO42–

H+ dan NO3–

H+ dan SO42–

Semua asid sulfurik mengion sepenuhnya dalam air kepada ion hidrogen dan ion sulfat .

CH3COOH (aq/ak )

Ethanoic acid partially ionises in water CH3COO (aq/ak ) + H (aq/ak ) into etahnoate ions and hydrogen ions. Some remain in the form of CH3COOH molecules . –

+

separa kepada ion Asid etanoik mengion etanoat dan ion hidrogen . Sebahagian lagi kekal dalam bentuk molekul CH3COOH.

H2CO3 (aq/ak ) 2H+ (aq/ak ) + CO32– (aq/ak )

2

NaOH (aq/ak )

Na (aq) + OH (aq) +

–

CH3COOH, CH3COO– and H+ CH3COOH, CH3COO– dan H+

H2CO3, H+ and Carbonic acid partially ionises in water into carbonate ions and hydrogen ion. Some CO32– + remain in the form of H CO molecules . H2CO3, H dan 3

CO32–

Sebahagian asid karbonik mengion dalam air kepada ion karbonat dan ion hidrogen. Sebahagian lagi kekal dalam bentuk molekul H2CO3.

Sodium hydroxide ionises completely in water into sodium ions and hydroxide ions.

Na+ and OH–

Potassium hydroxide ionises completely potassium ions and in water into hydroxide ions.

K+ and OH–

Na+

dan

OH–

Natrium hidroksida mengion sepenuhnya dalam air kepada ion natrium dan ion hidroksida .

KOH (aq/ak )

K+ (aq) + OH– (aq)

K+

dan

OH–

Kalium hidroksida mengion sepenuhnya dalam air kalium dan ion hidroksida . kepada ion

Ba(OH)2 (aq/ak )

Ba (aq) + 2OH (aq) 2+

–

Barium hidroksida, Ba(OH)2

Weak alkali

Zarah-zarah yang hadir

Semua asid nitrik mengion sepenuhnya dalam air kepada ion hidrogen dan ion nitrat .

Kalium hidroksida, KOH

Barium hydroxide, Ba(OH)2

Penerangan

Semua molekul hidrogen klorida melarut dalam air dan mengion sepenuhnya kepada ion hidrogen dan ion klorida .

Natrium hidroksida, NaOH

Potassium hydroxide, KOH

Particles present

Explanation

Barium hydroxide ionises completely in water into barium ions and hydroxide

Ba2+ and OH– Ba2+ dan OH–

ions.

Barium hidroksida mengion sepenuhnya dalam air kepada ion barium dan ion hidroksida .

NH3 (g)+ H2 O(l/ce) + NH4 (aq/ak ) + OH–(aq/ak )

Ammonia partially ionises in water into ammonium ions and hydroxide ions, some remain in the form of NH molecules . 3

separa dalam air kepada Ammonia mengion ion ammonium dan ion hidroksida , sebahagian lagi kekal dalam bentuk molekul NH .

NH3, NH4+ and OH– NH3, NH4+ dan OH–

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CONCENTRATION OF ACID AND ALKALI / KEPEKATAN ASID DAN ALKALI

A solution is a homogeneous mixture formed when a solute is dissolved in a solvent. For example copper(II) sulphate solution is prepared by dissolving copper(II) sulphate powder (solute) in water (solvent).

1

Larutan adalah campuran homogen yang terbentuk apabila bahan larut dilarutkan dalam pelarut. Contohnya larutan kuprum(II) sulfat disediakan dengan melarutkan serbuk kuprum(II) sulfat (bahan larut) di dalam air (pelarut). 2

Concentration of a solution the quantity of solute in a given volume of solution which is usually 1 dm3 of solution. Kepekatan sesuatu larutan ialah kuantiti bahan terlarut dalam isi padu larutan yang tertentu, biasanya isi padu 1 dm3 larutan.

3

Concentration can be expressed in two ways / Kepekatan boleh diwakili dengan dua cara : (a) Mass of solute in gram per 1 dm3 solution, g dm–3/ Jisim bahan larut dalam gram bagi setiap 1 dm3 larutan, g dm–3. Concentration of solution (g dm–3) =

Mass of solute in gram (g) / Jisim bahan larut dalam gram (g)

Volume of solution (dm3) / Isi padu larutan (dm3) Kepekatan larutan (g dm–3) (b) Number of moles of solute in 1 dm3 solution, mol dm–3 / Bilangan mol bahan larut dalam 1 dm3 larutan, mol dm–3. Concentration of solution (mol dm–3) =

Number of mole of solute (mol) / Bilangan mol bahan larut (mol)

Volume of solution (dm3) / Isi padu larutan (dm3) Kepekatan larutan (mol dm–3) The concentration in mol dm–3 is called molarity or molar concentration. The unit mol dm–3 can be represented by ‘M’.

4

Kepekatan dalam mol dm–3 dipanggil sebagai kemolaran atau kepekatan molar. Unit mol dm–3 boleh diwakili dengan‘M’.

Molarity = Kemolaran

Number of mole of solute (mol) / Bilangan mol bahan larut (mol)

Number of mole of solute (mol) = Molarity × Volume (dm3)

Kepekatan dalam mol dm–3 (kemolaran)

Kemolaran × Isi padu (dm3)

n = MV Mv n = 1 000

Bilangan mol bahan terlarut

M = Concentration in mol dm–3 (molarity)

Volume of solution (dm3) / Isi padu larutan (dm3)

Bilangan mol bahan larut (mol)

n = Number of moles of solute

V = Volume of solution in dm3 Isi padu larutan dalam dm3

v = Volume of solution in cm3 Isi padu larutan dalam cm3

The concentration of a solution can be converted from mol dm to g dm and vice versa. –3

5

–3

Kepekatan larutan boleh ditukar daripada mol dm–3 kepada g dm–3 dan sebaliknya. × molar mass of the solute / jisim molar bahan terlarut

mol dm–3

g dm–3 ÷ molar mass of the solute / jisim molar bahan terlarut

The pH value of an acid or an alkali depends on the concentration of hydrogen ions or hydroxide ions:

6

Nilai pH bagi asid atau alkali bergantung pada kepekatan ion hidrogen atau ion hidroksida:

The higher the concentration of hydrogen ions in acidic solution, the lower the pH value. Semakin tinggi kepekatan ion hidrogen dalam larutan berasid, semakin rendah nilai pH.

The higher the concentration of hydroxide ions in alkaline solution, the higher the pH value. Semakin tinggi kepekatan ion hidroksida dalam larutan beralkali, semakin tinggi nilai pH.

The pH value of an acid or an alkali is depends on / Nilai pH bagi asid atau alkali bergantung pada: (a) The strength of acid or alkali / Kekuatan asid atau alkali – the degree of ionisation or dissociation of the acid and alkali in water / darjah pengionan asid atau alkali dalam air. (b) Molarity of acid or alkali / Kemolaran asid atau alkali – the concentration of acid or alkali in mol dm–3 / kepekatan bahan terlarut dalam mol dm–3. (c) Basicity of an acid / Kebesan asid – the number ionisable of hydrogen atom per molecule of an acid molecule in an aqueous solution.

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bilangan atom hidrogen per molekul asid yang terion dalam larutan akueus.

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0.1 mol dm–3 HCl

1.21

2.98

0.01 mol dm–3 HCl

I

Bandingkan kepekatan ion hidrogen dan nilai pH

Nilai pH bagi asid hidroklorik 0.1 mol dm–3 lebih rendah daripada asid hidroklorik 0.01 mol dm–3.

hydrochloric acid.

– The pH value of 0.1 mol dm–3 of hydrochloric acid is lower than 0.01 mol dm–3 of

Kepekatan ion hidrogen dalam asid hidroklorik tinggi daripada asid 0.1 mol dm–3 lebih hidroklorik 0.01 mol dm–3.

0.01 mol dm–3 of hydrochloric acid.

–3

– Concentration hydrogen ion in 0.1 mol dm of hydrochloric acid is higher than

HCl H+ + Cl– –3 0.01 mol dm–3 0.01 mol dm

Asid hidroklorik 0.01 mol dm–3 mengion kepada 0.01 mol dm–3 ion hidrogen:

– 0.01 mol dm–3 of hydrochloric acid ionises to –3 form 0.01 mol dm hydrogen ion:


Asid hidroklorik 0.1 mol dm–3 mengion kepada 0.1 mol dm–3 ion hidrogen:

– 0.1 mol dm–3 of hydrochloric acid ionises to –3 form 0.1 mol dm hydrogen ion:

lengkap dalam air kepada ion hidrogen.

Compare – Hydrochloric acid is a strong acid ionises concentration completely in water to hydrogen ion. of H+ and pH kuat yang Asid hidroklorik adalah asid value mengion

Bacaan pH meter

pH meter reading

Eksperimen

Experiment

0.05 mol dm–3 HCl

2.25

diprotik .

Nilai pH bagi asid sulfurik 0.05 mol dm–3 lebih rendah daripada asid hidroklorik 0.05 mol dm–3.

hydrochloric acid.

– The pH value of 0.05 mol dm–3 of sulphuric acid is lower than 0.05 mol dm–3 of

Kepekatan ion hidrogen dalam asid sulfurik 0.05 mol dm–3 adalah dua kali ganda (lebih tinggi) daripada asid hidroklorik 0.05 mol dm–3.

0.05 mol dm–3 of hydrochloric acid.

– Concentration hydrogen ion in 0.05 mol dm–3 of sulphuric acid is double of (higher than)

H+ + Cl– –3 0.05 mol dm–3 0.05 mol dm

HCl

Asid hidroklorik 0.05 mol dm–3 mengion lengkap dalam –3 air menghasilkan 0.05 mol dm ion hidrogen:

– 0.05 mol dm–3 of ionises completely in water –3 to form 0.05 mol dm hydrogen ion:

Asid hidroklorik adalah asid kuat monoprotik .

– Hydrochloric acid is a strong monoprotic acid.

H2SO4 2H+ + SO42– –3 0.05 mol dm–3 0.1 mol dm

Asid sulfurik 0.05 mol dm–3 mengion lengkap kepada 0.1 mol dm–3 ion hidrogen:

hydrogen ion:

– 0.05 mol dm-3 of sulphuric acid ionises –3 completely in water to form 0.1 mol dm

Asid sulfurik adalah asid kuat

– Sulphuric acid is a strong diprotic acid.

0.05 mol dm–3 H2SO4

1.15

II

0.1 mol dm–3 CH3COOH

3.45

kuat yang mengion

lengkap dalam

H+ + CH3COO–(aq/ak ) less than/kurang dari 0.1 mol dm–3

0.1 mol dm–3

Nilai pH bagi asid hidroklorik 0.1 mol dm–3 lebih rendah daripada asid etanoik 0.1 mol dm–3.

– The pH value of 0.1 mol dm–3 of hydrochloric acid lower than of 0.1 mol dm–3 of ethanoic acid.

Kepekatan ion hidrogen dalam asid hidroklorik 0.1 mol dm–3 lebih tinggi daripada asid etanoik 0.1 mol dm–3.

ethanoic acid.

– Concentration hydrogen ion in 0.1 mol dm–3 of hydrochloric acid is higher than of 0.1 mol dm–3 of

0.1 mol dm–3

CH3COOH(aq/ak )

Asid etanoik 0.1 mol dm–3 mengion kurang daripada ion hidrogen:

– 0.1 mol dm–3 of ethanoic acid ionises to less than 0.1 mol dm–3 hydrogen ion:

Asid etanoik adalah asid lemah mengion separa dalam air menghasilkan kepekatan ion hidrogen yang lebih rendah .

– Ethanoic acid is a weak acid ionises partially in water to produce lower concentration hydrogen ion.


–3 Asid hidroklorik 0.1 mol dm–3 mengion lengkap kepada 0.1 mol dm ion hidrogen:

– 0.1 mol dm–3 of hydrochloric acid ionises to form 0.1 mol dm–3 hydrogen ion:

Asid hidroklorik adalah asid air kepada ion hidrogen.

– Hydrochloric acid is a strong acid ionises completely in water to hydrogen ion.

0.1 mol dm–3 HCl

1.21

III

Rajah di bawah menunjukkan bacaan pH meter untuk pelbagai jenis dan kepekatan asid. Tujuan eksperimen adalah untuk mengkaji hubungan antara kepekatan ion hidrogen dengan nilai pH. Bandingkan kepekatan ion hidrogen dan nilai pH untuk asid-asid yang berikut. Terangkan jawapan anda.

The diagram below shows the reading of pH meter for different types and concentration of acids. The aim of the experiment is to investigate the relationship between concentration of hydrogen ions with the pH value. Compare the concentration of hydrogen ions and the pH value of the following acids. Explain your answer.

Example / Contoh:


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PREPARATION OF STANDARD SOLUTION / PENYEDIAAN LARUTAN PIAWAI

Standard solution is a solution in which its concentration is accurately known.

1

Larutan piawai ialah larutan yang kepekatannya diketahui dengan tepat.

The steps taken in preparing a standard solution are:

2

Langkah-langkah yang diambil dalam menyediakan larutan piawai adalah:

(a) Calculate the mass of solute needed to give the required volume and molarity. Hitung jisim bahan larut yang diperlukan untuk menghasilkan isi padu dan kemolaran yang dikehendaki.

(b) The solute is weighed / Bahan larut ditimbang. (c) The solute is completely dissolved in distilled water and then transferred to a volumetric flask partially filled with distilled water. Bahan larut dilarutkan sepenuhnya dalam air suling dan dipindahkan kepada kelalang volumetrik yang sebahagiannya sudah diisi dengan air suling.

(d) Distilled water is added to the calibration mark of the volumetric flask and the flask is inverted to make sure thorough mixing. Air suling ditambah ke dalam kelalang volumetrik hingga tanda senggatan dan kelalang volumetrik ditelangkupkan beberapa kali untuk memastikan campuran sekata.

PREPARATION OF A SOLUTION BY DILUTION / PENYEDIAAN LARUTAN SECARA PENCAIRAN

Adding water to the standard solution lowered the concentration of the solution. Since no solute is added, the amount of solute in the solution before and after dilution remains unchanged: Penambahan air kepada larutan piawai merendahkan kepekatan larutan tersebut. Oleh kerana tiada bahan terlarut yang ditambah, kandungan bahan terlarut dalam larutan sebelum dan selepas pencairan tidak berubah:

Number of mol of solute before dilution = Number of mole of solute after dilution Bilangan mol bahan terlarut sebelum pencairan

M1V1

1 000

Therefore / Oleh itu,

Bilangan mol bahan terlarut selepas pencairan

=

M2V2 1 000

M1V1 = M2V2

M1 = Initial concentration of the solute / Kemolaran larutan awal V1 = Initial volume of the solution in cm3 / Isipadu larutan awal dalam cm3 M2 = Final concentration of the solute / Kemolaran larutan akhir V2 = Final volume of the solution in cm3 / Isipadu larutan akhir dalam cm3 Example / Contoh :

(a) What is meant by a standard solution / Apakah yang dimaksudkan dengan larutan piawai ?

1

Standard solution is a solution in which its concentration is accurately known. (b) (i)

You are given solid sodium hydroxide. Describe the procedure to prepare 500 cm3 of 1.0 mol dm–3 sodium hydroxide solution. [Relative atomic mass: H = 1; O = 16; Na = 23] Anda diberi pepejal natrium hidroksida. Huraikan kaedah untuk menyediakan 500 cm3 larutan natrium hidroksida 1.0 mol dm–3. [Jisim atom relatif: H = 1, O = 16, Na = 23]

Calculate the mass of sodium hydroxide / Hitung jisim natrium hidroksida :

– Molar mass of NaOH / Jisim molar NaOH = 23 + 16 + 1 = 40 g mol – Mol NaOH / Bilangan mol NaOH = 500 × 1.0/1 000 = 0.5 mol

–1

– Mass of NaOH / Jisim NaOH = mol / Bilangan mol × molar mass / Jisim molar –1 = 0.5 mol × 40 g mol = 20.0 g

Preparation of 500 cm3 1.0 mol dm–3 sodium hydroxide: Penyediaan 500 cm3 larutan natrium hidroksida 1.0 mol dm–3: Timbang

Publica

20.0 g

of NaOH accurately using a weighing bottle .

NaOH dengan tepat menggunakan

botol penimbang

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20.0 g

– Dissolve

20.0 g

Larutkan

distilled water

of NaOH in

air suling

NaOH dalam

in a beaker.

di dalam bikar.

– Transfer the content into a 500 cm volumetric flask . 3

Pindahkan kandungan ke dalam kelalang volumetrik 500 cm3.

– –

Rinse

the beaker with distilled water and transfer all the contents into the volumetric flask .

Bilas

bikar dengan

Distilled water

air suling

dan pindahkan semua kandungan ke dalam kelalang volumetrik .

is added to the volumetric flask until the calibration mark .

Air suling ditambah kepada kelalang volumetrik hingga

– The volumetric flask is closed tightly with homogeneous solution. Kelalang volumetrik ditutup dengan

penutup

tanda senggatan

stopper dan

and

.

inverted

ditelangkupkan

a few times to get

beberapa kali untuk mendapatkan larutan

yang homogen.

(ii) Describe how you would prepare 250 cm3 of 0.1 mol dm–3 sodium hydroxide from the above solution. Huraikan bagaimana anda menyediakan 250 cm3 larutan natrium hidroksida 0.1 mol dm–3 daripada larutan di atas.

Calculate the volume of 1 mol dm–3 sodium hydroxide used: Hitung isi padu natrium hidroksida 1 mol dm–3 yang digunakan:

– M1 × V1 = M2 × V2 – V1 =

M2 × V2 = 0.1 × 250 = 25 cm3 M1 1

Preparation of 250 cm3 1.0 mol dm–3 sodium hydroxide solution: Penyediaan 250 cm3 larutan natrium hidroksida 1.0 mol dm–3:

– A pipette is filled with

25 cm3

of 1.0 mol dm–3 sodium hydroxide solution.

Sebuah pipet diisi dengan

25 cm3

larutan natrium hidroksida 1.0 mol dm–3.

– –

25 cm3

of 1.0 mol dm–3 NaOH is transferred into 250 cm3 volumetric flask .

25 cm3

NaOH 1.0 mol dm–3 dipindahkan kepada kelalang volumetrik 250 cm3.

Distilled water

is added to the volumetric flask until the calibration mark .

Air suling ditambah kepada kelalang volumetrik hingga

– The volumetric flask is closed tightly with homogeneous solution. Kelalang volumetrik ditutup dengan

penutup

tanda senggatan

stopper dan

and

ditelangkupkan

.

inverted

a few times to get

beberapa kali untuk mendapatkan larutan

yang homogen.


The table below shows the pH value of a few substances / Jadual di bawah menunjukkan nilai pH bagi beberapa bahan. Substance / Bahan

pH value / Nilai pH

Ethanoic acid 0.1 mol dm / Asid etanoik 0.1 mol dm

3

Hydrochloric acid 0.1 mol dm–3 / Asid hidroklorik 0.1 mol dm–3

1

Glacial ethanoic acid / Asid etanoik glasial

7

–3

–3

(a) (i)

What is meant by weak acid and strong acid / Apakah yang dimaksudkan dengan asid lemah dan asid kuat ?

Weak acid / Asid lemah : An acid that partially ionises in water to produce low concentration of hydrogen ion, + H .

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Strong acid / Asid kuat : An acid that completely ionises in water to produce high concentration of hydrogen + ion, H .

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(ii) Between ethanoic acid and hydrochloric acid, which acid has the higher concentration of H+ ion? Explain your answer. Antara asid etanoik dengan asid hidroklorik, asid manakah mempunyai kepekatan ion H+ yang lebih tinggi? Terangkan jawapan anda.

higher

– Hydrochloric acid has

concentration of H+ than ethanoic acid. tinggi

Asid hidroklorik mempunyai kepekatan ion H+ yang lebih

berbanding dengan asid etanoik.

– Hydrochloric acid is a strong acid which ionises completely in water to produce concentration of H+: asid kuat

Asid hidroklorik ialah tinggi yang lebih

HCl (aq/ak )

:

sepenuhnya

yang mengion

lemah

:

CH3COOH (aq/ak )

dalam air untuk menghasilkan kepekatan ion H+

H+(aq/ak ) + Cl–(aq/ak )

– Ethanoic acid is a weak acid which ionises H+: Asid etanoik ialah asid rendah yang lebih

higher

partially

in water to produce lower concentration of

separa

yang mengion

dalam air untuk menghasilkan kepekatan ion H+

CH3COO– (aq/ak ) + H+ (aq/ak )

(iii) Why do ethanoic acid and hydrochloric acid have different pH value? Mengapakah asid etanoik dan asid hidroklorik mempunyai nilai pH yang berbeza?

– The concentration H+ in hydrochloric acid is tinggi

Kepekatan H dalam asid hidroklorik +

– The concentration H in ehanoic acid is +

Kepekatan H dalam asid etanoik +

rendah

higher

, the pH value is rendah

, nilai pH lebih

lower

, nilai pH lebih

.

.

, the pH value is tinggi

lower

higher

.

.

(b) Glacial ethanoic acid has a pH value of 7 but a solution of ethanoic acid has a pH value less than 7. Explain the observation. Asid etanoik glasial mempunyai nilai pH 7 tetapi asid etanoik mempunyai nilai pH yang kurang daripada 7. Terangkan pemerhatian tersebut.

– Glacial ethanoic acid molecules do not ionise . Glacial ethanoic acid consists of only CH3COOH molecules . The CH COOH molecules are neutral . No hydrogen ions present. The pH value of 3 glacial ethanoic acid is 7. mengion . Asid etanoik glasial hanya terdiri daripada molekul CH3COOH Molekul asid etanoik glasial tidak Molekul hidrogen CH COOH adalah neutral. Tiada ion hadir. Nilai pH asid etanoik glasial adalah 7. sahaja. 3

– Ethanoic acid ionises partially in water to produce ethanoate ions and hydrogen ions causes the solution to have acidic property. The pH value of the solution is less than 7. etanoat hidrogen dan ion Asid etanoik mengion separa dalam air untuk menghasilkan ion asid . Nilai pH bagi larutan tersebut adalah kurang daripada 7. larutan mempunyai sifat

yang menyebabkan

The table shows the pH value of a few solution / Jadual berikut menunjukkan nilai pH bagi beberapa jenis larutan berbeza.

2

Solution / Larutan

P

Q

R

S

T

U

pH

1

3

5

7

11

14

(a) (i)

Which solution has the highest concentration of hydrogen ion? Larutan manakah yang mempunyai kepekatan ion hidrogen yang paling tinggi?

Solution P

(ii) Which solution has the highest concentration of hydroxide ion? Larutan yang manakah mempunyai kepekatan ion hidroksida yang paling tinggi?

Solution U

m

0.01 mol dm–3 of hydrochloric acid / 0.01 mol dm–3 asid hidroklorik ?

Q

(i)

(ii) 0.01 mol dm–3 of ethanoic acid / 0.01 mol dm–3 asid etanoik ?

R

(iii) 0.1 mol dm–3 ammonia aqueous / 0.1 mol dm–3 larutan ammonia ?

T

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(b) Which is the following solution could be / Larutan manakah yang mungkin

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(iv) 1 mol dm–3 of hydrochloric acid / 1 mol dm–3 asid hidroklorik ?

P

(v) 1 mol dm–3 sodium hydroxide solution / 1 mol dm–3 larutan natrium hidroksida ?

