DLRC Chemistry 16 Comprehensive Samplex for Long Exam 2

Chemistry 16 Comprehensive Samplex for Long Exam 2 September 7, 2012 Instructions:

Answer as many items as you can. Read each question carefully and answer honestly. All information you need is in this paper. Good luck!

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MULTIPLE CHOICE. CHOICE. Choose the letter corresponding to the best answer. Which of the following set of quantum D. Same subshell and orbitals but different numbers is possible? shells A. n=1, l=0, m l=1, ms=1/2 3. For systems with positive changes in B. n=5, l=3, m l=-2, ms=-1/2 enthalpy and entropy, processes become C. n=2, l=1, m l=0, ms=0 spontaneous at __ D. None of these A. All temperatures Two electrons in the same atom which have B. Low identical values of quantum numbers l and C. High ml are said to be in ___ D. Absolute zero A. The same shell and subshell but in 4. Which of the following is a polar molecule? different orbitals A. RbCl B. Same shell but different subshells and B. BF3 orbitals C. CCl4 C. Same shell, subshell and orbitals D. PCl3

1|DLRC Chem 16 LE2 Samplex

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Which of the following is false about the structure of thiocyanate ion? A. The C-N bond is a double bond. B. There are no lone pairs on C. C. There is a lone pair of electrons on N. D. The C-H bond is a single bond. Of the following, which is a state function? A. ∆H B. q C. w D. All of the above The shorter the internuclear distance between two bonded atoms, the A. Less energy needed to break the bond B. Less multiple bonding character there is C. More resonance D. More energy needed to break the bond Which of the following is an unstable species and is bound not to exist? A. H2 B. He2+ C. Be2+ D. Be2 Which of the following pairs have similar molecular geometry? A. CH3+, CH3B. PF5, ClF5 C. H2O, NH2D. None of these Which of the following ions has the largest atomic radius? A. Na+ B. Mg+ C. O2D. FWhich of the following atoms has the largest first ionization energy? A. Rb B. Sr C. Al D. All have same magnitude of energy. What is the correct electronic configuration for a nickel atom? A. 1s22s22p64s23d23d23d23d13d1 B. 1s22s22p6 3s2 3p6 4s2 4p63d13d1 C. 1s22s22p63s23p63d10 D. none of these


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The Lewis structure for calcium hydride is A. [H:]-[Ca]2+[H:]B. [H:]-[Ca] + C. H-Ca-H D. Ca-H Which of the following is best represented by a set of resonance structures? A. Ammonium B. Methane C. Oxygen difluoride D. Sulfur dioxide What is the formal charge on N in nitrate ion? A. 2 B. 2C. 1 D. 1If there is a negative change in the Gibb’s energy, then the reaction is said to be ___. A. Reversible B. Spontaneous C. Exothermic D. Endothermic In the experiment on Corrosion, the portion of the nail that was oxidized was ___ in color and acted as the __. A. White, anode B. Pink, anode C. Blue green, cathode D. Blue green, anode In the experiment on Corrosion, zinc acted as the __ while the nail is the __ A. Protected Anode; Sacrificial Cathode B. Protected Cathode; Sacrificial Anode C. Sacrificial Cathode; Protected Anode D. Sacrificial Anode; Protected Cathode Which of the following reactions is nonspontaneous? A. Neutralization of NaOH and HCl B. Freezing water in room temperature C. Combustion of octane fuel D. None of the above Which of the following statements is false? A. Ferrous hydroxide is formed by adding soluble Fe3+ sample to NaOH or NH3. B. Calcium will not precipitate to limiting and excess ammonia.

C. A Prussian blue precipitate will form when Fe2+ is added to potassium ferricyanide. D. The formula of the brown ring in the nitrate test is [Fe(NO)] 2+. 21. Which of the following ionic compounds has the highest entropy? A. Calcium chloride B. Magnesium arsenate C. Sodium oxalate D. Tin(IV) oxide For numbers 22-26, refer to the structure of capsaicin below:

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The EGG, MGG, and hybridization of O 1 is: A. Tetrahedral; angular; sp 2 B. Trigonal planar; trigonal pyramidal; sp 3 C. Tetrahedral; angular; sp 3 D. Trigonal planar; angular; sp 2 The difference/s between C 2 and C5 is: A. Their EGG and hybridization B. Their MGG and hybridization C. Their EGG and MGG D. Their EGG, MGG, and hybridization How many σ and π bonds are in capsaicin? A. 43 ; 5 B. 42 ; 4 C. 42 ; 5 D. 43 ; 4 What is the hybridization overlap in O 4? A. sp2 – sp3 B. sp3 – sp2 C. sp2 – sp2 D. sp3 – sp3 The formal charge of N 3 is __ and the hybridization overlap in C – N3 – C is __. A. 0; sp2-sp3-sp2 B. 0; sp3-sp3-sp2 C. +1; sp3-sp3-sp2 D. +1; sp2-sp3-sp2 Which of the following is true? A. Manganese(II) cations are dominant in basic medium.


