Solutions Manual PDF

Solutions Manual: Module 1 The Chemical Earth

CHAPTER 1: THE EARTH: A MIXTURE OF CHEMICALS Review exercise 1.1 1

food preservative: analgesic: cleaning agent: fuel: fertiliser: pesticide: building material: antibiotic: textile: plastic:

sulfur dioxide salicylic acid carbon tetrachloride octane potassium phosphate pyrethrin cellulose (major constituent of wood) penicillin nylon polyvinyl chloride

2

Biochemistry is the study of the composition, structure, properties and reactions of chemicals concerned with living processes. Geochemistry is the study of the composition, structure, properties and reactions of chemicals concerned with the Earth. Astrochemistry is the study of the composition, structure, properties and reactions of chemicals concerned with the universe. Chemical engineering is the branch of engineering that deals with the design, construction and functioning of chemical plants and factories based on chemical processes, e.g. adhesive or paint factory.

Review exercise 1.2 1

Shape and volume: For a substance to have a definite shape, its particles must be unable to move relative to each other. For a substance to have a variable volume, its particles must be able to move away from the original source until confined by the walls of a container. Because the particles in a solid are packed tightly and cannot move past one another, the shape of the solid cannot change, and nor can its volume. The shape of liquids is variable because the particles are able to move relative to each other; however, the volume is unchangeable because the particles cannot move away from each other. In gases, the particles are moving rapidly and randomly and are confined only by the walls of the container, and so both shape and volume are variable. Compressibility: For a substance to be compressed, its particles must be able to be pushed closer together than they were originally. Therefore, solids are almost incompressible because the majority of particles are already as close together as they can be. Liquids can be very slightly compressed because there may be some empty space between the particles, depending on how well the particles are packed. Gases are easily compressed because the particles have a lot of empty space between them.

© Pearson Education Australia (a division of Pearson Australia Group Pty Ltd) 2006. This page from Chemistry Contexts 1 Teacher’s Resource second edition may be reproduced for classroom use.

Solutions Manual: Module 1 The Chemical Earth Diffusion: For a substance to be able to diffuse, its particles must be able to move away from their original source. Gases diffuse very quickly, as the particles are already moving very quickly and will move to occupy whatever space is available to them. Liquids will diffuse slowly as the particles move only relative to each other. Solids will not diffuse to any significant extent, because the particles do not move away from their original positions, although they do vibrate. Thermal expansion: When a substance expands when heated, the particles start moving more rapidly and away from each other. Remember that the particles themselves do not expand, just the spaces between them. Gas particles experience weak forces of attraction, and so heat is able to overcome these forces easily, and therefore the particles can move away from each other even more rapidly when heated. In liquids the particles are held together more tightly by stronger forces, which heat may be able to overcome partly, so liquids expand to some extent, until of course they turn into gases! In solids the particles are held together extremely tightly by very strong forces, which heat may not be able to overcome, and so solids expand to a very small extent, until of course they turn into liquids! 2

A smell is caused by a gaseous substance that reaches our nose and affects our olfactory sense (sense of smell). Because gases move quickly to occupy a space until confined by its walls (diffusion), any gas particles released during cooking will move into whatever part of the house is open to them.

3

The pressure of a gas is due to the number of collisions between the gas particles and the container walls, and the force of those collisions. The temperature of a gas is a measure of how fast the gas particles are moving. If the speed of the gas particles does not change (i.e. if the temperature doesn’t change) then the force of the collisions will not change. The number of collisions is related to the surface area provided by the volume of the container. A smaller surface area will mean more collisions. If the volume does not change then the surface area will not change and so the number of collisions will not change. Therefore if the number of collisions and their force do not change, the pressure will remain constant.

4

The reason that gases are easily compressed is that their particles have a lot of space between them, and this is reduced when the particles are pushed closer together. We observe this as the gas taking up less space, i.e. being compressed.

5

Gases spread to occupy the available space because their particles have very weak forces of attraction between each other and they are moving very rapidly. There is no force that makes them stay ‘together’.

6

In the gaseous state, the particles are widely spaced and exhibit random straight-line movement. Intermolecular forces are so small that they can be ignored. This means that the particles move away from each other very quickly and have a lot of empty spaces between them. So gases have variable shape and volume, they are highly compressible, they diffuse very quickly and are highly expandable when heat is applied.


Solutions Manual: Module 1 The Chemical Earth In the liquid state, the particles are packed together and can move relative to each other only. Intermolecular forces are fairly strong and cannot be ignored. This means that the particles move around each other and do not have very much empty space between them. So liquids have variable shape but a definite volume, they are very slightly compressible, they diffuse slowly and are moderately expandable when heat is applied. In the solid state, the particles are closely packed and vibrate in their positions only. Intermolecular forces are very strong and cannot be ignored. This means that the particles stay in their positions relative to each other and there is very little, if any, empty space between them. So solids have definite shape and volume, they are almost incompressible, diffuse negligibly and are only slightly expandable when heat is applied. Review exercise 1.3 1

2

a

An element is a pure substance that cannot be broken down into simpler substances. A compound is a pure substance made of two or more elements chemically combined in a fixed ratio.

b

A mixture is an impure substance that contains two or more substances that are not chemically combined. A homogeneous mixture has the same composition throughout, whereas a heterogeneous mixture has a variable composition throughout.

c

A compound is a pure substance made of two or more elements chemically combined in a fixed ratio. A homogeneous mixture is an impure substance made of two or more substances that are not chemically combined.

a

i

element

ii

compound

iii

element

iv

mixture

v

mixture

vi

mixture

i

The element has the symbolic representation of P, but its formula is P2.

ii

PQ3

iii

Q

iv

mixture of elements P (formula P2) and Q

v

mixture of element Q and compound PQ3

vi

mixture of elements Q, and P (formula P2) and compound PQ3

b


Solutions Manual: Module 1 The Chemical Earth 3

a

element

b

compound

c

compound

d

compound

e

heterogeneous mixture

f

element

g

compound

h

heterogeneous mixture

i

Depending on climate, location relative to water sources and pollution levels, the air may differ with respect to the amount of water and pollutants, i.e. it can be considered heterogeneous. However, its composition is still basically the same anywhere in the world with respect to the percentage of nitrogen and oxygen, i.e. it can be considered homogeneous.

j

homogeneous mixture


2

3

a

Biosphere: those parts of the Earth that support living things throughout their life cycles

b

Lithosphere: the solid crustal layer of the Earth

c

Atmosphere: the gaseous layer above the Earth’s surface

d

Hydrosphere: the parts of the Earth’s surface that are water-based, e.g. rivers, oceans, lakes, seas

a

oxygen and carbon

b

oxygen and carbon

c

water and oxygen

d

water and oxygen

Minerals are naturally occurring substances with a definite composition, e.g. quartz (silicon dioxide) and iron oxide.


a

C carbon

b

SiO2 silicon dioxide

c

H2O dihydrogen oxide (water)



The demand for drugs and dyes is so strong now that the chemicals could never be obtained in sufficient quantities from the natural world. So they are produced using technology that allows for cost-effective specialisation of particular chemicals that can be bought by drug or dye manufacturers, who can then package and market the products in the forms expected.

3

a

aluminium from bauxite

b

aviation fuel and petrol

c

sodium chloride

d

nitrogen (element) or carbon dioxide (compound)


evaporation, distillation and fractional distillation

2

a

Filtration—the sand and water would be poured through filter paper. The water would move through, while the sand would remain on the filter paper.

b

Distillation—by boiling the salt water solution, the water would boil at 100oC, leaving the salt behind. The gaseous water would be condensed back to a liquid.

c

Distillation—by boiling the wine, the alcohol would boil at 100oC, leaving the other liquids behind. The alcohol would be condensed back to a liquid.

d

Filtration and then evaporation—the sand, salt and water would be poured through filter paper. The salt water would move through, while the sand would remain on the filter paper. By boiling the salt–water solution, the water would escape as a gas, leaving the solid salt behind.

e

Crystallisation then filtration—dissolve both crystals in hot water. Then plunge the container into an ice bath. Filter the cooled mixture; the solid potassium dichromate will stay on the filter paper, while the ammonium chloride solution will pass through.

f

Chromatography—different drugs will adsorb or cling to the inert chromatographic substance with different strengths, and leave unique traces, which allow them to be identified as separate chemicals.


Volumetric and gravimetric analyses are both quantitative in nature. Volumetric analyses have measurements in volume (e.g. mL or L), whereas gravimetric analyses have measurements in mass (e.g. mg or g).

2

Initial mass = 247.5 g Mass of carbon left = 203.4 g 203.4 % carbon = × 100 247.5 = 82.18% © Pearson Education Australia (a division of Pearson Australia Group Pty Ltd) 2006. This page from Chemistry Contexts 1 Teacher’s Resource second edition may be reproduced for classroom use.


Careful analysis of the ore shows the exact amount of substance that is available for extraction. If the ore does not contain a high enough percentage of that substance, then the mine may not be commercially viable, i.e. it will cost more to extract the substance than the company will be able to sell it for.

Chapter 1—Application and investigation 1

Investigation

2

a

Solids have a definite shape, while liquids have a variable shape; solids are basically incompressible, while liquids can be compressed slightly; solids diffuse negligibly, while liquids can diffuse slowly. Liquids have a definite volume, while gases have a variable volume; liquids can be compressed slightly, while gases are easily compressible; liquids diffuse slowly, while gases diffuse quickly.

b

Solid: copper, wood, silicon dioxide. Liquid: blood, water, mercury. Gas: oxygen, carbon dioxide, nitrogen.

3

When the brake pedal is pushed, that pressure is transferred, via a system of tubes filled with liquid, to the brake pad/disc, which then pushes onto the moving wheel to slow it. Because the liquid is incompressible, it cannot take up less space in the tubes and so pushes onto the brake pad/disc.

4

a

Gas particles have a lot of empty space between them and so can move closer together if they are forced to. Liquid particles and solid particles do not have very much space between them at all, so cannot be pushed closer together.

b

Solid particles cannot move relative to each other—they can only vibrate in the same position; hence the shape of the solid does not change. Liquid and gas particles are able to move relative to each other and so move to (partly) take the shape of their container.

c

The forces between solid particles are very strong, and when they are heated these forces are somewhat overcome so the particles can move further apart (when the substance changes to a liquid). The forces between liquid particles are still relatively strong, but can be overcome by heat, so when the substance turns to a gas, the particles have enough energy to move away from each other. This means there is an increase in volume.

5

Assumptions 2, 3 and 4 (see coursebook pages 5 and 6).

6

Investigation

7

A pure substance has a chemical composition that is unchanging. This means that any sample of that substance will have the same composition. A mixture has a variable chemical composition because the substances that make it up are not chemically combined.



9

a

Magnesium and oxygen are elements, magnesium oxide is a compound.

b

The properties of a substance are a result of the atoms that make it up and how they are chemically combined. Because magnesium oxide is made up of two types of atoms chemically combined in a way that is different from that of either magnesium or oxygen, its properties are different.

10

Elements: N2, He, C, Cl2. Compounds: C2H5OH, KI, NH4NO2.

11

The Earth has a dense core of mostly iron and nickel. It has a thick, relatively homogeneous mantle composed mainly of magnesium and silicon oxides and magnesium and iron silicates. And it has a thin heterogeneous crust, consisting of oxygen, silicon, aluminium, iron, calcium, sodium, potassium, magnesium and less than 1 per cent of the many other elements. When the Earth was formed, it was molten and the heavier elements sank towards the centre, while lighter elements combined with other elements and stayed at different layers depending on their density and temperature. The Earth then cooled.

12

Element: gold. Compound: silicon dioxide. Heterogeneous mixture: granite.

13

Depending on climate, location relative to water sources and pollution levels, the air may differ with respect to the amount of water and pollutants, i.e. it can be considered heterogeneous. However, its composition is still basically the same anywhere in the world with respect to the percentage of nitrogen and oxygen, i.e. it can be considered homogeneous.

14

Investigation

15

If both solids are soluble, filtration will achieve very little, since both solutions will pass through the filter paper. Crystallisation involves then evaporating the solvent (water). Both solids will return in crystal form still mixed up together and they would still have to be separated by hand. Even if their solubilities were different at different temperatures, since we cannot be assured that the required temperature is consistent throughout the mixture, it is difficult to ensure that every part of each substance will either stay in solution or crystallise.

16

a

magnetic separation

b

evaporation

c

solubility

d

liquefaction and fractional distillation


Solutions Manual: Module 1 The Chemical Earth e

centrifuging

f

distillation

17

Gold: Petrol: Glass: Water: Oxygen: Milk: Wood: Ink:

used in naturally occurring form; element processed from crude oil; homogeneous mixture processed from silicon dioxide (sand); homogeneous mixture used in naturally occurring form; compound used in naturally occurring form; element used in (mostly) naturally occurring form; homogeneous mixture used in naturally occurring form; heterogeneous mixture processed; homogeneous mixture

18

Investigation

19

Investigation

20

% salt =

2.13 × 100 50 = 4.3%

CHAPTER 2: ELEMENTS OF THE EARTH Review exercise 2.1 1

Oxygen can be found: • as a gaseous element in the form of O2 • as a gaseous element in the form of O3 (ozone) • in compounds such as water H2O, which can be gaseous, liquid or solid • in gaseous compounds such as CO2 • in solid compounds such as silicon dioxide SiO2.

2

In the Earth’s crust, iron is combined with other elements to form minerals. The Earth’s core makes up a large proportion of the total Earth and is mostly iron and nickel.


A hydrogen molecule is made up of two hydrogen atoms joined together.

2

carbon (C); iron (Fe); bromine (Br2); calcium (Ca); fluorine (F); copper (Cu); silicon (Si)

3

a

helium, neon

b

oxygen, hydrogen

c

sulfur, carbon


Solutions Manual: Module 1 The Chemical Earth Review exercise 2.3 1

Elements in nature are surrounded by chemicals with which to react (e.g. oxygen and water) if the elements are reactive enough. If an element is found ‘free’ in nature it means that it is not very reactive. If it is found as a compound then its reactivity is greater. For example, a hydrogen molecule is made up of two hydrogen atoms joined together.

2

Reactive metal: sodium, calcium. Reactive non-metal: iodine. Non-reactive metal: gold. Non-reactive non-metal: helium, hydrogen.


Electrical conductivity. It is clear from the table that only metals have high electrical conductivity. Both metals and non-metals can have low density (e.g. aluminium and carbon), low thermal conductivity (mercury and carbon), and low melting point (bromine and mercury). This means that it is difficult to use those properties to distinguish metals from non-metals.

2

a

b

3

i

Electrical wiring in a house requires high electrical conductivity and high strength. V would be best.

ii

A bridge requires high strength. Z would be best.

iii

The frame of an ultralight aircraft needs reasonable strength but low density. Y would be best.

iv

Saucepans requires a high melting point and high thermal conductivity. V would be best.

i

Cost and abundance as well as toxicity, since it is inside a house.

ii

Cost, corrosion and aesthetics.

iii

How easily workable the material was (e.g. bending). Could its strength be increased by alloying it?

iv

Cost, corrosion, toxicity and abundance.

Property

Germanium

Typical of metal or non-metal

Melting point

938.3oC

Metal

Boiling point

5108oC

Metal

Electrical conductivity

Low (semi-conductor) –1

–1

–1

Non-metal

Heat conductivity

60 Js m K

Non-metal

Appearance

Greyish-white, lustrous, cubic crystal

Metal

Tensile strength

Low

Non-metal



Silicon chips are found in calculators, computers and microwave ovens.

5

Metals: Non-metals: Semi-metal:

6

The periodic table is set out in a way that shows trends in properties. The trend going across the periodic table is from metallic substances to non-metallic substances. Those substances that show some properties of both groups should be situated between the two groups, which is where the semi-metals are found.

Li, Zn, Ce Se, Br, Xe Sb


The elements can combine with each other in many different ways. Compounds can be made of two elements, or many elements, combined, and so the permutations and combinations are almost limitless.

2

The crust of the Earth is only a very small part of the whole Earth. The rest of the planet does not have a very high proportion of oxygen in it, so the overall figure is much less. The Earth is an insignificant part of the universe and so the amount of oxygen here is very small compared to the universe, which is mostly made up of hydrogen and helium.

3

Since the formula for carbon is the same as its elemental symbol, carbon must be either a monatomic molecule or a network structure. The subscript of 2 in the formula for nitrogen tells us that it is a diatomic molecule, i.e. two atoms of nitrogen joined together.

4

5

6

a

hydrogen and oxygen

b

carbon, hydrogen and oxygen

c

nitrogen and hydrogen

a

Potassium is a much more reactive metal than copper and so is normally found in compounds. The types of elements that potassium reacts with will not react with copper and so copper is found in its elemental state.

b

Bromine is a much more reactive non-metal than sulfur and so is normally found in compounds. The types of elements that bromine reacts with will not react with sulfur and so sulfur is found in its elemental state.



a

Tungsten is ductile and becomes white hot when electricity passes through it.

b

Lead is malleable and resistant to corrosion by water.

c

Steel is very strong and reasonably cheap.

d

Graphite is soft and rubs off easily.

e

Aluminium is light with good thermal conductivity.

f

Helium is light and a non-flammable gas.

g

Mercury expands when heated, has a shiny lustre and is liquid at the normal range of temperatures.

8

Non-metal, as it does not conduct electricity and has a low melting point. The shininess is a less important characteristic of metals than electrical conductivity.

9

Sr Ge Xe Mo H Se

10

Investigation

strontium germanium xenon molybdenum hydrogen selenium

metal semi-metal non-metal metal non-metal non-metal

CHAPTER 3: ATOMS COMBINE TO FORM COMPOUNDS Review exercise 3.1 1

a b

Postulates 5 and 6 were unique, although the ‘atomists’ did not define the differences between different elements as well as Dalton’s postulates 2 and 3. The revised postulates are: • postulate 1, as there are now many subatomic particles known • postulate 2, as many elements have different isotopes that have the same chemical properties but differ slightly in mass.

2

The nucleus of an atom contains protons and neutrons, while electrons move around outside the nucleus. In B-11, the nucleus contains 5 protons and (11 – 5) 6 neutrons. Since the atom is neutral overall, there must be 5 electrons around the outside of the nucleus.

3

Atomic number = number of protons Mass number = number of protons + number of neutrons



5

Protons

Neutrons

Electrons

a

3

4

3

b

10

10

10

c

17

18

17

d

47

60

47

e

82

126

82

f

7

7

7

a

56 26 Fe

b

28 14 Si

c

30 14 Si

d

39 19 K


lithium sulfur chlorine calcium

2

a

3

b

6

3

2, 1 2, 8, 6 2, 8, 7 2, 8, 8, 2

The noble gases are all chemically unreactive. Reactivity is a reflection of electron arrangement. If the noble gases do not react, then their electron arrangement must be stable—that is, they have full outer shells of eight electrons.


Cations are formed when atoms (usually metals) lose electrons, e.g. sodium has 11 protons and 11 electrons. When sodium loses one electron, it gains a charge of +1 and becomes the sodium ion Na+, which is a cation. Anions are formed when atoms (usually non-metals) gain electrons, e.g. sulfur has 16 protons and 16 electrons. When sulfur gains two electrons, it gains a charge of –2 and becomes the sulfide ion S2–, which is an anion.