U

(vi) 1 mol dm–3 potassium sulphate solution / 1 mol dm–3 larutan kalium sulfat ?

S

(c) (i)

State two solutions which react to form neutral solution. Nyatakan dua larutan yang bertindak balas untuk membentuk larutan neutral.

P/Q/R and T/U // Hydrochloric acid/ethanoic acid with ammonia aqueous/sodium hydroxide solution.

(ii) Which solutions will produce carbon dioxide gas when calcium carbonate powder is added? Larutan manakah menghasilkan gas karbon dioksida apabila ditambah serbuk kalsium karbonat?

P/Q // Hydrochloric acid/ethanoic acid 3

The molarity of sodium hydroxide solution is 2 mol dm–3. What is the concentration of the solution in g dm–3? [RAM: Na = 23, 0 = 16, H = 1] Kemolaran larutan natrium hidroksida ialah 2 mol dm–3. Apakah kepekatan larutan tersebut dalam g dm–3? [JAR: Na = 23, O = 16, H = 1]

Answer / Jawapan : 4

80 g dm–3

Calculate the molarity of the solution obtained when 14 g potassium hydroxide is dissolved in distilled water to make up 500 cm3 solution. [RAM: K = 39, H = 1, O = 16] Hitung kemolaran larutan yang diperoleh apabila 14 g kalium hidroksida dilarutkan dalam air suling untuk menyediakan larutan yang berisi padu 500 cm3. [JAR: K = 39, H = 1, O = 16]


0.5 mol dm–3

Calculate the molarity of a solution which is prepared by dissolving 0.5 mol of hydrogen chloride, HCl in distilled water to make up 250 cm3 solution. Hitung kemolaran larutan yang disediakan dengan melarutkan 0.5 mol hidrogen klorida, HCl dalam air suling untuk menyediakan larutan yang berisi padu 250 cm3.


2 mol dm–3

How much of sodium hydroxide in gram should be dissolved in water to prepare 500 cm3 of 0.5 mol dm–3 sodium hydroxide solution? [RAM: Na = 23, O = 16, H = 1] Berapakah jisim natrium hidroksida dalam gram yang patut dilarutkan dalam air untuk menyediakan 500 cm3 larutan natrium hidroksida 0.5 mol dm–3? [JAR: Na = 23, O = 16, H = 1]

10 g


300 cm3 water is added to 200 cm3 hydrochloric acid, 1 mol dm–3. What is the resulting molarity of the solution? Jika 300 cm3 air ditambah kepada 200 cm3 asid hidroklorik 1 mol dm–3, apakah kemolaran bagi larutan yang dihasilkan?

0.4 mol dm–3

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Answer / Jawapan :

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Calculate the volume of nitric acid, 1 mol dm–3 needed to be diluted by distilled water to obtain 500 cm3 of nitric acid, 0.1 mol dm–3.

8

Hitung isi padu asid nitrik 1 mol dm–3 yang diperlukan untuk dilarutkan oleh air suling bagi menghasilkan 500 cm3 asid nitrik 0.1 mol dm–3.

Answer / Jawapan :

50 cm3

(a) Compare the number of mol of H+ ions which are present in 50 cm3 of 1 mol dm–3 of sulphuric acid and 50 cm3 of 1 mol dm–3 of hydrochloric acid. Explain your answer.

9

Bandingkan bilangan mol ion H+ yang hadir dalam 50 cm3 asid sulfurik 1 mol dm–3 dengan 50 cm3 asid hidroklorik 1 mol dm–3. Terangkan jawapan anda. Acid Asid

50 cm3 of 1 mol dm–3 of sulphuric acid 50 cm3 asid sulfurik 1 mol dm–3

50 × 1 Calculate number of Number of mol of sulphuric acid = 1 000 hydrogen ion, H+ Hitung bilangan mol = 0.05 mol + ion hidrogen, H

H2SO4

2H + SO4 +

50 cm3 of 1 mol dm–3 of hydrochloric acid 50 cm3 asid hidroklorik 1 mol dm–3

HCl

2–

From the equation,

50 × 1 1 000 = 0.05 mol

Number of mol of hydrochloric acid =

H+ + Cl–

From the equation,

1 mol of H2SO4 : 2 mol of H

1 mol of HCl : 1 mol of H+

0.05 mol of H2SO4 : 0.1 mol of H+

0.05 mol of HCl : 0.05 mol of H+

+

Compare the number The number of H+ in 50 cm3 of 1 mol dm–3 of sulphuric acid is twice of the number of H+ in of hydrogen ions 50 cm3 of 1 mol dm–3 of hydrochloric acid. Bandingkan bilangan ion hidrogen

Explanation Penerangan

Sulphuric acid is diprotic acid whereas hydrochloric acid is monoprotic acid. 1 mol of sulphuric ionises to 2 mol of H+ whereas 1 mol of hydrochloric acid ionises to 1 mol of H+. The number of H+ in the same volume and concentration of both acids is doubled in sulphuric acid compared to hydrochloric acid.

(b) Suggest the volume of 1 mol dm–3 of hydrochloric acid that has the same number of H+ with 50 cm3 of 1 mol dm–3 of sulphuric acid. Cadangkan isi padu asid hidroklorik 1 mol dm–3 yang mempunyai bilangan ion H+ yang sama dengan 50 cm3 asid sulfurik 1 mol dm–3.

100 cm3

NEUTRALISATION / PENEUTRALAN Neutralisation is the reaction between an acid and a base to form only salt and water: Peneutralan ialah tindak balas antara asid dan bes untuk membentuk garam dan air sahaja:

1

Acid / Asid + Base / Bes

Salt / Garam + Water / Air

Example / Contoh : HCl (aq/ak ) + NaOH (aq/ak ) 2HNO3 (aq/ak ) + MgO (s/p)

NaCl (aq/ak ) + H2O (l/ce) Mg(NO3)2 (aq/ak ) + H2O (l/ce)

In neutralisation, the acidity of an acid is neutralised by an alkali. At the same time the alkalinity of an alkali is neutralised by an acid. The hydrogen ions in the acid react with hydroxide ions in the alkali to produce water:

2

Dalam peneutralan, keasidan asid dineutralkan oleh alkali. Pada masa yang sama, kealkalian alkali dineutralkan oleh asid. Ion hidrogen dalam asid bertindak balas dengan ion hidroksida dalam alkali:

H+ (aq/ak ) + OH– (aq/ak )

m

H2O (l/ce)

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Application of neutralisation in daily life / Aplikasi peneutralan dalam kehidupan seharian : Application

Example

Aplikasi

Agriculture Agrikultur

Contoh

1 Acidic soil is treated with powdered CaCO3)or ashes of burnt wood.

soda lime

(calcium oxide, CaO),

limestone

(calcium carbonate,

Tanah berasid dirawat dengan serbuk kapur tohor (kalsium oksida, CaO), batu kapur (kalsium karbonat, CaCO3) atau abu daripada kayu api.

2 Basic soil is treated with compost. The alkalis in basic soil. Tanah berbes dirawat dengan kompos. Gas dalam tanah berbes.

acidic berasid

gas from the decomposition of compost neutralises the yang terbebas daripada penguraian kompos

3 The acidity of water farming is controlled by adding

soda lime

meneutralkan

alkali

, (calcium oxide, CaO).

Keasidan air dalam pertanian dikawal dengan menambah kapur tohor (kalsium oksida, CaO).

Industries Industri

soda lime

1 Acidic gases emitted by industries are neutralised with are released into air.

, (calcium oxide, CaO) before the gases

Gas-gas berasid yang dibebaskan oleh kilang dineutralkan dengan kapur tohor (kalsium oksida, CaO), sebelum gas-gas tersebut dibebaskan ke udara.

2 Organic acid produced by bacteria in latex neutralises by

ammonia

and prevents coagulation.

Ammonia meneutralkan asid organik yang dihasilkan oleh bakteria dalam lateks dan mencegah penggumpalan.

Health Kesihatan

1 Excess acid in the stomach is neutralised with its anti-acids that contain bases such as aluminium hydroxide , calcium carbonate and magnesium hydroxide . Anti-asid mengandungi bes seperti aluminium hidroksida , meneutralkan asid berlebihan dalam perut.

2 Toothpastes contain mouth. Ubat gigi mengandungi mulut.

3

magnesium hidroksida

untuk

(such as magnesium hydroxide) to neutralise the acid produced by bacteria in (seperti magnesium hidroksida) untuk meneutralkan asid yang dihasilkan oleh bakteria dalam

(natrium hidrogen karbonat) digunakan untuk merawat sengatan lebah yang beralkali.

Vinegar (Ethanoic acid) is used to cure acidic wasp sting. Cuka

4

bes

dan

Baking powder (sodium hydrogen carbonate) is used to cure alkaline bee stings. Serbuk penaik

4

bases

kalsium karbonat

(asid etanoik) digunakan untuk merawat sengatan tebuan yang berasid.

An acid-base titration / Pentitratan asid-bes : (a) It is a technique used to determine the volume of an acid required to neutralise a fixed volume of an alkali with the help of acid-base indicator. Ianya adalah teknik yang digunakan untuk menentukan isi padu asid yang diperlukan untuk meneutralkan isi padu tertentu alkali dengan bantuan penunjuk asid-bes. Penunjuk yang biasa digunakan adalah fenolftalein dan metil jingga. (b) Steps taken are / Langkah-langkah yang diambil adalah :

(i)

An exact volume of alkali is measured with a pipette and poured into a conical flask.

Isi padu alkali yang tepat diukur dengan pipet dan dituang ke dalam kelalang kon. (ii) A few drops of indicator are added to the alkali / Beberapa titik penunjuk ditambah kepada alkali.

(iii) A burette is filled with an acid. An acid is added drop by drop into the alkali in the conical flask until the indicator changes colour, indicating the pH of neutral solution produced.

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Buret diisi dengan asid. Asid ditambah setitik demi setitik kepada alkali dalam kelalang kon sehingga warna penunjuk bertukar, menunjukkan pH larutan neutral telah dihasilkan. (c) When the acid has completely neutralised the given volume of an alkali, the titration has reached the end point. Apabila asid sudah lengkap meneutralkan isi padu alkali yang diberi, pentitratan telah mencapai takat akhir. (d) The end point is the point in the titration at which the indicator changes colour. Takat akhir ialah takat dalam pentitratan di mana penunjuk bertukar warna. (e) The commonly used indicators are phenolphthalein and methyl orange. Penunjuk yang biasa digunakan adalah fenolftalein dan metil jingga.

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The general steps used in any calculation involving neutralisation:

5

Langkah umum dalam penghitungan yang melibatkan peneutralan:

Step / Langkah 1 : Write the balanced equation / Tulis persamaan yang seimbang. Step / Langkah 2 : Write the information from the question above the equation. Tulis maklumat daripada soalan di atas persamaan.

Step / Langkah 3 : Write the information from the chemical equation below the equation (number of moles of substance involved). Tulis maklumat daripada persamaan kimia di bawah persamaan (bilangan mol bahan yang terlibat).

Step / Langkah 4 : Change the information to mole / Tukar maklumat kepada mol. Step / Langkah 5 : Use the relationship between the number of moles of the substances in Step 3. Guna hubungan di antara bilangan mol bahan-bahan dalam Langkah 3.

Step / Langkah 6 : Convert the number of mol to the required unit with the formula: Tukar bilangan mol kepada unit yang diperlukan dengan menggunakan formula:

n =

Mv 1 000

atau n = MV

n = Number of moles of solute / Bilangan mol bahan terlarut M = Concentration in mol dm–3 (molarity) / Kepekatan dalam mol dm–3 (kemolaran) V = Volume of solution in dm3 / Isi padu larutan dalam dm3 v = Volume of solution in cm3 / Isi padu larutan dalam cm3

EXERCISE / LATIHAN

50 cm3 of 1 mol dm–3 sodium hydroxide solution is neutralised by 25 cm3 of sulphuric acid. Calculate the concentration of sulphuric acid in mol dm–3 and g dm–3. [RAM: H = 1, S = 32, O = 16]

1

50 cm3 larutan natrium hidroksida 1 mol dm–3 dineutralkan oleh 25 cm3 asid sulfurik. Hitung kepekatan asid sulfurik dalam mol dm–3 dan g dm–3. [JAR: H = 1, S = 32, O = 16]

M = 1 mol dm–3 V = 50 cm3 2NaOH + H2SO4

M = ? V = 25 cm3 Na2SO4 + 2H2O

Number of mol of NaOH = 1 ×

n mol V dm3 0.025 mol = = 1 mol dm–3 25 dm3 1 000

Concentration of H2SO4 =

50 = 0.05 mol 1 000

From the equation, 2 mol of NaOH : 1 mol of H2SO4 0.05 mol of NaOH : 0.025 mol of H2SO4

Concentration of H2SO4 = 1 mol dm–3 × (2 × 1 + 32 + 16 × 4) g mol–1 = 98 g dm–3

Calculate the volume of 2 mol dm–3 sodium hydroxide needed to neutralise 100 cm3 of 1 mol dm–3 hydrochloric acid.

2

Hitung isi padu larutan natrium hidroksida 2 mol dm–3 yang diperlukan untuk meneutralkan 100 cm3 asid hidroklorik 1 mol dm–3.

M = 2 mol dm–3 V = ? cm3

M = 1 mol dm–3 V = 100 cm3

NaOH +

HCl

Number of mol of HCl = 1 ×

NaCl + H2O

100 = 0.1 mol 1 000

1 mol of HCl : 1 mol of mol NaOH 0.1 mol of HCl : 0.1 mol of mol NaOH n mol Volume of NaOH = M mol dm–3 0.1 mol = 2 mol dm–3 –3 = 0.05 dm = 50 cm3 m

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3

Experiment I / Eksperimen I 1 mol dm–3 of nitric acid is used to neutralise 100 cm3 of 1 mol dm–3 sodium hydroxide solution. Asid nitrik 1 mol dm–3 digunakan untuk meneutralkan 100 cm3 larutan natrium hidroksida 1 mol dm–3.

Experiment II / Eksperimen II 1 mol dm–3 of sulphuric acid is used to neutralise 100 cm3 of 1 mol dm–3 sodium hydroxide solution. Asid sulfurik 1 mol dm–3 digunakan untuk meneutralkan 100 cm3 larutan natrium hidroksida 1 mol dm–3.

Compare the volume of acids needed to neutralise 100 cm3 of 1 mol dm–3 sodium hydroxide solution in Experiment I and Experiment II. Explain your answer. Bandingkan isi padu asid yang diperlukan untuk meneutralkan 100 cm3 larutan natrium hidroksida 1 mol dm–3 dalam Eksperimen I dan Eksperimen II. Terangkan jawapan anda. Answer / Jawapan: Experiment Eksperimen

Balanced equation Persamaan kimia

Calculation Pengiraan

Experiment I

Experiment II

Eksperimen I

NaOH + HNO3

Eksperimen II

NaNO3 + H2O

2NaOH + H2SO4

Na2SO4 + 2H2O

100 1 000 = 0.1 mol From the equation / Daripada persamaan : 1 mol NaOH : 1 mol HNO

100 1 000 = 0.1 mol From the equation / Daripada persamaan : 2 mol NaOH / NaOH : 1 mol H SO

0.1 mol NaOH : 0.1 mol HNO 3 Mv Mol of HCl / Bilangan mol HNO3 = 1 000 M = Concentration of HNO3 / Kepekatan HNO3 v = Volume of HNO3 in cm3 / Isi padu HNO3 dalam cm3 1 mol dm–3 × v = 0.1 mol 1 000 3 v = 100 cm

0.1 mol NaOH / NaOH : 0.05 mol H SO 2 4 Mv Mol of H2SO4 / Bilangan mol H2SO4 = 1 000 M = Concentration of H2SO4 / Kepekatan H2SO4 v = Volume of H2SO4 in cm3 / Isi padu H2SO4 dalam cm3 1 mol dm–3 × v = 0.1 mol 1 000 3 v = 50 cm

Mol of KOH / Bilangan mol NaOH = 1 ×

Mol of KOH / Bilangan mol NaOH = 1 ×

3

2

Comparison and explanation

– The volume of acid needed in Experiment II is doubled of Experiment I.

Perbandingan dan penerangan

– Sulphuric acid is

4

Isi padu asid nitrik yang diperlukan adalah dua kali ganda dalam Eksperimen I dibandingkan dengan Eksperimen II.

diprotic

Asid sulfurik adalah asid

acid while nitric acid is monoprotic .

diprotik

– One mol of sulphuric ionises

manakala asid nitrik adalah asid

two

Satu mol asid sulfurik mengion kepada ion H+.

monoprotik

.

mol of H , one mol nitric acid ionises to

dua

+

one

mol of H+.

mol ion H manakala satu mol asid nitrik mengion kepada +

satu

mol

– The number of H+ in the same volume and concentration of both acids is doubled in sulphuric acid compared to hydrochloric acid. Bilangan ion H+ dalam isi padu dan kepekatan yang sama bagi kedua-dua asid adalah dua kali ganda dalam asid sulfurik dibandingkan dengan asid nitrik.

4

Diagram below shows the apparatus set-up for the titration of potassium hydroxide solution with sulphuric acid. Gambar rajah di bawah menunjukkan susunan radas bagi pentitratan larutan kalium hidroksida dengan asid sulfurik.

0.5 mol dm–3 sulphuric acid Asid sulfurik 0.5 mol dm–3

50 cm3 of 1 mol dm3 potassium hydroxide solution + methyl orange 50 cm3 larutan kalium hidroksida 1 mol dm3 + metil jingga

0.5 mol dm–3 sulphuric acid is titrated to 50 cm3 of 1 mol dm3 potassium hydroxide solution and methyl orange is used as indicator.

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Asid sulfurik 0.5 mol dm–3 ditambahkan kepada 50 cm3 larutan kalium hidroksida 1 mol dm3 dan metil jingga digunakan sebagai penunjuk.

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(a) (i)

Name the reaction between sulphuric acid and potassium hydroxide. Namakan tindak balas antara asid sulfurik dengan kalium hidroksida.

Neutralisation

(ii) Name the salt formed in the reaction / Namakan garam yang terbentuk dalam tindak balas tersebut. Potassium sulphate

(b) Suggest an apparatus that can be used to measure 25.0 cm3 of potassium hydroxide solution accurately. Cadangkan radas yang boleh digunakan untuk mengukur 25.0 cm3 larutan kalum hidroksida dengan tepat.

Pipette (c) What is the colour of methyl orange / Apakah warna metil jingga dalam (i) in potassium hydroxide solution / larutan kalium hidroksida? Red (ii) in sulphuric acid / asid sulfurik? Yellow (iii) at the end point of the titration / pada titik akhir pentitratan? Orange (d) (i) Write a balanced equation for the reaction that occurs / Tuliskan persamaan seimbang bagi tindak balas yang berlaku. 2KOH + H2SO4

K2SO4 + 2H2O

(ii) Calculate the volume of the 0.1 mol dm–3 sulphuric acid needed to completely react with 50 cm3 of 1 mol dm3 potassium hydroxide. Hitung isi padu asid sulfurik 0.1 mol dm–3 yang diperlukan untuk bertindak balas dengan lengkap dengan 50 cm3 larutan kalium hidroksida 1 mol dm–3.

Number of mol KOH = 1 × From the equation,

50 = 0.05 mol 1 000

2 mol of KOH : 1 mol of H2SO4 0.05 mol of KOH : 0.025 mol of H2SO4

n mol M mol dm–3 0.025 mol = 1 mol dm–3 = 0.025 dm3 = 25 cm3

Volume of H2SO4 =

(e) (i)

The experiment is repeated with 0.1 mol dm hydrochloric acid to replace sulphuric acid. Predict the volume of hydrochloric acid needed to neutralise 50.0 cm3 potassium hydroxide solution. –3

Eksperimen diulang dengan menggunakan asid hidroklorik 0.1 mol dm–3 untuk menggantikan asid sulfurik. Ramalkan isipadu asid hidroklorik yang diperlukan untuk meneutralkan 50.0 cm3 larutan kalium hidroksida.

50 cm3 // double the volume of sulphuric acid

(ii) Explain your answer in (e)(i) / Terangkan jawapan anda di (e)(i). – Hydrochloric acid is a monoprotic acid whereas sulphuric acid is a Asid hidroklorik ialah asid

monoprotik

manakala asid sulfurik ialah asid

diprotic diprotik

– The same volume and concentration of both acids, hydrochloric acid contains of mole of H+ ions as in sulphuric acid.

m

half separuh

the number bilangan

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Objective Questions / Soalan objektif 1

Which of the following substances changes blue litmus paper to red when dissolved in water?

5

The table below shows the concentration of hydrochloric acid and ethanoic acid. Jadual di bawah menunjukkan kepekatan asid hidroklorik dan asid etanoik.

Antara bahan berikut, yang manakah menukarkan warna kertas litmus merah kepada biru apabila dilarutkan dalam air?

A Sulphur dioxide

Acid

Sulfur dioksida B

Carbon dioxide Lithium oxide

Ethanoic acid

Sodium carbonate

Which of the following statements is true about both acids?

The table below shows the pH value of four acids which have the same concentration.

Antara berikut, yang manakah adalah betul tentang kedua-dua asid?

Jadual di bawah menunjukkan nilai pH empat larutan yang mempunyai kepekatan yang sama.

A Both are strong acids Kedua-duanya adalah asid kuat

Solution / Larutan

pH value / Nilai pH

B

Both acids are strong electrolyte

P

2

Q

7

C

The pH value of both acid are equal

R

12

S

13

Which of the following solutions has the highest concentration of hydroxide ion? Antara larutan berikut, yang manakah mempunyai kepekatan ion hidroksida paling tinggi?

A P B Q 3

A Sulphuric acid and copper(II) sulphate solution

6

A 20 B 40 7

Nitric acid and magnesium oxide Hydrochloric acid and sodium nitrate solution Asid hidroklorik dan larutan natrium nitrat Asid etanoik dan larutan natrium sulfat

Antara tindak balas berikut, yang manakah tidak akan membebaskan sebarang gas?

A Copper metal with sulphuric acid Logam kuprum dengan asid sulfurik

B

Zinc metal with hydrochloride acid Logam zink dengan asid hidroklorik

C

Ammonium chloride with calcium hydroxide Ammonium klorida dengan kalsium hidroksida

D Sodium carbonate hydrochloric acid

8

C 200 cm3 D 250 cm3

Which of the following solutions have the same concentration of hydrogen ions, H+, as in 0.1 mol dm–3 sulphuric acid, H2SO4? Antara asid berikut, yang manakah mempunyai kepekatan ion hidrogen, H+ yang sama dengan asid sulfurik 0.1 mol dm–3?

A 0.1 mol dm–3 hydrochloric acid Asid hidroklorik 0.1 mol dm–3

B

0.1 mol dm–3 carbonic acid Asid karbonik 0.1 mol dm–3

C

0.2 mol dm–3 ethanoic acid Asid etanoik 0.2 mol dm–3

D 0.2 mol dm–3 nitric acid Asid nitrik 0.2 mol dm–3

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Natrium karbonat dengan asid hidroklorik

What is the volume of 2.0 mol dm–3 potassium hydroxide solution needed to prepare 500 cm3 of 0.1 mol dm–3 potassium hydroxide solution?

A 100 cm3 B 150 cm3

D Ethanoic acid and sodium sulphate solution Which of the following reactions will not produce any gas?

C 80 D 120

Berapakah isi padu larutan kalium hidroksida 2.0 mol dm–3 diperlukan untuk menyediakan 500 cm3 larutan kalium hidroksida 1 mol dm–3?