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B. When organic chlorine and iodine are combined in a toluene-water mixture, there will be no visible reaction. C. A more acidic medium promotes corrosion of metals. D. Zinc, Nickel, and Cobalt can act as sacrificial anodes for Iron. What is the net heat change if 25.0 mL of 3.78 M sulfuric acid is neutralized by 50.0 mL of 1.90 M potassium hydroxide? A. -5.27 kJ B. 5.27 kJ C. 5.30 kJ D. -5.30 kJ What is the De Broglie wavelength and frequency of a photon with mass 1.08×10 -24 nanograms? A. 2.05×10 -7 m; 1.47×1014 s-1 B. 2.05×10 -6 m; 1.47×1015 s-1 C. 2.05×10 -6 m; 1.47×1014 s-1 D. 2.05×10 -7 m; 1.47×1015 s-1 If the uncertainty for a velocity of a photon is 3.14×10 -15 m/s, given that the mass of the photon is 7.09×10 -9 kg, what is the uncertainty of the position of the photon? A. 2.37×10 -12 m B. 2.63×10 -12 m C. 2.86×10 -12 m D. 2.45×10 -12 m Nitrogen gas and Hydrogen gas react to form ammonia gas. Given that ∆Hf of ammonia is -46.3 kJ, how many grams of ammonia are produced if 176.7 kJ of heat is released? A. 65.0 g B. 130.0 g C. 7.63 g D. 3.81 g Which ion/s can give an interfering false ring for nitrate test? A. Thiocyanate only B. Ammonium and thiocyanate C. Thiocyanate, Bromide, Iodide D. Sulfate, Bromide, Thiocyanate

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1. 2. 3. 4. 5.

MODIFIED TRUE OR FALSE. Write TRUE if the statement is true. Otherwise, determine what part of the statement made it false and offer corrections. 1. In the ozone layer, O 3(g) absorbs harmful UV radiation and forms oxygen gas and gaseous oxygen atom. The entropy change for this process is equal to zero. 2. The 4f orbitals can accommodate a maximum of 10 e-. 3. Hess’s Law is a consequence of the first law of thermodynamics. 4. On a humid day, we can observe water vapor to condense along the outer wall of cold drink bottles. The formation of the vapor is an endothermic process. 5. The form of carbon with a ∆H o formation value of zero is diamond. 6. The azimuthal quantum number tells us how far from the nucleus an electron is. 7. The oxidation state assigned to P in the phosphate ion is +5. 8. In the aluminum atom, the outermost electron has the quantum numbers n=3, l=1, m l=1, and ms=+1/2. 9. qsoln is still part of q cal even if the volumes of the solution used for calibration and calculation are not constant. 10. XeO2F2 has a trigonal bipyramidal electron group geometry, and a square pyramidal molecular group geometry. 11. Naphthalene, C10H8, is made up of two benzene rings “stacked” or “glued” together. The number of dominant resonance structures of naphthalene is two. 12. The order of hydrohalidic acids’ ionic character in increasing order is: I < Br < Cl < F. 13. Prussian blue precipitate from potassium ferricyanide indicates the presence of Fe 3+ ions, while from potassium ferrocyanide indicates the presence of Fe 2+ ions. 14. The net dipole moment of water explains sodium chloride’s miscibility in it. 15. In excess ammonia, Zn2+ will form a colorless complex [Zn(NH3) 2]2+, while in excess NaOH, Zn 2+ will form a colorless complex [Zn(OH) 4]2-. 16. Nitrate ion does not exhibit pi bond delocalization. 17. Adding potassium ferricyanide to Zn 2+ will form yellow precipitate KZn[Fe(CN) 6], while adding potassium ferrocyanide will form white precipitate KZn 3[Fe(CN)6]. 18. There will be an exothermic change in heat if the system gives off heat to the surroundings, and the surroundings does work to the system. 19. An exothermic reaction from low entropy to high entropy will only proceed at high temperatures. 20. If a reaction is endothermic from high entropy to low entropy, the reverse reaction is non-spontaneous. CAUSE AND EFFECT. Determine the effect of the cause stated in the first column on the parameter in the second assuming all other variables are held constant. Effects: Increase, Decrease, No Effect Cause Actual molarity of the acid is greater than the recorded value. The cylinder used for measuring the base solution was wet. The student pulled the thermometer out of the set-up to take temperature readings. There is a tiny hole in the cork, exposing the inside of the calorimeter. Actual molarity of the base is greater than the recorded value.


Parameter

Magnitude of calculated ∆H

Magnitude of calculated Ccal

Effect

Effects: Becomes present, Becomes absent, No Effect (stays the same) Cause Parameter 6. Mossy Zinc and Aluminum metal were accidentally coiled together with iron Corroded Zinc nail. 7. Gelatin mixture was accidentally Appearance of Prussian blue added with concentrated NaOH (after 1 day) 8. KMnO4 was accidentally added in Persistence of Pink Color excess. 9. Litmus paper was accidentally dipped Indication of gaseous ammonia into unknown solution with NaOH. 10. Copper(II) nitrate was added with Dissolution to deep blue NaOH, then with excess ammonia. complex 11. Bromide sample was accidentally Pink-red toluene layer contaminated with iodide sample for Yellow toluene layer KMnO4 elimination test. IV.