2

To have a charge of –2, two electrons must have been gained by the oxygen atom (the ion now has two more electrons than the number of protons).

3

a

Cations are usually formed from metals, e.g. Na+, Ca2+, Al3+.

b

Anions are usually formed from non-metals, e.g. F–, O2–, N3–.



K+, P3–, Ca2+, Al3+, I–

5

SO42–, OH–, NO3–, NH4+


a

b

2

a

b

c

d

3

a

b

c


Solutions Manual: Module 1 The Chemical Earth d

4

5

a

1 pair

b

2 pairs

c

3 pairs

a

b

c

d


They exist as diatomic molecules.

2

a

NH3

b

NO2

c

H2S

d

H2O2


Solutions Manual: Module 1 The Chemical Earth

3

4

5

e

SF4

f

SF6

g

N2O5

h

NI3

a

carbon monoxide

b

carbon dioxide

c

hydrogen fluoride

d

carbon tetrahydride or methane

e

sulfur trioxide

f

phosphorus trichloride

g

dinitrogen trioxide

h

bromine pentafluoride

a

NaNO3

b

KI

c

Hg(CN)2

d

Al(OH)3

e

Ca(MnO4)2

f

(NH4)2SO4

g

Fe2O3

h

Mg3(PO4)2

a

potassium permanganate

b

calcium hydrogencarbonate

c

magnesium hydrogensulfate

d

sodium carbonate-10-water

e

lead(II) oxide

f

sodium hydrogensulfite

g

copper(I) oxide

h

iron(III) hydroxide



a

(NH4)2S

b

KNO2

c

CaSO4.2H2O

d

PbF2

e

Cu(NO3)2

f

SnCO3

g

Mg3N2

h

Al2(SO4)3


2

a

i

A, C and D are isotopes.

ii

C and F each have the same number of electrons and protons.

iii

A, D and E all have more protons than electrons.

iv

B has more electrons than protons.

b

A, C and D represent calcium. B represents chlorine. E represents potassium. F represents argon.

c

A: Ca2+ B: Cl– C: Ca D: Ca2+ E: K+ F: Ar Symbol

16 8

O

17 8

O 2−

23 11

Na +

37 17

Cl −

9 4

Be 2+

Protons

8

8

11

17

4

Neutrons

8

9

12

20

5

Electrons

8

10

10

18

2

Net charge

0

–2

+1

–1

+2

Atomic number

8

8

11

17

4

Mass number

16

17

23

37

9



a

2

b

2, 8, 3

c

2, 8

d

2, 5

e

2, 4

f

2, 8, 8, 1

4

5

Ne

F

C

O

a

group

VIII

VII

IV

VI

b

electron configuration

2, 8

2, 7

2, 4

2, 6

c

valence electrons

8

7

4

6

d

charge on simple ion

ion not formed

1–

ion not formed

2–

a

b

c

d

6

a

b

c


Solutions Manual: Module 1 The Chemical Earth d

e

f

7

Investigation

8

a

The symbol for an element is the one or two letters used to identify it. A formula for an element gives some information about its structure in its standard state, e.g. monatomic, diatomic.

b

Co is the symbol for the element cobalt. CO is the formula for the compound carbon monoxide.

a

Atoms are the smallest particle of matter that can take part in a chemical reaction. Molecules are groups of atoms chemically bonded together. Simple ions are single charged atoms (either positive or negative). Polyatomic ions are groups of atoms chemically combined together that have an overall net charge (either positive or negative).

b

Elements and compounds are both pure substances. Elements cannot be broken down into simpler substances, whereas compounds can, as they are made of two or more elements chemically joined together.

9

10

Aspirin Sucrose

11

a

2

b

8

c

6

d

9

12

C9H8O4 C12H22O11

Monatomic: Diatomic: Polyatomic:

Ne CO, Cl2, HCl, F2 NH3, CH4, HNO3



14

15

16

17

a

I2

b

P4

c

CH4

d

HBr

e

PCl3

f

PCl5

g

CCl4

h

ClF3

a

fluorine

b

sulfur dioxide

c

hydrogen iodide

d

iodine

e

hydrogen chloride

f

diphosphorus pentoxide

g

nitrogen trifluoride

h

carbon tetrafluoride

a

magnesium chloride

b

potassium hydrogensulfate

c

iron(III) oxide-3-water

d

ammonium sulfite

e

calcium hydrogencarbonate

f

lead(IV) oxide

a

HgCl

b

HgNO3.H2O

c

AlPO4

d

(NH4)2CO3

e

KMnO4

f

CuCO3

Investigation


Solutions Manual: Module 1 The Chemical Earth CHAPTER 4: CHEMICAL CHANGE


a

physical

b

chemical

c

physical

d

chemical

e

chemical

f

physical

g

physical

h

chemical

2

A physical change does not produce new substances, e.g. when freezing liquid water to ice, the first substance is H2O, as is the second substance. In a chemical change, the substance is actually changed and no longer exists, e.g. in burning coal, carbon is changed into CO and CO2 by combining with oxygen.

3

i

ii

a

Physical process

b

Water molecules that were far apart and moving rapidly in straight-line motion (since the substance was in a gaseous state) are brought closer together and slowed down until the substance is in the liquid state.

c

Energy is released, much less than the energy involved in part ii.

a

Chemical process

b

The hydrogen molecules break apart, as do the oxygen molecules. The hydrogen and oxygen atoms then rearrange themselves so that H2O is formed.

c

Energy is released and it is a much larger amount of energy than in part i.


The law of conservation of mass (and energy).

2

When gaseous methane burns, it chemically combines with gaseous oxygen to form carbon dioxide gas and water vapour.

3

a

NH3(g) + HCl(g) → NH4Cl(s)

b

CaCO3(s) → CaO(s) + CO2(g)

c

AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)



2

a

Cl2(g) + H2(g) → 2HCl(g)

b

2Cl2(g) + O2(g) → 2Cl2O(g)

c

CaO(s) + H2O(l) → Ca(OH)2(aq)

d

Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)

e

4Fe(s) + 3O2(g) → 2Fe2O3(s)

f

2C2H6(g) + 7O2(g) → 4CO2(g) + 6H2O(g)

a

C(s) + O2(g) → CO2(g)

b

2C(s) + O2(g) → 2CO(g)

c

2Al(s) + 3Cl2(g) → 2AlCl3(s)

d

2KClO3(s) → 2KCl(s) + 3O2(g)

e

2Al(s) + 6HCl(aq) → 2AlCl3(aq) + 3H2(g)

f

2H2S(g) + 3O2(g) → 2SO2(g) + 2H2O(g)


Chemical decomposition is a chemical change in which a substance is broken down into simpler substances, often all the way to the elements that make it up.

2

a

Before water is boiled, it exists as water molecules and is in the liquid state. After water is boiled, it exists as water molecules and is in the gaseous state. Before water is electrolysed, it exists as water molecules and is in the liquid state. After electrolysis, the water molecules no longer exist as they have been broken down and the atoms rearranged into hydrogen and oxygen molecules.

b

Boiling is a physical change, as water molecules existed before and after. Electrolysis is chemical decomposition, as the substances after the process are not the same as before and are simpler than the original substance.

3

When sodium hydrogencarbonate is heated, it decomposes and one of the products is CO2 gas. The bubbles of CO2 gas try to escape from the cake mixture and push the mixture up as they travel upwards. Some gas bubbles get trapped and remain in the cake mixture, so the cake rises.

4

a

chemical decomposition

b

physical change

c

physical change

d




Extraction is the process by which the substance desired (usually a metal) is removed from the compound in which it is found. Synthesis is the process by which a chemical is manufactured from simpler substances in a chemical reaction. Decomposition can be thought of as the reverse of chemical synthesis. Decomposition is the breaking down of a complex chemical into simpler ones by a chemical reaction.

2

An element is not made up of other chemicals, so you cannot put chemicals together to make an element. Elements can be synthesised by nuclear reactions (which rearrange particles in nuclei) but not by chemical reactions (which rearrange electrons between atoms).

3

Heat: extracting mercury from mercury oxide Light: decomposing silver chloride into silver atoms Electricity: decomposing water into hydrogen and oxygen

4

a

Light energy is absorbed.

b

Heat energy is released.

c

Electrical energy is released.

d

Light energy is released.


Covalent: sharing of electrons between atoms. Ionic: electrostatic attraction between oppositely charged ions.

2

The stronger the force holding two things together, the more energy is required to overcome that force. If a compound requires a lot of energy to decompose it into its elements, then the elements must have had very strong forces holding them together in the compound originally.


a

physical

b

chemical

c

chemical

d

physical

e

chemical

f

physical



Lead: physical change. It would appear that the grey-silver solid has melted, as it occurred at a defined temperature and there is no evidence to suggest that a new liquid, or any other product, formed. Zinc iodide: chemical decomposition. Two products appeared at the end of the process (it is a chemical change of some sort). Iodine as an element is a purple solid and zinc is a grey metal, so we can assume that decomposition has taken place. Lead bromide: chemical decomposition. Two products appeared at the end of the process (it is a chemical change of some sort). Bromine at 500ºC as an element is a brown gas and lead is a silver liquid, so we can assume that decomposition has taken place.

3

The white solid appears to be a compound undergoing chemical decomposition. There is no other chemical available to react with and so, since new products are made, we cannot say that the original substance was an element reacting with another substance to produce two new products. To separate a mixture, physical methods should be sufficient.

4

i

release light and heat energy; large; chemical change

ii

absorbing energy; small; physical change

iii

releasing energy; small; physical change

iv

release light and heat energy; small; chemical change

5

i

ii

6

7

a

When methane gas is burnt in gaseous oxygen, carbon dioxide gas and water vapour are formed.

b

When solid copper is placed in a solution of silver nitrate, solid silver is precipitated and the remaining solution is copper nitrate.

a

4Al(s) + 3O2(g) → 2Al2O3(s)

b

2C2H2(g) + 5O2(g) → 4CO2(g) + 2H2O(g)

c

BaCl2(aq) + Na2SO4(aq) → BaSO4(s) + 2NaCl(aq)

d

Ca(s) + 2H2O(l) → Ca(OH)2(s) + H2(g)

e

2NaHCO3(s) → Na2CO3(s) + H2O(g) + CO2(g)

f

2C7H5N3O6(s) → 3N2(g) + 5H2O(g) + 7CO(g) + 7C(s)



a

2Na(s) + Cl2(g) → 2NaCl(s)

b

Na2O(s) + H2O(l) → 2NaOH(aq)

c

MgCO3(s) → MgO(s) + CO2(g)

d

Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g)

e

CaCO3(s) + 2HCl(aq) → CO2(g) + H2O(l) + CaCl2(aq)

f

2C6H6(l) + 15O2(g) → 12CO2(g) + 6H2O(g)

g

2Al(s) + 3CuSO4(aq) → 3Cu(s) + Al2(SO4)3(aq)

h

2NH4NO3(s) → 2N2O(g) + 4H2O(g)

9

Some of the products of wood burning are gaseous and escaped into the air. If the mass of the gaseous products and the ash could be combined, they would have the same mass as the original wood + oxygen.

10

If nine molecules of nitrogen dioxide are used, the ratio multiplier for all chemicals in the equation is 3.

11

12

a

2 × 3 = 6 molecules of oxygen and octane reacted

b

3 × 1 = 3 molecules of water produced

If 96 molecules of carbon dioxide are produced, the ratio multiplier for all chemicals in the equation is 6. a

2 × 6 = 12 molecules of octane reacted 25 × 6 = 150 molecules of oxygen reacted

b

18 × 6 = 108 molecules of water produced

a


b


c

chemical synthesis

d

chemical synthesis

13

Investigation

14

a

The intermolecular forces must be very weak because it does not take very much energy to overcome them and separate the molecules from each other.

b

The larger energy required to break the ethanol molecule apart suggests that the forces within the molecule, holding the atoms together, are quite strong. These forces are covalent bonds.


Solutions Manual: Module 1 The Chemical Earth CHAPTER 5: BONDING AND STRUCTURE DETERMINE PROPERTIES


Paper: Physical properties: tears easily, solid at room temperature. Chemical property: burns in oxygen to produce carbon (ash) as one product. Petrol: Physical properties: liquid at room temperature, evaporates easily. Chemical property: burns to release a large amount of gaseous by-products (carbon monoxide, carbon dioxide and water vapour). Water: Physical properties: tasteless, odourless, m.p. 0°C, b.p. 100°C. Chemical properties: solvent for many salts, neutral (neither acidic nor basic). Natural gas: Physical properties: gas at room temperature, colourless. Chemical properties: burns in air to produce large amounts of heat.

2

Copper sulfate forms a blue solution, i.e. colour. Sodium chloride is salty, i.e. taste (taste should never be used to identify a substance!). Oxygen is a gas at room temperature, i.e. boiling point. Water is a non-conductor of electricity, i.e. electrical conductivity. Iron is magnetic. Sulfur is a yellow solid at room temperature with a characteristic smell, i.e. colour, melting point and odour.


a

b

Water is a colourless, odourless liquid at room temperature. Oxygen is a colourless, odourless gas at room temperature that will support combustion. Hydrogen is a colourless, odourless gas at room temperature that will explode in the presence of a flame.

c

Water is a compound composed of two parts of hydrogen to one part of oxygen. Oxygen and hydrogen are diatomic elements.



The law of constant proportions states that all pure samples of the same compound will contain the same proportion by mass of the constituent elements. In the 10 g sample, the ratio of oxygen to copper is 2 : 8 = 1 : 4. Thus in the 30 g sample, the ratio must be the same, and so it must be 15 oxygen = 6 g.


Charged species (electrons, cations or anions) that are mobile are necessary for a substance to conduct an electric current.

2

If the m.p. is high then the forces between particles that are separated on melting are very strong. Obviously if the m.p. is low then the forces between particles that are separated on melting are not very strong.

3

A: B: C: D:

covalent molecular metallic ionic covalent network

4

a

ionic

b

covalent molecular

c

metallic

d

ionic

e

covalent molecular

f

covalent network

g

covalent molecular

h

covalent network

5

The forces that hold the atoms together in a chlorine molecule are covalent bonds and are very strong. So it requires very high temperatures to break each molecule up into its constituent atoms. The forces between chlorine molecules are very weak intermolecular forces, so the molecules move apart from each other (i.e. the substance melts) at a low temperature.

6

To melt a covalent molecular substance, individual molecules must be separated from each other. The forces that hold these molecules together are weak intermolecular forces, and so the melting point is low. Note that the forces holding the atoms together within each molecule remain undisturbed. To melt a covalent network substance, individual atoms must be separated from each other. The atoms are held together by strong covalent bonds operating in three dimensions from each atom. Breaking covalent bonds requires lots of energy and so the m.p. and b.p. are very high.



a

Both solid and molten KCl contain ions. However, in solid KCl these ions are not free to move relative to each other as they are held in a fixed ionic lattice, and so the substance cannot conduct electricity. In molten KCl the ions have been separated from each other and are free to move, and so it can conduct electricity.

b

The forces of attraction between the ions (ionic bonds) are strong and so a lot of energy is required to overcome them.

8

Malleable: The layers of cations can slide past one another if a force is applied, as the electrons between layers will move if enough energy is applied. Once in their new position, the layers are held in place by the delocalised electrons. Ductile: The layers of cations can slide past one another to create a more end-to-end structure if enough force is applied. Once in their new position, the end-to-end layers are held in place by the delocalised electrons. Conduct electricity: The electrons are free to move throughout the lattice, so it is able to conduct electricity. Thermal conductivity: The kinetic energy supplied by the heat can be transferred from one electron to the next along the length of the metal, because the electrons are not attached to any one particular cation and can move and collide with neighbouring electrons.

9

a

ionic lattice: NaF

b

metallic: Cu

c

covalent network: SiC

d

covalent molecular: N2


a

Iron: high mechanical strength, so used in construction. Copper: high electrical conductivity, so used in domestic wiring. Lead: highly malleable, so used in flashing in construction. Alcohol: evaporates easily, so used as a cooling ‘rub’. Helium: less dense than air, so used in ballooning.

b

Petrol: burns efficiently to produce gaseous products, so used in engines to move pistons. Oxygen: combines with glucose to provide chemical energy for life processes. Carbon dioxide: will not support combustion, so used in fire extinguishers.

2

Investigation

3

Investigation


Solutions Manual: Module 1 The Chemical Earth Metallic bonding: metal cations are held together in a three-dimensional lattice by delocalised mobile electrons. This leads to the class of metallic substances. Ionic bonding: cations and anions are held together by electrostatic attraction. This leads to the class of ionic substances. Covalent bonding: electrons are shared between atoms. There are two classes of covalently bonded substances: covalent molecular, in which individual molecules (groups of atoms) are separate from each other but held loosely together by intermolecular forces; and covalent network, where each atom is held in a threedimensional lattice by covalent bonds.

4

a

5

To have an odour, a substance must be evaporating (so that gas molecules can travel to peoples’ noses!). To evaporate from a liquid (or solid), the particles must be only held loosely together so the energy at room temperature is enough to separate them. This suggests weak intermolecular forces in a covalent molecular structure.

6

From strongest to weakest: H2O > SO2 > NH3 > CH4

7

It is either a covalent network or an ionic substance. Both have a lattice or threedimensional structure. (It cannot be molecular, since it has a high m.p., and can’t be metallic because it is not a conductor.) If, on melting, it conducts electricity, then the substance can be assumed to be ionic. If it does not conduct in the molten state then it is covalent network.

8

CO2 is covalent molecular and in order to change from the solid state, only weak intermolecular forces need to be overcome. To melt and then boil SiO2, strong covalent bonds must be broken because it has a covalent network structure.

9

Investigation

10

Solid sodium chloride has charged particles but they are not free to move in the solid state. Water does not have a significant number of charged particles and so will not conduct electricity very well at all. When sodium chloride dissolves in the water, its charged particles (Na+ and Cl–) are free to move and so it can conduct electricity.

11

Metals are malleable because the layers of cations can slide past one another if a force is applied but are still held together in their new location (still layered) by delocalised electrons. When an ionic substance has force applied to it and the layers slip past one other, they can line up in such a way that cations are next to cations and anions next to anions. This provides a repulsive electrostatic force and so the substance is brittle.

12

a

Metallic bonds: Cu, sterling silver, Pb, bronze

b

Ionic bonds: MgBr2, Na2S, K2O, F2O, strontium chloride, MgH2, AsH3, CsCl

c

Covalent bonds: SiH4, sulfur dioxide, P4O10, SiC, B2O3, diamond



14

15

16

17

18

a

cations of sodium and delocalised electrons

b

cations of sodium and anions of chlorine

c

atoms of carbon

d

molecules of a carbon atom bonded to two oxygen atoms

a

intermolecular forces

b

covalent bonds

c

ionic bonds

d

metallic bonds

a

Na+ and I–

b

electrons and copper ions

c

electrons

d

Mg+ and NO3–

e

Al3+ and NO3–

a

Copper has delocalised mobile electrons able to conduct electricity. The layers can be pushed past each other and still be held together by the delocalised electrons. Diamond does not have charged species in its structure and so cannot conduct. It is extremely hard, as the covalent bonds that hold it together act in three dimensions around each atom.

b

Naphthalene is a covalent molecular species and its molecules are held together by weak intermolecular forces. This means that it has a low melting point and the molecules can easily be pushed around each other, i.e. it is soft. Quartz is extremely hard, as the covalent bonds that hold it together act in three dimensions around each atom. These covalent bonds are very strong and require a lot of energy to break, and so it is hard and has a high m.p.

a

covalent network

b

covalent molecular

c

metallic

d

ionic

e

metallic

a

covalent molecular

b

covalent molecular

c

ionic

d

metallic



20

a

gold

b

potassium bromide

c

In solid gold, the conduction occurs because of electrons. In molten potassium bromide, the conduction occurs because of potassium cations and bromide anions.

d

Silicon carbide, because it is a covalent network substance and has the strongest type of bonds (covalent) holding it together in a three-dimensional lattice.