Asid nitrik dan magnesium oksida

4

The molarity of sodium hydroxide solution 0.5 mol dm–3. What is the concentration of the solution in g dm–3? [Relative atomic mass: H = 1, O =16, Na = 23] Kemolaran larutan natrium hidroksida adalah 0.5 mol dm–3. Apakah kepekatan larutan itu dalam g dm–3? [Jisim atom relatif: H = 1, O =16, Na = 23]

Asid sulfurik dan larutan kuprum(II) sulfat

C

Nilai pH kedua-dua asid adalah sama

50 cm3 setiap asid memerlukan 50 cm3 larutan natrium hidroksida 0.1 mol dm–3 untuk dineutralkan

C R D S

Antara pasangan bahan tindak balas berikut, yang manakah akan menghasilkan tindak balas?

Kedua-duanya adalah elektrolit yang kuat

D 50 cm3 of each acid need 50 cm3 of 0.1 mol dm–3 of sodium hydroxide to be neutralised

Which of the following pairs of reactants would result in a reaction?

B

0.1

Asid etanoik

Natrium karbonat

2

0.1

Asid hidroklorik

Litium oksida

D

Kepekatan / mol dm–3

Hydrochloric acid

Karbon dioksida

C

Concentration / mol dm–3

Asid

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9

Which of the following sodium hydroxide solutions have concentration of 1.0 mol dm–3? [Relative atomic mass: H=1, O=16, Na =23] Antara larutan natrium hidroksida berikut, yang manakah mempunyai kepekatan 1.0 mol dm–3? [JAR: H = 1, O = 16, Na = 23]

I

10 The diagram below shows 25.0 cm3 of 1.0 mol dm–3 of sulphuric acid and 50.0 cm3 of 1.0 mol dm–3 of sodium are added hydroxide solution to form solution A. Rajah di bawah menunjukkan 25.0 cm3 asid sulfurik 1.0 mol dm–3 dan 50.0 cm3 larutan natrium hidroksida 1.0 mol dm–3 ditambah bersama untuk menghasilkan larutan A.

5 g of sodium hydroxide dissolved in distilled water to make 250 cm3 of solution.

50 cm3 of 1 mol dm–3 of hydroxide solution 50 cm3 larutan natrium hidroksida 1 mol dm–3

5 g natrium hidroksida dilarutkan dalam air suling menjadikan 250 cm3 larutan.

II 20 g of sodium hydroxide dissolved in distilled water to make 500 cm3 of solution. 20 g natrium hidroksida dilarutkan dalam air suling menjadikan 500 cm3 larutan.

III 250 cm of 2 mol dm sodium hydroxide solution is added to distilled water to make 1 dm3 of solution. 3

–3

250 cm3 larutan natrium hidroksida 2 mol dm ditambah air suling menjadikan 1 dm3 larutan.

25 cm3 of 2.0 mol dm–3 sulphuric acid 25 cm3 asid sulfrik 2.0 mol dm–3

Solution A / Larutan A

–3

IV 500 cm3 of 2 mol dm–3 sodium hydroxide solution is added to distilled water to make 1 dm3 of solution. 500 cm3 larutan natrium hidroksida 2 mol dm–3 ditambah air suling menjadikan 1 dm3 larutan.

A I and III only I dan III sahaja

B

II and III only

C

II and IV only

II dan III sahaja II dan IV sahaja

Which of the following is true about the solution A? Antara berikut, yang manakah adalah benar tentang larutan A?

A The solution has a pH value of 7 Larutan itu menpunyai nilai pH 7

B

The solution will react with any acid

C

The solution turns a red litmus paper blue

Larutan itu boleh bertindak balas dengan sebarang asid Larutan itu menukarkan warna kertas litmus merah kepada biru

D The solution will react with zinc to produce hydrogen gas Larutan itu bertindak balas dengan zink untuk menghasilkan gas hidrogen

D I, II, III and IV

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SALT

7

GARAM PREPARATION OF SALTS / PENYEDIAAN GARAM

• THE MEANING OF SALTS / MAKSUD GARAM – To write the meaning of salts and the formulae for all types of salt that are commonly found in this topic. Menyatakan maksud garam dan menulis formula semua jenis garam yang biasa ditemui dalam tajuk ini.

• THE SOLUBILITY OF SALTS / KETERLARUTAN GARAM – To determine the solubility of nitrate, sulphate, carbonate and chloride salts. Menentukan keterlarutan semua garam nitrat, sulfat, karbonat dan klorida.

• EXPERIMENTS FOR THE PREPARATION OF SALTS BASED ON SOLUBILITY EKSPERIMEN PENYEDIAAN GARAM BERDASARKAN KETERLARUTAN

– To determine the suitable methods for the preparation of salts based on solubility: Menentukan kaedah yang sesuai bagi penyediaan garam berdasarkan keterlarutan:

–

i. Acid + metal / Asid + logam ii. Acid + metal oxides / Asid + oksida logam iii. Acid + alkali / Asid + alkali iv. Acid + metal carbonate / Asid + karbonat logam v. Double decomposition reaction / Tindak balas penguraian ganda dua To describe the experiments for each method of preparation and explain the rationale for each step. Menghuraikan eksperimen bagi setiap jenis kaedah penyediaan serta menerangkan rasional setiap langkah.

CALCULATION ON QUANTITY OF REACTANTS/PRODUCTS [ QUANTITATIVE ANALYSIS ] PENGHITUNGAN KUANTITI BAHAN/HASIL [ ANALISIS KUANTITATIF]

• CONTINUOUS VARIATIONS METHODS / KAEDAH PERUBAHAN BERTERUSAN – To describe the methods of experiment to determine the formulae of insoluble salts. Menghuraikan eksperimen bagi kaedah penentuan formula garam tak larut.

• SOLVING VARIOUS PROBLEMS RELATING TO QUANTITY OF REACTANTS/PRODUCTS IN SOLID, LIQUID AND GAS FORMS MENYELESAIKAN PELBAGAI MASALAH BERKAITAN KUANTITI BAHAN DALAM BENTUK PEPEJAL, LARUTAN DAN GAS – Using the formula / Menggunakan formula:

i. n =

MV 1 000

ii. Mole / Mol =

Mass / Jisim RAM/RMM/RFM / JAR/JMR/JFR

iii. The molar volume of gas at room temperature and s.t.p / Isi padu molar gas pada suhu bilik dan s.t.p

IDENTIFICATION OF IONS [ QUALITATIVE ANALYSIS ] / MENGENAL ION [ ANALISIS KUALITATIF ] • ACTION OF HEAT ON SALTS / KESAN HABA KE ATAS GARAM – To state the colour of the residue of lead(II) oxide, zinc oxide and copper(II) oxide. Menyatakan warna baki bagi plumbum(II) oksida, zink oksida dan kuprum(II) oksida.

– To state the confirmatory tests for carbon dioxide and nitrogen dioxide. Menyatakan ujian pengesahan bagi gas karbon dioksida dan nitrogen dioksida.

– To write the equations of the decomposition of carbonate and nitrate salts. Menulis persamaan penguraian semua garam karbonat dan nitrat.

• CONFIRMATORY TEST CATIONS AND ANIONS / UJIAN PENGESAHAN KATION DAN ANION – To state the confirmatory tests for all cations using sodium hydroxide and ammonia solution. Menghuraikan ujian pengesahan semua kation menggunakan natrium hidroksida dan larutan ammonia.

– To state the confirmatory tests to differentiate Al3+ and Pb2+. Menghuraikan ujian untuk membezakan Al3+ dan Pb2+.

– To state the confirmatory tests for anions of sulphate, nitrate, carbonate and chloride.

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Menghuraikan ujian pengesahan anion sulfat, nitrat, karbonat dan klorida.

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PREPARATION OF SALT / PENYEDIAAN GARAM A salt is a compound formed when the hydrogen ion in an acid is replaced with metal ion or ammonium ion. Example: Sodium chloride, copper(II) sulphate, potassium nitrate and ammonium sulphate.

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Garam ialah sebatian ion yang terhasil apabila ion hidrogen daripada asid diganti oleh ion logam termasuk ion ammonium. Contoh: natrium klorida, kuprum(II) sulfat, kalium nitrat dan ammonium sulfat.

Write the formulae of the salts in the table below by replacing hydrogen ion in sulphuric acid, hydrochloric acid, nitric acid and carbonic acid with metal ions or ammonium ion.

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Tuliskan formula kimia garam berikut dengan menggantikan ion hidrogen dalam asid sulfurik, asid hidroklorik, asid nitrik dan asid karbonik dengan ion logam atau ion ammonium: Metal ion Ion logam

Sulphate salt (from H2SO4) Garam sulfat (dari H2SO4 )

Chloride salt (from HCl) Garam klorida (dari HCl)

Nitrate salt (from HNO3) Garam nitrat (dari HNO3 )

Carbonate salt (from H2CO3) Garam karbonat (dari H2CO3 )

Na+

Na2SO4

NaCl

NaNO3

Na2CO3

K+

K2SO4

KCl

KNO3

K2CO3

Mg2+

MgSO4

MgCl2

Mg(NO3 )2

MgCO3

Ca2+

CaSO4

CaCl2

Ca(NO3 )2

CaCO3

Al3+

Al2(SO4 )3

AlCl3

Al(NO3 )3

Al2(CO3 )3

Zn2+

ZnSO4

ZnCl2

Zn(NO3 )2

ZnCO3

Fe2+

FeSO4

FeCl2

Fe(NO3 )2

FeCO3

Pb2+

PbSO4

PbCl2

Pb(NO3 )2

PbCO3

Cu2+

CuSO4

CuCl2

Cu(NO3 )2

CuCO3

Ag+

Ag2SO4

AgCl

AgNO3

Ag2CO3

NH4+

(NH4 )2SO4

NH4Cl

NH4NO3

(NH4 )2CO3

Ba2+

BaSO4

BaCl2

Ba(NO3 )2

BaCO3

Solubility of salts in water: / Keterlarutan garam dalam air: (a) All K+, Na+ and NH4+ salts are soluble. / Semua garam K+, Na+ dan NH4+ larut. (b) All nitrate salts are soluble. / Semua garam nitrat larut. (c) All carbonate salts are insoluble except K2CO3, Na2CO3 and (NH4)2CO3. Semua garam karbonat tak larut kecuali K2CO3, Na2CO3 dan (NH4 )2CO3. (d) All sulphate salts are soluble except CaSO4, PbSO4 and BaSO4.

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Semua garam sulfat larut kecuali CaSO4, PbSO4 dan BaSO4.

(e) All chloride salts are soluble except PbCl2 and AgCl. / Semua garam klorida larut kecuali PbCl2 dan AgCl. * Based on the solubility of the salts in water, shade the insoluble salts in the above table. * Berdasarkan keterlarutan garam dalam air, lorekkan garam yang tak larut dalam jadual di atas.

Method used to prepare salt depends on the solubility of the salt.

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Kaedah penyediaan garam bergantung pada keterlarutan garam tersebut. Soluble salts are prepared from the reactions between an acid with a metal/ base/ metal carbonate: Garam terlarut disediakan melalui tindak balas antara asid dengan logam/bes/karbonat logam: i. Acid + metal / Asid + logam salt + hydrogen / garam + hidrogen Acid + *base salt + water ii. Acid + metal oxide / Asid + oksida logam salt + water / garam + air Asid + *bes garam + air iii. Acid + alkali / Asid + alkali salt + water / garam + air iv. Acid + metal carbonate / Asid + karbonat logam salt + water + carbon dioxide / garam + air + karbon dioksida

* Most bases are metal oxide or metal hydroxide. / Kebanyakan bes adalah oksida logam atau hidroksida logam. * All metal oxides and hydroxides are insoluble in water except Na2O, K2O, NaOH and KOH. Semua oksida logam dan hidroksida logam tidak larut dalam air kecuali Na2O, K2O, NaOH dan KOH.

* Alkali is a base that soluble in water and ionises to hydroxide ion.

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Salts are prepared based on their solubility as shown in the flow chart below:

Asid + Alkali Garam + Air (Tindak balas Peneutralan)

– Acid + Alkali Salt + Water (Neutralisation Reaction)

Garam ini disediakan melalui kaedah pentitratan di antara asid dan alkali dengan menggunakan penunjuk.

The salt is prepared by titration method of acid and alkali using an indicator.

Method I / Kaedah I

Salts / Garam K+, Na+, NH4+

Soluble salt Garam larut

Isi padu asid yang sama juga ditambah kepada isi padu alkali yang sama tanpa penunjuk untuk mendapatkan garam yang tulen dan neutral.

– The same volume of acid is then added to the same volume of alkali without any indicator to obtain pure and neutral salt solution.

Pentitratan dijalankan dengan menentukan isi padu asid yang diperlukan untuk meneutralkan alkali yang isi padunya sudah ditetapkan dengan menggunakan penunjuk.

Salt + Hydrogen (Displacement reaction)

Garam + Air + Karbon dioksida

Keringkan baki dengan menekankan antara kertas turas.

– Dry the residue by pressing it between filter papers.

Bilas baki dengan air suling.

– Rinse the residue with distilled water.

Turas dengan corong turas.

– Filter using filter funnel.

Kacau dengan rod kaca.

– Stir with glass rod.

Campur dua larutan yang mengandungi kation dan anion garam tak larut.

– Mix two solutions containing cations and anions of insoluble salts.

Garam ini disediakan melalui kaedah pemendakan. (Tindak balas penguraian ganda dua).

The salt is prepared by precipitation method. (Double decomposition reaction)

Method III / Kaedah III

Insoluble salt Garam tak larut

Turas dan keringkan hablur garam dengan menekan antara kertas turas.

– Cooled at room temperature / Biarkan sejuk pada suhu bilik. – Filter and dry the salt crystals by pressing them between filter papers.

Celupkan dengan rod kaca, jika hablur terbentuk dengan serta merta, larutan adalah tepu.

– Evaporate the filtrate until it becomes a saturated solution/ Sejatkan hasil turasan hingga larutan tepu. – Dip in a glass rod, if crystals are formed, the solution is saturated.

Turas campuran tersebut untuk mengeluarkan pepejal logam/oksida logam/karbonat logam yang berlebihan.

– Filter the mixture to remove excess metal/metal oxide/metal carbonate

Tambah serbuk logam/oksida logam/karbonat logam ke dalam isi padu tetap asid yang dihangatkan sehingga berlebihan.

volume of the heated acid

– Add metal/metal oxide/metal carbonate powder until excess into a fixed

Asid + Karbonat logam

Salt + Water + Carbon Dioxide

Garam + Air (Tindak balas Peneutralan)

– Acid + Metal carbonate

Asid + Oksida bes

Salt + Water (Neutralisation Reaction)

Garam + Hidrogen (Tindak balas penyesaran)

– Acid + Metal oxide

Asid + Logam

– Acid + Metal

Garam ini disediakan melalui tindak balas antara asid dengan logam/oksida logam/ karbonat logam yang tak larut:

The salt is prepared by reacting acid with insoluble metal/metal oxide/ metal carbonate:

Method II / Kaedah II

Other than / Garam selain K+, Na+, NH4+

Preparation of salt / Penyediaan garam

Garam disediakan berdasarkan keterlarutannya sebagaimana yang ditunjukkan pada carta aliran di bawah:

– A titration is conducted to determine the volume of acid needed to neutralise a fixed volume of an alkali with the aid of an indicator.

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PREPARATION OF SOLUBLE AND INSOLUBLE SALT / PENYEDIAAN GARAM LARUT DAN GARAM TAK LARUT


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50 – 100

yang berlebihan.

3

0.5 – 2

–3

Hasil turasan ialah larutan garam

The filtrate is salt solution

logam/ Baki adalah logam oksida/ logam karbonat

Heat

Panaskan

Acid Asid

Salt crystals

hablur garam dengan Keringkan menekan antara kertas turas.

salt crystals by • Dry the pressing them between filter papers.

serbuk logam/ oksida logam/ karbonat logam pada isi padu asid yang Tambahkan tetap sambil dihangatkan perlahan-lahan .

.

.

Panaskan

Heat

Sejatkan larutan sehingga terbentuk.

Hablur

are terbentuk.

Baki adalah hablur garam

Residue is salt crystals

Turaskan campuran tersebut untuk mengasingkan hablur garam .

garam

Salt crystals

hablur garam

• Filter the mixture to separate the salt crystals .

Sejukkan pada suhu bilik sehingga

crystals salts

tepu

Larutan garam

salt solutions

Saturated

larutan tepu

Larutan garam dituangkan dalam mangkuk penyejat .

salt solution is poured into evaporating dish .

• Evaporate the salt solution until saturated solution is formed.

• The

• Cool it at room temperature until formed.

.

The residue is metal /metal oxide /metal carbonate .

Turas campuran tersebut untuk mengasingkan bahan berlebihan iaitu logam/oksida logam/karbonat logam larutan garam dengan .

• Filter the mixture to separate metal /metal oxide excess /metal carbonate with the salt solution .

cm of mol dm of any acid • Measure and pour and pour into a beaker. 50 – 100 Sukat dan tuangkan cm3 sebarang asid berkepekatan –3 0.5 – 2 mol dm dan tuangkan ke dalam bikar. • Add metal/metal oxide/ metal carbonate powder into the acid and heat gently .

Panaskan

Heat

Logam/oksida logam/ karbonat logam

Excess of metal/ metal oxide/ metal carbonate

Tambah serbuk logam / oksida logam / karbonat logam kepada asid sehingga berlebihan .

• Add metal/metal oxide / metal powder carbonate to the acid excess until .

campuran dengan Kacau rod kaca menggunakan .

• Stir the mixture with a glass rod .

Method II:/Kaedah II: Soluble salt except K+, Na+ and NH4+ / Garam larut selain K+, Na+ dan NH4+

2 Steps to Prepare Soluble Salt/Langkah Penyediaan Garam Larut Method I:/Kaedah I:

1 mol dm–3 sebarang asid dititratkan kepada alkali sehingga neutral menggunakan penunjuk. Isi padu asid yang digunakan dicatat.

1 mol dm–3 of any acid is titrated to the alkali until neutral by using an indicator. The volume of acid used is recorded.

Alkali Alkali

Acid Asid

Ulang titratan tanpa penunjuk untuk mendapatkan larutan garam yang tulen dan neutral .

• Repeat the titration without the indicator to get pure and neutral salt solution.

•

Sukat dan tuangkan 50 cm3 sebarang alkali berkepekatan 1 mol dm–3 ke dalam kelalang. Tambah beberapa titis fenolftalein.

• Measure and pour 50 cm3 of 1 mol dm–3 any alkali into a conical flask. Add a few drops of phenolphthalein.

Garam larut K+, Na+ dan NH4+

Soluble salt of K+, Na+ and NH4+


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Steps to Prepare Insoluble Salt / Penyediaan Garam Tak Larut Insoluble salts are prepared by the precipitation method through double decomposition reactions. Garam tak larut disediakan dengan cara pemendakan melalui tindak balas penguraian ganda dua. (i) In this reaction, the precipitate of insoluble salt is formed when two different solutions that contain the cation and anion of the insoluble salt are mixed. Dalam tindak balas ini, mendakan garam tak larut terbentuk apabila dua larutan berbeza yang mengandungi kation dan anion garam tak terlarut dicampurkan.

(ii) The insoluble salt is obtained as a residue of a filtration. Garam tak terlarut tersebut diperoleh daripada baki penurasan. Method III: Preparation of Insoluble XnYm Salt by Double Decomposition Reaction Kaedah III: Penyediaan Garam Tak Larut XnYm Melalui Tindak balas Penguraian Ganda Dua 1) Measure and pour 50 – 100 cm3 0.5 – 2 of mol dm–3 of aqueous solution contains X m+ cation.

2) Measure and pour 50 – 100 cm3 0.5 – 2 mol dm–3 of of aqueous solution contains Yn– anion into another beaker.

Sukat dan tuangkan 50 – 100 cm3 0.5 – 2 mol dm–3 larutan berkepekatan mengandungi kation Xm+ ke dalam bikar.

Precipitate of salt is formed.

XnYm

Sukat dan tuangkan 50 – 100 cm3 0.5 – 2 larutan berkepekatan mol dm–3 mengandungi anion Yn– ke dalam bikar yang lain.

3) Mix both solutions and stir the glass rod .

Campur dan kacaukan rod kaca .

XnYm

Mendakan garam terbentuk.

The residue is salt.

XnYm

Mendakan adalah garam XnYm .

mixture

campuran

with

menggunakan

4)

Filter the with salt.

mixture

distilled water

and

rinse

the precipitate

. The residue is

XnYm

Turas campuran dan bilas mendakan itu menggunakan air suling. Baki ialah garam XnYm.

Salt

XnYm

5)

Press the precipitate

between

filter papers to dry it.

Tekankan mendakan antara kertas turas untuk mengeringkannya.

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Garam

XnYm

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Complete the following table: Lengkapkan jadual berikut: X m+

Yn–

XnYm

Pb2+ [Pb(NO3)2]

I– [KI]

PbI2

Ba2+[ BaCl2 ]

SO4 [ Na2SO4 ]

BaSO4

Ag+ [AgNO3]

Cl– [NaCl]

AgCl

Ag+ + Cl–

Ca2+ [Ca(NO3)2]

CO32– [Na2CO3]

CaCO3

Ca2+ + CO32–

2–

Ion equation/Persamaan ion

Pb2+ + I–

Ba2+ + SO42–

PbI2

BaSO4 AgCl CaCO3

Complete the following table by writing “S” for soluble salts and “IS” for insoluble salts. Write all the possible chemical equations to prepare soluble salts and two chemical equations for insoluble salts.

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Lengkapkan jadual berikut dengan menulis “L” bagi garam larut dan “TL” bagi garam tak larut. Tuliskan semua persamaan kimia dalam penyediaan garam larut dan dua persamaan kimia bagi garam tak larut. Salt

Garam

Zinc chloride Zink klorida

“S” / “IS”

Chemical equations

“L” / “TL”

Persamaan kimia

ZnCl + H 2 2 ZnCl + CO + H O ZnCO3 + 2HCl 2 2 2 ZnCl + H O ZnO + 2HCl Zn + 2HCl

S

2

Sodium nitrate Natrium nitrat

Silver chloride Argentum klorida

Copper(II) sulphate Kuprum(II) sulfat

Lead(II) sulphate Plumbum(II) sulfat

Aluminium nitrate Aluminium nitrat

Lead(II) chloride Plumbum(II) klorida


Potassium chloride Kalium klorida

Lead(II) nitrate Plumbum(II) nitrat

Barium sulphate

m

IS

S

IS

NaNO + H O 3 2

NaOH + HNO3

AgCl + HNO

AgNO3 + HCl

3

AgCl + NaNO 3

AgNO3 + NaCl

CuSO + H O 4 2 CuSO + CO + H O CuCO3 + H2SO4 4 2 2

CuO + H2SO4

Pb(NO3)2 + H2SO4 Pb(NO3)2 + Na2SO4

PbSO + 2HNO 4 3 PbSO + 2NaNO 4

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2Al(NO ) + 3H 3 3 2 2Al(NO ) + 3H O Al2O3 + 6HNO3 3 3 2 2Al(NO ) + 3CO + 3H O Al2(CO3)3 + 6HNO3 3 3 2 2 2Al + 6HNO3

S

IS

Pb(NO3)2 + 2HCl Pb(NO3)2 + 2NaCl

PbCl + 2HNO 2 3 PbCl + 2NaNO 2

3

Mg(NO ) + H 3 2 2 Mg(NO ) + H O MgO + 2HNO3 3 2 2 Mg + 2HNO3

S

MgCO3 + 2HNO3

Mg(NO3)2 + CO2 + H2O

KCl + H O 2

S

KOH + HCl

S

Pb(NO ) + H O 3 2 2 Pb(NO ) + CO + H O PbCO3 + 2HNO3 3 2 2 2

IS

PbO + 2HNO3

BaCl2 + H2SO4 BaCl2 + Na2SO4

BaSO + 2HCl 4 BaSO + 2NaCl 4

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Barium sulfat

S

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The diagram below shows the set-up of apparatus to prepare soluble salt Y. Rajah di bawah menunjukkan susunan radas bagi menyediakan garam larut Y.