Investigation

Module 1 REVIEW 1

C

2

C

3

C

4

B

5

C

6

D

7

A

8

B

9

C

10

B

11

B

12

a

b 13

14

i

7

ii

8

14 7 X

a

The particles in ice are held together in a lattice by intermolecular forces. The particles are vibrating slowly but are not free to move relative to each other. As heat is applied to the ice, the particles start to move more rapidly and as they start being able to move past one another, we say that the ice is melting.

b

Water (H2O) molecules

a

CaO is solid, SiO2 is solid and H2O is liquid.

b

CaO is ionically bonded, SiO2 is covalently bonded (in a network structure), H2O is covalently bonded (in a molecular structure).



Add water and filter. The sodium chloride would dissolve in the water and travel through the filter paper. The two solids would be left on the paper. Take the filtrate and evaporate to dryness to collect the sodium chloride. Take the filter paper and dry it. Use a sieve to separate the fine sulfur powder from the coarser sand particles.

16

A: B: C: D: E:

SiO2 HI Mg I2 KCl

17

i

a

Intermolecular forces are being broken.

b

Molecules are being separated.

a

Covalent bonds are being broken.

b

One atom in each molecule is being separated from the other atom.

a

Covalent bonds are being broken.

b

Groups of atoms held together in a lattice are being pulled apart from other atoms held together in the same lattice.

a

Ionic bonds are being broken.

b

Groups of cations and anions held together in a lattice are being pulled apart from other cations and anions held together in the same lattice.

a

Metallic bonds are being broken.

b

Groups of cations held together in a lattice are being pulled apart from other cations held together in the same lattice.

ii iii

iv

v

18

a

Liquid substances often evaporate, e.g. simmering a sauce to reduce its volume. Energy is absorbed during this process.

b

A decomposition reaction of sodium hydrogen carbonate to carbon dioxide gas occurs when cakes rise during cooking.


Solutions Manual: Module 2 Metals

CHAPTER 6: THE EXTRACTION AND USES OF METALS Review exercise 6.1 1

Copper, gold and silver are often found in their elemental form, i.e. not chemically combined with other substances, and so could be used in the form in which they were found. The technology required to extract other metals from their ores took time to develop.

2

The chemical bonds that hold metal atoms to the other atoms in the mineral are very strong and require energy to break or rearrange them.

3

Both copper sulfide and aluminium oxide have ionic bonds holding the metal ions of copper and aluminium close to the oppositely charged ions of sulfur and oxygen respectively.

4

The extraction of metals from their ores and the ability to form alloys both increased the range of metallic materials available for human use.

5

Copper was the first metal to be extracted because the two ingredients necessary for the reactions to occur were readily available in campfires and pottery kilns, that is, heat and carbon. The addition of tin to copper forms bronze. Tin is between copper and iron in the activity series and requires more energy for extraction than copper, but not so much that the technology of furnaces was needed for its extraction. Iron is higher in the activity series than copper and requires higher temperatures for extraction. It was not until improvements to furnaces allowed increased temperatures to be achieved that iron was able to be extracted from its ore.

6

Copper is used for: • electrical wires because of its high electrical conductivity and its ductility • pipes in plumbing because it is resistant to corrosion in water • pipes used to dissipate heat (e.g. in air conditioners) because of its thermal properties.

7

The addition of tin to copper to form bronze allowed for a durable, strong and corrosion-resistant metal. Bronze could be used in weaponry and so allowed groups at war an advantage over enemies who did not have bronze weapons.

8

Iron(III) oxide

9

a

Bronze was used for tools and weapons where sharpness and durability were very important. It was discovered that steel could be made even harder and sharper by alternating heating and sudden cooling. This simple process meant that bronze was eventually replaced by iron in many applications.

b

Iron is the major constituent of steel, and steel is a very widely used building material. Iron is the second most abundant metal in the Earth’s crust.


Solutions Manual: Module 2 Metals c

Iron in its pure form is easily corroded when exposed to air and moisture. The rust that forms is weak and brittle.


The aluminium atoms are held very tightly in the Al2O3.3H2O complex. Large amounts of electrical energy are needed to extract it, so it was not commercially viable until early in the twentieth century, when large amounts of electricity were available at reasonably cheap rates.

2

Aluminium is strong enough for a roof but it is much lighter than steel, so it is a more structurally sound material to use.

3

Titanium is relatively unreactive and so is resistant to corrosion in salt water.


a

Alloys are substances made up of two or more metals combined in some way.

b

At its simplest level, alloys are made by melting the components together to form a ‘mixture’ of the metals.

c

Alloys are important in society because they can be manufactured so that the mixtures of properties from the constituent metals suit particular purposes.

2

Metals can have all of their atoms lined up in ‘sheets’ or ‘layers’ that can slip past one another. This makes the metal soft. By placing other atoms in these sheets or layers, the bonds created reduce the slippage and so the alloy is stronger than the original metal.

3

a

Different compositions mean different properties and so the uses of iron can be greatly increased.

b

Different heat treatments produce different-sized crystals, which changes the strength and ductility of the alloy.

4

Bronze (Cu and Sn) is easily cast and sonorous (used in church bells). Brass (Cu and Zn) is easy to machine and polishes well (used in decorative door knobs). Cupronickel (Cu and Ni) is hard-wearing and a silver colour (used in coinage).

5

Carbon steel: mild steel (nails), structural steel (girders), tool steel (tool blades). Alloy steel: stainless steel (cutlery), tungsten (grinding tools), silicon (electromagnets).


If the mineral is present in sufficient quantity to make the mining and extraction of the metal economically viable, it is called an ore.

2

Extraction of the metal from the silicate is difficult, whereas extraction from the oxide or sulfide is easier in terms of energy needed and the chemical processes involved.

3

Although it is only a small part of the lithosphere, when it is found, nickel is usually in very concentrated ore deposits.


Solutions Manual: Module 2 Metals Review exercise 6.5 1

Mining (extracting the ore from its natural environment), milling (concentrating or purifying the ore by physical means), smelting (extracting the metal from its ore using chemical processes) and refining (purifying the crude metal by heat or electrical processes).

2

The frothing agent produces bubbles. The chemical ore adheres to the bubbles as they rise to the top and are then skimmed off. The physical property is adhesion.

3

a

The iron is removed by roasting the ore in oxygen. This removes the iron from the ore and converts it to iron oxide. This oxide then is combined with silicon dioxide to form an iron silicate, which is easily removed physically.

b

In the last two smelting processes, gaseous sulfur dioxide forms. This gas causes bubbles in the molten copper and appears as blisters on the surface of the copper as it solidifies.

4

Slabs of impure blister copper create the anode (positive electrode) of an electrolytic cell. The cathode is a thin sheet of pure copper and the electrolyte is copper(II) sulfate solution, acidified with sulfuric acid. All the metals in the slab go into solution. A small voltage is applied so that only copper deposits as a solid on the cathode, while the ions of the more reactive metals remain in solution and the less reactive metals fall to the bottom. Both impurities are removed from time to time.


Gold is expensive because it is sought after due to its attractive appearance, its ability to be shaped easily to make different objects, its very high conductivity and reflectivity and its resistance to chemical attack. Although easy to extract, it is a rare metal.

2

Sodium extraction requires large amounts of energy, so although it is abundant in the Earth’s crust and this should mean a low price, its extraction costs outweigh this factor and the price remains high.

3

a

Iron is found in high-grade ores and is abundant in the Earth’s crust, it is reasonably cheap to produce in large scale and it can be alloyed to many other substances to produce a wide range of specialist products.

b

Extraction costs for aluminium are much higher than for iron. Aluminium extraction requires large amounts of electricity, whereas iron is extracted from its ores in blast furnaces, which run continuously, thereby reducing production costs.

c

Although its extraction is relatively easy, mercury is very rare in the Earth’s crust.


The aluminium is sorted according to its alloy type. It is then sent to a smelter where it is melted in specially designed furnaces. The molten aluminium is then analysed and its composition adjusted before it is cast into ingots, which are sent to manufacturers.


Solutions Manual: Module 2 Metals 2

Recycling aluminium requires less than 5% of the energy needed to produce the same amount of aluminium from its ore.

3

Once molten Al2O3 is produced from the bauxite, it is electrolysed. This requires vast amounts of electricity. It costs money to transport such high amounts of power and so the smelters are situated near power stations.

4

a

The rate of recycling will depend on how much of the original product is given up for recycling. Many of the aluminium products used domestically are small and light, and are used for consumables (e.g. drinks). Domestic objects of iron and steel are heavier and are used for longer (e.g. cars). It is thus easier to ‘give up’ aluminium products for recycling. Also, of course, advertising and financial incentives have been made available to consumers.

b

Fossil fuels are commonly used to create electricity (coal-fired power stations). Electricity is used in vast amounts in the smelting of aluminium from its ores. Recycling processes use only 5% of this amount of energy and so, by recycling, less electricity is used and fossil fuels are conserved somewhat.


Investigation

2

Investigation

3

Iron alloys were stronger and harder than bronze. Iron ores were also more abundant than copper and tin ores.

4

Iron in its elemental form is soft and rusts easily. When combined with other elements these poor properties are reduced dramatically, e.g. stainless steel does not rust, structural steel is very strong, and tool steel can be sharpened.

5

Investigation

6

Metals can have all of their atoms lined up in ‘sheets’ or ‘layers’, which can slip past one another, making the metal soft. By placing other atoms in these sheets or layers, the slippage is reduced and so the alloy is stronger than the original metal. For example, bronze is far stronger than copper. Bronze is copper with one in every ten atoms (10%) replaced by a tin atom.

7

a

Copper: good electrical conductivity; used exclusively for electrical wiring in every home, office and industry in industrialised countries.

b

Iron: ability to form alloy with carbon; cheap to produce large quantities of steel for building construction.

c

Aluminium: lightweight but strong; has replaced the heavier steel in many building applications to reduce costs associated with structures to support the steel.


Solutions Manual: Module 2 Metals d

Titanium: ability to withstand corrosion in a marine environment. Very few metals can survive salt water conditions and yet many industries are dependent on components that spend long periods in salt water, e.g. propellers in boats.

8

Investigation

9

Investigation

10

We now have more sophisticated methods of extraction and alloy production. This is because more energy is available on a low-cost basis and our knowledge of chemistry is greatly advanced, allowing higher temperatures to be used in some processes and electrolysis (which uses large amounts of electricity) to be used in others.

11

A metal is an element, e.g. copper. A mineral is a naturally occurring compound, e.g. chalcopyrite CuFeS2. An ore is a substance in which the mineral is present in sufficient quantity to make the mining and extraction of the metal economically viable.

12

Copper was the first metal to be extracted because the two ingredients necessary for the reactions to occur were readily available in campfires and pottery kilns, that is, heat and carbon. Aluminium is held very tightly in its mineral and vast quantities of electricity are required for its extraction. Electricity production on this scale was not available until the early twentieth century.

13

Copper is concentrated using a method called froth flotation. In this process, air is bubbled through a suspension of the pulverised ore in water containing a flotation or frothing agent. The desired copper mineral particles adhere to rising bubbles and are skimmed off as froth, leaving the unwanted silicate minerals to settle out. Thus the copper ore is separated from the gangue.

14

Use the impure copper as an anode, a copper solution as the electrolyte and a sheet of pure copper as the cathode. As an electrical current is applied, copper will dissolve from the anode and be deposited on the cathode in a more pure form.

15

2CuO(s) + C(s) → 2Cu(s) + CO2(g)

16

It reduces the temperature required to keep the Al2O3 molten. This saves money by saving the electricity needed to maintain such a high temperatures. © Pearson Education Australia (a division of Pearson Australia Group Pty Ltd) 2006. This page from Chemistry Contexts 1 Teacher’s Resource second edition may be reproduced for classroom use.


a

Because the extraction and refining of aluminium is so expensive with respect to the amount of electricity used.

b

The final stage of refining the aluminium electrolytically uses very high currents and is therefore very expensive. Situating a smelter near a power station reduces the costs of transporting such high currents of electricity. This cost would be higher than the cost of transporting the ore from its location to a site near a power station.

18

Titanium has very special properties and so is in demand. It is tough, durable and resistant to corrosion in marine environments.

19

Investigation

CHAPTER 7: CHEMICAL REACTIONS OF METALS Review exercise 7.1 1

a

Disadvantage: the rusting of iron can cause weaknesses in structures made of steel (an alloy of Fe).

b

Advantage: when the corroded layer is impervious to oxygen or moisture, it protects the metal underneath, e.g. Al2O3.

2

Aluminium, chromium, zinc

3

Sodium and potassium react with oxygen in the air and vigorously with moisture in the air to produce large amounts of heat. Therefore they are stored in paraffin oil to prevent oxygen or water coming in contact with the metal.

4

a

2Ca(s) + O2(g) → 2CaO(s)

b

4Al(s) + 3O2(g) → 2Al2O3(s)


2

a

Potassium and water react to produce potassium hydroxide and hydrogen. 2K(s) + 2H2O(l) → 2KOH(aq) + H2(g)

b

Calcium and water react to produce calcium hydroxide and hydrogen. Ca(s) + 2H2O(l) → Ca(OH)2(s) + H2(g)

c

Aluminium reacts with steam to produce aluminium oxide and hydrogen. 2Al(s) + 3H2O(g) → Al2O3(s) + 3H2(g)

Iron reacts with water to produce iron oxide and hydrogen. The container would rust and hydrogen gas would also be produced. Hydrogen is explosive in the presence of an open flame.



Cu, Ag and Pt

2

a&b Mg(s) + 2H+(aq) → Mg2+(aq) + H2(g) Bubbles of gas form on the magnesium, the magnesium dissolves and the container becomes warm.

3

Magnesium, aluminium, zinc, chromium, iron, cadmium, cobalt, nickel and tin.

4

a

2Al(s) + 12H+(aq) + 6SO42–(aq) → 2Al3+(aq) + 6(SO4)2–(aq) + 3SO2(g) + 6H2O(g)

b

2Al(s) + 6H+(aq) → 2Al3+(aq) + 6Cl–(aq) + 3H2(g)


The activity series is a list of metals ranked according to their reactivity.

2

The negatively charged valence electron is held by the electrostatic attraction to the positively charged nucleus. This force must be overcome if the electron is to leave the atom—that is why energy is needed.

3

The valence electron is more tightly held in iron than in calcium and in potassium. It therefore requires more energy to remove that valence electron in iron than in calcium and potassium.

4

Sodium (group 1) then magnesium (group 2) and then zinc (transition metal). Order of first ionisation energies: Zn > Mg > Na Order of reactivity: Na > Mg > Zn


2

a

Zn oxidised

O reduced

b

C oxidised

Pb reduced

c

Mg oxidised

S reduced

d

C oxidised

O reduced

e

Zn oxidised

H+ reduced

f

Fe oxidised

Cu2+ reduced

a

Cl2 is reduced and is ∴ an oxidising agent. Br– is oxidised and is ∴ a reducing agent.

b

Cl2 is reduced and is ∴ an oxidising agent. Mg is oxidised and is ∴ a reducing agent.

c

Ag+ is reduced and is ∴ an oxidising agent. Cu is oxidised and is ∴ a reducing agent.

d

Br2 is reduced and is ∴ an oxidising agent. I– is oxidised and is ∴ a reducing agent.



4

a

Mg(s) → Mg2+(aq) + 2e–

b

Na(s) → Na+(aq) + e–

c

2H+(aq) + 2e– → H2(g)

d

Cl2(g) + 2e– → 2Cl–(aq)

a

2Na(s) → 2Na+(s) + 2e– F2(g) + 2e– → 2F–(s)

b

2Ca(s) → 2Ca2+(s) + 4e– O2(g) + 4e– → 2O2–(s)

c

Mg(s) → Mg2+(aq) + 2e– 2H+(aq) + 2e– → H2(g)

d

Cu(s) → Cu2+(aq) + 2e– 2Ag+(aq) + 2e– → 2Ag(s)


a

Gold can be used in jewellery, which is in close contact with moist, slightly acidic skin, without tarnishing.

b

While it can conduct electricity very well, gold will not react with anything around it and so the electrical conductivity property remains unaltered.

2

Iron will corrode slowly in hot water, whereas copper will not.

3

Aluminium corrodes more readily in salt water than the steel that makes up the gas platform. The aluminium is a sacrificial anode.

4

Steel is an alloy of iron. Since copper is less reactive than iron it would not be a suitable sacrificial anode, as the iron would corrode before the copper.


Aluminium reacts readily in air to form Al2O3, which is dull grey and masks the shiny aluminium.

2

a

no reaction

b

no reaction

c

2Zn(s) + O2(g) → 2ZnO(s)

d

Fe(s) + H2SO4(aq) → FeSO4(aq) + H2(g)

e

no reaction

3

Observe for any reactions of titanium with cold water, hot water, cold dilute acid, warm dilute acid, and compare with reactions of other known metals. It may be necessary to find some measure of the strength of the reactions by timing how long a given piece takes to dissolve compared to the same mass of another metal. © Pearson Education Australia (a division of Pearson Australia Group Pty Ltd) 2006. This page from Chemistry Contexts 1 Teacher’s Resource second edition may be reproduced for classroom use.


a

X>Z>Y>W

b

X could be K, Na or Ca. Z could be Mg. Y could be any metal in the activity series from Al to Ni. W could be Cu, Ag, Hg, Pt or Au.

5

Investigation

6

Most metals will react with oxygen, water or other substances given enough time. Heat is often also available in the Earth’s crust, allowing the more unreactive metals to react.

7

a

2Al(s) + 6HCl(aq) → 2AlCl3(aq) + 3H2(g)

b

2Al(s) + 6H2SO4(aq) → Al2(SO4)3(aq) + 3SO2(g) + 6H2O(g)

8

9

10

Exchange species

Oxidation

Reduction

a

Oxygen

gained

lost

b

Electrons

lost

gained

An oxidising agent is the species undergoing reduction. a

Fe2O3

b

O2

c

Cl2

a

2Zn(s) → 2Zn2+(aq) + 4e– O2 + 4e– → 2O2–(aq)

oxidation reduction

b

Mg(s) → Mg2+(aq) + 2e– 2H+(aq) + 2e– → H2(g)

oxidation reduction

c

2H2(g) → 4H+(aq) + 4e– 4e– + Sn4+(aq) → Sn(s)

oxidation reduction

11

Iron will react with hot water, whereas copper will not.

12

Magnesium is more reactive than iron. Magnesium is the sacrificial anode and will react first, leaving the iron in the steel uncorroded.

13

Investigation

14

Investigation

CHAPTER 8: THE PERIODIC TABLE Review exercise 8.1 1

The chemical and physical properties of carbon, nitrogen and oxygen are very different: for example, the formulas of their chlorides are different (CCl4, NCl3, OCl2), while carbon is a solid, and nitrogen and oxygen are both colourless gases at room © Pearson Education Australia (a division of Pearson Australia Group Pty Ltd) 2006. This page from Chemistry Contexts 1 Teacher’s Resource second edition may be reproduced for classroom use.

Solutions Manual: Module 2 Metals temperature. In constructing his table, Döbereiner paid more attention to the trends in the properties than the trend in the atomic masses. 2

Newlands conceived of the periodic nature of elements, and was convinced that the groupings should occur in octaves. To make all the known elements fit into this octave pattern, some dissimilar elements were placed together, while others occupied the same place in the ‘table’ of elements.

3

a

Mendeleev noticed that there were inconsistencies in atomic masses and trends, i.e. a trend would suggest a certain set of properties for the next element. However, the next known element did not fit the trend, so he left a gap.

b

By studying the properties of elements surrounding it above and below, a trend could be discerned. The new element, ‘eka-silicon’, would then have properties that fitted the trend between silicon and tin.

a

The noble gases were not known about at the time, and the trend from one period to the next was explainable, so there were no gaps that needed to be filled.

b

The noble gases are inert and form a very small percentage of the atmosphere. Also, they are colourless.

4

5

‘When the elements are listed in order of increasing atomic number, similar chemical properties occur periodically.’


a

Br, F, Cl

b

Na, Al, S, Cl

c

Ag, Cu, Fe,

2

metals: Li, Zn, Ce non-metals: Se, Br, Xe semi-metals: Te

3

Ca and Ba. Both elements are in Group 11, which means that they have two valence electrons. Since valence electrons determine chemical and physical properties, these two elements are most similar.

4

a

Cl

b

Be

c

Te

d

Ag

e

Hg

5

In some periodic tables, hydrogen is placed above lithium because it has some properties similar to group I elements. In other periodic tables, hydrogen is placed above fluorine because it has properties similar to group VII elements. In yet other © Pearson Education Australia (a division of Pearson Australia Group Pty Ltd) 2006. This page from Chemistry Contexts 1 Teacher’s Resource second edition may be reproduced for classroom use.

Solutions Manual: Module 2 Metals periodic tables, hydrogen is placed by itself to indicate that it does not readily fit into any group. Review exercise 8.3 1

2

a

Ionisation energy increases across the period. The number of protons increases and the number of electrons going into the same energy level increases. Each electron is as tightly bound to the nucleus as any other in the same shell. As the electrostatic force increases between the more positive nuclei and the valence electron, ionisation energies increase.

b

As the ionisation energy increases, the metallic properties decrease. Metals (e.g. Li, Na) easily lose electrons and so have low ionisation energies. Non-metals do not readily lose electrons and so have high ionisation energies. The trend is from metals, through semi-metals to non-metals.

a

The bonding changes from metallic for metals such as Na to covalent for nonmetals such as P. At the end, the noble gases are monoatomic, i.e. atoms of the element do not bond to each other.

b

The melting points increase to Si. As the charge on the metal cation increases, there are more electrostatic attractions to overcome to melt the solid. Si is a covalent network compound, and so requires a large amount of energy to break the strong covalent bonds. The melting points decrease from here as the only forces to be overcome are intermolecular forces between polyatomic elements (P4 and S8), then diatomic molecules (Cl2) and then singular atoms (Ar).

3

Moving across a period, electrical and thermal conductivity decreases. Metals on the left conduct heat and electricity as they have delocalised electrons. Non-metals on the right do not conduct electricity, as there are no electrons free to move.

4

There is an increase in combining power from 1 to 4 and then a decrease. Cl tends to gain one electron. Na has one valence electron that it loses to one Cl. Mg loses two valence electrons to two Cl atoms and Al loses three valence electrons to three Cl atoms. Si shares four electrons and so needs four Cl atoms to obtain the four extra electrons. P needs an extra three electrons and so shares with three Cl atoms; S needs two electrons and so shares with two Cl atoms. Cl atoms each need one extra electron and so share with one other Cl atom. Ar does not need any more electrons as it has a filled shell and so does not form a compound with Cl.

5

a

Atomic radius decreases.

b

Metallic character decreases.

c

Ionisation energy increases.

d

Electronegativity increases.

e

The number of oxygen (the ratio of O to the other atom) in the formula increases to carbon and then decreases.



Chemical properties are related to the number of valence electrons. Elements in the same group have the same number of valence electrons.

7

As you go down a group, the first ionisation energy decreases; that is, it is easier to remove the electron. As you go down a group, the electron in the outer shell being removed is further from the nucleus and the force of attraction is less (even though the nucleus is more positive).

8

a

PH3

b

AsH3

9

Bi as there is an increase in metallic character going down a group.


Investigation

2

3

CaCl2, CsCl, GaCl3, GeCl4, PCl3 or PCl5, BaCl2

4

The elements’ boiling points do show periodicity and have their highest values in the middle groups of the periodic table. Within a period, as atomic number increases there © Pearson Education Australia (a division of Pearson Australia Group Pty Ltd) 2006. This page from Chemistry Contexts 1 Teacher’s Resource second edition may be reproduced for classroom use.

Solutions Manual: Module 2 Metals is a sharp rise in boiling point for the first four elements, then a sharp decline and then a small decline. 5

Li, Na and K. They are all highly reactive metals.

6

The valence electrons for elements further down a group are located at lower and lower energy levels and are further and further away from the nucleus. They are held less tightly and so are more easily lost. This gives these elements an increasingly metallic character. In group IV, C and Si are non-metals, while Ge could be considered a semi-metal. Sn and Pb are both metals.

7

a

At2

b

Hat

c

MgAt2

d

covalent

8

The noble gases have a decreasing trend in ionisation energy (IE). Since argon has an IE of approximately 1500 kJ mol–1, it would be expected that the IE for krypton would be lower still.

9

a

Cs

b

Cs and F

c

i

Al

ii

Ar

10

11

12

d

Al as it does not have a complete outer shell. Argon does and thus will be very unreactive.

e

Cl

f

Na and Cs

a

Metallic character increases going down a group, so Te will be most metallic.

b

Metallic character decreases going across a period, so Al will be most metallic.

c

Metallic character increases going down a group, so Tl will be most metallic.

a

As

b

K

c

O

d

Cr

a

The size of an atom is determined by the distance of the outermost electron from the nucleus. Potassium has eight more electrons than sodium and so its outermost electron is further from the nucleus than sodium’s.


Solutions Manual: Module 2 Metals

13

b

Bromine and iodine both have seven electrons in their outermost shell. An element’s properties are governed by the number of electrons in the outer shell, so Br and I have similar chemical properties as they have similar outer shells.

a

Order by atomic mass alone may give some indication of sequential order. The order is: R A E Q M D G L X J B T Z. Order given by state at room temperature: A E G J T B L Q D M R X Z. Order given by conductivity: E B A G J D L M Q R T X Z. Order given by reactivity: D R A E G J L Q T Z B M X. So the suggested table could be: R A

E

Q

M

D

G

*

L

X

#

J

B

T

Z

&

b

Gaps appear as shown above. The first missing element (between E and B) would be expected to be a solid, golden in colour, with very high conductivity and medium reactivity. It would probably have an atomic mass between 16 and 20. The second missing element (after R and D) would be a colourless gas, with low conductivity and high reactivity. Its atomic mass would be probably between 24 and 28. The missing element would be similar to #.

c

i

The new element of atomic mass 25.8 is most probably the element that would fit into the second gap # described above.

ii

It would be a colourless gas, low electrical conductivity and high reactivity.

CHAPTER 9: MEASURING THE AMOUNTS OF SUBSTANCES: THE MOLE Review exercise 9.1 1

1 of C-12 and 1 None, as each scale still represents one subatomic particle, i.e. 12 19 of F-19 are the same.

2

Mg-24. The atomic mass is calculated using weighted proportions. Since the outcome of that calculation is closer to 24 than 25 (or 26), the most abundant isotope, i.e. the one mathematically contributing most to the relative atomic mass, must be Mg-24.

3

20 × 10 + (80 × 11) 100 = 10.80 Not exactly the same, as the 20% and 80% figures were approximate only. Ar(B) =



2

3

a

Mr(O2) = 2 × Ar(O) = 2 × 16.00 = 32.00

b

Mr(CO2) = Ar(C) + 2 × Ar(O) = 12.01 + 2 × 16.00 = 44.01

c

Mr(P4) = 4 × Ar(P) = 4 × 30.97 = 123.9

d

Mr (glucose) = 180.16

a

Mr(NaCl) = Ar(Na) + Ar(Cl) = 22.99 + 35.45 = 58.44

b

Mr(Ba(NO3)2) = Ar(Ba) + 2 × (Ar(N) + 3 × Ar(O)) = 137.3 + 2 × (14.01 + 3 × 16) = 261.3

c

Mr (calcium sulfate-2-water) = 172.2

d

Mr (magnesium sulfate) = 262.9

C3H8 propane, Mr = 44.09 and CO2 carbon dioxide, Mr = 44.01


The three ‘relative mass’ terms apply to the mass of an atom, a molecule and a formula unit of an ionic substance compared to 1/12th of the mass of a C-12 atom. They have no units. The term ‘molar mass’ is the actual mass of one mole of a substance in grams.

2

Ar(N) M(N) Mr(N2) M(N2)

3

a

12.01 g mol–1

b

44.01 g mol–1

c

M(Fe2(CO3)3) = 2 × Ar(Fe) + 3(Ar(C) + 3 × Ar(O)) = 291.7 g mol–1

d

M(UF6) = 6 × Ar(U) + 6(Ar(F) = 352.0 g mol–1

= = = =

14.01 14.01 g mol–1 28.02 28.02 g mol–1



2

3

a

N = 3.5 × NA = 2.108 × 1024

b

n =

a

n(Fe3+) n(CO32–) n(C) n(O)

b

1 mol (NH4)2HPO4 contains 9 mol hydrogen atoms. Therefore 2 mol hydrogen atoms will be in 2/9 mol (NH4)2HPO4.

3.10 × 10 22

6.022 × 10 23 = 0.05 = = = =

2 × 6 = 12 3 × 6 = 18 3 × 6 = 18 3 × 3 × 6 = 54

1 mol Ca(HCO3)2 has 6 mol oxygen atoms. Therefore, 0.5 mol of Ca(HCO3)2 has 3 mol oxygen atoms.


2

3

4

a

1 mol of magnesium atoms = 24.3 g ∴ 3.5 mol = 24.3 × 3.5 = 85.05 g

b

1 mol of CO2 molecules = 44.01 g ∴ 0.750 mol = 44.01 × 0.750 = 33.01 g

c

7.154 g

a

n(O2) =

64 =2 32

b

n(Ar) =

19.95 = 0.5 39.9

c

14.98

d

5.01 × 10–3

20.4 = 0.2 moles 101.96 There are 0.2 × 2 mol aluminium present = 0.4 × 26.98 = 10.79 g 20.4 g of aluminium oxide represents

15 = 1.07 moles 14.01 There are 2 mol nitrogen for every 1 mole of urea. ∴ 1.07 mol nitrogen must be in 0.535 mol urea = 0.535 × 60.062 = 32.13 g 15 g of nitrogen represents



M(CH4) = 16.042 12.01 × 100 %C = 16.042 = 74.87%

a

%H b

M(H2C2O4.2H2O) %C

%H %O 2

= 100% – 74.87% = 25.13% = 126.068 12.01 × 2 × 100 = 126.068 = 19.05% 1.008 × 6 × 100 126.068 = 4.797% =

= 100% – 19.05% – 4.797% = 76.15%

% N in NH4NO3 = 35% % N in (NH4)2SO4 = 21.2% % N in NH3 = 82.25% For each gram of compound, ammonia has the highest nitrogen level and would make the best nitrogenous fertiliser.


Elements % Mass (in 100 g) Mole ratio

C 9.9 9.9

F 31.4 31.4

1.5 58.7 58.7

9.9 12.01

:

31.4 19.00

:

58.7 35.45

= 0.8243 = 1 Empirical formula is CF2Cl2.

: :

1.6526 2

: :

1.6559 2



Elements Mass ratio Mole ratio

= =

C 1.86 1.86 12.01 0.1549 1.997 2

≈ Empirical formula is C2H5Cl. 3

H 0.39 0.39 1.008 0.3869 4.99 5

: : : :

Relative empirical formula mass

=

Relative molecular mass

= =

: : : :

Cl 2.75 2.75 35.45 0.0776 1 1

5 × 12.01 + 7 × 1.008 + 14.01 81.12 162

∴ Molecular formula is twice the value of the empirical formula and is C10H14N2. 4

Elements Mass (in 100 g) Mole ratio

C 52.2

= =

O 34.8

H 13.0

52.2 12.01

:

34.8 16.00

:

13.0 1.008

4.3464 1.998 2

: : :

2.175 1 1

: : :

12.897 5.93 6

≈ Empirical formula is C2H6O Relative empirical formula mass Relative molecular mass ∴ Empirical formula

= = =

46.1 g 46.068 g molecular formula C2H6O


a

b

2H2(g) volume ratio 2 200 L i.e. 200 L of H2 needed

+

O2(g) 1 100 L

→

2H2O(g) 2 200 L

200 L of H2O produced



a

b

2NH3(g) volume ratio 2 20 L i.e. 20 L of NH3 decomposed

→

N2(g) 1 10 L

+

3H2(g) 3 30 L

30 L of H2 produced

3

4

volume ratio

4NH3(g) 4 4×5 4

=

+

5O2(g) 5 5×5 4

a

5 L of NH3 and 6.25 L of O2 needed

b

7.5 L of H2O produced

→

4NO(g) 4 5

+

6H2O(g) 6 6×5 4


a

1 mole CH3OH will be produced for every 2 mol H2 consumed.

b

2 moles Fe will be produced for every 3 mol CO consumed.

c

4 moles HCl will be produced for every 1 mole of TiCl4 consumed.

d

7 moles O2 will be consumed for every 2 mol C2H6 consumed.

2

Mole ratio

2 Mg(s) 2 8

a

4 mol O2

b

8 moles MgO

+

O2(g) 1 4

→

2MgO(s) 2 8


Solutions Manual: Module 2 Metals 4HNO3(aq) 4 12

3

mole ratio

4

→

4NO2(g) 4 12

a

12 moles HNO3

b

12 moles NO2

a

19.25 moles O2

b

11 moles CO2 and 16.5 moles H2O

+

2H2O(l) 2 6

+

O2(g) 1 3

16.3 = 0.2 moles 81.39 0.2 moles ZnO gives 0.2 moles Zn (1:1 mole ratio from equation) 0.2 moles Zn = 0.2 × 65.39 = 13.10 g

5

16.3 g ZnO =

6

25.4 g Cu = 0.4 moles 0.4 moles Cu comes from 0.2 mol Cu2S (2:1 mole ratio) 0.2 moles Cu2S = 0.2 × 159.17 = 31.83 g

7

91.2 g C8H18 = 0.798 moles a

0.798 moles C8H18 use 9.98 moles O2 = 319.4 g

b

0.798 moles C8H18 produce 6.384 moles CO2 = 281.0 g

c

0.798 moles C8H18 produce 7.182 moles H2O = 129.5 g


A limiting reagent is one that is completely used up in a reaction by other reagents that are in excess.

2

mole ratio a i b i c i d i 3

2Al 2 Cl2 neither Al Cl2

+ ii ii ii ii

3Cl2 3 4 moles 6 moles 6 moles 4 moles

→ iii iii iii iii

2AlCl3 2 2 moles none 3 moles 5 moles

2.48 g NaOH = 0.062 moles, 1.30 g H2S = 0.038 moles a

Limiting reagent is NaOH.

b

0.062 moles NaOH produce 0.031 moles Na2S = 2.42 g

c

H2S is in excess by 0.038 – 0.031 moles = 0.2386 g



1 kg CuFeS2 = 5.448 moles, 1 kg SiO2 = 16.64 moles a

Limiting reagent is CuFeS2.

b

5.448 moles CuFeS2 produces 5.448 moles Cu = 346.2 g

c

5.448 moles CuFeS2 produces 10.896 moles SO2 = 698.1 g


Ar(Ag) = 108

2

Ar(Ne) =

3

a

Mr(O3) = 48

b

Mr(PCl3) = 137.3

c

Mr(Ca(HCO3)2) = 162.1

d

Mr(Fe3(PO4)2) = 357.5

a

1 mole

b

2 moles

c

1 mole

d

2 moles

e

9 moles

f

10 moles

4

90.9 × 20 + 0.3 × 21 + 8.8 × 22 100 = 20.18

5

Investigation

6

a

2.6498 g

b

126.036 g

c

5.134 g

d

25.047 mol of Fe = 1399 g

e

0.002616 mol of Cl = 0.2823 g

a

14.2 g = 0.2 mol of chlorine molecules ∴ 0.4 mol of chlorine atoms

b

20.53 mol of NaCl formula units

c

0.0851 mol SO2 molecules ∴ 0.0851 mol of sulfur atoms and 0.1701 mol of oxygen atoms

d

0.002839 mol

7



9

a

% Ni = 64.67%; % S = 35.33%

b

% H = 3.09%; % P = 31.60%; % O = 65.31%

c

% C = 45.89%; % H = 2.75%; % N = 7.65%; % O = 26.20%; % S = 17.51%

a

CuO

b

CF2

c

U3O8

10

empirical formula is C3H8O3 relative empirical mass = 92.094 relative molecular mass = 92.0 ∴ empirical formula is the molecular formula

11

empirical formula is CH3O3 relative empirical mass = 63.034 relative molecular mass = 126 ∴ the molecular formula is twice the empirical formula, i.e. C2H6O6. However, we are told it is a dihydrate (two water). So the correct formula is C2H2O4.2H2O.

12

empirical formula is CH3O relative empirical mass = 31.034 relative molecular mass = 62.07 ∴ molecular formula is twice the empirical formula, i.e. C2H6O2

13

14

2.024 − 0.989 × 100 2.024 = 51.14% 51.14 relative molecular mass water = So 100 relative formula mass of compound 18 x 51.14 = 120.38 + 18 x 100 x=7 Molecular formula is MgSO4.7H2O, so there are 7 molecules of water per formula unit of magnesium sulfate.

% H2O =

a b

15

1 3 1 6

a

22 kg of CuFeS2 = 119.9 moles 119.9 moles CuFeS2 produce 119.9 mol Cu = 7619.6 g

b

119.9 mol CuFeS2 produce 119.9 × 2 moles SO2 = 15364.0 g

c

119.9 mol CuFeS2 consume 119.9 × 2.5 moles O2 = 9592.0 g



14 g TiCl4 = 0.074 moles a

0.074 moles TiCl4 produce 0.074 mol TiO2 = 5.896 g

b

0.074 moles TiCl4 consume 2 × 0.074 mol H2O = 2.664 g

c

0.074 moles TiCl4 produce 4 × 0.074 mol HCl = 10.764 g

17

2.85 g of Fe = 0.051 moles Assuming 100% purity, we would expect 0.0255 mol Fe2O3. 0.0250 × 100 = 98.22% We only achieve 4 g = 0.0250 moles ∴ purity = 0.0255

18

300 mg aspirin = 0.001665 mol. ∴ 0.001665 mol salicylic acid needed (1:1 mole ratio from equation) = 0.22997 g

19

5.25 g of CO2 = 0.1193 moles

20

a

0.1193 mol of ethanol must be made (1:1 mole ratio from equation) = 5.496 g

b

1/2 × 0.1193 mol of glucose used = 10.748 g

a

diagram ii

b

The diagram in the text shows 12 H2 and 6 N2, and the ratio of H2:N2 from the equation is 3:1. So the limiting reagent is H2 (2 N2 left).