25 cm3 of 1 mol dm–3 potassium hydroxide solution + phenolphthalein 25 cm3 larutan kalium hidroksida 1 mol dm-3 + fenolftalein

Phenolphthalein is used as an indicator in a titration between nitric acid and sodium hydroxide solution. 25 cm3 of nitric acid completely neutralises 25 cm3 of 1 mol dm–3 potassium hydroxide solution. The experiment is repeated by reacting 25 cm3 of 1 mol dm–3 potassium hydroxide solution with 25 cm3 nitric acid without phenolphthalein. Salt Y is formed from the reaction. Fenolftalein digunakan sebagai penunjuk dalam pentitratan antara asid nitrik dengan larutan kalium hidroksida. 25 cm3 asid nitrik meneutralkan 25 cm3 larutan kalium hidroksida 1 mol dm–3. Eksperimen ini diulang dengan menindakbalaskan 25 cm3 larutan kalium hidroksida 1 mol dm–3 dengan 25 cm3 asid nitrik tanpa fenolftalein. Garam Y terbentuk daripada tindak balas ini.

(a) Name salt Y. Nyatakan nama garam Y.

Potassium nitrate (b) Write a balanced equation for the reaction that occurs. Tuliskan persamaan seimbang bagi tindak balas yang berlaku.

HNO3 + KOH

KNO3 + H2O

(c) Calculate the concentration of nitric acid. Hitungkan kepekatan asid nitrik tersebut.

1 = 0.025 mol 1 000 From the equation, 1 mol of KOH : 1 mol of HNO3 0.025 mol of KOH : 0.025 mol of HNO3 Concentration of HNO3, M 25 0.025 = M × 1 000 M = 1 mol dm–3 Mol of NaOH = 1 ×

(d) Why is the experiment is repeated without phenolphthalein? Mengapakah eksperimen ini diulang tanpa menggunakan fenolftalein?

To get pure and neutral salt solution Y. (e) Describe briefly how a crystal of salt Y is obtained from the salt solution. Huraikan secara ringkas bagaimana hablur garam Y diperoleh daripada larutan garamnya.

– The salt solution is poured into an evaporating dish. – The solution is heated to evaporate the solution until one third its original volume// a saturated solution formed. – The saturated solution is allowed to cool until salt crystals Y are formed. – The crystals are filtered and dried by pressing them between filter papers. (f) Name two other salts that can be prepared with the same method. Namakan dua garam lain yang boleh disediakan dengan kaedah yang sama.

Potassium/sodium/ammonium salt. Example: potassium nitrate, sodium sulphate. (g) State the type of reaction in the preparation of the salts. Nyatakan jenis tindak balas dalam penyediaan garam ini. n io

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Neutralisation

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The following is the steps in the preparation of dry copper(II) sulphate crystals.

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Berikut adalah langkah-langkah dalam penyediaan hablur garam kuprum(II) sulfat kering.

Step I:

Copper(II) oxide powder is added a little at a time with constant stirring to the heated 50 cm3 of 1 mol dm–3 sulphuric acid until some of it no longer dissolve.

Langkah I: Serbuk kuprum(II) oksida ditambahkan, sedikit demi sedikit sambil dikacau ke dalam 50 cm3 asid sulfurik 1 mol dm-3 yang dipanaskan sehingga serbuk itu tidak boleh larut lagi.

Step II:

The mixture is filtered.

Langkah II: Campuran dituras.

Step III:

The filtrate is poured into an evaporating dish and heated to evaporate the solution until one third of its original.

Langkah III: Hasil turasan dipanaskan di dalam mangkuk penyejat sehingga isi padunya menjadi satu pertiga daripada isi padu asal.

Step IV:

The salt solution is allowed to cool at room temperature for the crystallisation to take place.

Langkah IV: Hasil turasan itu dibiarkan sejuk ke suhu bilik sehingga penghabluran berlaku.

Step V:

The crystals formed are filtered and dried by pressing them between filter papers.

Langkah V: Hablur yang terbentuk dituraskan dan dikeringkan dengan menekan antara kertas turas.

(a) (i)

State two observations during Step I. Nyatakan dua pemerhatian pada Langkah I.

– Black solid dissolve – Colourless solution turns black

(ii) Write a balance chemical equation for the reaction that occur in Step I. Tuliskan persamaan kimia seimbang bagi tindak balas yang berlaku dalam Langkah I.

CuO + H2SO4

CuSO4 +H2O (iii) State the type of reaction in the preparation of the salts. Nyatakan jenis tindak balas yang berlaku dalam penyediaan garam.

Neutralisation (b) Why is copper(II) oxide powder added until some of it no longer dissolve in Step I? Mengapakah serbuk kuprum(II) oksida ditambah pada larutan tersebut sehingga ia tidak boleh melarut lagi dalam Langkah I?

To make sure that all sulphuric acid has reacted. (c) What is the purpose of heating in Step III? Apakah tujuan pemanasan dalam Langkah III?

To evaporate the water and copper(II) sulphate solution becomes saturated (d) What is the colour of copper(II) sulphate? Apakah warna kuprum(II) sulfat?

Blue (e) What is the purpose of filtration in

Apakah tujuan penurasan dalam (i) Step II? / Langkah II?

– To remove the excess copper(II) oxide. – To obtain copper(II) sulphate solution as a filtrate

(ii) Step V? / Langkah V?

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(f) Draw the a labelled diagram to show the set-up of apparatus used Step II and Step III. Lukiskan gambar rajah berlabel untuk menunjukkan susunan alat radas yang digunakan dalam Langkah II dan Langkah III. Filter paper

Excess of copper(II) oxide


Heat Copper(II) sulphate solution (g) Can copper powder replace copper(II) oxide in the experiment? Explain your answer. Bolehkah serbuk kuprum digunakan untuk menggantikan kuprum(II) oksida dalam eksperimen ini? Terangkan jawapan anda. Cannot. Copper is less electropositive than hydrogen in the electrochemical series, copper cannot displace hydrogen from the acid. (h) Name other substance that can replace copper(II) oxide to prepare the same salt. Write a balance chemical equation for the reaction that occur. Namakan sebatian lain yang dapat menggantikan kuprum(II) oksida dalam penyediaan garam yang sama. Tuliskan persamaan kimia yang seimbang bagi tindak balas yang berlaku.

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Copper(II) carbonate

Substance / Garam larut :

Balance equation / Persamaan seimbang :

CuCO3 + H2SO4

CuSO4 + H2O + CO2

The diagram below shows the flow chart for the preparation of lead(II) nitrate and lead(II) sulphate through reaction I and II. Rajah di bawah menunjukkan carta aliran bagi penyediaan plumbum(II) nitrat dan plumbum(II) sulfat melalui tindak balas I dan II. Reaction I Tindak balas I

Lead(II) carbonate

Lead(II) nitrate

Plumbum(II) karbonat

(a) (i)

Reaction II Tindak balas II

Plumbum(II) nitrat

Lead(II) sulphate Plumbum(II) sulfat

What is meant by salt? Apakah maksud garam?

Salts are ionic compounds produced when hydrogen ion from acid is replaced with metal ion including ammonium ion.

(ii) Based on the flow chart above, classify the above salt to soluble salt and insoluble salt. Berdasarkan carta aliran di atas, kelaskan garam-garam tersebut kepada garam larut dan garam tak larut.

Soluble salt / Garam larut : Lead(II) nitrate

Insoluble salt / Garam tak larut : Lead(II) carbonate, Lead(II) sulphate

(b) (i)

Describe how lead(II) nitrate solution is obtained in reaction I. Terangkan bagaimana larutan plumbum(II) nitrat diperoleh daripada tindak balas I.

–

Measure Sukat

and pour sebanyak

– Lead(II) carbonate Serbuk

– Stir the

cm3 of 1 mol dm–3 cm asid 3

nitrik

nitric

acid in a beaker.

1 mol dm dan tuangkan ke dalam bikar. -3

is added to the acid in the beaker until

plumbum(II) karbonat ditambahkan kepada asid di dalam bikar sehingga

mixture

Campuran

– The

powder

50 50

mixture

Campuran

.

berlebihan

.

with a glass rod.

tersebut dikacau dengan rod kaca.

in the beaker is filtered. dituraskan.

– The filtrate is lead(II) nitrate larutan

solution

.

plumbum(II) nitrat. n io

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Hasil turasan ialah

excess

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(ii) Write a balanced chemical equation for the reaction that occur. Tuliskan persamaan kimia yang seimbang bagi tindak balas yang berlaku.

(c) (i)

PbCO3 + HNO3

Pb(NO3)2 + H2O + CO2

Describe how to prepare pure and dry lead(II) sulphate in reaction II. Huraikan bagaimana cara menyediakan plumbum(II) sulfat yang tulen dan kering dalam tindak balas II.

–

50 cm3

1 mol dm–3 lead(II) nitrate solution is added to of sodium sulphate solution in a beaker. 50 cm3 larutan plumbum(II) nitrat 1 mol dm–3 ke dalam bikar.

– The

mixture

– The

mixture

Campuran

mol dm–3 ditambahkan kepada

of

50 cm3

1

mol dm–3

larutan natrium sulfat

is stirred with glass rod. tersebut dikacau dengan rod kaca.

Campuran

is filtered. The white precipitate of lead(II) sulphate is collected as the residue. dituraskan. Mendakan putih plumbum(II) sulfat dikumpulkan sebagai baki.

– The precipitate is rinsed with Mendakan tersebut dibilas dengan

– The precipitate is Mendakan tersebut

1

50 cm3

distilled water air suling

. .

pressed

between sheets of

ditekan

antara

kertas turas

filter papers to dry it. .

(ii) Write an ionic equation the reaction that occur. Tuliskan persamaan ion bagi tindak balas yang berlaku.

2+ 2– Pb + SO4

PbSO4

(iii) Name the type of reaction that occur in reaction II. Namakan jenis tindak balas yang berlaku dalam tindak balas II.

Double decomposition reaction

(iv) What is the step taken to make sure that pure lead(II) sulphate in reaction II is pure? Apakah langkah yang diambil untuk memastikan plumbum(II) sulfat dalam tindak balas II tulen?

(d) (i)

The precipitate is rinsed with distilled water. Can lead(II) sulphate be prepared by adding excess of lead(II) nitrate to calcium(II) sulphate followed by filtration. Explain your answer. Bolehkah plumbum(II) sulfat disediakan dengan menambahkan plumbum(II) nitrat berlebihan kepada kalsium(II) sulfat dan diikuti dengan penurasan? Terangkan jawapan anda.

– Cannot. – Calcium sulphate is insoluble salt, it cannot form a solution and there are no free moving ions. – Double decomposition reaction cannot occur.

(ii) Can lead(II) sulphate be prepared by adding excess of lead(II) oxide to sulphuric acid. Explain your answer. Bolehkah plumbum(II) sulfat disediakan dengan menambahkan plumbum(II) oksida berlebihan kepada asid sulfurik? Terangkan jawapan anda.

– Cannot. – Lead(II) sulphate and lead(II) oxide are insoluble, both cannot be separated by filtration.

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– The insoluble lead(II) sulphate will prevent lead(II) oxide to undergo further reaction with sulphuric acid.

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4

The diagram below shows the flow chart for the preparation of zinc carbonate and zinc sulphate through reactions I and II. Rajah di bawah menunjukkan carta aliran bagi penyediaan garam zink karbonat dan zink sulfat melalui tindak balas I dan tindak balas II.

Zinc nitrate

Reaction I Tindak balas I

Zinc carbonate

Zink nitrat


Zinc sulphate

Zink karbonat

Zink sulfat

(a) Based on the flow chart above, classify the above salt to soluble salt and insoluble salt. Berdasarkan carta aliran di atas, kelaskan garam di atas kepada garam larut dan garam tak larut.

Soluble salt / Garam larut : Zinc nitrate, zinc sulphate

Insoluble salt / Garam tak larut : Zinc carbonate

(b) (i)

State the reactant for the preparation of zinc carbonate from zinc nitrate in reaction I. Nyatakan bahan tindak balas untuk penyediaan zink karbonat dalam tindak balas I.

Sodium carbonate solution / potassium carbonate solution / ammonium carbonate solution

(ii) State the type of reaction the occurs in reaction I. Nyatakan jenis tindak balas yang berlaku dalam tindak balas I.

Double decomposition (iii) Describe the preparation zinc carbonate from zinc nitrate in the laboratory through reaction I. Huraikan penyediaan zink karbonat dari zink nitrat melalui tindak balas I.

– 50 cm3 of 1 mol dm–3 zinc nitrate solution is added to 50 cm3 of 1 mol dm–3 sodium carbonate solution in a beaker. – The mixture is stirred with a glass rod and a white solid, ZnCO3 is formed. – The mixture is filtered and the residue is rinsed with distilled water. – The white precipitate is dried by pressing it between filter papers. (iv) Write the chemical equation for the reaction in (b)(iii). Tuliskan persamaan kimia untuk tindak balas yang berlaku dalam (b)(iii).

Zn(NO3)2 + Na2CO3 (c) (i)

ZnCO3 + 2NaNO3

State the reactant for the preparation of zinc sulphate from zinc carbonate in reaction II. Nyatakan bahan tindak balas bagi penyediaan zink sulfat dari zink karbonat dalam tindak balas II.

Sulphuric acid (ii) Describe laboratory experiment to prepare zinc sulphate from zinc carbonate through reaction II. Huraikan eksperimen dalam makmal untuk menyediakan zink sulfat dari zink karbonat melalui tindak balas II. – 50 cm3 of 1 mol dm–3 of sulphuric is measured and poured into acid in a beaker. – The white precipitate from reaction I/ zinc carbonate powder is added to the acid until in excess. – The mixture is stirred with a glass rod. – The excess white precipitate is filter out.

– The filtrate is poured into an evaporating dish.

– The salt solution is gently heated until saturated.

– The hot saturated salt solution is allowed to cool for crystals to form.

– The crystals formed are filtered and dried by pressing it between sheets of filter papers. (iii) Write the chemical equation for the reaction in (c)(ii). Tuliskan persamaan kimia untuk tindak balas yang berlaku dalam (c)(ii).

ZnSO4 + H2O + CO2 n io

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ZnCO3 + H2SO4

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Constructing Ionic Equation for the Formation of Insoluble Salt Membina Persamaan Ion bagi Pembentukan Garam Tak Larut

The ionic equation for the formation of insoluble salt can be constructed if the number of moles of anion and cation to form 1 mol of insoluble salt are known.

1

Persamaan kimia untuk pembentukan garam tak terlarut dapat ditulis jika bilangan mol anion dan kation untuk membentuk 1 mol garam tak larut diketahui.

The number mol of cation and anion which combined to form 1 mol of insoluble salt is determined experimentally by a continuous method:

2

Bilangan mol kation dan anion yang bergabung untuk membentuk 1 mol garam tak terlarut dapat ditentukan secara eksperimen menggunakan kaedah perubahan berterusan:

(a) A fixed volume of a solution A contains cations, Xm+ of the insoluble salt reacts with increasing volume of another solution B contains the anions, Yn– of the insoluble salt. Isi padu tetap larutan A mengandungi kation, X m+ daripada garam tak terlarut bertindak balas dengan isi padu yang meningkat larutan B yang mengandungi anion, Y n– daripada garam tak terlarut.

(b) The volume of solution B needed to completely react with fixed volume of solution A is determined. Isi padu larutan B yang diperlukan untuk bertindak balas dengan isi padu larutan A yang ditetapkan ditentukan.

(c) The number of mol of Xm+ react with Yn– is calculated based on the result of the experiment. Bilangan mol X m+ yang bertindak balas dengan Y n– dihitung berdasarkan keputusan eksperimen.

(d) The simplest ration of mol of Xm+: mol of Yn– is calculated. Nisbah di antara bilangan mol X m+: bilangan mol Y n– dihitung.

(e) Use the ratio to construct ionic equation. Gunakan nisbah tersebut untuk membina persamaan ion.

Example: / Contoh: 5.0 cm³ of 0.5 mol dm–3 copper(II) sulphate solution is poured to 8 test tubes with the same size. Different volume of 1.0 mol dm–3 potassium hydroxide solution are added to each test tube. The test tubes are stoppered and shaken well. The test tubes are left for 30 minutes. The height of precipitate formed in each test tube is measured. The graph below is obtained when the height of precipitate is plotted against the volume of potassium hydroxide solution.

3

5.0 cm³ larutan kuprum(II) sulfat 0.5 mol dm–3 dituang ke dalam setiap 8 tabung uji yang mempunyai saiz yang sama. Larutan kalium hidroksida 1.0 mol dm–3 yang berlainan isi padu ditambah kepada setiap tabung uji. Tabung uji tersebut digoncangkan dan dibiarkan selama 30 minit. Tinggi mendakan yang terbentuk dalam setiap tabung uji diukur. Graf di bawah diperoleh apabila ketinggian mendakan diplot melawan isi padu larutan kalium hidroksida. Height of precipitate (cm) / Tinggi mendakan (cm)

5

0 (a) (i)

Volume of potassium hydroxide /cm3 1

2

3

4

5

6

7

8

9

Isi padu kalium hidroksida /cm3

Name the precipitate formed. Nyatakan nama mendakan yang terbentuk.

Copper(II) hydroxide

(ii) What is the colour of the precipitate? Apakah warna mendakan?

Blue (b) Based on the above graph, what is the volume of potassium hydroxide solution needed to completely react with copper(II) sulphate solution? Berdasarkan graf di atas, apakah isi padu larutan kalium hidroksida yang diperlukan untuk bertindak balas dengan larutan kuprum(II) sulfat secara lengkap?

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(c) (i)

Calculate the number of moles of copper(II) ions in 5.0 cm³ of 0.5 mol dm–3 copper(II) sulphate solution. Hitung bilangan mol ion kuprum(II) dalam 5.0 cm³ larutan kuprum(II) sulfat 0.5 mol dm–3.

CuSO4

Cu2+ + SO42– 5 × 0.5 = 0.0025 mol Mol of CuSO4 = 1 000 From the equation, 1 mol CuSO4 : 1 mol Cu2+ 0.0025 mol CuSO4 : 0.0025 mol Cu2+

(ii) Calculate the number of mol of hydroxide ions needed to react with 5.0 cm³ of 0.5 mol dm–3 copper(II) sulphate solution. Hitung bilangan mol ion hidroksida yang diperlukan untuk bertindak balas dengan 5.0 cm³ larutan kuprum(II) sulfat 0.5 mol dm–3.

KOH

K+ + OH– Mol of KOH = 5 × 1.0 = 0.005 mol 1 000 From the equation, 1 mol KOH : 1 mol OH– 0.005 mol KOH : 0.005 mol OH– (iii) How many moles of hydroxide ions react with one mole of copper(II) ions to form a precipitate? Berapakah bilangan mol ion hidroksida yang bertindak balas dengan satu mol ion kuprum(II) untuk membentuk mendakan?

0.0025 mol Cu2+ : 0.005 mol OH– 1 mol Cu2+ : 2 mol of OH– (d) Calculate the number of mol of hydroxide ions needed to react with 5.0 cm³ of 0.5 mol dm–3 copper (II) sulphate solution. Tuliskan persamaan ion bagi pembentukan mendakan.

Cu2+ + 2OH–

Cu(OH)2

Solving Numerical Problems Involving the Salt Preparation Penghitungan Pelbagai Masalah Melibatkan Penyediaan Garam Mass in gram Jisim dalam gram

÷ (RAM/RMM/RFM) g mol–1 ÷ (JAR/JMR/JFR) g mol–1

Solution concentration in mol dm–3 (M) and volume in cm3 (V) Kepekatan larutan dalam mol dm–3 (M) dan isi padu dalam cm3 (V)

MV n = 1000

× (RAM/RMM/RFM) g mol–1 × (JAR/JMR/JFR) g mol–1 × 24 dm3 mol–1/22.4 dm3 mol–1

Number of mol (n)

Volume of gas in dm3

Bilangan mol (n)

Isi padu gas dalam dm3 ÷ 24 dm mol /22.4 dm mol 3

–1

3

–1

Gas occupies the volume of 24 dm3 at room temperature and 22.4 dm3 at s.t.p (standard temperature and pressure). 1 mol sebarang gas menempati isipadu 24 dm3 pada suhu bilik dan 22.4 dm3 pada s.t.p (suhu dan tekanan piawai). Calculation steps: / Langkah-langkah pengiraan:

S1 Write a balanced equation. L1

Tuliskan persamaan seimbang.

S2 Write the information from the question above the equation. L3

Tuliskan maklumat daripada soalan di atas persamaan tersebut.

S3 Write the information from the chemical equation below the equation (the number of moles of reactants/products). L3

Tuliskan maklumat daripada persamaan kimia di bawah persamaan tersebut (bilangan mol bagi bahan/hasil tindak balas).

S4 Change the information in S2 into moles by using the method shown in the chart below. L4

Tukar maklumat dalam L2 menjadi mol dengan menggunakan kaedah yang ditunjukkan dalam carta di atas.

S5 Use the relationship between number of moles of substance involved in S3 to find the answer. L5

Gunakan perhubungan bilangan mol bahan terlibat dalam L3 untuk mendapatkan jawapan.

S6 Change the information to the unit required using the chart below. Tukar maklumat tersebut kepada unit yang dikehendaki mengikut carta di atas. n io

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EXERCISE / LATIHAN

50 cm3 of 2 mol dm–3 sulphuric acid is added to an excess of copper(II) oxide powder. Calculate the mass of copper(II) sulphate formed in the reaction. [Relative atomic mass: H = 1, O = 16, Cu = 64, S = 32]

1

50 cm3 asid sulfurik 2 mol dm–3 ditambah kepada serbuk kuprum(II) oksida berlebihan. Hitungkan jisim kuprum(II) sulfat yang terbentuk dalam tindak balas itu. [Jisim atom relatif: H = 1, O = 16, Cu = 64, S = 32]

M = 2 mol dm–3 V = 50 cm3 CuO(aq) + H2SO4(aq)

? g CuSO4(ak) + 2H2O(l) 2 × 50 Number of moles of sulpuric acid = = 0.1 mol 1 000 From the equation, 1 mol CuO : 1 mol CuSO4 0.1 mol CuO : 0.1 mol CuSO4 Mass of CuSO4 = 0.1 mol × [64 + 32 + (16 × 4)] g mol–1 = 16 g

27.66 g of lead(II) iodide is precipitated when 2.0 mol dm–3 of aqueous lead(II) nitrate solution is added to an excess of aqueous potassium iodide solution. Calculate the volume of aqueous lead(II) nitrate solution used. [Relative atomic mass: I = 127, Pb = 207]

2

27.66 g plumbum(II) iodida termendak apabila 2.0 mol dm–3 larutan plumbum(II) nitrat akueus ditambahkan kepada larutan kalium iodida akueus berlebihan. Hitungkan isi padu plumbum(II) nitrat yang digunakan. [Jisim atom relatif: I = 127, Pb = 207]

M = 2 mol dm–3 V = ? cm3 Pb(NO3)2(aq) + 2KI(aq)

Mol of PbI2 =

From the equation,

n mol 0.06 mol = = 0.03 dm3 = 30 cm3 Volume of Pb(NO3)2 = M mol dm–3 2 mol dm–3

3

Zinc oxide powder is added to 100 cm3 of 2 mol dm–3 nitric acid to form zinc nitrate. Calculate

25 g PbI2(s) + 2KNO3(aq)

27.66 = 0.06 mol (207 + 2 × 127) 1 mol PbI2 : 1 mol Pb(NO3)2 0.06 mol PbI2 : 0.06 mol Pb(NO3)2

Serbuk zink oksida ditambahkan kepada 100 cm3 asid nitrik 2 mol dm–3 untuk membentuk zink nitrat. Hitungkan

(i) the mass of zinc oxide that has reacted. jisim zink oksida yang bertindak balas.