21

mole ratio we have reacting left over

2SO2 2 3 2 1

+

O2 1 1 1 0

+

2H2O 2 5 2 3

→

2H2SO4 2 – – 2

1 mol SO2 in excess no O2 left over (limiting reagent) 3 mol H2O in excess 2 mol H2SO4 produced 22

23

2.17 g of MnO2 = 0.0250 moles 2.74 g of HCl = 0.0752 moles mole ratio from equation of MnO2: HCl = 1 : 4, and we have a ratio of 1 : 3 a

All HCl is used, so HCl is the limiting reagent.

b

0.0188 mol Cl2 is produced = 1.333 g

c

(0.0250 – (0.0752/4)) mol MnO2 in excess = 0.539 g

a

15 g NH3 = 0.881 moles, and so 0.4405 mol ammonium sulfate expected = 58.21 g

b

85% × 58.21 = 49.482 g


Solutions Manual: Module 2 Metals Module 2 REVIEW 1

C

2

D

3

A

4

B

5

A

6

A

7

D

8

A

9

B

3.85 g = 0.024 moles 2FeCl3 → 2Fe + 3Cl2 ratio FeCl3 : Cl2 = 1 : 1.5

10

A

0.4 mol oxygen

11

a

Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g)

b

5000 g of ore

a

Ca, Sr and Ba are all metals. They all form dihalides (e.g. BaCl2) and monoxides (e.g. BaO).

b

i

They are arranged according to increasing atomic number

ii

Mendeleev arranged them according to increasing atomic mass (although he realised that there were anomalies and would rearrange according to chemical trends rather than atomic mass).

12

13

14

= 0.9 × 5000 g of magnetite = 4500 g 4500 4500 g Fe3O4 = = 19.434 moles 231.55 19.434 moles Fe3O4 gives 3 × 19.434 moles Fe = 3256 g

a

Coke (carbon) is oxidised to carbon dioxide. It is the reductant for iron ore that is reduced from the oxide to the metal.

b

The limestone gets rid of P4O10 and SiO2 impurities to form slag.

c

It oxidises the impurities, i.e. C, S, P and Si are converted to CO2, SO2, P4O10 and SiO2 , e.g. C + O2 → CO2

a

3Cu + 8HNO3 → 3Cu(NO3)2 + 2NO + 4H2O 2.5 g Cu = 0.0393 moles mass of copper nitrate= 0.0393 × (63.55 + (2 × 14.01) + (6 × 16)) = 7.378 g


Solutions Manual: Module 2 Metals b

If 0.0393 mole Cu reacts then = 0.0262 mol mass = n × Mr = 0.786 g

15

16

17

0.0393 × 2 moles NO form 3

c

Cu is oxidised to Cu2+, so something else must be reduced. Nitrogen is reduced by gaining 3 electrons to go from NO3– to NO.

a

The general trend is for increasing ionisation energy as the atomic radius decreases and the outer electron is held more tightly by the increasing nuclear charge. More energy is needed to remove tightly held electrons.

b

The general trend is for decreasing ionisation energy as the outer electron is shielded from the nucleus by one extra complete electron shell as you descend the group and so it is held less tightly.

Answer should make note of: Döbereiner

early 1800s

Triads of substances with similar physical and chemical properties

Newlands

mid 1800s

Arranged substances according to atomic mass, and recognised a pattern of properties for every eighth element; ‘law of octaves’

Mendeleev

late 1800s

Organised a ‘table’ with rows and columns; left gaps where elements did not fit the pattern

a

Al > Fe > Cu Al must be most reactive as the energy required is so high that only electricity will achieve extraction. Fe is extracted by chemical means. Cu can be extracted by less extreme chemical means.

b

The conditions for extraction of Cu were to be found in early kilns and fires. The conditions necessary for Al were not available until electricity was available in mass production.

c

Aluminium metal is more expensive than iron as aluminium is more expensive to extract from the mineral. Aluminium is extracted by electrolysis, which requires large amounts of electricity, and this is very costly. Iron is extracted by heating in a furnace, which does not require as much power. The equipment required for the aluminium extraction is also more complicated and expensive.


Solutions Manual: Module 3 Water

CHAPTER 10: WATER ON EARTH Review exercise 10.1 1

a

liquid

b

liquid

c

gas

2

Oceans, lakes, rivers

3

When some compounds crystallise, they do so with molecules of water as part of the lattice structure. The water is not chemically bonded to the compound itself, but it is held within the structure with some force. Heat can usually remove the water of crystallisation from a crystal. Magnesium sulfate can crystallise with seven molecules of water per formula unit (MgSO4.7H2O) in the crystal structure. If these crystals are heated, anhydrous (without water) magnesium sulfate (MgSO4) results.


A compound is a pure substance made up of two or more elements combined in a fixed ratio. A mixture can be made up of many substances in a range of compositions. A solution is an example of an homogeneous mixture (which has a fairly definite composition), although the parts of the solution are not chemically bonded together.

2

a

Unsaturated (could dissolve 6 g more).

b

Saturated (2 g more than it can normally hold).

c

Saturated (equivalent to 36 g NaCl in 100 g of water).


Living things that are multicellular need to be able to transport nutrients from where they enter the body to individual cells and then takes wastes from those cells to areas where they are excreted. Water acts as a solvent for oxygen and many nutrients to organs, as well as a solvent to remove wastes such as urea and carbon dioxide. For plants, water is an essential reactant in the production of glucose via photosynthesis. For both multicellular and unicellular organisms, water facilitates the transfer of gases into cells. The high specific heat capacity of water means that it is an efficient transporter of heat around body systems.

2

The heat capacity of water means that it acts as a heat sink during the day and a heat source at night. It therefore moderates temperatures and creates air currents (onshore and offshore breezes). Thus the climates of areas in close proximity to large bodies of water are moderated.


Solutions Manual: Module 3 Water 3

Water can cause the physical breakdown of rocks by wearing them away as it falls on them, e.g. waterfalls. In desert areas, particularly where there is a large difference in day and night temperatures, water can reside inside cracks in rocks. As it freezes at night it expands and causes cracks. Water can also cause the chemical breakdown of rocks: as chemicals such as carbon dioxide, some pollutants and other minerals dissolve in water, they create slightly acidic solutions.


As temperature increases, particles have more energy and move more quickly and move away from each other as intermolecular forces are overcome. This means that most substances expand, i.e. their volume increases. Density is defined as mass divided by volume, so as the substance is heated, the volume increases but the mass stays the same, and so density decreases.

2

1 cm–3 is the same as 1 mL and 1 dm–3 is the same as 1 L. There are 1000 mL in 1 L and so the mass would be 1000 × 0.917 = 917 g.

3

a

As water cools to 4ºC it contracts (as expected), i.e. its density increases. Cooling from 4ºC causes the density to decrease as the water expands, which is different from the behaviour of most other compounds.

b

Ice is less dense than liquid water and so floats on top, allowing aquatic life to exist under the ice. The ice forms an insulating layer, so the water below remains in a liquid state. If the ice was more dense than the liquid water, it would sink to the bottom, exposing more liquid water to the same freezing conditions, so that eventually all the water would freeze and the life forms would perish.

c

Water is most dense at 4ºC.


Investigation. Some examples of terms are: aquatic: usually meaning a water environment aquamarine: a description of a particular colour blue (presumably as in a marine environment) aquapuncture: puncture of the skin with a fine jet of water under pressure aqueduct: a human-made channel to transport water from one place to another

2

Investigation

3

All answers are approximate as there is no grid to the graph. a b

i

36 g in 100 g water

ii

93 g in 100 g water

i

34ºC

ii

56ºC



4

c

At 0ºC ammonium chloride has a lower solubility than lithium sulfate. At 100ºC the reverse is true.

d

36 g in 100 g is saturated at 50ºC, so 72 g would be needed to saturate 200 g of water.

e

From 100ºC to just over 53ºC, the solution would remain unsaturated and so a colourless solution would be observed. From 53ºC, as the solution cooled it would either:

a

b

•

be supersaturated, in which case a colourless solution would still be observed, OR

•

the ammonium chloride would start coming out of solution, i.e. white crystals would form in the solution and slowly settle at the bottom of the container.

If the student put the crystal into the solution and it: •

dissolved, then the solution was originally unsaturated

•

remained the same, i.e. did not dissolve, and no other crystals came out of solution, then the original solution was saturated

•

remained the same, i.e. did not dissolve and some of the dissolved solute came out of solution, then the solution was originally supersaturated.

The temperature and pressure would need to be kept constant. Stirring or agitation of the solution when the crystal was added would also need to be kept constant.

5

Investigation

6

Investigation

7

Investigation

8

Notable features are the extreme drop in density at 0ºC compared to the slow drop from 100ºC to 0ºC.



The water will always be warmer than the air, since it takes longer to cool (higher specific heat capacity). The warmest water is the most dense and so it sinks to the bottom. Cooler water has expanded, so its density is lower, and so it rises.

10

Water is essential to carbon-based life forms because it acts as a solvent for nutrients and wastes. Both unicellular and multicellular life forms need to be able to exchange gases, and water allows for that exchange.

CHAPTER 11: THE MOLECULAR STRUCTURE OF WATER AND HYDROGEN BONDING Review exercise 11.1 1

The atoms in molecules will arrange themselves in order to decrease the repulsion between negative charges. The arrangement of atoms in a molecule gives the molecule its shape.

2

a

i

four

ii

one

iii

linear H—Cl

i

two

ii

two

iii

linear Cl—Be—Cl

i

three

ii

three

b

c


Solutions Manual: Module 3 Water iii

triangular planar

d

i

four

ii

four

iii

tetrahedral

four

ii

three

iii

pyramidal

four

ii

two

iii

bent

e

i

f

i


Solutions Manual: Module 3 Water g

four

ii

four

iii

tetrahedral

i

four

ii

four

iii

tetrahedral

a

bent

b

tetrahedral

c

bent

d

trigonal pyramidal

i

h

3


a

difference in electronegativities = 1.0 ∴ polar

b

difference in electronegativities = 0.0 ∴ not polar

c


d


e


f

difference in electronegativities = 0.0 ∴ not polar

g


h




3

4

a

A polar bond is a covalent bond in which the shared pair of electrons are more likely to be nearer the atom with the higher electronegativity. Thus one end of the bond is slightly negative and the other end is relatively positive. A polar molecule is one in which one part of the molecule is slightly negative relative to the other end of the molecule.

b

If the polarities of the bonds ‘cancel each other out’, i.e. the polar bonds are symmetrical around the molecule, then the molecule will be non-polar.

c

There must be at least one polar bond; the molecule must be asymmetrical.

a

The Be-Cl bond is polar and the molecule is bent ∴ a polar molecule.

b

The C-F bond is polar and the bonds are symmetrically placed around the tetrahedral structure ∴ a polar molecule.

c

The O-Cl bond is polar and the molecule is bent ∴ a polar molecule.

d

The F-Cl bond is polar and the molecule is linear ∴ a polar molecule.

e

The Br-H bond is polar and the molecule is linear ∴ a polar molecule.

f

The C-F bond is polar, while the C-H is less polar. The bonds are not symmetrically placed around the tetrahedral structure ∴ a polar molecule.

g

The C-Cl bond is polar and the C-F bond is more polar and the molecule is tetrahedral. The bonds are not symmetrically placed around the tetrahedral structure ∴ a polar molecule.

h

The P-F bond is polar and the molecule is trigonal planar. The bonds are symmetrically placed around the structure ∴ a non-polar molecule.

a

ionic

b

pure covalent

c

polar covalent

d

ionic

e

polar covalent

f

polar covalent

g

ionic

h

polar covalent


dispersion forces < dipole–dipole interactions < hydrogen bonds

2

a

The only forces acting between these molecules are dispersion forces: the bonding within each molecule is pure covalent and so the molecules are nonpolar. As the mass of the molecule increases (from fluorine to iodine), the electron


Solutions Manual: Module 3 Water cloud gets larger and so the temporary dipoles are stronger as more electrons are displaced. This results in stronger intermolecular forces and therefore increased melting points.

3

4

b

The only forces acting between these molecules are dispersion forces. As the mass of the molecule increases (from carbon tetrafluoride to carbon tetraiodide), the electron cloud gets larger and so the temporary dipoles are stronger as more electrons are displaced. This results in increased melting points. Carbon tetrachloride decomposes rather than melts, since the intermolecular forces are sufficiently strong that they are of similar strength to the intramolecular bonds, so bonds between molecules and bonds within molecules all start breaking.

a

dispersion forces

b

hydrogen bonds

c

hydrogen bonds

d

dispersion forces

e

hydrogen bonds

f

hydrogen bonds

g

dispersion forces

h

dispersion forces

i

dispersion forces

j

hydrogen bonds

a

Nitrogen (N2). Both substances have only dispersion forces attracting particles to each other. In the case of helium, the particles are atoms. In the case of nitrogen, the particles are diatomic molecules. In nitrogen, the mass of the particle is greater, so the electron cloud is bigger and the electric dipole will be stronger, resulting in a higher boiling point than that for helium.

b

Methanol (CH3OH). Methanol will have hydrogen bonds as well as dispersion forces holding the molecules next to each other. These are stronger than the dispersion forces holding methane molecules together.

c

Hydrogen fluoride (HF) as hydrogen bonds act between HF molecules, whereas only dipole–dipole forces act between HCl molecules. Since hydrogen bonds are stronger than dipole–dipole forces, HF will have a higher boiling point.


acetone < water < glycerol

2

Cohesive forces are those that act between particles in the liquid. In the case of water, these cohesive forces are very strong (hydrogen bonds) and cause the molecules at the surface to be pulled inwards towards the bulk of the water. In the case of petrol (mostly octane), cohesive forces are weaker and so molecules at the surface of the petrol do not © Pearson Education Australia (a division of Pearson Australia Group Pty Ltd) 2006. This page from Chemistry Contexts 1 Teacher’s Resource second edition may be reproduced for classroom use.

Solutions Manual: Module 3 Water have a net force pulling them inwards. The adhesive forces between the air and petrol molecules are similar in strength to the cohesive forces. 3

Hydrogen bonds act between H2O molecules, whereas only dipole–dipole forces act between H2S molecules. As hydrogen bonds are stronger than dipole–dipole forces, H2O has higher melting and boiling points.

4

A hydrogen bond occurs between the hydrogen of one molecule and the non-bonding pair of electrons of oxygen in another water molecule. Each oxygen atom in water has two pairs of non-bonding electrons, so it can form two hydrogen bonds with hydrogen from other water molecules. Each hydrogen atom in the water molecule has a hydrogen bond with oxygen in a neighbouring water molecule. This makes four hydrogen bonds per molecule of water.


2

3

a

trigonal planar

b

tetrahedral

c

trigonal planar

d

bent

e

tetrahedral

f

trigonal pyramidal

a

The ammonia molecule and the methane are both tetrahedral in shape.

b

Carbon is surrounded by four hydrogen atoms, and the nitrogen is surrounded by three hydrogen atoms with a lone electron pair at the apex of the tetrahedron. This lone pair repels the N-H to a greater extent than the C-H bonds repel each other.

a

A bond is polar if there is a difference in electronegativities between the atoms that are covalently bonded. The deciding factor is difference in electronegativity.

b

A molecule is polar if the effects of the polar bonds do not eliminate each other— i.e. due to asymmetry of the bonds, the molecule has one end that is relatively more positive or negative than the other end. The factors are bond polarity, placement of atoms in the structure, and symmetry of the structure.



5

Polar bonds?

Polar molecule?

a

BCl3

Yes

trigonal planar—No

b

C2H4

C-H Yes, C-C No

No

c

PH3

Yes

trigonal pyramidal—Yes

d

CH3Cl

C-H & C-Cl Yes, C-C No

tetrahedral—Yes

e

SO3

Yes

trigonal planar—No

f

SiH4

Yes

tetrahedral—No

g

I2

No

No

h

CO2

Yes

linear—No

van der Waals force

Conditions

Strength

Dispersion

All covalently bonded molecules will have dispersion forces, but they are the only forces acting in pure covalently bonded molecules, e.g. I2

Weakest

Dipole–dipole

The covalently bonded molecule must be polar. There should not be an H on one molecule, which can interact with the non-bonding pair of electrons in N, O or F in another molecule, e.g. HI

Medium

Hydrogen bonds

The covalently bonded molecule must be polar and there should be an H on one molecule which can interact with the non-bonding pair of electrons in N, O or F in another molecule, e.g. HF

Strongest

Dispersion forces are present between all covalent molecules. This is the weakest type of van der Waals force. Dipole–dipole interactions are present between polar molecules. These are stronger than dispersion forces but not as strong as hydrogen bonds. Hydrogen bonds exist between the hydrogen of one molecule and the lone electron pair of a fluorine, nitrogen or oxygen atom on another molecule.



a

Covalent molecular

b

Both the B-F bond and the N-F bond would be polar but the molecules themselves would be differently shaped. The BF3 would be trigonal planar, while the NF3 would be trigonal pyramidal, since nitrogen also has a non-bonding pair of electrons that contribute to the shape. BF3 would therefore not be a polar molecule and its intermolecular forces would be dispersion only. The NF3 molecule would be polar and so dipole–dipole forces would exist between molecules.

c

The melting point of NF3 would be higher than that of BF3 because dipole–dipole forces are stronger than simple dispersion forces.

7

CO2 is a covalent molecular substance with dispersion forces between its molecules. These forces are weak and so it is a gas at room temperature. SiO2 is a covalent network substance and in order for it to melt or boil, strong covalent bonds have to be broken between atoms in a three-dimensional array.

8

a

CH4 < NH3 < SO2 < H2O

b

CH4 has only dispersion forces acting between its non-polar molecules. These forces are weak. Both NH3 and SO2 are polar and so dipole–dipole forces must be overcome. Although ammonia also has hydrogen bonding between its molecules, these must be weaker than the dipole strength of sulfur dioxide (S-O electronegativity difference is 1.2, while N-H is 0.8). Finally, water molecules have four strong hydrogen bonds between molecules. This is the strongest of the intermolecular forces.

a

In carbon dioxide, the intramolecular bonding (within the molecule) is polar covalent double bonds. The bonding between molecules is simple dispersion forces as the molecule itself has no dipole.

b

In ammonia, the bonding within a molecule is polar covalent single bonds. Between ammonia molecules is hydrogen bonding, with the hydrogen of one molecule strongly attracted to the lone electron pair of nitrogen in another ammonia molecule.

c

In hydrogen sulfide, the bonding within a molecule is polar covalent double bonds. The bonding between molecules is dipole–dipole, as the polar bonds are not distributed symmetrically around the bent molecule, hence it is polar.

a

Both O2 and H2 have dispersion forces. O2 will have the higher melting point as it is the larger molecule and so has a larger electron cloud. This produces a stronger temporary dipole.

b

Both molecules are non-polar and therefore have only dispersion forces between them. The greater molecular mass of C4H10 and therefore greater number of electrons will result in greater dispersion forces and therefore higher melting point.

9

10


Solutions Manual: Module 3 Water c

HF would have hydrogen bonds between molecules, while HCl would have dipole–dipole interactions. Since hydrogen bonds are stronger than dipole–dipole forces, HF would have a higher boiling point.

11

Investigation. The adhesive forces between the glass molecules and the water molecules are very strong. This causes any water molecule near the edge of the glass container to be drawn up and cling to the glass. Water molecules nearer the centre are still drawn to each other through cohesive forces and so a sloped surface appears around the edge of the container.

12

Investigation

13

Investigation

CHAPTER 12: WATER AS A SOLVENT Review exercise 12.1 1

2

3

Water is a polar solvent. a

polar

∴ soluble

b

polar

∴ soluble

c

non-polar ∴ insoluble

d

polar

∴ soluble (in fact, it dissociates in water)

e

polar

∴ soluble

f

non-polar ∴ insoluble

Carbon tetrachloride is a non-polar solvent. ∴ insoluble

a

ionic

b

non-polar ∴ soluble

c

ionic

∴ insoluble

d

polar

∴ insoluble

e

non-polar ∴ soluble

f

polar

∴ insoluble

Yes, solubility is a matter of relative strengths of forces within the solute and the forces within the solvent. ‘Like dissolves like’ is a rule of thumb only—each individual solute and solvent is a unique combination of forces.


Solutions Manual: Module 3 Water Review exercise 12.2 1

2

3

a

Copper sulfate is ionic and would be soluble in a polar solvent like water. The forces that would form would be ion–dipole attraction, with the negative end of water molecules surrounding the Cu2+ ion and the positive ends of water molecules surrounding the SO42– ion.

b

Sulfur dioxide is polar and would be soluble in water. The forces that would form would be dipole–dipole, with the negative end of water molecules surrounding the positive end of the SO2 molecule and the positive ends of water molecules surrounding the negative end of SO2.

c

Glucose would be expected to be soluble and the OH groups would form hydrogen bonds with the water molecules.

d

Octane is non-polar, so it would not be soluble in water.

e

Diamond is covalent network and so would not be soluble in water.

f

PVC is a polymer and so would not be soluble in water.

4

Ionisation occurs when a previously non-ionic substance produces ions when it reacts with water. Dissociation is the separation of ions from a crystal lattice away from each other as it goes into solution.

5

The left-hand structure is vitamin A, with a large non-polar component. The right-hand structure is vitamin C, with many water-soluble polar groups. © Pearson Education Australia (a division of Pearson Australia Group Pty Ltd) 2006. This page from Chemistry Contexts 1 Teacher’s Resource second edition may be reproduced for classroom use.

Solutions Manual: Module 3 Water Chapter 12—Application and investigation 1

Solutes that have similar intermolecular forces to the solvent are likely to dissolve in that solvent.

2

Urea would exist as molecules surrounded by polar water molecules. Sodium nitrate would dissolve by its ions dissociating from the lattice as they become surrounded by polar water molecules.

3

For a substance to be removed, it must dissolve in the ‘removing’ substance. Oil and grease are non-polar solutes and will dissolve in similar non-polar solvents such as kerosene or petrol. Water is a polar solvent and the non-polar oil and grease would not dissolve in it.

4

a

Ammonia is polar and methane is not, so ammonia dissolves in polar water.

b

Polar substances are surrounded by polar water molecules when they dissolve and the cations and anions in an ionic substance dissociate themselves from the crystal as they become surrounded by polar water molecules.

c

5

6

7

When sodium chloride is dissolved in water: • the ionic bonds between sodium cations and chlorine anions break • the hydrogen bonds between water molecules break • ion–dipole attractive forces between polar water molecules and the ions are set up.

a

CH3OH would be more soluble, because although both compounds have a polar OH group on the end, CH3CH2CH2CH2OH has a long non-polar hydrocarbon chain which is insoluble and which may overwhelm the effect of the soluble OH end.

b

C6H13OH would be more soluble as it has a polar OH group and so has the potential to form hydrogen bonds in water. C6H14 is non-polar and so would not be soluble.

c

MgCl2 would be more soluble as it is ionic and could dissociate in water. Cl2 is covalent molecular and is non-polar.

a

insoluble

b

soluble: ion–dipole attraction

c

soluble: N2O is linear and polar so it is dipole–dipole attraction

d

insoluble

e

insoluble

f

soluble: both N2H4 and H2O can form hydrogen bonds with each other

Investigation



Hydrogen chloride is a covalent molecular substance that contains no charged species. When dissolved in water, however, it ionises and so the solution is an electrolyte.

9

Hydrogen chloride is a covalent molecular substance which, when dissolved in water, ionises as follows: HCl(l) → H+(aq) + Cl–(aq) Sodium chloride is already an ionic species and when it dissolves in water the lattice simply breaks apart or dissociates into separate ions: NaCl(s) → Na+(aq) + Cl–(aq)

10

Investigation

11

Investigation

CHAPTER 13: SOLUBLE AND INSOLUBLE SALTS Review exercise 13.1 1

2

3

These are full ionic equations. a

Cu2+(aq) + SO42–(aq) + 2Na+(aq) + CO32–(aq) → CuCO3(s) + 2Na+(aq) + SO42–(aq)

b

Fe3+(aq) + 3Cl–(aq) + 3K+(aq) + 3OH–(aq) → 3K+(aq) + 3Cl–(aq) + Fe(OH)3(s)

c

2Ag+(aq) + 2NO3– (aq) + 2Na+(aq) + S2– (aq) → Ag2S(s) + 2Na+(aq) + 2NO3– (aq)

d

no precipitate

e

3Ca2+(aq) + 6NO3–(aq) + 6K+(aq) + PO43–(aq) → Ca3(PO4)2(s) + 6K+(aq) + 6NO3–(aq)

f

no precipitate

a

H+(aq) + CO32–(aq) → H2O(l) + CO2(g)

b

Ca2+(aq) + CO32–(aq) → CaCO3(s)

c

Pb2+(aq) + H2S(g) → PbS(s) + 2H+(aq)

d

CaCO3(s) + CO2(g) + H2O(l) →Ca2+(aq) + 2HCO3–(aq)

a

2K+(aq) + CO32–(aq) + Ba2+(aq) + 2Cl–(aq) → BaCO3(s) + 2K+(aq) + 2Cl–(aq)

b

Pb2+(aq) + 2NO3–(aq) + 2NH4+(aq) + CO32–(aq) → PbCO3(s) + 2NH4+(aq) + 2NO3–(aq)

c

Mg(s) + 2H+(aq) + 2Cl–(aq) →Mg2+(aq) + 2Cl–(aq) + H2(g)

d

2H+(aq) + 2Cl–(aq) + CaCO3(s) → Ca2+(aq) + 2Cl–(aq) + H2O(l) + CO2(g)

e

Ca2+(aq) + 2HCO3–(aq) → CaCO3(s) + CO2(g) + H2O(l)


Solutions Manual: Module 3 Water f

3Ba2+(aq) + 6NO3–(aq) + 6NH4+(aq) + PO43–(aq) → Ba3(PO4)2(s) + 6NH4+(aq) + 6NO3–(aq)


At 20°C, the rate of solution and crystallisation would be equal and 4 g of NaCl would remain in the solid state. As the temperature increased, the rate of solution would increase and the rate of crystallisation would decrease until the temperature was at 50°C and only 3 g of NaCl would be in the solid state. At this new temperature, a new equilibrium will be established and the rate of solution and crystallisation would again become equal.

2

N2(g) + 3H2(g) ↔ NH3(g)


5g = 5 × 10–3 g = 5 × 10–3 × 400 g =2g

2

a b

4.5 × 100 = 1.8% 254.5 4.5 × 10 3

mg 250 × 10 −3 kg = 18 000 ppm

3

n = cV = 0.215 × 25 × 10–3 = 0.0054 mol

4

n =

= c = = =

in 1 L in 1 × 10–3 L (i.e. 1 mL) in 400 × 1 × 10–3 L (i.e. 400 mL)

1.32 (46 + 12 + 48) 0.0125 mol n/V 0.0125/0.25 0.05 mol L–1

5

n = cV = 0.525 × 20 × 10–3 = 0.0105 mol 1 mol CuSO4.5H2O = 249 g ∴ 0.0105 mol = 2.615 g

6

a

concentration Fe(NO3)3 = 0.45 mol L–1 ∴ concentration Fe3+ = 0.45 mol L–1 ∴ concentration NO3– = 0.45 × 3 = 1.35 mol L–1


Solutions Manual: Module 3 Water b

n = cV = 0.45 × 50 × 10–3 = 0.0225 mol Fe(NO3)3 ∴ 0.0225 mol Fe3+ ∴ 0.0225 × 3 = 0.0675 mol NO3–


2

n = = = V = =

cV 2.5 × 5 12.5 mol n/c = 0.694 L

12.5/18

The number of mol remains constant, i.e. n1 = n2, so c1V1 = c2V2 c V ∴ c2 = 1 1 V2 =

4.5 × 5 × 10 −3

100 × 10 −3 = 0.225 mol L–1 3

n1 = 100 × 10–3 × 2 = 0.2 n2 = 80 × 10–3 × 2.5 = 0.2 So total number of mol = 0.4 in 180 mL c = n/V 0.4 = 180 × 10 −3 = 2.2 mol L–1


2

a

n = 50 × 10–3 × 0.25 = 0.0125

b

0.0125

c

1 mol BaSO4 = 233.37 g ∴ 0.0125 mol = 2.92 g

d

number of mol K2SO4 needed = 0.0125 V = n/c = 0.03125 L

a

1 mol Al(OH)3 = 78.004 g n = 0.0064

b

3 × 0.0064 = 0.0192

c

V = n/c = 0.128 L



3

d

0.0064 mol

e

c = n/V = 0.0064/0.5 = 0.0128 mol L–1

a

0.12

b

0.0799 (ratio of 3:2)

c

0.32 (ratio of 8:3)

d

V = n/c = 0.08 L


a

b

c

white precipitate + Pb2+(aq) + →PbSO4(s)

2NO3–(aq) 2NH4+(aq)

+ +

2NH4+(aq) 2NO3–(aq)

+

SO42–(aq)

blue precipitate + Cu2+(aq) →Cu(OH)2(s) +

2Cl–(aq) Ba2+(aq)

+ +

Ba2+(aq) 2Cl–(aq)

+

2OH–(aq)

pale yellow precipitate + NO3–(aq) Ag+(aq) →AgBr(s) + Na+(aq)

+ +

Na+(aq) NO3–(aq)

+

Br–(aq)

+ +

2K+(aq) 2NO3–(aq)

+

S2–(aq)

d

no reaction

e

white precipitate + Zn2+(aq) →ZnS(s) +

2NO3–(aq) 2K+(aq)

f

copper carbonate dissolves, bubbles produced and blue solution + 2H+(aq) + 2Cl–(aq) CuCO3(s) + 2Cl–(aq) + H2O(l) + CO2(g) →Cu2+(aq)

g

bubbles produced, Mg dissolves and colourless solution + SO42–(aq) Mg(s) + 2H0+(aq) + SO42–(aq) + H2(g) →Mg2+(aq)

2

If the system was at equilibrium there would be no colour change.

3

A saturated solution of copper (II) sulfate at equilibrium would be a constant colour and there would be no change in the absence of precipitate.

4

a

I2(C2H5 OH) ↔ I2(s)

b

As the temperature was lowered, all particles would lose kinetic energy (KE). As this occurred, fewer and fewer of the solid particles would have the energy to escape into solution, and more and more particles in solution would precipitate out. Thus the original equilibrium would be disturbed and the rate of the forward reaction (as written) would start increasing and the rate of the backward reaction would start decreasing and a new equilibrium would be established. More crystals of iodine would be observed.



51.2 g 375 × 10

−3

= 136.53 g L–1

6

0.9 × 10–2 × 500 = 4.5 g

7

1.35 × 10 3 = 450 ppm 3

8

a

n = cV = 3 × 2.5 = 7.5 mol

b

n = 0.105 × 15.5 × 10–3 = 0.00163 mol

9

10

11

12

a

14 53.492 c = n/V = 1.05 mol L–1 n =

b

0.1 mol L–1

c

0.068 mol L–1

a

n

b

n = 8 mass = 319.984 g

c

n = 0.025 mass = 6.24 g

a

density = mass ÷ volume ∴ mass = 1 × 250 = 250 g 0.51% = 1.275 g

b

n = 0.0277 mol c = 0.11 mol L–1

= = mass = =

250 × 10–3 × 0.05 0.0125 0.0125 × 105.99 1.325 g

glucose none sucrose none NaCl [Na+] = 1 mol L–1 NH4NO3 [NH4+] = 1 mol L–1 Na2SO4 [Na+] = 2 mol L–1 FeCl3 [Fe+] = 1 mol L–1

[Cl–] = 1 mol L–1 total ions = 2 mol L–1 [NO3–] = 1 mol L–1 total ions = 2 mol L–1 [SO42–] = 1 mol L–1 total ions = 3 mol L–1 [Cl–] = 3 mol L–1 total ions = 4 mol L–1



14

a

[H+] = [Cl–] = 0.105 mol L–1

b

[Na+] = 4 mol L–1

[SO42–] = 2 mol L–1

c

[Al3+] = 0.108 mol L–1

[SO42–] = 0.162 mol L–1

a

n1 = n2

so V1 = =

= 15

16

17

18

c 2V2 c1 6 × 0.5 16 187.5 mL

b

277.8 mL

a

n = = c = =

b

n1 = 0.002625; n2 = 0.012; c = n/V = 0.14625 mol L–1

a

nMgCl2 = 0.005 ∴ 0.005 mol of Mg(OH)2 makes = 0.2916 g

b

0.0417 L

a

6.5 g KMnO4 = 0.041 mol ∴ mol HCl = 8 × 0.041 = 0.329 V = n/c = 0.0548 L

b

7.278 g

15 × 10–3 × 0.5 0.0075 mol n/V 0.075 mol L–1 ntotal = 0.01463

Investigation

CHAPTER 14: THE HEAT CAPACITY OF WATER Review exercise 14.1 1

∆H = = = =

2

C =

mC ∆ T 45 000 × 4.18 × (27.5 – 18.0) 1 700 000 J 1.7 MJ

∆H m∆T = 441 J kg–1 K–1 or 0.441 J g–1 K–1


Solutions Manual: Module 3 Water Review exercise 14.2 1

a

b

c

d

2

i

reactants

ii

exothermic

iii

released

i

products

ii

endothermic

iii

absorbed

i

products

ii

endothermic

iii

absorbed

i

reactants

ii

exothermic

iii

released

Although energy is put into the system to start it off (e.g. lighting a match), more energy is given out as heat and light.


a

i

ii

exothermic

iii

increase


Solutions Manual: Module 3 Water b

c

2

i

ii

endothermic

iii

decrease

i

ii

exothermic

iii

increase

a

KBr has ionic bonds holding cations and anions together in a crystal lattice. Water has hydrogen bonds holding water molecules together in a liquid form. When KBr dissolves, the ionic bonds break down, and the hydrogen bonds are overcome. The cations and anions are surrounded by water molecules and the new forces are ion–dipole.

b

∆ H for the process is positive, which means that energy has been absorbed as KBr dissolves. KBr dissolves because the strength of the ion–dipole attraction (between ions and the water molecules) is greater than the strength of the original hydrogen bonds in the solvent and the ionic bonds in the lattice. Thus dissolved KBr is at a higher energy level than crystal KBr and liquid water—this is why the value for ∆ H for the process is positive.



∆H = =

mC ∆ T 26 668.4 J for 2.4 g 40 ∴2675.2 × = 44 586.7 J for one mol 2.4 = 44.59 kJ

4

nHCl = 50 × 10–3 × 2 = 0.1 nNaOH = 100 × 10–3 × 1 = 0.1 The enthalpy change for 0.1 mol = 5.7 kJ 5.7 = 150 × 10–3 × 4.18 × 103 × (Tfinal – 18) (Tfinal – 18) = 9.09° So the final temperature is 27.09°C.


At 15°C the solubility of O2 is about 1.6 mmol/L, compared to a considerably lower value of about 1.1 mmol/L at 40°C. Therefore less oxygen is available to aquatic animals at higher temperatures.

2

These are bubbles of dissolved gas (mixture of gases from the atmosphere including oxygen) coming out of solution. As the beaker is warmed, the solubility of dissolved gases decreases (i.e. the equilibrium between gases in solution and gases in the gaseous state shifts to the right).

3

Gas solubility decreases because all molecules gain KE as the temperature increases. The dissolved gaseous molecules have enough energy to overcome the intermolecular forces that keep them surrounded by water molecules and so are able to leave the liquid state. The forces that keep them in solution are dispersion and dipole–dipole forces, depending on the gas. These forces are weaker than the forces holding water molecules together and so the gases come out of solution before the water boils.


2

3

a

685 520 J = 685.5 kJ

b

m = 126 g; ∆ H = 2680 J; c = 2.42 J g–1 K–1 ∆T = ∆H mc = 8.79° = 2680 = 2680 ∴ initial temp = 30.0 + 8.79 = 38.79°C

a

0.1255 J g–1 K–1

b

No, as the value for the specific heat capacity is different. Aluminium has a higher specific heat capacity and will produce a higher temperature change for the same amount of heat.

Investigation © Pearson Education Australia (a division of Pearson Australia Group Pty Ltd) 2006. This page from Chemistry Contexts 1 Teacher’s Resource second edition may be reproduced for classroom use.


a

b

c

5

6

i

reactants

ii

exothermic

iii

released

i

products

ii

endothermic

iii

absorbed

i

reactants

ii

exothermic

iii

released

a

H2O(g) ↔ H2O(l) is exothermic so ∆ H is negative

b

positive

c

negative

d

positive

a

exothermic

b

endothermic

c

exothermic

d

endothermic

7

4Fe(s) + 3O2(g) → 2Fe2O3(s) + 3296 kJ

8

a

endothermic

b

The products are at a higher energy level as the reactants absorbed energy from the environment (area of sporting injury) to create products.

c

9

50 g = 1.25 moles ∴ ∆ H = 1.25 × 44500 J ∆ T = 133º © Pearson Education Australia (a division of Pearson Australia Group Pty Ltd) 2006. This page from Chemistry Contexts 1 Teacher’s Resource second edition may be reproduced for classroom use.


a

3762 J

b

nHCl = 0.1 nMg = 0.008 So the HCl is in excess and all of the magnesium is used up. 0.008 mol produces 3762 J, so 1 mol of Mg produces 47 0250 J = 470.3 kJ.

11

Changing the temperature of the water can alter the environment negatively for living creatures, e.g. if temperature of the water increases, available oxygen decreases. Factories often use water for cooling their manufacturing processes, and the water that leaves the factory is often warmer than the water entering the factory. This is a form of pollution and must be monitored so that the negative impact on the ecosystem is minimised.