(ii) the mass of zinc nitrate produced. [Relative atomic mass: H = 1, O = 16, Cl = 35.5, Zn = 65]

m

(i) 2HNO3(aq) + ZnO(s)

Zn(NO3)2(aq) + H2O(l)

(ii) From the equation, 2 mol of HNO3 : 1 mol of Zn(NO3)2 0.2 mol of HNO3 : 0.1 mol of Zn(NO3)2 Mass of Zn(NO3)2 = 0.1 mol × [65 +[14 + (16 × 3)] × 2] g mol–1 = 0.1 × 189 = 18.9 g

100 × 2 = 0.2 mol 1 000 From the equation, 2 mol of HNO3 : 1 mol of ZnO 0.2 mol of HNO3 : 0.1 mol of ZnO Mass of ZnO = 0.1 × [65 + 16] = 8.1 g Number of moles of HNO3 =

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jisim zink nitrat yang terhasil. [Jisim atom relatif: H = 1, O = 16, Cl = 35.5, Zn = 65]

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4

200 cm3 of 1 mol dm–3 barium chloride solution reacts 100 cm3 of 1 mol dm–3 silver nitrate solution. Calculate the mass of precipitate produced. [Relative atomic mass Ag = 108, Cl = 35.5] 200 cm3 larutan barium klorida 1 mol dm–3 bertindak balas dengan 100 cm3 larutan argentum nitrat 1 mol dm–3. Hitungkan jisim mendakan yang terbentuk. [Jisim atom relatif: Ag = 108, Cl = 35.5]

M = 0.1 mol dm–3 V = 100 cm3

2AgNO3 2AgCl + Ba(NO3)2 1 × 200 Mol of barium chloride = = 0.2 mol (excess) 1 000 1 × 100 = 0.1 mol Mol of silver nitrate = 1 000 From the equation, 1 mol of BaCl2 : 2 mol of AgNO3 : 2 mol of AgCl 0.2 mol of BaCl2 (lebih) : 0.1 mol of AgNO3 : 0.1 mol of AgCl Mass of AgCl = 0.1 mol × [108 + 35.5] g mol–1 = 14.35 g

BaCl2

M = 0.2 mol dm–3 V = 100 cm3, ? g

+

Qualitative Analysis of Salts / Analisis Kualitatif Garam 1

Qualitative analysis of a salt is a chemical technique to identify the ions present in a salt. Analisis kualitatif garam ialah suatu teknik dalam kimia yang digunakan untu mengenal pasti ion-ion yang hadir dalam garam.

2

The qualitative analysis consists of the following steps: Analisis kualitatif terdiri daripada langkah-langkah berikut: (a) Observe the physical properties on salt. Perhatikan sifat-sifat fizik garam. (b) The action of heat on salts. Kesan haba ke atas garam. (c) Prepare aqueous solution of salts and conduct confirmatory test for cation and anion present. Sediakan larutan akueus garam dan menjalankan ujian pengesahan untuk kation dan anion yang hadir.

Physical Properties of Salt Sifat-Sifat Fizik Garam 1

Physical properties such as colour and solubility indicate the possibility of the presence of certain cations, anions or metal oxide. Sifat-sifat fizikal seperti warna dan keterlarutan menunjukkan kemungkinan kehadiran kation, anion atau oksida logam tertentu. Pepejal

Larutan akueus

Aqueous

Salts/ Cation/Metal oxide

White Putih

Colourless Tanpa warna

K+, Na+, Ca2+, Mg2+, Al3+, Zn2+, Pb2+, NH4+

Green Hijau

Insoluble Tak larut

CuCO3

Light green Hijau muda

Light Green Hijau muda

Fe2+, contoh: FeSO4, FeCl2, Fe(NO3)2

Blue Biru

Blue Biru

CuSO4, Cu(NO3 )2 dan CuCl2

Brown Perang

Brown Perang

Fe3+

Black Hitam

Insoluble Tak larut

CuO

Yellow when hot, white when cold Kuning apabila panas, putih apabila sejuk

Insoluble Tak larut

ZnO

Brown when hot, yellow when cold Perang apabila panas, kuning apabila sejuk

Insoluble Tak larut

PbO

Garam/Kation/Oksida logam

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Action of Heat on Salt / Kesan Haba ke atas Garam

Some salts decompose when they are heated:

1

Beberapa jenis garam terurai apabila dipanaskan:

Salt

metal oxide

Garam

oksida logam

+

gas gas

Common Gas Identification: / Pengesahan Gas yang biasa:

2

Gas Gas

Observation/ Test Pemerhatian/Ujian

Nitrogen dioxide, NO2 Nitrogen dioksida, NO2

Oxygen,O2 Oksigen,O2

Carbon dioxide, CO2 Karbon dioksida, CO2

Inference Inferens

– Brown gas. Wasap perang. – Place a moist blue litmus paper at the mouth of the boiling tube, blue litmus paper turns red. Letakkan kertas litmus biru lembap pada mulut tabung didih, kertas litmus biru bertukar menjadi merah.

– Nitrogen dioxide gas is produced by heating nitrate salt. Nitrogen dioksida terhasil apabila garam nitrat dipanaskan. – Nitrate ion, NO3– present. Ion nitrat, NO3– hadir.

– Colourless gas. Gas tanpa warna. – Put a glowing wooden splinter near to the mouth of a boiling tube, the glowing wooden splinter is relighted. Dekatkan kayu uji berbara ke mulut tabung didih, kayu uji berbara menyala.

– Oxygen gas is produced by heating nitrate or chlorate(V) salt. Gas oksigen terhasil apabila garam nitrat atau klorat(V) dipanaskan. – Nitrate ion, NO3– present or ClO3– ion present. Ion nitrat, NO3– atau ion ClO3– hadir.

– Colourless gas. Gas tanpa warna. – Pass the gas through lime water, lime water turns chalky. Lalukan gas pada air kapur, air kapur menjadi keruh. – Draw the set-up of apparatus to conduct the test: Lukiskan susunan radas untuk menjalankan ujian:

– Produced by heating carbonate salt. Terhasil apabila garam karbonat dipanaskan. – Carbonate ion, CO3– present. Ion karbonat, CO3– hadir.

Calcium carbonate Heat Lime water

Ammonia, NH3 Ammonia, NH3

– Colourless gas with pungent smell. Gas tanpa warna dengan bau yang sengit. – Place a moist red litmus paper at the mouth of the boiling. tube, red litmus paper turns blue. Letakkan kertas litmus merah lembap pada mulut tabung didih, kertas litmus merah bertukar menjadi biru.

– Produced by heating ammonium salt with alkali. Terhasil apabila garam ammonium dipanaskan dengan alkali. – Ammonium ion NH4+ present. Ion ammonium NH4+ hadir.

Action of heat on nitrate and carbonate salts.

3

Kesan haba ke atas garam nitrat dan garam karbonat. Cation Kation

Nitrate (NO3–) / Nitrat ( NO3–) Decompose to oxygen gas and metal nitrite when heated Terurai kepada gas oksigen dan logam nitrit apabila dipanaskan

K+

m

2KNO2 + O2

2NaNO3

2NaNO2 + O2

White solid White solid Pepejal putih Pepejal putih

Does not decompose when heated Tidak diuraikan apabila dipanaskan

–

–

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Na+

2KNO3

White solid White solid Pepejal putih Pepejal putih

Carbonate (CO32–) / Karbonat (CO32–)

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Decompose to oxygen gas, nitrogen dioxide gas and metal oxide when heated Terurai kepada gas oksigen, gas nitrogen dioksida dan oksida logam apabila dipanaskan Ca2+

Mg2+

Al3+

Zn2+

2Ca(NO3)2

Decompose to carbon dioxide gas and metal oxide when heated Terurai kepada gas karbon dioksida dan oksida logam apabila dipanaskan CaCO3

2CaO + 4NO2 + O2

White solid White pepejal Brown fume Pepejal putih Pepejal putih Wasap perang 2Mg(NO3)2

MgCO3

2MgO + 4NO2 + O2

White solid White pepejal Brown fume Pepejal putih Pepejal putih Wasap perang 4Al(NO3 )3

CaO + CO2

White solid White solid Turn lime water chalky Pepejal putih Pepejal putih Air kapur menjadi keruh MgO + CO2

White solid White solid Turn lime water chalky Pepejal putih Pepejal putih Air kapur menjadi keruh 2Al2 (CO3)3

2Al2O3 + 12NO2 + O2

White solid White solid Turn lime water chalky Pepejal putih Pepejal putih Air kapur menjadi keruh

2Zn(NO3)2

ZnCO3

2ZnO + 4NO2 + O2

ZnO + CO2

White solid Yellow when hot Turn lime water chalky Pepejal white when cold Air kapur menjadi keruh Putih Kuning apabila panas,

White solid Yellow when hot Brown gas Pepejal white when cold Gas perang putih Kuning apabila panas, putih apabila sejuk

2Pb(NO3)2 Pb2+

putih apabila sejuk

2PbO + 4NO2 + O2

PbCO3 PbO + CO2 White solid Brown when hot Turn lime water chalky Pepejal Yellow when cold Air kapur menjadi keruh Putih Perang apabila panas,

White solid Brown when hot Brown fume Pepejal yellow when cold Wasap perang Putih Perang bila panas,

kuning apabila sejuk

kuning apabila sejuk

Cu2+

4

2Al2O3 + 6CO2

White solid White pepejal Brown fume Pepejal putih Pepejal putih Wasap perang

2Cu(NO3)2

2CuO + 4NO2 + O2

CuO + CO2 CuCO3 Green solid Black solid Turn lime water chalky Pepejal hijau Pepejal hitam Air kapur menjadi keruh

Blue solid Black solid Brown fume Pepejal biru Pepejal hitam Wasap perang

Sulphate salts are more stable, they are not easily decompose when heated. Garam sulfat lebih stabil kerana ia tidak terurai dengan mudah apabila dipanaskan.

5

Chloride salts do not decompose except NH4Cl: NH4Cl(s)

Garam klorida tidak terurai kecuali NH4Cl: NH4Cl(p) 6

NH3(g) + HCl(g)

NH3(g) + HCl(g)

Complete the following table: Lengkapkan jadual berikut: Observation Pemerhatian A white salt is heated.

Inference/conclusion Inferens/kesimpulan

Gas

Garam berwarna putih dipanaskan.

– Brown gas is released, the gas turns moist blue litmus paper red. Gas perang dibebaskan, menukar kertas litmus biru lembap kepada merah.

– Residue is yellow when hot and white when cold.

Garam berwarna hijau dipanaskan.

– Colourless gas released, the gas turns lime water chalky. Gelembung gas dibebaskan, ia menukar air kapur menjadi keruh.

– Residue is black

zinc

zinc nitrate

–

zink nitrat

karbon dioksida

– The residue is Baki ialah

kuprum(II)

– The green salt is Garam hijau ialah

Carbonate

dibebaskan. Ion

copper(II)

hadir.

.

.

Carbon dioxide gas released. Gas

hadir.

ion present. zink

oksida. Ion

– The white salt is Garam putih ialah

Zinc

oxide.

zink

Baki ialah

nitrat

dibebaskan. Ion

oxide.

Copper(II)

copper(II) carbonate

hadir.

ion present.

kuprum(II)

oksida. Ion

kuprum(II) karbonat

ion present karbonat

hadir.

.

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Baki berwarna hitam.

gas released. Nitrate ion present.

nitrogen dioksida

– The residue is

Baki berwarna kuning apabila panas dan putih apabila sejuk

A green salt is heated.

Nitrogen dioxide

–

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A white salt is heated.

–


Carbon dioxide Gas

– Colourless gas released, the gas turns lime water chalky.

Baki ialah

Baki berwarna perang apabila panas dan kuning apabila sejuk.

oksida. Ion

zinc zink

Baki ialah

– Residue is yellow when hot and white when cold.

Garam putih ialah

hadir.

. .

zinc

oxide.

ion present.

zink

zinc carbonate zink karbonat

hadir.

ion present.

plumbum(II)

oksida. Ion

– The white salt is

Gelembung gas dibebaskan, ia menukar air kapur menjadi keruh.

Lead(II)

plumbum(II) karbonat

Garam putih ialah

– The residue is

– Colourless gas released, the gas turns lime water chalky.

ion present karbonat

lead(II) carbonate

A white salt is heated. Garam berwarna putih dipanaskan.

oxide.

plumbum(II)

– The white salt is

– Residue is brown when hot and yellow when cold.

Carbonate

dibebaskan. Ion

lead(II)

– The residue is

Gelembung gas dibebaskan, ia menukar air kapur menjadi keruh.

gas released.

karbon dioksida

hadir.

. .

Baki berwarna kuning apabila panas dan putih apabila sejuk.

A blue salt is heated.

– Nitrogen dioxide gas released. Nitrate ion present.

Garam berwarna biru dipanaskan.

Gas nitrogen dioksida dibebaskan. Ion nitrat hadir.

– Brown gas is released, the gas turns moist blue litmus paper red.

– The residue is copper(II) oxide. Copper(II) ion present.

Gas perang terbebas menukar warna kertas limus biru menjadi merah.

– Residue is black.

Baki ialah kuprum(II) oksida. Ion kuprum(II) hadir.

– The blue salt is

Baki berwarna hitam.


Garam biru ialah

copper(II) nitrate

.

kuprum(II) nitrat

.

– Nitrogen dioxide gas released. Nitrate ion present.


Gas nitrogen dioksida dibebaskan. Ion nitrat hadir.

– Brown gas is released, the gas turns moist blue litmus paper red.

– The residue is lead(II) oxide. Lead(II) ion present.

Gas perang terbebas menukar warna kertas limus biru menjadi merah.

– Residue is brown when hot and yellow when cold. Baki berwarna perang apabila panas dan kuning apabila sejuk.


Baki ialah plumbum(II) oksida. Ion plumbum(II) hadir.

– The blue salt is

lead(II) nitrate

plumbum(II) nitrat

Garam putih ialah

– Carbon dioxide gas released.


– Colourles gas released, the gas turns lime water chalky. Gelembung gas dibebaskan, ia menukar air kapur menjadi keruh.

karbon dioksida

Gas

. .

Carbonate dibebaskan. Ion

ion present. karbonat

hadir.

– The possible residue are ZnO/PbO/MgO/Al2O3 Baki yang mungkin adalah CaOl/MgO/Al2O3.

– Residue is white Baki berwarna putih.

– From the above table, action of heat on heat on salt can be used to identify lead(II) nitrate , lead(II) carbonate , zinc nitrate , zinc carbonate , copper(II) nitrate and copper(II) carbonate . Daripada jadual di atas, kesan haba ke atas garam boleh digunakan untuk mengenal garam zink nitrat

,

zink karbonat

,

kuprum(II) nitrat

dan

plumbum(II) nitrat

kuprum(II) karbonat

, plumbum(II) karbonat ,

.

– Confirmatory test for other cations and anions is carried out by Confirmatory Tests for Anions and Cations Ujian pengesahan untuk kation dan anion lain boleh dijalankan dengan menggunakan Ujian Pengesahan Anion dan Kation. Confirmatory Tests for Cations Ujian Pengesahan bagi Kation

Chemical tests is conducted for confirmation of cations in aqueous form.

1

Ujian-ujian kimia dijalankan bagi pengesahan kation dalam bentuk akueus.

Confirmatory test is carried out by adding a small amount of sodium hydroxide solution / ammonia solution followed by excess sodium hydroxide / ammonia solution to the solution contains the cation.

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Ujian pengesahan dijalankan dengan menambah sedikit larutan natrium hidroksida / larutan ammonia diikuti dengan larutan natrium hidroksida / larutan ammonia berlebihan kepada larutan yang mengandungi kation.

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Sodium hydroxide solution Larutan natrium hidroksida

Cations Kation

small amount sedikit

K+ Na+ Ca2+

No change

No change

No change

Tiada perubahan

Tiada perubahan

Tiada perubahan

No change

No change

No change

No change

Tiada perubahan

Tiada perubahan

Tiada perubahan

Tiada perubahan

White precipitate

Insoluble in excess

No change

No change

Mendakan putih

Tak larut dalam berlebihan

Tiada perubahan

Tiada perubahan

Insoluble in excess

White precipitate

Soluble in excess

White precipitate

White precipitate

Soluble in excess

White precipitate

Soluble in excess

Mendakan putih

Larut dalam berlebihan

Mendakan putih


White precipitate

Soluble in excess

White precipitate

Soluble in excess

Mendakan putih


Mendakan putih


Green precipitate

Insoluble in excess

Green precipitate

Soluble in excess

Mendakan hijau


Mendakan hijau


Brown precipitate

Insoluble in excess

Brown precipitate

Soluble in excess

Mendakan perang


Mendakan perang


Blue precipitate

Insoluble in excess

Blue precipitate

Soluble in excess

Mendakan biru


Mendakan biru


Zn2+

White precipitate

Mendakan putih Mendakan putih

Fe2+ Fe3+ Cu2+ NH4+

excess berlebihan

No change

White precipitate

Pb2+

small amount sedikit

excess berlebihan

Tiada perubahan

Mg2+

Al3+

Ammonia solution Larutan ammonia


Mendakan putih


Insoluble in excess


Soluble in excess

Mendakan putih


No change

No change

No change

No change

Tiada perubahan

Tiada perubahan

Tiada perubahan

Tiada perubahan

(a) Reaction with small amount until excess of sodium hydroxide solution: (refer to the above table) Tindak balas dengan larutan natrium hidroksida sedikit demi sedikit sehingga berlebihan: (rujuk jadual di atas) Pungent smell, moist red litmus paper turn to blue Bau sengit, menukarkan kertas litmus merah lembap kepada biru

Solution contains: Larutan mengandungi:

Heat Add a little sodium hydroxide solution

K+, Ca2+, Mg2+, Al , Zn , Pb , 3+

2+

2+

Fe2+, Fe3+, Cu2+, NH4+

NH4+

Tambahkan sedikit larutan natrium hidroksida

K , NH4 +

Panaskan

+

No precipitate Tiada mendakan

No changes

K+

Tiada perubahan

Precipitate formed Mendakan terbentuk

Cu2+ (blue), Fe2+ (green), Fe3+ (brown)

Coloured precipitate Mendakan berwarna Add excess sodium hydroxide solution White precipitate Mendakan putih

Pb2+, Al3+, Zn2+, Ca2+, Mg2+

Soluble Larut

Tambahkan larutan natrium hidroksida berlebihan

Insoluble

Zn2+, Al3+, Pb2+

Ca2+, Mg2+

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(b) Reaction with small amount until excess of ammonia solution: Tindak balas dengan larutan ammonia sedikit demi sedikit sehingga berlebihan:

Solution contains: Larutan mengandungi:

Add a little solution of ammonia Tambah sedikit larutan ammonia

No precipitate

Cu2+ (blue), Fe2+ (green), Fe3+ (brown)

Tiada mendakan

K+, Na+, Ca2+, Mg2+, Al3+, Zn2+, Pb2+, Fe2+, Fe3+, Cu2+

Add excess aqueous ammonia

Ca2+, K+, Na+

Precipitate formed Mendakan terbentuk

Tambahkan larutan ammonia berlebihan

Soluble Larut

Cu2+ Fe2+, Fe3+

Insoluble Coloured precipitate

Add excess aqueous ammonia

Mendakan berwarna

White precipitate Mendakan putih

Pb , Al , Zn2+, Mg2+ 2+

3+

Tambahkan larutan ammonia berlebihan

Tak larut

Soluble Larut

Insoluble

Zn2+ Mg2+, Al3+, Pb3+

Tak larut

(c) Conclusion of the confirmatory test for colourless/white cations: Kesimpulan ujian pengesahan bagi kation tanpa warna/putih:

(i) Zn2+: White precipitqte, soluble in excess of sodium hydroxide and ammonia solution (ii) Mg2+: White precipitate, insoluble in excess of sodium hydroxide and ammonia solution (iii) Al3+: White precipitate, soluble in excess of sodium hydroxide and insoluble in excess ammonia solution (iv) Ca2+: White precipitate insoluble in excess of sodium hydroxide and no precipitate with ammonia solution (v) NH4+: No precipitate with sodium hydroxide solution and pungent smell released when heated (d) Conclusion of the confirmatory test for coloured cations. Kesimpulan untuk ujian pengesahan bagi kation berwarna.

(i) Cu2+: Blue precipitate insoluble in excess of sodium hydroxide solution and soluble in excess ammonia solution (ii) Fe2+: Green precipitate, insoluble in excess of sodium hydroxide and ammonia solution (iii) Fe3+: Brown precipitate, insoluble in excess of sodium hydroxide and ammonia solution (e) All cations can be identified with confirmatory test using sodium hydroxide solution and ammonia solution except Al3+ and Pb2+. Semua kation boleh dikenal pasti dengan ujian pengesahan menggunakan larutan natrium hidroksida dan larutan ammonia kecuali Al3+ dan Pb2+.

(f) To differentiate between Al3+ and Pb2+: Untuk membezakan Al3+ dengan Pb2+:

– Al3+ and Pb2+ are differentiated by double decomposition reaction. An aqueous solution containing SO42–/ Cl–/ I– anion is used to detect the presence of Al3+ and Pb2+. Al3+ dan Pb2+ boleh dibezakan dengan menggunakan tindak balas pernguraian ganda dua. Larutan akueus yang mengandungi anion SO42–/ Cl– / I– digunakan untuk mengesan kehadiran Al3+ dan Pb2+.

– Precipitate is formed when solution containing SO42–/ Cl–/ I– added to Pb2+. Mendakan terbentuk apabila larutan mengandungi SO42–/ Cl–/ I– ditambah kepada Pb2+. – No precipitate when solution containing SO42–/ Cl– / I– added to Al3+.

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Tiada mendakan terbentuk apabila larutan mengandungi SO42–/ Cl–/ I– ditambah keepada Al3+.

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(g) Write the ionic equations for the formation of precipitates: Tuliskan persamaan ion bagi pembentukan mendakan:

Al3+ and Pb2+ Al3+ dan Pb2+

Add sodium sulphate solution

Add potassium iodide solution

Tambahkan larutan natrium sulfat

No changes

Add sodium chloride solution

White precipitate

Tiada perubahan

Mendakan putih

Pb2+ + SO42–

No changes Tiada perubahan

Tambahkan larutan natrium klorida

Pb2+

Al3+

Tambahkan larutan kalium iodida

Yellow precipitate Mendakan kuning

Pb2+

Al3+ Pb2+ + 2I–

PbSO4 No changes

PbI2

White precipitate

Tiada perubahan

Mendakan putih

Al3+

Pb2+ Pb2+ + 2Cl–

PbCl2

Confirmatory tests for Anions Ujian Pengesahan untuk Anion Anion/Anion

Procedure/Prosedur

Remark/Catatan

– 2 cm3 of dilute hydrochloric acid / nitric acid /sulphuric acid is added to 2 cm3 of carbonate salt. 2 cm3 asid nitrik/asid sulfurik cair ditambah kepada 2 cm3 garam karbonat.

– The gas given off is passed through lime water: Gas yang terbebas dilalukan air kapur.

Carbonate ion, CO32–

Draw a labelled diagram to conduct the test:

Ion karbonat, CO32–

Lukiskan gambar rajah berlabel untuk menjalankan ujian:

Acid Carbonate salt

Pembuakan berlaku dan air kapur menjadi keruh.

Inference: / Inferens: The gas is carbon dioxide. Gas tersebut ialah karbon dioksida.

Ionic equation: / Persamaan ion: H2O + CO2 CO32– + 2H+

Lime water

– 2 cm3 of dilute nitric acid is added to 2 cm3 solution of chloride ions followed by 2 cm3 of silver nitrate solution. 2 cm3 asid nitrik cair ditambah kepada 2 cm3 larutan ion klorida diikuti dengan 2 cm3 larutan argentum nitrat.

Chloride ion, Cl–

Observation: / Pemerhatian: Effervescence occurs and lime water turns chalky.

Ion klorida, Cl–

Observation: / Pemerhatian: A white precipitate is formed. Mendakan putih terbentuk.

Inference: / Inferens: The precipitate is silver chloride Mendakan ialah argentum klorida.

Ionic equation: / Persamaan ion: AgCl Ag+ + Cl– – 2 cm3 of dilute hydrochloric / nitric acid is added to 2 cm3 of sulphate solution followed by 2 cm3 of barium chloride solution / barium nitrate solution. Sulphate ion, SO4 Ion sulfat SO4

2–

2–

2 cm3 asid sulfurik asid/asid nitrik cair ditambah kepada 2 cm3 larutan sulfat diikuti dengan 2 cm3 larutan barium klorida/larutan barium nitrat.

Observation: / Pemerhatian: A white precipitate is formed. Mendakan putih terbentuk.

Inference: / Inferens: The precipitate is barium sulphate Mendakan tersebut ialah barium sulfat.

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Ionic equation: / Persamaan ion: BaSO4 Ba2+ + SO42–

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– 2 cm3 of dilute sulphuric acid is added to 2 cm3 solution of nitrate ions followed by 2 cm3 of iron(II) sulphate solution. 2 cm3 larutan ion nitrat ditambah kepada 2 cm3 asid sulfurik cair diikuti dengan 2 cm3 larutan ferum(II) sulfat.

Nitrate ion, NO3

–

Ion nitrat, NO3–

– – –

The mixture is shaken. Campuran digoncang.

The test tube is slanted and held with a test tube holder.

Observation: / Pemerhatian: A brown ring is formed between two layers. Cincin perang terbentuk di antara dua lapisan.

Inference: / Inferens: Nitrate ion present. Ion nitrat hadir.

Tabung uji dicondongkan dan diapit dengan pemegang tabung uji.

A few drops of concentrated H2SO4 acid is dropped along the wall of the test tube and is held upright. Beberapa titis H2SO4 pekat dititiskan melalui dinding tabung uji dan ditegakkan.

EXERCISE / LATIHAN

(a) Substance A is white in colour. When A is strongly heated, a brown gas, B and gas C are released. These gases lighted a glowing wooden splinter. Residue D which is yellow in colour when hot and white when cold is formed.

1

Bahan A berwarna putih. Apabila A dipanaskan dengan kuat, gas berwarna perang B dan gas C dibebaskan. Gas C menyalakan kayu uji berbara. Baki D yang berwarna kuning apabila sejuk dan putih apabila sejuk terbentuk.

(i)

Name substances A, B, C and D. Namakan bahan A, B, C dan D.

A:

Zinc nitrate

B:

Nitrogen dioxide

C:

(ii) Write the chemical equation when substance A is heated. Tuliskan persamaan kimia apabila bahan A dipanaskan. 2Zn(NO3)2 2ZnO + 4NO2 + O2

Oxygen

D:

Zinc oxide

(b) Write the chemical equation when substance E is heated. Larutan tanpa warna E memberi keputusan berikut apabila dijalankan beberapa siri ujian:

S1 – Add sodium hydroxide solution, a white precipitate is formed. The precipitate is soluble in excess sodium hydroxide solution. L1 – Apabila ditambah dengan larutan natrium hidroksida, mendakan putih terbentuk. Mendakan ini larut apabila ditambah natrium hidroksida berlebihan.

S2 – Add ammonia solution, a white precipitate is formed. The precipitate is insoluble in excess ammonia solution. L2 – Apabila ditambah larutan ammonia, mendakan putih terbentuk dan mendakan ini tidak larut dalam larutan ammonia berlebihan.

S3 – Add potassium iodide solution, a yellow precipitate F, is formed. L3 – Apabila ditambah dengan larutan kalium iodida, mendakan kuning F terbentuk.

(i)

What are the possible cations present in substance E as a result of S1 test? Apakah kation-kation yang mungkin hadir dalam bahan E hasil ujian L1?

Pb2+, Al3+ and Zn2+

(ii) What are the possible cations present in solution E as a result from S1 and S2 tests? Apakah kation yang mungkin hadir dalam larutan E hasil ujian L1 dan L2? Pb2+ and Al3+

m

2+ Ion present /Ion hadir : Pb

Ionic equation/Persamaan ion : Pb2+ + 2I–

PbI2

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(iii) What is the ion present in E after S3 test has been done? Write an ionic equation for the formation of substance F. Apakah ion yang disahkan hadir dalam E setelah dilakukan ujian L3? Tulis persamaan ion bagi pembentukan bahan F.

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2

The diagram below shows the flow chart for Test I and Test II on colourless solution P. Rajah di bawah menunjukkan carta aliran bagi ujian I dan ujian II ke atas larutan tanpa warna P. Gas Q with a pungent smell is released and turns moist red litmus paper blue.

Test I Ujian I

Gas Q berbau sengit terbebas dan menukarkan warna kertas litmus merah lembap kepada biru.

Test II

Effervescence occurs and gas S is released

Colourless solution P

Ujian II

Larutan tanpa warna P

Add dilute hydrochloric acid

Pembuakan berlaku dan membebaskan gas S.

Tambah asid hidroklorik cair

(a) Identify gas Q and state its chemical properties. Kenal pasti gas Q dan nyatakan sifat kimia yang ditunjukkan oleh gas Q.

Ammonia, alkaline gas (b) State the reagent used in test I and state how the test is carried out. Nyatakan bahan uji yang digunakan dalam ujian I serta huraikan bagaimana ujian dilakukan.

Add sodium hydroxide solution, heat it. Name gas S and write the ionic equation that occurred in Test II:

(c) (i)

Namakan gas S dan tuliskan persamaan ion bagi tindak balas yang berlaku dalam ujian II:

Gas S/Gas S : Carbon dioxide

+ 2– Ionic equation/Persamaan ion: CO3 + 2H

H2O + CO2

(ii) Explain how you confirmed gas S. Terangkan bagaimana anda mengesahkan gas S.

Pass the gas through lime water, lime water turns chalky.

(iii) Name salt P based on the results of tests I and II. Namakan garam P berdasarkan keputusan ujian I dan II.

Ammonium carbonate 3

The table below shows the colour of five solutions labelled A, B, C, D and E added with small amount until excess of ammonia solution and sodium hydroxide solution. Jadual di bawah menunjukkan warna lima larutan berlabel A, B, C, D dan E yang ditambah dengan larutan natrium hidroksida dan larutan ammonia sedikit demi sedikit sehingga berlebihan. Solution

Colour

Larutan

With sodium hydroxide solution

Warna

A B C

Dengan larutan natrium hidroksida

With ammonia solution Dengan larutan ammonia

Blue

Blue precipitate insoluble in excess

Blue precipitate soluble in excess

Biru

Mendakan biru tidak larut dalam berlebihan

Mendakan biru larut dalam berlebihan

Colourless

White precipitate soluble in excess


Tanpa warna

Mendakan putih larut dalam berlebihan


Light green

Green precipitate

Dirty green precipitate

Hijau muda

Mendakan hijau kotor

Mendakan hijau kotor

D

Colourless


White precipitate insoluble in excess

Tanpa warna


Mendakan putih tidak larut dalam berlebihan

E

Colourless



Tanpa warna



(a) What are the cations present in Apakah kation yang terdapat dalam A: Cu

2+

2+ B: Zn

2+ C: Fe

2+ E: Mg

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(b) State another test to identify C. Nyatakan satu lagi ujian bagi mengenali C.

Add potassium hexacyanoferrate(II) solution, light blue precipitate formed (c) What are the possible cations present in solution D? Apakah kation-kation yang mungkin terdapat dalam larutan D?

Al3+, Pb2+ (d) Describe briefly a test that can differentiate the cations present in solution D. Terangkan secara ringkas satu ujian yang boleh digunakan untuk membezakan kation-kation yang hadir dalam larutan D.

– Add a few drops of potassium iodide / sodium chloride / sodium sulpahte solution into 1 cm3 of solution D. – Yellow/white precipitate formed, lead(II) ion / Pb2+ present – No precipitate, aluminium ion / Al3+ present. You are given lead(II) carbonate, zinc(II) carbonate and copper(II) carbonate. Without using any reagents, describe how you can differentiate the three substances in the laboratory.

4

Anda diberi plumbum(II) karbonat, zink(II) karbonat dan kuprum(II) karbonat. Tanpa menggunakan sebarang bahan uji, terangkan bagaimana anda membezakan ketiga-tiga bahan tersebut di dalam makmal.

•

Heat

strongly

kuat

Panaskan dengan bakinya:

boiling tube

one spatula of each salt in a

satu spatula setiap jenis garam dalam

and observe the residue: tabung didih

dan perhatikan baki-

– If the residue is yellow when hot and white when cold, then zinc oxide is formed. The salt is zinc carbonate . Jika baki berwarna kuning apabila panas dan putih apabila sejuk, maka zink karbonat . adalah

– If the residue is black, then

copper(II) oxide

Jika baki berwarna hitam, maka

zink oksida

terbentuk. Garam tersebut

is formed. The salt is copper(II) carbonate .

kuprum(II) oksida

kuprum(II) karbonat

terbentuk. Garam tersebut adalah

– If the residue is brown when hot and yellow when cold, then lead(II) carbonate . Jika baki berwarna perang apabila panas dan kuning apabila sejuk, maka plumbum(II) karbonat . tersebut adalah

lead(II) oxide

.

formed. The salt is

plumbum(II) oksida

terbentuk. Garam

The diagram below shows the flow chart of changes that took place beginning from solid M. Solid M is a zinc salt. When solid M is heated strongly, it decomposes into solid Q which is yellow when hot and white when cold.

5

Rajah di bawah menunjukkan carta aliran bagi perubahan yang berlaku bermula daripada pepejal M. Pepejal M adalah suatu garam bagi zink. Apabila pepejal M dipanaskan dengan kuat, ia terurai kepada suatu pepejal Q yang berwarna kuning apabila panas dan putih apabila sejuk. Reaction I Tindak balas I

Solid M Pepejal M


Add dilute nitric acid/Tambah asid nitrik cair

Panaskan Heat

Solid Q + carbon dioxide gas Pepejal Q + gas karbon dioksida

Solution S Larutan S

+

Carbon dioxide gas Gas karbon dioksida

+

Water Air

Reaction III + Magnesium Tindak balas III + Magnesium

Zinc metal + Magnesium nitrate solution / Logam zink + Larutan magnesium nitrat

(a) (i)

Berikan satu ujian kimia bagi gas karbon dioksida.

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Passed the gas through lime water, lime water turns chalky

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(ii) Draw a diagram of the apparatus set-up to carry out reaction I. Lukiskan gambar rajah susunan radas untuk menjalankan tindak balas I. Solid M Heat

Lime water

(b) Name solids M and Q. Nyatakan nama pepejal M dan Q.

M : Zinc carbonate

Q: Zinc oxide

(c) State the observations made when excess ammonia solution is added to solution S. Nyatakan pemerhatian yang dibuat apabila larutan ammonia berlebihan ditambahkan kepada larutan S.

White precipitate, soluble in excess of ammonia solution (d) (i)

Write the chemical equation for reaction II. Tuliskan persamaan kimia bagi tindak balas II.

ZnCO3 + 2HNO3

Zn(NO3)2 + H2O + CO2

(ii) For reaction II, calculate the volume of carbon dioxide gas released at room condition if 12.5 g solid M decomposes completely. [Relative atomic mass: C =12, O =16, Zn = 65, 1 mole of gas occupies 24 dm3 at room condition] Bagi tindak balas II, hitungkan isi padu gas karbon dioksida yang dibebaskan pada keadaan bilik, jika 12.5 g pepejal M terurai dengan lengkap. [Jisim atom relatif: C = 12, O = 16, Zn = 65, 1 mol gas menempati 24 dm3 pada suhu bilik]

12.5 = 0.1 mol 125 From the equation, 1 mol M : 1 mol CO2 0.1 mol M : 0.1 mol CO2 Volume of CO2 = 0.1 mol × 24 dm3 mol–1 = 2.4 dm3 Mol of solid M =

(e) Name reaction III. Namakan tindak balas III.

Displacement reaction (f) Describe a chemical test to determine the presence of anion in the magnesium nitrate solution. Huraikan ujian kimia untuk menentukan kehadiran anion dalam larutan magnesium nitrat.

– About Masukkan

2

cm3 of magnesium nitrate solution is poured into a test tube.

2

cm3 larutan magnesium nitrat ke dalam tabung uji.

– 2 cm3 of dilute sulphuric acid is added to the solution followed by 2 cm3 of 2 cm

asid sulfurik

3

shaken

– The mixture is Campuran

digoncang

– The test tube is Tabung uji

cair ditambah kepada larutan diikuti dengan larutan

iron(II) sulphate

ferum(II) sulfat

solution.

.

. .

slanted

and held with a test tube holder.

dicondongkan

dan dipegang dengan pemegang tabung uji.

– A few drops of concentrated sulphuric acid is dropped along the wall of the test tube and is held upright. pekat

Beberapa titis asid sulfurik

– A

dititiskan melalui dinding tabung uji dan ditegakkan.

brown ring is formed between two layers. Gelang perang

– Anion present is

terbentuk antara dua lapisan.

nitrate

ion.

nitrat

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Anion yang hadir adalah ion

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The diagram below shows list of chemical substances.

6

Rajah di bawah menunjukkan senarai bahan-bahan kimia. Hydrochloric acid, 1.0 mol dm–3

Barium chloride solution, 1.0 mol dm–3

Larutan asid hidroklorik, 1.0 mol dm

Larutan barium klorida, 1.0 mol dm–3

Iron(II) sulphate solution, 1.0 mol dm–3

Solid copper(II) oxide

Solid calcium carbonate

Larutan ferum(II) sulfat, 1.0 mol dm–3

Pepejal kuprum(II) oksida

Pepejal kalsium karbonat

–3

(a) (i)

Choose two solutions that can be used to prepare insoluble salts. Pilih dua larutan yang digunakan untuk menyediakan garam tak terlarutkan.

Barium chloride and iron(II) sulpahate

(ii) What is the type of reaction for the preparation of the salt in (a)(i)? Apakah jenis tindak balas bagi penyediaan garam di (a)(i)?

Double decomposition reaction

(iii) Write the ionic equation for the production of the salt in (a)(i). Tulis persamaan ion bagi penghasilan garam di (a)(i).

Ba2+ + SO42–

BaSO4

(iv) Describe how to collect the pure salt produced. Huraikan bagaimana anda mendapatkan pepejal garam tulen yang terhasil.

Filter the mixture and rinse with distilled water (b) State the observations when sodium hydroxide solution is added in small amount until in excess into iron(II) sulphate solution./ Nyatakan pemerhatian anda apabila larutan natrium hidroksida ditambah sedikit sehingga berlebihan kepada larutan ferum(II) sulfat.

Green precipitate formed, insoluble in excess of sodium hydroxide solution (c) (i)

Choose two chemical substances that can react to produce carbon dioxide gas. Pilih dua bahan yang boleh bertindak balas untuk menghasilkan gas karbon dioksida.

Calcium carbonate and hydrochloric acid

(ii) Write a balanced chemical equation for the reaction in (c)(i). Tulis persamaan kimia seimbang bagi tindak balas di (c)(i).

CaCO3 + 2HCl

CaCl2 + H2O + CO2

You are given zinc chloride crystals. Describe how you would conduct a chemical test in the laboratory to identify the ions presence ions in zinc chloride crystals./ Anda diberi hablur zink klorida. Huraikan bagaimana anda boleh menjalankan ujian kimia di

7

dalam makmal untuk mengenal pasti ion-ion yang hadir dalam hablur zink klorida.

Dissolve

–

1 spatula zinc chloride crystals in 10 cm3 of 2

distilled

water.

1 spatula hablur zink klorida di dalam 10 cm3 air suling . 2 Larutan solution is poured in three test tubes./ tersebut dituang ke dalam tiga tabung uji.

Larutkan

– The

– Add a few drops sodium hydroxide solution are added to zinc chloride precipitate soluble in excess of sodium hydroxide solution. natrium hidroksida Tambahkan beberapa titik larutan ke dalam Mendakan putih natrium hidroksida larut dalam larutan

larutan

solution

until excess. A white

zink klorida sehingga

berlebihan

.

berlebihan.

ammonia solution solution are added to another zinc chloride until excess. A white – Add a few drops excess ammonia zinc ions precipitate soluble in of solution. Ions present are . ammonia Tambahkan beberapa titik larutan Mendakan putih larut dalam larutan

ke dalam ammonia

larutan

zink klorida yang lain sehingga ion zink berlebihan. Ion yang hadir adalah

berlebihan

.

.

nitric acid is added to 2 cm3 solution of chloride ions followed by 2 cm3 of silver nitrate – About 2 cm3 of dilute solution. White precipitate formed. Ions present are chloride ions.

m

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asid nitrik 2 cm3 cair ditambahkan kepada 2 cm3 larutan ion klorida diikuti dengan 2 cm3 larutan Mendakan putih terbentuk. Ion yang hadir adalah ion klorida.

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8

The diagram below shows the formation of zinc nitrate and the changes to other compounds. Rajah berikut menunjukkan pembentukan zink nitrat dan perubahannya kepada sebatian lain. Heat

+ Substance X Zinc oxide

+ Bahan X

Zink oksida

Panaskan

Zinc nitrate Zink nitrat

Brown gas Gas perang

+ Potassium carbonate solution/ + Larutan kalium karbonat Precipitate Z + Potassium nitrate Mendakan Z

(a) (i)

Kalium nitrat

Zinc oxide reacts with substance X to form zinc nitrate. State the name of substance X. Zink oksida bertindak balas dengan bahan X untuk membentuk zink nitrat. Namakan sebatian X.

Nitric acid

(ii) Write the chemical equation for the reaction in (a)(i). Tuliskan persamaan kimia untuk tindak balas dalam (a)(i).

(b) (i)

ZnO + HNO3 → Zn(NO3)2 + H2O State the name of the brown gas formed. Namakan gas perang yang terbentuk.

Nitrogen dioxide

(ii) Write the chemical equation for the reaction in (b)(i). Tuliskan persamaan kimia untuk tindak balas dalam (b)(i).

2Zn(NO3)2 → 2ZnO + 2NO2 + O2 (c) When potassium carbonate solution added to zinc nitrate solution, precipitate Z and potassium nitrate formed. Apabila larutan kalium karbonat ditambah kepada larutan zink nitrat, mendakan Z dan kalium nitrat terbentuk.

(i)

State the type of reaction occurs. Namakan jenis tindak balas yang berlaku.

Precipitation

(ii) Write the ionic equation for the formation of compound Z. Tulis persamaan ion untuk pembentukan sebatian Z.

Zn2+ + CO32– → ZnCO3

(iii) State how the precipitate Z separated from potassium nitrate. Nyatakan bagaimana mendakan Z diasingkan daripada kalium nitrat.

Filtration

(d) Excess of zinc nitrate solution is added to 100 cm3 of 1 mol dm–3 potassium carbonate. Calculate the mass of zinc carbonate formed. [Relative atomic mass: Zn = 65, C = 12, O = 16] Larutan zink nitrat berlebihan ditambah kepada 100 cm3 larutan kalium karbonat 1 mol dm–3. Hitungkan jisim zink karbonat yang terbentuk. [Jisim atom relatif: Zn = 65, C = 12, O = 16]

Zn(NO3)2 + K2CO3 → ZnCO3 + 2KNO3 100 Mol of K2CO3 = 1× = 0.1 mol 1 000 From the equation, 1 mol K2CO3 : 1 mol ZnCO3 0.1 mol K2CO3 : 0.1 mol ZnCO3 Mass of ZnCO3 = 0.1 mol × 125 g mol–1 = 12.5 g

(e) Sodium hydroxide solution is added until excess to zinc nitrate solution. State the observation that can be made. Larutan natrium hidroksida ditambah sedikit demi sedikit hingga berlebihan kepada larutan zink nitrat. Nyatakan pemerhatian yang dapat dibuat.

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White precipitate soluble in excess of sodium hydroxide solution.

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Objective Questions / Soalan Objektif Which of the following is a salt?

1

5

Antara berikut, yang manakah adalah garam? A Lead(II) oxide Plumbum(II) oksida B Calcium hydroxide Kalsium hidroksida C Barium sulphate Barium sulfat D Tetrachloromethane Tetraklorometana

Antara tindak balas berikut, yang manakah akan menghasilkan kuprum(II) klorida?

I

Kuprum dan asid hidroklorik Kuprum(II) oksida dan asid hidroklorik

III Copper(II) carbonate and hydrochloric acid Kuprum(II) karbonat dan asid hidroklorik

IV Copper(II) sulphate and sodium chloride Kuprum(II) sulfat dan natrium klorida

Antara garam berikut, yang manakah larut dalam air? A Iron(II) sulphate Ferum(II) sulfat B Silver chloride Argentum klorida C Calcium carbonate Kalsium karbonat D Lead(II) bromide Plumbum(II) bromida

Which of the following salts can be prepared by double decomposition reaction?

3

Antara garam berikut, yang manakah boleh disediakan dengan kaedah pemendakan? A Copper(II) chloride Kuprum(II) klorida B Lead(II) nitrate Plumbum(II) nitrat C Barium sulphate Barium sulfat D Zinc sulphate Zink sulfat

Which pair of substances represented by the following formulae react to produce salt?

4

Antara pasangan bahan tindak balas berikut, yang manakah dapat bertindak balas menghasilkan garam?