12

Investigation

Module 3 REVIEW 1

C

2

B

3

C

4

C

5

B

6

Dispersion forces: arise between all molecules as the electron cloud around them is constantly changing shape and may become temporarily polar. This may then induce a temporary dipole in another molecule. Attractive forces then exist temporarily between oppositely charged ends of these molecules. Dipole–dipole forces: arise between polar molecules where permanent attractive forces exist between oppositely charged ends of molecules. Hydrogen bonds: arise between the hydrogen of one molecule and lone electron pairs in oxygen, nitrogen or fluorine of another molecule.

7

The solution process involves forces between solvent particles, as well as forces between solute particles, to be overcome by the stronger force between solvent and solute particles. This usually only happens when the solvent–solvent and solute–solute forces are of the same type (‘like’ each other).

8

a

Full ionic equation is 2K+(aq) + SO42–(aq) + 2Ag+(aq) + 2NO3–(aq) →Ag2SO4(s) + 2K+(aq) + 2NO3–(aq)

b

No, although there are charged particles in silver(I) sulfate as it is ionic, the ions are held rigidly in a lattice structure and are unable to move.



9

10

c

Yes, this solution contains charged particles (K+ and NO3–), which are able to move and so conduct electricity.

a

Molarity is the concentration of the solution measured in mol of solute dissolved per litre of solution.

b

C = n/V

c

2 mol L–1

a

CuSO4(aq) + 2NaOH(aq) → Cu(OH)2(s) + Na2SO4(aq)

b

number of mol Cu(OH)2 precipitated = 0.024 number of mol of CuSO4 available = 0.05 To produce 0.024 mol of copper(I) hydroxide, only 0.024 mol of copper sulfate was needed ∴ the CuSO4 is in excess. Only 0.024 mol will react, leaving 0.026 mol unreacted.

11

2C2H5COOH(aq) + Na2CO3(s) → 2NaC2H5COO(aq) + CO2(g) + H2O(l) nC2H5COOH = 0.25 ∴ nNa2CO3 = 0.125 mass = 13.25 g This is the reaction between an acid and a carbonate, and so the reaction will be complete when no more CO2 is produced, i.e. no more bubbles can be seen.

12

a

13

b

For each water molecule, four hydrogen bonds can form as each oxygen atom has two lone pairs of electrons, and each hydrogen atom has one hydrogen bond to another water molecule. Because of the shape of ammonia molecules, there are fewer hydrogen bonds per molecule. (The nitrogen has only one lone pair and the three hydrogens are not all involved in bonding with the nitrogen of another molecule.)

a

0.1599 mol

b

0.3198 mol L–1

c

7.995 × 10–3 mol

d

Na2CO3(aq) 1

+

2HCl(aq) → 2NaCl(aq) 2

2

→

H2O 1

+

CO2 1

–3

n Na2CO3 = 7.995 × 10 ∴ nHCl = 2 × 7.995 × 10–3 = 1.599 × 10–2 1.599 × 10 −2 VHCl = = 1.599 L 0.010 © Pearson Education Australia (a division of Pearson Australia Group Pty Ltd) 2006. This page from Chemistry Contexts 1 Teacher’s Resource second edition may be reproduced for classroom use.


15

16

a

The sodium carbonate may have some water of crystallisation attached. Since the only substance that is required to react with the HCl is pure Na2CO3, the water of crystallisation needs to be removed.

b

i

500 mL volumetric flask

ii

25 mL pipette

c

Na2CO3(s) + 2HCl(aq) → 2NaCl(aq) + CO2(g) + H2O(l)

d

Average titration volume used = 12.925 mL (first and last readings appear to be inaccurate) n(sodium carbonate) weighed out = 0.0496 moles concentration of the Na2CO3 solution in the 500 mL volumetric flask 0.0496 c Na 2 CO 3 = = 0.0992 mol L–1 0.5 moles Na2CO3 in conical flask = 2.48 × 10–3 moles HCl needed = 2 × 2.48 × 10–3 = 4.96 × 10–3 c = n/V = 0.384 mol L–1

a

n = cV = 0.50 × 0.1 = 0.050 mol

b

Full ionic equation 2Na+ (aq) + 2Cl– (aq) + Pb2+(aq) + 2NO3–(aq) → PbCl2(s) + 2Na+(aq) + 2NO3–(aq)

c

Conductivity is initially high as there are Pb2+ and NO3– ionic species in solution. As the NaCl solution is added, solid PbCl2 is formed, removing lead and chloride ions from solution. When all the Pb ions have been precipitated and excess NaCl(aq) is added, conductivity increases again as the number of free ions increases.

a

The food chemist needs to isolate the sugar from the rest of the liquid and so will need to evaporate it to dryness. Therefore she needs to know the weight of the solution before and then the weight of the sugar separately afterwards.

b

The units will be % composition (g of sugar per g of solution). The mass of sugar in a known mass of solution is easy to measure, and the unit is also meaningful for the consumers of the product.


Solutions Manual: Module 4 Energy

CHAPTER 15: SOME COMPOUNDS ARE IMPORTANT SOURCES OF ENERGY Review exercise 15.1 1

Chlorophyll traps the light energy that is used to chemically combine carbon dioxide and water to form carbohydrates and energy. The bonds in the carbohydrates store the energy from the sun as chemical energy.

2

The energy trapped in the bonds of the carbohydrates is released or transformed when animals (herbivores and omnivores) eat the plants. These animals may be eaten by other animals (omnivores and carnivores), which will use the energy again, and so on up the food chain until the animal material is decomposed and returned to the air, water and soil.

3

a

carbon dioxide + water + energy → glucose + oxygen 6CO2(g) + 6H2O(l) + energy → C6H12O6(aq) + 6O2(g)

b

oxygen + glucose → carbon dioxide + water + energy 6O2(g) + C6H12O6(aq) → 6CO2(g) + 6H2O(l) + energy


Oil, natural gas and coal.

2

Sunlight is the ultimate source of energy. Fossil fuels are made up of the remains of animals and plants. Plants convert sunlight energy into chemical energy. This chemical energy is transferred to any animals that eat the plants and so on up the food chain.

3

4

Organic matter

Peat

Brown coal (lignite)

Bituminous coal

Anthracite

a

Coal formed in environments where there was accumulated plant material and limited oxidation, such as swamps, marshes and mangroves.

b

Petroleum and natural gas formed in environments where the remains of marine organisms (especially bacteria and plankton) had been trapped and experienced high pressure and temperature.

5

The oil and natural gas are formed at depth. Because the oil and gas are less dense than the rocks around them, they will seep upwards until met by a barrier of impervious rock.

6

Initially an absence of oxygen prevented decay. This was followed by long periods of time with high pressure and temperature.


Solutions Manual: Module 4 Energy Chapter 15—Application and investigation 1

Investigation

2

Rabbits are herbivores, i.e. they eat plant material only. The plants they eat contain carbohydrates whose bonds have stored chemical potential energy. This energy originally came from the sun and was necessary for the plants to synthesis the carbohydrates from carbon dioxide and water. The chemical potential energy in the plants is used and stored by the rabbit and passed to the dingo as it eats the rabbit.

3

They are fuels because on combustion they release large amounts of heat energy. They are ‘fossil’ fuels because they are made of the remains of organisms that lived millions of years ago.

4

Solar energy is energy from the sun. Fossil fuels are made up of the remains of living organisms, and the chemicals that make up those remains contain chemical potential energy. If the remains are from a plant, this chemical potential energy was transformed from sunlight energy in the process of photosynthesis. If the remains are from animals, this chemical energy came from other animals from plants that were eaten. Ultimately all the chemical energy can be traced back to plants, and their energy is traced back to the sun.

5

Investigation

6

Investigation

7

Investigation

CHAPTER 16: THE VARIETY OF CARBON COMPOUNDS Review exercise 16.1 1

Organic: glucose, methane and petrol. Inorganic: carbon dioxide, sodium carbonate, ammonium hydrogencarbonate and carbon disulfide.

2

a

Nuclear composition: all isotopes have six protons. The major isotope, C-12, has six neutrons.

b

Electron configuration: six electrons, which in their lowest energy levels have the configuration 2, 4.

c

Valence electrons: four.

d

Position in the periodic table: group IV and period 2.


An allotrope of an element contains the atoms of that element structured in such a way that they have different physical properties from other allotropes of that element. Their chemical properties are identical. © Pearson Education Australia (a division of Pearson Australia Group Pty Ltd) 2006. This page from Chemistry Contexts 1 Teacher’s Resource second edition may be reproduced for classroom use.

Solutions Manual: Module 4 Energy 2

Physical properties are a reflection of the structure, bonding and forces that exist within the substance. Therefore if the structure and forces are different in different allotropes, so too are their physical properties.

3

Diamond: each carbon atom is covalently bonded to four other carbon atoms in a covalent network arrangement. The basic shape is a tetrahedron. Graphite: each carbon atom is covalently bonded to three other carbon atoms in layers that consist of hexagons as the basic shape. The fourth valence electron is delocalised and able to move between layers. It is a covalent network substance. Buckminsterfullerene: these are molecules of 60 carbon atoms covalently bonded to each other (each carbon atom forms three bonds). There are 12 pentagons and 20 hexagons in the shape of a ‘soccer ball’.

4

Graphite has delocalised electrons able to move between the layers of carbon atoms. These delocalised electrons arise because each carbon atom only forms three covalent bonds. Diamond has no delocalised electrons or mobile electrons. This is because each carbon atom is covalently linked to four other carbon atoms, i.e. all four valence electrons are used in bonding.

5

In order to melt diamond and graphite, covalent bonds must be broken, as each allotrope is a covalent network substance. Covalent bonds are the strongest bonds of all and so require lots of energy to break.


Silicon belongs to group IV, as does carbon. This means it has four valence electrons, as does carbon, and so forms four covalent bonds.

2

a

b

c

There are four regions of negative charge corresponding to the four covalent bonds (three C–Cl bonds and one C–H bond). The shape that offers maximum distance between these bonds is the tetrahedron.

a

b

3


Solutions Manual: Module 4 Energy c

The double bond between the two carbon atoms makes the molecule planar. The C–H and C–Cl bonds that come from that double bond are found at 120° from the carbon atoms and in the same plane.


2

a

b

c

d

a

b

c

d


Solutions Manual: Module 4 Energy e

f


Carbon easily forms a large number of bonds around it (4) and the nucleus is sufficiently small that it can be constructed into many shapes with a variety of other atoms bound to it. But most importantly, it can bond with other carbon atoms to form chains and rings of almost any length.

2

Investigation

3

In diamond, each carbon atom is covalently bonded to four other carbon atoms in a covalent network arrangement. The basic shape is a tetrahedron. These bonds are very strong and so diamond has a high m.p. The structure is also very strong and because there are no delocalised electrons to form new bonds in a different shape, it does not deform. Therefore it can be used as an abrasive. In graphite, each carbon atom is covalently bonded to three other carbon atoms in layers that consist of hexagons as the basic shape. These bonds are very strong and so graphite has a high m.p. The fourth valence electron is delocalised and able to move between layers. It is a covalent network substance, but the layers can slide past one another easily, as the delocalised electrons can make new temporary bonds between new positions on the layers. Therefore it can be used as a lubricant.

4

Investigation

5

There are four regions of negative charge corresponding to shared electron pairs in the four covalent bonds. The electrostatic repulsion between these electron pairs is minimised by forming a tetrahedral arrangement.

6

A single bond consists of one shared pair of electrons, the bond length is 0.154 nm and the arrangement of atoms is a tetrahedron around each C atom (the other three corners taken up with the other three bonds of that C atom). A double bond consists of two shared pairs of electrons, the bond length is 0.134 nm and the arrangement of atoms is planar around each C atom. A triple bond consists of three shared pairs of electrons, the bond length is 0.120 nm and the arrangement of atoms is linear.

7

The structural formula of ethyne is H–C≡C–H. It has a linear structure. There are three regions of negative charge: the two C–H bonds and the C≡C bond. Minimum electrostatic repulsion is achieved along a single line.



a

b

c

d

e

f

g

h

or

9

i

ii

a

tetrahedral

b

tetrahedral around each carbon

iii



10

c

planar

d

planar

e

O=C=O

linear

O=C=O

f

H—C≡N

linear

H—C≡N

a

The valency of carbon is shown as 5, rather than 4. One of the hydrogen, bromine or chlorine atoms must go for the formula to be correct, e.g. CHBr2Cl or CH2BrCl or CH2Br2.

b

The valency of the first carbon is shown as 5, but it should be 4, i.e. CH2=CBr2.

c

The valency of oxygen is shown as 2, hydrogen as 1 and carbon as 4. There is no way this structure will allow all of these valences to be correct. Possible solutions are CH3CH2OH (where a hydroxyl group is bonded to the second carbon through oxygen) or CH3CHO (where the oxygen is double bonded to the carbon).

d

Hydrogen cannot have a valency of 2. A possible solution is CCl3H or CCl2H2.

e

The chlorine atom and the hydroxy group are incorrectly oriented relative to each other.

f

This is a planar molecule and all the hydrogen atoms are in the same plane.


Solutions Manual: Module 4 Energy CHAPTER 17: HYDROCARBONS Review exercise 17.1 1

Hydrocarbons are a, c, e.

2

Both alkenes and alkynes have multiple carbon–carbon bonds. Alkenes have at least one double bond, and alkynes have at least one triple bond. If these bonds were broken and changed to single bonds, the associated carbons could have more atoms attached to them, so the groups are considered unsaturated.

3

a

3

b

2

c

8

d

6

e

1

f

4


two carbons: ethane

2

a

CnH2n

b

C5H10 and C7H14

c

Pentenes:

four: butane

six: hexane

eight: octane


Solutions Manual: Module 4 Energy Heptenes:

3

If an organic molecule can have a functional group, e.g. a methyl group or a double bond, in a number of places then those molecules are considered position structural isomers.

e.g. CH3—CH2—CH=CH2 (1-butene) and CH3—CH=CH—CH3 (2-butene) 4

5

a

1-butene

b

3-hexene

a

b

6

a

2-pentene

b

3-hexene


a

Crude oil is a mixture of liquids of different boiling points. Fractional distillation is used to separate these liquids from the mixture.

b

The crude oil is heated to about 400°C to form a hot liquid/vapour mixture. As the vapours rise through the fractionating tower, they condense at different levels according to their boiling point (lower b.p. fractions rise furthest.) Each fraction is a distinct substance that has a distinct set of uses.



a

The main use of the gasoline fraction is in motor fuels. Pentane and hexane.

b

The main use of the refinery gas fraction is as starting material for the manufacture of plastics. Methane and propane.

3

C8H18

4

Natural gas, liquefied petroleum gas (LPG), petrol fuel, diesel fuel and aviation fuel.

5

Lubricating oils, waxes for wrapping/cooking paper, tar for roads, asphalt for roads.


Alkanes are non-polar and so the intermolecular forces are dispersion forces. The force of attraction is between temporary dipoles set up in two molecules when electrons move randomly and create a temporary dipole in a molecule by their uneven distribution.

2

The trend is that boiling points increase as molecular mass increases. As the carbon chain gets longer, there is a larger electron cloud and a larger surface area over which these intermolecular dispersion forces can act.

3

Butane because there is greater surface contact between the chain-like butane molecules than between the branched methyl propane molecules.

4

Grease is a long chain hydrocarbon and has dispersion forces acting between its molecules—that is, it is non-polar. It is more likely to dissolve in a non-polar solvent such as kerosene or petrol than a polar solvent like water. This is because the dispersion forces that would act between water and the grease molecules are not as strong as the hydrogen bonds that already exist between water molecules and so the grease never breaks apart the water’s structure, i.e. the grease never dissolves into the water.


2

a

As the temperature increases, the average molecular velocity of the molecules also increases.

b

As the temperature increases, the average kinetic energy of the molecules also increases.

a

Evaporation is the process of particles leaving the liquid phase and entering the gaseous phase. This can occur at any temperature. Boiling only occurs at a specific temperature (for a given pressure).

b

Evaporation occurs from the surface of the liquid. Boiling occurs as particles enter the gaseous phase throughout the body of the liquid.



As the water particles in the wet clothes are exposed to the higher temperature of the air around them, they start to move faster. Their KE is likely to be sufficient to allow them to escape from the liquid phase and enter the gaseous phase: this makes the clothes dry. On a cold day, not as many water particles in the liquid phase will have sufficient KE to escape at the same rate as on the hot day.

4

a

Boiling point is defined as the temperature at which vapour pressure is equal to atmospheric pressure. As vapour pressure increases with increasing temperature, you would expect pentane to reach atmospheric pressure (101.3 kPa) first and so have the lower boiling point. Water will have the higher b.p.

b

The lower vapour pressure of water indicates that water molecules have much stronger intermolecular forces than pentane, since far fewer water particles have sufficient kinetic energy to escape the liquid state to become a vapour. The magnitude of the difference (2.34 kPa compared to pentane 56.6 kPa) suggests hydrogen bonding in water with weak dispersion forces in pentane.

5

The boiling point of trichloromethane (chloroform) would be just over 60°C.

6

Boiling point is defined as the temperature at which vapour pressure is equal to atmospheric pressure. If the pressure in the ‘atmosphere’ above the water is more than 101.3 kPa, then it will require a higher temperature to get the vapour pressure higher. A higher vapour pressure means that the molecules have more kinetic energy. This is a reflection of the higher temperature. The cooking of food requires bonds within the food to be broken and this requires both time and heat energy. If the heat energy is greater, then more bonds will be broken in a shorter time. Hence cooking times are decreased.


Flash point is the temperature at which an explosive mixture exists if a flame is added. Ignition temperature is the temperature at which a fuel/air mixture will spontaneously ignite.

2

This fuel’s flash point is below room temperature, so at room temperature any spark or flame will cause an explosion. So there is a danger.

3

a

LPG: use in a well-ventilated area, no naked flames or sparks. Store in sturdy pressurised containers, which must be pressure checked, by law, every nine years or discarded.

b

Petrol: no naked flames or electric sparks, use in a well-ventilated area. Store in sturdy metal or anti-static plastic containers with a narrow neck (to reduce the surface area for evaporation) and tight-fitting lids. Only minimal quantities of petrol should be stored.


Solutions Manual: Module 4 Energy Chapter 17—Application and investigation 1

coal, crude oil, natural gas

2

Alkanes (CnH2n+2): C12H26, C8H18, C20H42 Alkenes (CnH2n): C7H14, C5H10, C6H12 Alkynes (CnH2n–2): C9H16, C32H62, C12H22, C15H28

3

a

1-butene

b

2-pentene

c

hexane

d

3-heptene

4

A condensed structural formula gives you structural isomeric information. That is, the structural formula tells you where in the molecule the functional groups are positioned, although it is longer to write.

5

a

b

c

d

6

a

C5H12

CH3CH2CH2CH2CH3

b

C4H8

CH2CHCH2CH3 CH3CHCHCH3



8

C2H6 is an alkane in which both carbon atoms are said to be saturated as they have a single bond between them, i.e. one shared pair of electrons. C2H4 is an alkene in which carbon atoms are said to be unsaturated as they have a double bond between them, i.e. two shared pair of electrons. C2H2 is an alkyne in which carbon atoms are said to be unsaturated as they have a triple bond between them, i.e. three shared pair of electrons.