I II III IV

HNO3(aq) + NaOH(aq) HCl(aq) + NaCl(aq) H2SO4(aq) + MgSO4(aq) H2CO3(aq) + KOH(aq) A I and II only I dan II sahaja B

I and IV only

C

I, II and IV only

I dan IV sahaja I, II dan IV sahaja

D

I, II, III and IV

m

A

I and II only

B

II and III only

C

III and IV only

D

I, II, III and IV

I dan II sahaja II dan III sahaja III dan IV sahaja I, II, III dan IV

6

If 0.2 mole of calcium carbonate is heated until no further change, what is the mass of calcium oxide produced? [Relative atomic mass of C=12, O=16, Ca=40] Jika 0.2 mol kalsium karbonat dipanaskan sehingga tiada perubahan, berapakah jisim kalsium oksida, CaO yang terhasil? [Jisim atom relatif: C = 12, O = 16, Ca = 40] A 5.6 g B 11.2 g C 16.8 g D 22.4 g

7

The diagram below shows observations when white solid X heated strongly. Rajah di bawah menunjukkan pemerhatian apabila pepejal X dipanaskan dengan kuat. White solid X / Pepejal putih X Heat strongly/Panaskan dengan kuat – Brown gas is released/ Gas perang terbebas – Residue is a solid which is yellow when hot and white when cold/ Baki perang apabila panas dan kuning apabila sejuk.

Which of the following substance is X? Antara berikut, yang manakah adalah bahan X? A Zinc nitrate Zink nitrat B Zinc carbonate Zink karbonat C Lead(II) nitrate Plumbum(II) nitrat D Lead(II) carbonate Plumbum(II) karbonat

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I, II, III dan IV

Copper and hydrochloric acid

II Copper(II) oxide and hydrochloric acid

Which of the following salts is soluble in water?

2

Which of the following reactions will produce copper(II) chloride?

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8

The diagram below shows a series of tests carried out on solution Y. Rajah di bawah menunjukkan satu siri ujian kimia ke atas larutan Y. Solution Larutan

Y

Sodium hydroxide solution

Green precipitate

Larutan natrium hidroksida

Mendakan hijau

10 The diagram below shows the reaction between 20 cm3 of 0.5 moldm–3 of sodium chloride solution is and to 20 cm3 of 1.0 moldm–3 silver to produce silver chloride precipitate and solution X. Rajah di bawah menunjukkan tindak balas antara 20 cm3 larutan natrium klorida 0.5 mol dm–3 dengan 20 cm3 larutan argentum nitrat 1.0 mol dm–3 untuk menghasilkan mendakan argentum klorida dan larutan X. 20 cm3 of 1.0 moldm–3 silver nitrate solution

Dilute nitric acid followed by silver nitrate solution

20 cm3 argentum nitrat 1.0 mol dm–3

Asid nitrik cair diikuti dengan larutan argentum nitrat

White precipitate/Mendakan putih

Which of the following is solution Y? Antara berikut, yang manakah adalah bahan Y? A Iron(II) chloride C Copper(II) chloride Ferum(II) klorida Kuprum(II) klorida B Iron(II) sulphate D Copper(II) carbonate Ferum(II) sulfat Kuprum(II) karbonat

9

The diagram below shows two bottles of aqueous solutions. Rajah di bawah menunjukkan dua botol mengandungi larutan garam aluminium nitrat dan larutan plumbum(II) nitrat.

20 cm3 of 0.5 moldm–3 of sodium chloride solution 20 cm3 larutan natrium klorida 0.5 mol dm–3

Solution X Larutan X

Silver chloride precipitate Mendakan argentum klorida

Which of the following ions are present in the solution X? Antara ion berikut, yang manakah yang hadir dalam larutan X?

Aluminium nitrate solution Larutan aluminium nitrat

Lead(II) nitrate solution Larutan plumbum(II) nitrat

Which of the following substances can be used to differentiate between and aluminium nitrate solution and lead(II) nitrate solution? Antara bahan berikut, yang manakah dapat digunakan untuk membezakan larutan aluminium nitrat dan larutan plumbum(II) nitrat?

A

Sodium hydroxide solution

B

Ammonia solution

C

Potassium chloride solution

D

Barium nitrate solution

I II III IV

Na+ Ag+ NO3– Cl– A I and III only I dan III sahaja B

II and III only

C

I, II and III only

D

I, II, and IV only

II dan III sahaja I, II dan III sahaja I, II dan IV sahaja

Larutan natrium hidroksida Larutan ammonia Larutan kalium klorida

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Larutan barium nitrat

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8

MANUFACTURED SUBSTANCES IN INDUSTRY BAHAN KIMIA DALAM INDUSTRI

• SULPHURIC ACID/ASID SULFURIK – Write an equation for Contact process and Haber process, stating the temperature, pressure and catalyst required. Menulis persamaan untuk Proses Sentuh dan Proses Haber, menyatakan suhu, tekanan dan mangkin yang diperlukan.

• AMMONIA/AMMONIA – List the uses of sulphuric acid and ammonia. Menyenaraikan kegunaan asid sulfurik dan ammonia.

– Explain how sulphur dioxide causes environmental pollution. Menerangkan bagaimana sulfur dioksida menyebabkan pencemaran alam.

• – – – –

ALLOY/ALOI State the meaning of an alloy. / Menyatakan maksud aloi. Draw the arrangement of atoms in metals and alloys. / Melukis susunan atom di dalam aloi dan logam. Explain why an alloy is stronger than its pure metal. / Menerangkan mengapa aloi lebih kuat daripada logam tulennya. Design an experiment to investigate the hardness of a material and its alloy. Mereka bentuk eksperimen untuk mengkaji kekerasan aloi dan logam tulennya.

– List the examples of alloys, compositions and properties of alloys. / Menyenaraikan contoh aloi, komposisi dan sifat aloi. – Relate properties of alloys to their uses. / Mengaitkan sifat aloi dengan kegunaannya.

• – – – –

POLYMERS/POLIMER Sate the meaning of polymers. / Menyatakan maksud polimer. List naturally occurring polymers and synthetic polymers. / Menyenaraikan polimer semula jadi dan polimer sintetik. State the uses of synthetic polymers. / Menyatakan kegunaan polimer sintetik. Explain the effect of environmental pollution caused by the disposal of synthetic polymers. Menghuraikan kesan pembuangan polimer sintetik ke atas pencemaran alam sekitar.

– Ways to reduce pollution caused by synthetic polymers. / Cara-cara mengurangkan pencemaran yang disebabkan polimer sintetik.

• – – –

GLASS AND CERAMICS/KACA DAN SERAMIK List uses of glass and ceramics. / Menyenaraikan kegunaan kaca dan seramik. List types of glass and their properties. / Menyenaraikan jenis-jenis kaca dan kegunaannya. State properties of ceramics. / Menyenaraikan sifat-sifat seramik.

• COMPOSITE MATERIALS/BAHAN KOMPOSIT – State the meaning of composite materials. / Menyatakan maksud bahan komposit. – List examples of composite materials and their components and uses. Menyenaraikan contoh-contoh bahan komposit dan komponen dan kegunaannya.

– Compare and contrast properties of composite materials with those of their original component Membanding dan membezakan sifat bahan komposit dengan bahan asalnya.

– Design an experiment to produce composite materials.

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Mereka bentuk eksperimen untuk menghasilkan bahan komposit.

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Sulphuric Acid / Asid Sulfurik 1

Sulfuric acid is manufactured through the Contact Process. This process consists of three stages. Asid sulfurik dihasilkan melalui Proses Sentuh. Proses ini terdiri daripada tiga peringkat. Sulphur Sulfur

Sulphur dioxide SO2

Sulphur trioxide SO3

Oxygen

Sulfur dioksida SO2

Sulfur trioksida SO3

Oleum H 2 S2 O7

Asid sulfurik H2SO4

Oleum H2 S2 O7

Oksigen

Stage I/Peringkat I

Sulphuric acid H2SO4

Stage II/Peringkat II

Stage III/Peringkat III Concentrated sulphuric acid Asid sulfurik pekat

Waste gas Gas terbuang

Molten sulphur Sulfur lebur

SO3

Dry air Udara kering

Burner Pembakar

SO2 + O2

H2S2O7 (Oleum)

Catalytic converter Bekas mangkin

H2S2O7 (Oleum)

Water/Air H2SO4

Stage I/Peringkat I 2

Stage II/Peringkat II

Stage III/Peringkat III

Based on the above diagram, explain each stage and state the conditions required. Include all the balanced chemical equations involve in each stage. Berdasarkan rajah di atas, terangkan setiap peringkat serta keadaan yang diperlukan. Sertakan semua persamaan kimia yang seimbang yang terlibat dalam setiap peringkat. Stage

Explanation/Equation

Peringkat

Stage I: / Peringkat I: Production of sulphur dioxide Penghasilan

sulfur dioksida

Stage II: / Peringkat II: Production of sulphur trioxide Penghasilan

sulfur trioksida

Penerangan/Persamaan kimia

– Molten sulphur is burnt in dry air to produce sulphur dioxide. Sulfur lebur dibakar dalam udara kering untuk menghasilkan sulfur dioksida. Balanced equation: / Persamaan seimbang:

S + O2

SO2

– In a converter, sulphur dioxide and excess oxygen are passed through vanadium(V) oxide . Di dalam bekas mangkin, sulfur dioksida dan oksigen dialirkan melalui Balanced equation: / Persamaan seimbang:

2SO2 + O2

vanadium(V) oksida

.

2SO3

– Optimum conditions for maximum amount of product are: Keadaan optimum untuk penghasilan sulfur trioksida yang maksimum adalah:

Temperature / Suhu:

450 – 500 °C 2 – 3 atm

Catalyst / Mangkin:

vanadium(V) oxide, V2O5

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Pressure / Tekanan:

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Stage III: / Peringkat III: sulphuric acid Production of Penghasilan

–

Sulphur trioxide Sulfur trioksida

asid sulfurik

is dissolved in concentrated sulphuric acid to form oleum.

Balanced equation: / Persamaan seimbang:

SO3 + H2SO4

Oleum

–

Oleum

.

H2S2O7

is diluted in water to produce concentrated sulphuric acid. asid sulfurik pekat

dilarutkan dalam air untuk menghasilkan

Balanced equation: / Persamaan seimbang:

H2O + H2S2O7

*

oleum

dilarutkan dalam asid sulfurik pekat untuk menghasilkan

.

2H2SO4

Note that directly dissolving SO3 in water is impractical due to the highly exothermic nature of the reaction. Acidic vapour or mists are formed instead of a liquid.

Melarutkan sulfur dioksida dalam air secara terus tidak dapat dilakukan kerana pembebasan haba yang sangat banyak. Ini kerana tindak balas tersebut adalah eksotermik. Asid yang terhasil adalah dalam bentuk wap air dan bukannya cecair.

State five main uses of sulphuric acid.

3

Nyatakan lima kegunaan utama asid sulfurik.

(i) To manufacture detergents

(iv) As electrolyte in car batteries

(ii) To manufacture fertilizers

(v) To manufacture synthtetic fibers

(iii) To manufacture paints

Sulphur dioxide and environmental pollution:

4

Sulfur dioksida dan pencemaran alam:

(a) Major sources of sulphur dioxide in the air is combustion of fuel in power station or factories. Punca utama kehadiran sulfur dioksida di udara adalah pembakaran bahan bakar di stesen janakuasa dan kilang.

(b) Sulphur dioxide dissolve in rainwater to form sulphurous acid which will cause acid rain, balanced equation: Sulfur dioksida larut dalam air hujan untuk membentuk asid sulfurus yang menghasilkan hujan asid, persamaan seimbang:

SO2 + H2O

H2SO3

Oxidation of sulphurous acid in the air will produce sulphuric acid which will also cause acid rain. Pengoksidaan asid sulfurus di udara akan menghasilkan asid sulfurik yang juga merupakan penyebab kepada hujan asid.

(c) Effect of acid rain: Kesan hujan asid:

corrodes building, monuments and statues made from marble (calcium carbonate) because – Acid rain calcium carbonate react with acid to produce salt, water and carbon dioxide, balanced equation: mengkakis Hujan asid bangunan, monumen dan tugu yang diperbuat daripada marmar (kalsium karbonat) kerana kalsium karbonat bertindak balas dengan asid menghasilkan garam, air dan karbon dioksida, persamaan seimbang:

CaCO3 + H2SO4

CaSO4 + H2O + CO2

corrodes structures of the buildings or bridges which are made from – Acid rain iron rusts faster with the presence of sulphuric acid.

metal

. The

mengkakis Hujan asid struktur bangunan-bangunan dan jambatan-jambatan yang diperbuat daripada logam. Besi berkarat lebih cepat dengan kehadiran asid sulfurik.

– Acid rain Hujan asid

– Acid rain Hujan asid

increases

the acidity of lakes and river that causes aquatic organism to die.

meningkatkan

increases

keasidan tasik-tasik dan sungai-sungai yang menyebabkan kematian hidupan akuatik.

the acidity of soil. Acidic soil is not suitable for the growth of plants.

meningkatkan

keasidan tanah. Tanah yang berasid tidak sesuai untuk pertumbuhan tanam-tanaman.

(d) Ways to reduce production of sulphur dioxide and effect of acid rain: Cara-cara mengurangkan penghasilan sulfur dioksida dan kesan-kesan hujan asid:

– Gas released from power station and factories are sprayed with powdered limestone ( calcium carbonate ). Gas yang dilepaskan dari stesen janakuasa dan kilang boleh disembur dengan serbuk batu kapur (

– Add lime (

calcium oxide

m

).

) and limestone ( calcium carbonate ) to the lake or river.

kalsium oksida

) dan batu kapur (

kalsium karbonat

) ke tasik atau sungai.

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Menambahkan kapur (

kalsium karbonat

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Ammonia / Ammonia 1

In industry, ammonia is manufactured through the Haber Process: Dalam industri, ammonia dihasilkan melalui Proses Haber.

N2 + 3H2

Balanced equation of reaction / Persamaan seimbang tindak balas: Ferum

400 – 500°C

Catalyst / Mangkin : Temperature / Suhu : Pressure / Tekanan : 2

2NH3

200 atm

Ammonia is used in the manufacture of: Ammonia digunakan dalam pembuatan:

(a) Synthetic fertilizer such as ammonium sulphate, ammonium nitrate, ammonium phosphate and urea Baja sintetik seperti ammonium sulfat, ammonium nitrat, ammonium fosfat dan urea.

(b) Nitric acid in Ostwald Process. Asid nitrik dalam Proses Ostwald.

(c) Synthetic fiber and nylon. Gentian kaca sintetik dan nilon.

(d) Liquid form of ammonia is used as cooling agent in refrigerators. Cecair ammonia digunakan sebagai penyejuk dalam peti sejuk.

(e) Prevent coagulation of latex. Mencegah penggumpalan lateks. 3

Ammonia is a colourless gas with pungent smell and very soluble in water. Ammonia adalah gas yang tidak berwarna dengan bau yang sengit dan sangat larut di dalam air.

4

Chemical properties of ammonia: Sifat-sifat kimia ammonia: Property Sifat

Dissolve in water to form weak alkali Larut di dalam air membentuk alkali lemah

Effect on moist red litmus paper Kesan ke atas kertas litmus merah

Neutralise any acid to form ammonium salt Meneutralkan asid untuk membentuk garam ammonium

Chemical equation / Observation Persamaan kimia / Pemerhatian

NH3(g) + H2O(ce)

NH4+(ak) + OH –(ak)

The presence of hydroxide ions causes aqueous solution of ammonia to become alkaline. Kehadiran ion hidroksida menyebabkan larutan ammonia akueus menjadi alkali.

Turn moist red litmus paper to blue Ammonia reacts with sulphuric acid to form ammonium sulphate salt. Ammonia bertindak balas dengan asid sulfurik untuk membentuk garam ammonium sulfat. Balanced equation: / Persamaan seimbang:

2NH3 + H2SO4

(NH4)2SO4

Alloy / Aloi 1

Complete the following table: Lengkapkan jadual di bawah: Questions Soalan

1 What is the meaning of alloy? Apakah maksud aloi?

Facts / Elaboration / Drawing Fakta / Penerangan / Lukisan

mixture elements of two or more with a certain Alloy is a fixed/specific composition. The major component in the mixture is a metal. tetap

.

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campuran unsur dua atau lebih dengan komposisi yang Aloi ialah logam Komponen utama dalam campuran tersebut ialah .

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2 Relate the arrangement of atoms in pure metals to their ductile and malleable properties. Nyatakan hubungan antara susunan atom dalam logam tulen dengan sifat mulur dan mudah ditempa.

Force/Daya

Pure metals/Logam tulen atoms

.

Pure metal is made up of one type of

atom

Logam tulen terbentuk daripada satu jenis

Atoms in pure metals are all the same The same

size

layers saiz

Atom-atom dalam logam tulen mempunyai

.

. yang sama.

atoms are orderly arranged in layers. saiz

Atom-atom yang mempunyai

lapisan

yang sama ini tersusun dalam

force is applied to the pure metal, layers of atoms When easily over one another. daya Apabila sama lain.

3 Draw the arrangement of atoms in Lukiskan susunan atom dalam

(a) (b)

dikenakan ke atas logam tulen, lapisan atom

menggelongsor

.

slide

di antara satu

(b) Steel / Keluli

(a) Bronze / Gangsa

Bronze (90% copper and 10% tin) Gangsa (90% kuprum dan 10% timah)

Carbon

Steel (99% iron and 1% of carbon) Keluli (99% besi dan 1% karbon)

[Relative atomic mass: Cu = 64, Sn = 119, Fe = 56; C = 12] [Jisim atom relatif: Cu = 64, Sn = 119, Fe = 56, C = 12]

4 Explain why an alloy is stronger than its pure metal in terms of the arrangement of atoms in metals and alloys. Terangkan mengapa aloi lebih kuat daripada logam tulen dari segi susunan atom dalam logam dan aloi.

Iron

Copper

Tin

Atoms of other element added to the pure metal to make an alloy are different in size. Atom-atom unsur lain yang ditambah dalam logam tulen membentuk aloi yang terdiri daripada atom-atom berlainan saiz. yang

These atoms Atom-atom ini

disrupts

the orderly arrangement of atoms in pure metal.

mengganggu

susunan atom yang teratur dalam logam tulen.

force is applied to an alloy, the presence of added other atoms When prevent layers of atoms from sliding . daya dikenakan ke atas aloi, kehadiran atom-atom asing ini Apabila menggelongsor atom-atom ini daripada .

5 State three reason why pure metals are alloyed before used. Nyatakan tiga sebab mengapa logam tulen dialoikan sebelum digunakan.

strength

(a) To increase the

kekuatan

Meningkatkan

and dan

(b) To increase the resistance to Mencegah

kakisan

(c) To improve the Membaiki

lapisan

hardness of pure metals.

kekerasan

corrosion

logam tulen.

of a pure metals.

logam tulen.

appearance

rupa

menghalang

of a pure metal.

logam tulen.

Experiment to compare the hardness of brass and pure copper.

5

Eksperimen untuk membandingkan kekerasan loyang dengan kuprum tulen.

(a) Hypothesis: / Hipotesis: Brass is harder than copper (b) Manipulated variable: / Pemboleh ubah dimanipulasi: Copper and brass block (c) Responding variable: / Pemboleh ubah bergerak balas: Hardness of the copper and brass block (d) Fixed variable: / Pemboleh ubah dimalarkan:

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(e) Apparatus: / Alat radas: Retort stand and clamp, 1 kg weight, string, metre ruler.

Materials: / Bahan-bahan: Steel ball, copper block, brass block

(f) Procedure: / Prosedur: 1. A steel ball bearing is tapped onto a copper block. Set-up of the apparatus: / Susunan alat radas:

Satu bola keluli dilekatkan di atas sebuah bongkah kuprum.

2. A 1 kg weight is hung at a height of 50 cm above the copper block as shown in the diagram. Sebiji pemberat 1 kg digantung setinggi 50 cm di atas bongkah kuprum seperti yang ditunjukkan.

Retort stand

3. Drop the 1 kg weight on the steel ball. Pemberat 1 kg dijatuhkan ke atas bebola keluli.

String

4. Measure the diameter of the dent formed on the copper block with a ruler.

1 kg weight

Diameter lekuk yang terbentuk di atas bongkah kuprum diukur dengan pembaris.

5. Repeat the experiment three times on the other part of the copper block.

Steel ball

Eksperimen diulang tiga kali, pada ruang berbeza pada bongkah kuprum yang sama.

Cellophane tape

6. Steps 1 to 5 are repeated using a brass block to replace the copper block.

Copper block

Langkah 1 hingga 5 diulang dengan menggunakan bongkah loyang, menggantikan bongkah kuprum.

(g) Results: / Keputusan: Experiment

Average diameter/cm

Eksperimen

1

2

3

Diameter of dent on copper block/cm

a

b

c

a + b + c = x 3

Diameter of dent on brass block/cm

d

e

f

d + e + f = y 3

Diameter purata / cm

(h) Discussion: / Perbincangan: The average diameter of dent on copper, x is larger than the average diameter of dent on brass, y. (i) Conclusion: / Kesimpulan:

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Brass is harder than copper// alloy is harder than pure metal.

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Flow chart shows the composition, properties and uses of some alloys. Carta aliran di bawah menunjukkan komposisi, sifat-sifat dan kegunaan aloi-aloi. ALLOY / ALOI Major component / Komponen utama

COPPER / KUPRUM

IRON / FERUM Type of alloy/Jenis aloi

Type of alloy/Jenis aloi

BRONZE/GANGSA (90% Cu, 10% Sn) – Hard and strong,

does not corrode, (shiny surface)

–

BRASS/LOYANG (70% Cu, 30% Zn) – Hard and strong.

STEEL/KELULI (99% Fe, 1% C) – Hard and strong.

Keras dan kuat. Uses: / Kegunaan:

–

Keras dan kuat. Tidak berkarat (permukaan bersinar) Uses: / Kegunaan:

–

Musical instrument and Kitchenware

KELULI TAHAN KARAT

(74% Fe, 8% C, 18% Cr) – Shiny, strong and does

Keras dan kuat. Uses: / Kegunaan:

not rust

Construction of building and bridge and railway tracks

Alat muzik dan perkakas dapur

Building statue or monuments, medal, swords and artistic materials

STAINLESS STEEL

–

Bersinar, kuat dan tidak berkarat. Uses: / Kegunaan:

Making cutlery and surgical instrument

Pembinaan bangunan dan jambatan serta landasan keretapi.

Membuat sudu, garpu dan alat-alat pembedahan.

Pembuatan tugu atau monumen pingat, pedang dan bahan hiasan

ALUMINIUM / ALUMINIUM Type of alloy Jenis aloi

CUPRONICKEL KUPRONIKEL

(75% Cu, 25% Ni) – Shiny, hard and does

–

TIN / TIMAH Type of alloy Jenis aloi

DURALUMIN DURALUMIN

not corrode

(93% Al, 3% Cu & 1% Mn) – Light and strong

Making coins

– Uses: / Kegunaan: Building body of aeroplane and bullet train.

Bersinar, keras dan tidak berkarat. Uses: / Kegunaan: Membuat duit syiling

Ringan dan kuat.

Membuat rangka kapal terbang dan keretapi laju.

PEWTER / PEWTER (96% Sn, 3% Cu, 1% Sb) – Luster, shiny and strong

–

Berkilau, bersinar dan kuat. Uses: / Kegunaan:

Making souvenirs. Membuat cenderamata.

SYNTHETIC POLYMERS / POLIMER SINTETIK Polymer is a long chain molecules made up of a monomer.

1

large

number of small repeating

Polimer ialah molekul berantai panjang yang terbentuk daripada gabungan monomer. 2

Monomer is small identical

repeating

Monomer adalah unit kecil yang

berulang

banyak

unit kecil yang

identical sama

unit of dipanggil

units in the polymer. dalam polimer.

Polymers can be naturally occurring or synthetic.

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Polimer boleh didapati secara semula jadi atau sintetik.

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Example of naturally occurring polymers and their monomers are: Contoh polimer semula jadi dan monomernya:

5

Synthetic Polymer / Polimer

Monomer / Monomer

Protein / Protein

Amino Acid / Asid amino

Starch / Kanji

Glucose / Glukosa

Rubber / Getah

Isoprene / Isoprena

Synthetic polymers are made polymers. The monomers are usually obtained from petroleum after refining and cracking process. Polimer sintetik adalah polimer buatan. Monomer biasanya adalah daripada petroleum yang telah mengalami penyulingan dan peretakan.

6

Example of synthetic polymers, their monomers and uses: Contoh polimer sintetik, monomernya dan kegunaannya: Synthetic polymer

Monomer

Polimer sintetik

Example of uses

Monomer

Contoh kegunaan

Ethene, C2H4

Polythene Politena

Polypropene Polipropena

Plastic bags, shopping bags, plastic containers and plastic toys

Etena, C2H4

Beg plastik, beg membeli belah, bekas plastik dan permainan plastik

Propene, C3H6

Plastic bottles, plastic tables and chairs, car batteries casing and ropes Botol plastik, meja dan kerusi plastik, bekas bateri kereta dan tali

Propena, C3H6

Waterproof materials such as rain clothes, bags, shoes, artificial leather. Polyvinylchloride (PVC) Polivinil klorida (PVC)

Bahan kalis air seperti baju hujan, beg, kasut dan kulit tiruan.

Chloroethene, C2H3Cl

Insulation for electric wiring. Bahan penebat pendawaian wayar elektrik.

Kloroetena, C2H3Cl

Making water pipes because it does not rust. Paip air sebab ia tidak berkarat.

Styrene, C2H3C6H5

Polystyrene Polistirena

Packaging materials, disposable cups and plates

Stirena, C2H3C6H5

Bahan pembungkus, cawan dan pinggan pakai buang.

Perspex

Methylmetacrylate

Safety glass, car lamps and lens

Perspeks

Metil metakrilat

Kaca keselamatan, lampu kereta dan kanta

Hexane-1, 6-diol Terylene (polyester) Terilena (poliester)

Heksana-1, 6-diol

Clothing, sails, sleeping bags, ropes and fishing net

Benzene-1, 4-dicarboxylic acid

Pakaian, kain layar, tali dan jala

Benzena-1, 4-dikarboksilik asid

7

joining

Polymerisation is the process of Pempolimeran ialah proses

penggabungan

together the large number of monomers to form a polymer. monomer-monomer untuk membentuk polimer.

Example: / Contoh: (a) Polymerisation of ethene: Pempolimeran etena:

n

H C H

H = C H

H – C – H n, n is large number up to a few thousands

Polythene

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Ethene / Etena

H – C H

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(b) Polymerisation of propene:

(c) Polymerisation of chloroethene:

Pempolimeran propena:

n

H C H

Pempolimeran kloroetena:

CH3 = C H

H – C H

Propene / Propena

CH3 – C – H n

Polypropene

n

H C H

H = C Cl

H – C H

Chloroethene / Kloroetena

Polyvinylchloride

Complete the following table related to issues of the use of polymers in everyday life.

8

H – C – Cl n

Lengkapkan jadual di bawah berkaitan isu penggunaan polimer sintetik dalam kehidupan seharian. Advantages of synthetic polymers

Environmental pollution from synthetic polymers Pencemaran alam sekitar dari penggunaan polimer sintetik

Kebaikan polimer sintetik

(a) Very stable and do not corrode . Sangat stabil dan tidak berkarat .

(b) Inert to chemical reaction.

(a) Disposal of synthetic polymers such as plastic bottles and containers cause blockage of drainage systems and river thus causing flash floods . Pembuangan polimer sintetik seperti botol plastik dan bekas tersekat yang menyebabkan sistem saliran dan sungai banjir kilat mengakibatkan .

Reducing pollution of synthetic polymers

Pengurangan pencemaran dari polimer sintetik

recycle and (a) Reduce, reuse the synthetic polymers. Mengurangkan, mengitar semula dan mengguna semula polimer sintetik.

biodegradable Lengai terhadap tindak balas (b) Open burning of polymers will release acidic and poisonous (b) Using polimer. kimia gas that will cause air pollution: . Menggunakan polimer Pembakaran polimer sintetik secara terbuka membebaskan gas strong . (c) Light and berasid dan beracun yang menyebabkan pencemaran udara: terbiodegradasi . kuat . Ringan dan – Burning most of the synthetic polymers will produce: Pembakaran kebanyakan polimer sintetik menghasilkan: (c) On-going research to produce (d) Cheap.

cheap biodegradable polymers. (i) carbon dioxide gas which cause green house effect . Penyelidikan berterusan kesan rumah hijau . karbon dioksida yang menyebabkan untuk menghasilkan polimer (ii) carbon monoxide which is poisonous . terbiodegradasi yang murah. beracun . karbon monoksida yang (d) Disintegrate plastics by – Burning of PVC will release hydrogen chloride gas which pyrolysis : Plastic can will cause acid rain . be disintegrated by heating at Pembakaran PVC membebaskan gas hidrogen klorida yang temperature between hujan asid . menyebabkan 400 – 800°C without oxygen. – Burning of synthetic polymers contains carbon and Penguraian plastik secara pirolisis : Plastik boleh diuraikan nitrogen such as nylon will produce highly poisonous dengan pemanasan pada suhu gas such as hydrogen cynide .

Murah.

shaped (e) Easily and coloured.

dibentuk Mudah dan diwarnakan.

Pembakaran polimer sintetik mengandungi karbon dan nitrogen seperti nilon membebaskan gas sangat beracun seperti hidrogen sianida .

antara 400 – 800 °C tanpa oksigen.

(c) Plastic containers that are left in open area collect rain water will become breeding ground for mosquito which will cause diseases such as dengue fever. Bekas plastik yang ditinggalkan di tempat terbuka menakung air nyamuk yang menyebabkan hujan menjadi tempat pembiakan penyebaran penyakit seperti demam denggi.

Glass / Kaca 1 Name the element which forms the major component of glass. Namakan unsur yang membentuk komponen utama kaca.

2 List the property of glass. Senaraikan sifat-sifat kaca.

Silicon dioxide

, SiO2 which exist naturally in

sand

Silikon dioksida

, SiO2 yang boleh didapati secara semula jadi di dalam

. pasir

.

Properties: / Sifat-sifat: Transparent, hard but brittle, non-porous, heat insulator, electric insulator, resistant to

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chemical, easy to clean, can withstand compression

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Complete the table below. Lengkapkan jadual di bawah. Types of glass Jenis kaca

Soda-lime glass Kaca soda kapur

Composition Komposisi Silicon dioxide, sodium carbonate or calcium calcium carbonate

Special properties Sifat istimewa

Borosilicate glass Kaca borosilikat

Silikon dioksida, boron dioksida, natrium oksida, aluminium oksida

durability kimia

Tahan kakisan bahan

High

–

tinggi

Pekali pengembangan haba haba Tidak tahan .

chemical

– Good

Low

mirrors, glass containers

.

kimia

.

Making cookware and laboratory

thermal expansion. rendah

Pekali pengembangan haba

heat

haba

.

glassware such as boiling tube and

when heated to

– Resistant to high temperature. Tahan tinggi.

Making flat glass, electrical bulbs,

durability

Tahan kakisan bahan

–

.

termal expansion but does not heat .

withstand

Silikon dioksida, natrium karbonat, kalsium karbonat

Silicion dioxide, boron dioxide, sodium oxide, aluminum oxide

chemical

– Good

Uses Kegunaan

beakers.

apabila dipanaskan pada suhu

– Optically transparent. Lut sinar.

chemical

– Good Fused glass Kaca silika terlakur

durability kimia

Tahan kakisan bahan

Silicon dioxide

– Low thermal expansion

Silikon dioksida

Pekali pengembangan haba

.

Laboratory glassware, lenses,

rendah

telescope mirrors, optical fibres.

.

high temperature – Can be heated to and resistance to thermal shock. tinggi Boleh dipanaskan pada suhu yang tahan terhadap pertukaran suhu yang cepat.

Lead glass Kaca plumbum

Silicon dioxide, sodium oxide, lead(II) oxide Silikon dioksida, natrium oksida, plumbum(II) oksida

– High

refractive index and

Indeks

biasan

Glittering

–

Kelihatan

dan

density

ketumpatan

appearance.

berkilat

,

.

yang tinggi

Tableware, crystal glass ware and decorative glassware.

.

Ceramics / Seramik 1

Name the elements found in ceramic. Namakan unsur-unsur yang terkandung dalam seramik.

Aluminium, silicon, oxygen and hydrogen 2

Ceramics are made from clay. Name the main component of clay. Seramik dibuat daripada tanah liat. Namakan komponen utama tanah liat.

which is rich in

hydrated aluminium silicate

Kaolin

yang mengandungi

aluminium silikat terhidrat

, Al2O32SiO2.2H2O. , Al2O32SiO2.2H2O.

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Complete the following table for the properties and uses of ceramic.

3

Lengkapkan jadual berikut untuk menunjukkan sifat-sifat dan kegunaan seramik. Property/Sifat

Uses/Kegunaan

Hard and strong.

Building materials such as

Keras dan kuat.

Bahan binaan seperti

Chemically inert and non-corrosive.

cement

simen

tiles

, jubin

,

– Kitchenware such as cooking pots and plates. periuk

Perkakas dapur seperti

pinggan

dan

Tidak reaktif secara kimia dan tidak mudah menghakis.

– Decorative items such as vases and pottery.

Have high melting point and good insulator of heat, remain stable under high temperature.

Insulation such as parts.

lining

Penebat haba seperti enjin bahagian

melapik

Mempunyai takat lebur yang tinggi dan penebat haba yang baik serta stabil dalam suhu yang tinggi.

Penebat elektrik yang baik.

.

Barang hiasan seperti pasu dan lain-lain.

of furnace, wall of nuclear reactor and dinding

dinding relau,

engine

bagi reaktor nuklear dan

.

Electric insulator in electrical items such as electric cables .

Good insulator electric.

, bricks, roof and toilet bowl.

, batu-bata, atap dan tandas.

Penebat elektrik bagi alat-alat elektrik seperti kabel elektrik .

electric plugs

plug elektrik

,

oven

and

ketuhar

,

dan

Medical dental and apparatus such as orthopedic joint replacement, dental restoration and bone implants.

Non compressible.

perubatan Alat-alat palsu dan pemindahan tulang.

Tidak boleh dimampatkan.

pergigian

dan

seperti penukaran sendi ortopedik, gigi

Composite Materials / Bahan Komposit (a) Composite materials are structural materials that are formed by combining two or more different substances such as metal alloys ceramic glass polymer , , , and .

1

Bahan-bahan komposit adalah bahan yang diperbuat daripada gabungan dua atau lebih bahan berbeza seperti aloi seramik kaca polimer , , dan .

superior

(b) Composite materials have properties that are Bahan-bahan komposit mempunyai sifat-sifat yang

logam

,

than those of the original components.

lebih baik

berbanding dengan komponen-komponen asal.

Complete the table below:

2

Lengkapkan jadual di bawah: Types of composite materials

Components

Special properties

Komponen

Sifat istimewa

Example of uses Contoh kegunaan

Jenis bahan komposit

Superconductors Super konduktor

Reinforced concrete Konkrit yang diperkukuhkan

Copper(II) oxide, barium carbonate and Yttrium oxide heated to form a type of ceramic known as perovoskyte

Conduct electricity with no resistance when it is cooled at low temperature.

Kuprum(II) oksida, barium karbonat dan natrium oksida dipanaskan membentuk sejenis seramik dipanggil perovoskit

amat rendah.

Concrete ( cement , sand and pebbles) reinforced with steel and polymer fibers

m

Very

strong

moulded

and can be

into any shape.

devices(MRI), generators, transformers, computer parts and bullet train

Construction of building, bridges and oil platforms

kuat dan boleh Sangat dibentuk menjadi pelbagai bentuk.

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Konkrit ( simen , pasir dan batu kerikil) diperkukuhkan dengan keluli dan polimer gentian.

Boleh mengalirkan arus elektrik tanpa rintangan pada suhu yang

Used in medical magnetic-imaging

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High tensile strength, low density, easily moulded in

Plastic reinforced with glass fiber .

Fibre glass Plastik yang diperkukuhkan dengan kaca

Plastik yang diperkukuhkan dengan gentian kaca .

thin

Kaca fotokromik

boats, helmets

layers.

regangan tinggi, ketumpatan Daya rendah , mudah dibentuk menjadi nipis

lapisan

Photochromic glass

Making water storage tanks,

.

Darken

Photochromic substance like silver chloride embedded in glass/transparent polymers

when exposed to bright clear when light and becomes exposed to dim light.

Bahan fotokromik seperti argentum klorida digabungkan

gelap apabila dikenakan Menjadi cerah cahaya cerah dan menjadi dalam cahaya malap.

dengan kaca atau polimer lut sinar.

Making optical lens, car wind shield light intensity meters


The diagram below shows the reaction involve in the production of fertilizer Z in industry. Rajah berikut menunjukkan tindak balas yang terlibat dalam pembuatan baja Z dalam industri.

Ammonia

Process X Proses X

Ammonia

Process Y

Sulphuric acid

Compound Z Sebatian Z

Asid sulfurik

Proses Y

(a) (i)

Reaction P Tindak balas P

Name Process X and Process Y. Namakan Proses X dan Proses Y.

Haber process

Process X / Proses X:

(ii) Complete the following table related to process X and Y.

Process Y / Proses Y:

Contact process

Lengkapkan jadual berikut yang berkaitan dengan proses X dan Y. Process Proses

Catalyst Mangkin

Process X

Iron

Proses X

Besi

Process Y

Vandaium(V) oxide

Proses Y

Vanadium(V) oksida

Temperature/°C Suhu/°C

Pressure/ atm

Tekanan / atm

Balanced equation for the reaction that Involve a catalyst

Persamaan kimia tindak balas yang melibatkan mangkin

400 – 500

200

N2 + 3H2

2NH3

450 – 500

2 – 3

2SO2 + O2

2SO3

(b) Ammonia react with sulphuric acid through reaction P to produce compound Z. Ammonia bertindak balas dengan asid sulfurik melalui tindak balas P menghasilkan sebatian Z.

(i)

Write a balance equation for reaction P. Tuliskan persamaan seimbang bagi tindak balas P.

NH3 + H2SO4

(NH4 )2SO4

(ii) What is the type of reaction that takes place? Apakah jenis tindak balas yang berlaku?

Neutralisation

(iii) State one important use of compound Z. Nyatakan satu kegunaan penting sebatian Z.

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(iv) Calculate the percentage by mass of nitrogen in compound Z. [Relative atomic mass: N = 14, S = 32, O = 16, H = 1] Hitungkan peratusan jisim nitrogen dalam sebatian Z. [Jisim atom relatif: N = 14, S = 32, O = 16, H = 1]

%N =

2 × 14 × 100% = 21.2% 2(14 + 4 × 1) + 32 + 4 × 16

The table shows the examples and component of four types of manufactured substances in industry.

2

Jadual berikut menunjukkan contoh-contoh dan komponen bagi empat jenis bahan buatan dalam industri. Type of manufactured substances

Example

Jenis bahan buatan

P

Component

Contoh

Komponen

Cement, sand, small pebbles and steel

Reinforced concrete Konkrit yang diperkukuhkan

Q

Bronze / Gangsa

Polymer / Polimer

R

Glass / Kaca

S

Simen, pasir, batu kecil dan keluli

Copper and tin / Kuprum dan stanum Chloroethene / Kloroetena Silicon dioxide, sodium carbonate, calcium carbonate Silikon dioksida, natrium karbonat, kalsium karbonat

(a) State the name of P, Q, R and S. Namakan P, Q, R dan S.

P: Composite materials

Q: Alloy

R: Polyvinyl chloride

S: Soda-lime glass

(b) (i)

State two uses of reinforced concrete. Nyatakan dua kegunaan konkrit yang diperkukuhkan.

To make framework of buildings and bridges.

(ii) What is the advantage of using reinforced concrete compared to concrete? Apakah kelebihan konkrit yang diperkukuhkan berbanding dengan konkrit?

Reinforced concrete can withstand higher pressure/support heavier loads/ stronger/ higher tensile strength than concrete. (c) (i)

Draw the arrangement of particles in Lukis susunan atom dalam

Pure copper / Kuprum tulen

Bronze / Gangsa Copper

Copper

Tin

(ii) Bronze is harder than pure copper. Explain. Gangsa lebih keras daripada kuprum. Terangkan. – Atoms of pure copper metal are the of same size, they arranged orderly in layers. – Layers of atoms are easily slide over each other when external force is applied on them. – The size of tin atoms which are bigger than copper in bronze disrupt the orderly arrangement of copper atoms.

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– Layers of metal atoms are prevented from sliding each other when external force is applied.

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12/9/2011 5:54:32 PM


(d) The diagram shows the structure of R. / Rajah berikut merupakan struktur bagi R.

(i)

H C H

H – C C1

n

Draw the structural formula for monomer R. / Lukiskan formula struktur bagi monomer R.

H H C = C H C1

(ii) State one use of polymer R.

Nyatakan satu kegunaan polimer R.

Pipe / wire cables / bags / footwear

(iii) State two ways how R causes environmental pollution.

Nyatakan dua cara R menyebabkan pencemaran alam.

– R is non biodegradable, it can cause blockage of drainage system and flash flood. – Burning of R produces hydrogen chloride gas which is poisonous and acidic. (e) (i)

Explain why glass containers are more suitable for storing acid in the laboratory. Terangkan mengapa bekas kaca lebih sesuai digunakan untuk menyimpan asid di dalam makmal.

Glass is chemically inert/ glass is non-reactive

(ii) Soda-lime glass cannot withstand high temperature. State the name of another type of glass that is more heat resistant.

Kaca soda kapur tidak tahan suhu yang tinggi. Namakan jenis kaca lain yang lebih tahan haba.

Borosilicate glass


Which of the following are the uses of sulphuric acid? Antara berikut, yang manakah adalah kegunaan asid sulfurik?

I Detergent II

Cat

Fertiliser

IV Synthetic fiber

Baja

A B 2

Gentian sintetik

C

I and II only I dan II sahaja

D

III and IV only III dan IV sahaja

I, II dan IV sahaja

I, II, III and IV I, II, III dan IV

SO2

II

SO3

III

H2S2O7

IV

Fe

2NH3

Which of the following is the function of iron, Fe in the process? Antara berikut, yang manakah adalah fungsi besi, Fe dalam proses itu?

Rajah di bawah menunjukkan peringkat I, II, III dan IV dalam Proses Sentuh.

I

N2 + 3H2

I, II and IV only

The diagram below shows the stages I, II, III and IV in the Contact Process.

S

The equation below shows chemical equation to produce ammonia in Haber Process. Persamaan tindak balas berikut menunjukkan persamaan kimia untuk menghasilkan ammonia dalam Proses Haber.

III Paint

Detergen

3

H2SO4

Which of the following stages requires the use of a catalyst?

A B C D

To lower the pressure required for the process. Merendahkan tekanan yang diperlukan untuk proses itu.

To lower the temperature required for the process. Merendahkan suhu yang diperlukan untuk proses itu.

To increase the rate of production of ammonia. Untuk meningkatkan kadar pengeluaran ammonia.

To increase the percentage of production of ammonia. Untuk meningkatkan peratus penghasilan ammonia.

Antara peringkat berikut, yang manakah memerlukan mangkin?

I II

C D

III IV n io

Sdn. B

m

181

. hd

Publicat

A B

Nila


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4

The diagram below shows the arrangement of atoms in alloy X.

8

Rajah di bawah menunjukkan susunan atom dalam aloi X.

Which of the following are the characteristics of synthetic polymers that causes environmental pollution? Antara berikut, yang manakah adalah ciri-ciri polimer sintetik yang menyebabkan pencemaran alam sekitar?

I

Copper/Kuprum

Polymers are non biodegradable Polimer adalah tidak terbiodegradasi

II

Zinc/Zink

Polymers dissolve in water and increase pH of water Polimer larut dalam air dan meningkatkan pH air

III Burning of polymers release toxic gas Pembakaran polimer membebaskan gas beracun

What is alloy X?

IV Disposal of polymers promote excessive growth of algae

Apakah aloi X?

A B 5

C

Brass Loyang

D

Bronze Gangsa

Pembuangan polimer meningkatkan pertumbuhan alga berlebihan

Cupronickel

A

Kupronikel

Duralumin

B

Duralumin

C

An alloy Y is used to make a body of an aeroplane. Which of the following is alloy Y and its major component?

D

Aloi Y digunakan untuk membuat badan kapal terbang. Antara berikut, yang manakah adalah aloi Y dan komponen utamanya?

Alloy Y

Major component

Duralumin

Magnesium

Duralumin

Magnesium

Duralumin

Aluminium

Duralumin

Aluminium

Bronze

Copper

Gangsa

Kuprum

C

Cupronickel

Copper

Kupronikel

Kuprum

D

Aloi Y

A B C D 6

Komponen utama

Which type of glass is suitable for making beakers and test tubes that can be used for heating? Kaca yang manakah adalah sesuai untuk membuat bikar dan tabung uji yang boleh digunakan untuk pemanasan?

A B 7

Lead glass Kaca plumbum


C D

9

A B

What is glass X?

Hard and strong Keras dan kuat

Good insulator electric Penebat elektrik yang baik

Remain stable under high temperature Kekal stabil pada suhu tinggi

Chemically inert and non corrosive Lengai terhadap bahan kimia dan tidak terkakis

Maklumat berikut adalah berkaitan dengan bahan Z yang digunakan dalam keretapi laju.

Conducts electricity with no resistance at low temperature.

Photochromic glass

Apabila kaca X dipanaskan dengan kuat dan seterusnya dimasukkan ke dalam air sejuk, kaca itu tidak pecah.

II, III and IV only II, III dan IV sahaja

10 The following information is about substance Z which is used in bullet train.

Mengkonduksi elektrik tanpa rintangan pada suhu rendah.

Kaca fotokromik

When the glass X is heated to a high temperature and plunged into cold water, the glass does not crack.

I, III and IV only I, III dan IV sahaja

Seramik digunakan untuk membuat dinding reaktor nuklear. Antara berikut, yang manakah adalah ciri seramik untuk penggunaan itu?

Borosilicate glass

Maklumat di bawah menunjukkan sifat kaca X.

II and III only II dan III sahaja

Ceramic is used to make wall of reactor nuclear. Which of the following is the characteristic of ceramic for the usage?

Kaca borosilikat

The information below shows the property of a glass X.

I and III only I dan III sahaja

What is substance Z? Apakah bahan Z?

A B C D

Fiber glass Duralumin

Superconductors Super konduktor

Polyvinylchloride Polivinil klorida

Fibre glass Plastik yang diperkukuhkan dengan kaca

Apakah kaca X?

A

m


C D

Fused glass Kaca silika terlakur

Borosilicate glass Kaca borosilikat

Publica

n Sdn.

182

tio

Nil a

B

Lead crystal glass Kaca plumbum

d. Bh


12/9/2011 5:54:33 PM

Nilam Publication module Chemistry Form 4 answer

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