9

Investigation

10

a

Boiling point

b

Lower molecular mass components have higher volatilities, i.e. lower boiling points, and are more likely to be gases. The top of the fractionating column will be cooler than the bottom and so these low melting points will remain as gases until they reach the top levels.

11

Investigation

12

Hydrocarbons release vast amounts of energy when they burn, i.e. the energy input to break the C–H and O=O bonds is much less than the energy released when CO2 and H2O are produced.



a, b

c

There is a general increase in melting and boiling point with increasing length of the carbon chain for both alkanes and alkenes (although this increase is not totally regular). The general trend can be explained in terms of stronger dispersion forces with increasing molecular size. The efficient crystal packing of the small molecules of CH4 and C2H6 means relatively more energy has to be put into breaking down the crystal lattice when they are solids and thus accounts for the high melting points of these alkanes.

14

Grease is a long-chain hydrocarbon and has dispersion forces acting between its molecules, i.e. it is non-polar. It is more likely to dissolve in a non-polar solvent such as kerosene or petrol than a polar solvent like water. This is because the dispersion forces that would act between water and the grease molecules are not as strong as the hydrogen bonds that already exist between water molecules, so the grease never breaks apart the water’s structure, i.e. the grease never dissolves into the water.

15

In hot weather the vapour pressure increases as more molecules in the liquid phase have sufficient KE to leave the liquid phase and enter the gaseous phase. The increased pressure caused by gas particles colliding with the container walls may cause it to bulge.

16

The process of leaving the liquid phase and entering the gas phase (whether by evaporation or boiling) requires that the particles can escape the forces holding them in the structure of the liquid. Substances that have low intermolecular forces have high volatility and low boiling points as the particles in the liquid phase are able to escape into the gaseous state. Substances that have strong intermolecular forces have low volatility and high boiling points, because they require a lot of energy to overcome their intermolecular forces.



a

ethanoic acid < ethanol < acetone < pentane

b

pentane < acetone < ethanol < ethanoic acid

c

The intermolecular forces in ethanoic acid must be the strongest of these four compounds as its b.p. is highest. The intermolecular forces of pentane are dispersion forces only and they must be the weakest of the four compounds since pentane has the lowest b.p.

d

18

a

The risk is much greater for pentane and octane than decane, although all three liquids should be transported carefully without vapour loss or exposure to naked flames. Both pentane and octane will ignite if a flame is introduced at room temperature, since their flash points are below 25°C. Decane will not combust with a flame until 46°C is reached.

b

The vapour pressure and volatility of the liquids in order from most volatile (most vapour pressure) to least is: pentane > octane > decane. The boiling points will increase from pentane to octane and then decane.

c

Decane will have the strongest intermolecular forces and pentane will have the weakest. All three only have dispersion forces operating and these will increase in strength as the molecular mass increases.


Solutions Manual: Module 4 Energy CHAPTER 18: COMBUSTION OF HYDROCARBONS Review exercise 18.1 1

There could be a change in temperature and/or odours (new gaseous chemicals) could be given off.

2

C5H12(g) + 8O2(g) → 5CO2(g) + 6H2O(l)

3

Incomplete combustion causes the black deposit. The kerosene does not have sufficient oxygen to burn completely to carbon dioxide. One of the products of incomplete combustion is solid carbon, also known as soot.

4

Combustion of the gas uses up oxygen from the air around it, so it is important to continue to provide sufficient oxygen for both the heater (for complete combustion) and for any living creatures in the room. Additionally some unburned fuel and carbon monoxide might be produced and a constant source of fresh air dilutes these toxic chemicals.


2

a

A hydrogen molecule is formed of two hydrogen atoms bonded together. Since bond formation releases energy, the enthalpy of the two hydrogen atoms must be higher than that of the hydrogen molecule.

b

A hydrogen fluoride molecule is formed of a fluorine and a hydrogen atom bonded together. Since bond formation releases energy, the enthalpy of the fluorine and the hydrogen atom must be higher than that of the hydrogen fluoride molecule.

a

bonds broken: O=O, C–H and C=C bonds formed: C=O, O–H

b

bonds broken: N–H, N–O, N=O, ionic bond between NH4+ and NO2– bonds formed: N≡N, O–H

3

Since the reaction is exothermic, energy is released. Energy is released when forming bonds and absorbed when breaking bonds. Since overall energy is released when hydrogen and chlorine react to form HCl, the energy released when the H–Cl bonds formed must be in total greater than the energy absorbed to break the H–H and Cl–Cl bonds.

4

Activation energy is the energy barrier that must be overcome to allow a reaction to occur.


Solutions Manual: Module 4 Energy Review exercise 18.3 1

2

a

Global warming: The Earth’s atmosphere normally acts as a greenhouse trapping infra-red radiation (heat) from the sun. There are certain gases (principally CO2) responsible for this trapping of heat and because these gases are increasing in concentration (due to burning of fossil fuels), the amount of heat trapped is increasing, i.e. warming the globe.

b

Acid rain: Certain oxides of sulfur, nitrogen and carbon can react with water to produce mild acids. These oxides are released into the atmosphere when fossil fuels are burnt, where they react with water vapour to form acid rain.

c

Photochemical smog is a form of pollution produced under the influence of light. Sunlight allows oxygen atoms to be created from the brown-coloured gaseous pollutant nitrogen dioxide. These atoms react with oxygen molecules to produce ozone and a wide range of toxic organic molecules.

a

Carbon monoxide results from insufficient oxygen in the combustion process, i.e. incomplete combustion. CO prevents haemoglobin picking up oxygen and transporting it around the body.

b

Carbon dioxide is produced by any combustion of hydrocarbons. It is a greenhouse gas.

c

Particulates: soot comes from incomplete combustion of hydrocarbons and ash from inorganic impurities in the fossil fuels. Both can lodge in the respiratory system and reduce the efficiency of gaseous exchange.

d

Sulfur oxides: most coals have sulfur in them. When the coal is burned, sulfur oxides form. These can react with water to produce acid rain. They are also choking toxic fumes.

e

Nitrogen oxides are formed when nitrogen reacts with oxygen at the high temperatures found in internal combustion engines. These can cause photochemical smog and acid rain.

f

Hydrocarbons are released into the environment through incomplete combustion of the hydrocarbon itself and from their use as solvents. They are usually toxic at high concentrations and can react with oxygen atoms and ozone to form a range of other toxic organic compounds.


2

a

2C6H14(l) + 19O2(g) → 12CO2(g) + 14H2O(g)

b

A number of answers are possible, e.g. C6H14(l) + 8O2(g) → CO(g) + C(s) + 4CO2(g) + 7H2O(g)

Investigation



a

bonds broken: H–H and Cl–Cl

b

bonds formed: 2 H–Cl bonds

c

∆ H = –184 kJ tells us that energy is released. Bond breaking absorbs energy and bond making releases energy. Since there is energy released, there must be more energy involved in making the bonds than is needed to break them.

4

When liquid water is vaporised, the only ‘bonds’ that need to be broken are hydrogen bonds. While these are the strongest intermolecular forces, they are nowhere near as strong as the covalent bonds between oxygen and hydrogen in a water molecule. It is these bonds that must be broken when liquid water is decomposed.

5

In order for kerosene to ignite, the vapour/air mixture must be in the correct ratio. This does not happen until the temperature is at 48°C (flash point), at which point enough vapour is being produced to cause the kerosene to ignite if there is a source of ignition. The kerosene at room temperature will not ignite with the match, but will at 80°C, which is above the flash point of 48°C.

6

Ignition temperature is the temperature at which a fuel/air mixture will spontaneously ignite. The increased energy the reactants have at this higher temperature is a direct measure of the activation energy, since they have the energy required for the reaction to proceed.

7

Investigation

8

Investigation

9

Investigation

CHAPTER 19: RATES OF CHEMICAL REACTIONS Review exercise 19.1 1

Zn(s) + H2SO4(aq) → H2(g) + ZnSO4(aq) • Measure the volume of H2 produced per unit time. • Measure the mass of Zn dissolved per unit time.

2

C7H16(l) + 11O2(g) → 7CO2(g) + 8H2O(l) • Measure the mass of C7H16 reacting per unit time. • Measure the volume of CO2 produced per unit time.


a and d If a reaction does not involve bonding rearrangements, it is likely to be rapid at room temperature, as no complex bond-breaking and bond-forming processes are required.

2

If the volume is decreased, the probability increases that gas molecules with enough energy to react will collide, as there is less room in which they can move around. © Pearson Education Australia (a division of Pearson Australia Group Pty Ltd) 2006. This page from Chemistry Contexts 1 Teacher’s Resource second edition may be reproduced for classroom use.


Many metal catalysts are used in the form of a fine wire mesh as this increases surface area. Many metal catalysts work by supplying a surface area on which the reactant particles can settle and react. By increasing the surface area, the number of reactant particles that have room to settle and then react is increased.

4

a

It is usually when there is dust above the grain that there is an explosion. This is because the grain has an increased surface area to react with oxygen in the air. This lowers the activation energy for the wheat-burning reaction.

b

Deterioration will be slowed at lower temperatures as reactant particles are less likely to have enough energy to overcome the activation energy barrier.

c

Air is only 20% oxygen. By reacting with pure oxygen, the proportion of collisions between reactant particles increases, and so the reaction is faster.

5

Rate might be increased by: • increasing temperature (more particles will have energy to overcome the activation energy barrier) • decreasing volume (more particles colliding) • increasing concentration of a reactant, e.g. NH3 or O2 (more reactant collisions).


Factors affecting whether a particular collision between reactant molecules will lead to a reaction are: • orientation of the particles to each other • energy of the colliding particles.

2

They ‘bounce’ off each other if they have sufficient energy to overcome intermolecular forces.

3

Activation energy for the phosphorus reaction is very much lower than that for the rusting of iron. Indeed, the energy of activation for the phosphorus reaction is small enough to be amply provided for by the heat available at room temperature.

4

The activation energy for natural gas would appear to be higher than that for kerosene, as natural gas needs to have more heat supplied before the reactant particles have enough energy to react with oxygen.



a

b

6

i

Fastest forward reaction is reaction Y (as per table). Slowest forward reaction is reaction Z.

ii

Fastest reverse reaction is reaction Z. Slowest reverse reaction is reaction X.

N2(g) + 3H2(g) ↔ 2NH3(g)


The fraction of collisions may be the same but it is the same fraction of a larger number (as pure oxygen has many more oxygen molecules than air).

2

At the higher temperature (T2) a greater fraction of reactant molecules have sufficient kinetic energy to supply the activation energy needed for reaction. As a result, the ‘souring’ reaction will occur more rapidly at room temperature (T2) than in a refrigerator (T1). (See following graphs.)



3

The particles in the vapour state have a higher surface area and a greater energy than particles in the liquid state, so collision rates will be greater. The number with enough energy to overcome the activation energy barrier will also be greater.

4

A catalyst can reduce the activation energy by providing an alternative pathway for the reaction. This lower activation energy means that of the same number of collisions, more will have the required amount of energy.

5

a

Reactant particles will combine to form products at the flash point if an external source of energy is provided, i.e. a spark or flame. This external source of energy allows the activation energy to be overcome. At the ignition temperature, the activation energy has already been overcome by the kinetic energy of the particles themselves.

b

Small boats: trapped petrol fumes Silos and coal mines: fine particles with a large surface area

c

Explosions occur as the result of a rapid expansion of gases. The products of combustion for butane are both gases, CO2 and H2O, whereas the product of combustion of phosphorous is a solid. So although combustion may occur, it is not an explosive combustion.


Rate of reaction could be measured by measuring the: • volume of gas appearing and • mass of magnesium dissolving per unit time.



a

b

The rate of reaction is rapid at first and then decreases with time.

c

The reaction slows due to the decreasing concentration of the reactant, C4H9Cl.

3

The fast reaction does not require any bonds to be broken, and has both reactants in the same state, i.e. they are aqueous solutions. This allows for collisions between reactant particles to occur reasonably easily. In the slower reaction, many bonds must be broken before the products can begin to be formed. The slower reaction has reactants in different states, which allows for very little common contact area, i.e. the surface of the liquid (unless of course the oxygen is bubbled through the solution).

4

a

More oxygen would mean more collisions and so the numbers that were successful would increase, thus increasing the forward rate.

b

More oxygen would mean more collisions and so the numbers that were successful would increase, thus increasing the rate.

c

No change, as oxygen plays no part in this reaction.

d

More oxygen would mean more collisions and so the rate of the reverse reaction would increase.

5

Powdered zinc would cause an increased rate of reaction because of the larger surface area provided to the HCl to react with. Given that zinc is already quite reactive with HCl, the powdered zinc could produce a violent explosive reaction, producing a lot of heat very quickly and perhaps igniting the hydrogen gas.

6

The time taken would be much less, as there are more reactant particles. There is no reason to suppose that because the concentration is doubled, the reaction rate is halved, but it would certainly be less.

7

a

This energy is required to break the bonds in the molecules of the reactants. The minimum energy is the activation energy.

b

The particles also need to be orientated the right way to each other and the collision needs to occur for a long enough time for rearrangement to occur.



9

a

Endothermic

b

25 kJ mol–1

c

Forward reaction activation energy = 60 kJ mol–1(approx) Reverse reaction activation energy = 35 kJ mol–1(approx)

d

Forward reaction activation energy = 50 kJ mol–1(approx) Reverse reaction activation energy = 25 kJ mol–1(approx)

e

The rate of both the forward and reverse reactions is increased in the presence of a catalyst.

a

i

ii

b

i

Ea (reverse) = 173 kJ

ii

Ea (reverse) = 10 kJ



a

b

11

a

A greater concentration or pressure of reactants leads to an increased number and rate of collision between reactant particles and consequently a greater reaction rate.

b

When the degree of sub-division of a solid or liquid reactant is increased, the number of collisions between reactant particles increases, as does the total surface area of the reactants. Hence the reaction rate increases.

c

At a higher temperature, a greater fraction of molecular collisions have sufficient energy to overcome the activation energy barrier. (See following graphs.)



12

Without the platinum gauze catalyst, the activation energy for this reaction is very high, so at room temperature the chance of molecules colliding with sufficient energy to react is small. However, the activation energy when the catalyst is used is much lower, so at room temperature many more collisions will result in a reaction. 13

Catalysts certainly speed up a reaction, but they do take part in it—it is difficult to think how they could affect the reaction without taking part! Many create intermediate products that have a lower activation energy before they decompose to form the products and re-form the catalyst. The important point about catalysts is that they speed up the reaction without being used up in the reaction.

14

a

Three processes that use transition metals as catalysts: • hydrogenation of vegetable oils to produce margarine (uses nickel or platinum) • manufacture of ammonia (uses iron) • production of nitric acid (uses platinum).


Solutions Manual: Module 4 Energy b

Chlorine atoms are used in the first reaction and are produced in the second reaction. If this set of reactions always occurs as a sequence, and the rate of ozone depletion is faster with the chlorine atoms than without, then chlorine is acting as a catalyst.

15

Investigation

16

Investigation

Module 4 REVIEW 1

A

2

C

3

B

4

C

5

D

6

A

7

C

8

B

9

A

10

C

11

i

Zinc powder + hydrochloric acid → zinc chloride + hydrogen gas

ii

Hydrogen peroxide → oxygen + water (catalyst is manganese dioxide)

iii

Carbon dioxide + water + light → glucose + oxygen

a

Per mole values are converted to per gram values by dividing by the molar mass (M).

12

M(C) = 12 g

M(C3Hg) = 44 g

M(C9H20) = 128 g

∆H carbon g

∆H C3H8 g

∆H C9H20 g

393 12 = 32.75 kJ g–1

2220 44 = 50.45 kJ g–1

=

=

8125 128 = 47.85 kJ g–1

=

∴ propane (C3H8) has the highest energy output per gram.



13

b

Carbon is very safe to transport as a solid. However, build-up of carbon dust in the air should be avoided. Propane, being a liquefiable gas, is normally stored under pressure in metal cylinders kept cool and restrained from moving. Large pressurised tankers can transport propane or, alternatively, gas pipelines can transport propane. Nonane (C9H20) has a relatively low flashpoint but can safely be transported as a liquid in tanks provided flames and sparks are kept away and the tanks or drums are sealed.

a

Hexane has a flashpoint of –23°C. Therefore it ignites easily, whereas hexadecane, having a flashpoint in the order of +400°C, does not readily ignite.

b

There is insufficient concentration of oxygen in air to fully combust hexane molecules and so some carbon (soot) is produced in the combustion. One possibility is: 2C6H14(g) + 15O2(g) → 2C(s) + 4CO(g) + 14H2O(g) + 6CO2(g)

14

As the length of the carbon chains in alkanes increases, the total of the dispersion forces acting also increases. This is reflected in the increasing amount of energy (temperature) required to boil the alkanes as their chain length increases.

15

a

Petrol is one of the constituents of a fossil fuel. Fossil fuels are made from the remains of plants and animals. The molecules that made up these plants and animals were synthesised originally by plants using the process of photosynthesis: 6CO2(g) + 6H2O(l) + light energy → C6H12O6(aq) + 6O2(g)

b

Petrol is obtained from petroleum by a process of fractional distillation. A fractionating tower has petroleum supplied at the bottom of the tower. It is heated and volatile components move up the tower, recondensing at various points. Each of these levels has a structure that allows recondensed liquids to be syphoned off.

c

i

The activation energy of a fuel is the energy that must be supplied before a reaction (combustion) takes place. This energy may be in the form of a spark or flame, or for fuels with low activation energies it may be the heat available at room temperature. In this case the fuel will combust spontaneously. Petrol’s activation energy is high enough so that it is safe at room temperature and needs a spark to set off the combustion process.

ii

If a fuel is very volatile then its transport and storage can be a problem, as the vapour is the state in which a fuel reacts. Petrol is quite volatile and so storage, transport and delivery require safety precautions.



a

N2O4(g) → 2NO2(g)

∆ H = –24 kJ mol–1

b

17

18

c

24 + 55 = 79 kJ

a

CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) 2CH4(g) + 3O2(g) → 2CO(g) + 4H2O(l) CH4(g) + O2(g) → C(s) + 2H2O(l)

b

A blue flame is most efficient and, as the products are carbon dioxide and water, we know that complete combustion took place. More energy is released per mole (or per g) of CH4.

a

2C8H18(g) + 25O2(g) → 16CO2(g) + 18H2O(l)

∆ H = –804 kJ mol–1 ∆ H = –521 kJ mol–1 ∆ H = –410 kJ mol–1

b

c

Hexadecane is C16H34. M = 16 × 12 + 34 = 226 g ∴ ∆ H = 47.3 × 226 = 10 690 kJ mol–1

d

As the length of the carbon chain increases, so too does the strength of dispersion forces acting between molecules. Stronger dispersion forces require higher temperatures to overcome them. Thus there is an increase in boiling points as chain length increases.



a

Burning coal produces: carbon dioxide, a contributor to the greenhouse effect; sulfur dioxide, a contributor to acid rain; soot, a contributor to photochemical smog; carbon monoxide, a poison and a contributor to the greenhouse effect; and an extensive number of aromatic and hydrocarbon compounds.

b

CO2 release cannot be avoided. Sufficient oxygen will avoid much emission of CO and the aromatics and hydrocarbons. Electrostatic precipitation reduces soot emission. Use of low-sulfur coal avoids excess SO2 generation.


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