Effect

PROBLEM SOLVING. Read (without underestimation) the problems carefully, and solve for what is asked. There are ten problems in this part of the review.

Problem 1: Lead-Acid Battery The lead-acid battery, invented in 1959 by French physicist Gaston Planté, is the oldest model of rechargeable batteries. Lead-acid batteries are normally used in automobile engines, because it can generate large amounts of current for short periods of time. When a lead-acid battery is fully charged, the battery contains sulfuric acid as the electrolyte. Two grids, lead grid and lead(IV) oxide grid, serve as the electrodes of the battery. During discharge, the grids are converted to lead(II) sulfate, and when the battery is fully discharged, the sulfuric acid is almost diluted to water. Assume that sulfuric acid does not fully ionize during discharge; it will only give one H + during ionization. a) WRITE the balanced half-reactions for the anode and the cathode. INDICATE the reducing and oxidizing agents. b) WRITE the overall balanced equation for lead-acid battery discharge. c) CALCULATE the number of electrons loss/gained after full discharge if 150.0 g of solid lead, 150.0 g lead(IV) oxide, and 3.00 L of 4.2 M sulfuric acid was present at full charge. The problem with lead-acid batteries is its environmental risks. Even though lead-acid batteries are rechargeable, lead can still escape from the battery, especially during battery explosion. A battery explosion happens when water molecules hydrolyze to hydrogen and oxygen gas during recharging. d) WRITE the balanced equation for the hydrolysis of water. An explosion can spray battery acid and release harmful lead compounds in the environment. Lead mostly accumulates in bodies of water in the form of lead(II) ions. To detect concentration of lead(II) ions in bodies of water, chemists use cerimetry, a redox titration by using cerium(IV) ions as the titrant.


50 mL of a wastewater sample near a lead-acid battery explosion site was acidified with 25 mL 6 M HCl, and was titrated with 32.6 mL of cerium(IV) sulfate solution. The cerium(IV) sulfate solution was prepared dissolving 250 g of the solid in 800 mL distilled water. e) If lead(II) and cerium(IV) becomes lead(IV) and cerium(III) during titration, WRITE the balanced equation of the redox reaction. f) CALCULATE the concentration of lead(II) ions in the wastewater sample.

Problem 2: Bomb Calorimetry A more sophisticated calorimeter than the constant-pressure (coffee cup) calorimeter is the bomb calorimeter. The bomb calorimeter uses an insulated sealed vessel where the reaction (with the sample) takes place. The vessel is pressurized with excess oxygen gas and it is placed inside the calorimeter full of water. Electric current will run through the vessel, igniting the sample. A thermometer will then measure the temperature change by contact with the water inside the calorimeter. Bomb calorimetry is used mostly in calculating energy needed to burn food or fuel. a) The bomb calorimeter is also known as “constant -volume” calorimeter, because react ions are carried out at constant volume. EXPLAIN why ∆E corresponds to the transferred heat in bomb calorimeter rather than ∆H. The bomb calorimeter is calibrated by burning 25.0 g of solid glucose (C 6H12O6) in the sealed vessel, and the temperature rose from 25.00 °C to 59.50 °C. The heat of reaction for the combustion of one mole of solid glucose is -2803 kJ. b) WRITE the balanced equation for the combustion of solid glucose. c) CALCULATE the heat capacity of the bomb calorimeter. A certain milk tea was found to contain fructose (C 6H12O6), gallic acid and palmitic acid. Gallic acid (MW = 170.12 g/mol) is an antioxidant found in most Chinese herbs, while palmitic acid (C 16H32O2) is an unsaturated fat found in dairy products. 100.0 mL of the sample (filtered, only containing the desirable compounds) was burned in the sealed vessel of the calorimeter in an initial temperature of 25.00 °C. Several experiments were conducted to find the percent volumes of some of the compounds and the amount of products formed. The following data were collected before and after the tests: Final Temperature Percent volume of fructose Percent volume of palmitic acid Percent volume of gallic acid Grams of carbon dioxide Grams of water

27.24 °C Density of fructose 0.400% Density of palmitic acid 0.200% Density of gallic acid 0.300% Heat of combustion per mole of fructose 2.385 g Heat of combustion per mole of gallic acid 0.7604 g

d) CALCULATE the molecular formula of gallic acid. e) CALCULATE the molar heat of combustion of palmitic acid. f) WRITE the balanced equations for the combustion of palmitic acid and gallic acid. 6|DLRC Chem 16 LE2 Samplex

1.694 g/mL 0.853 g/mL 1.7 g/mL -2812 kJ/mol -2655 kJ/mol

[Problem 3]: DLRC-Land In the year 21XX, scientists discovered a peculiar planet outside Earth, and they named it “DLRC Land”. What makes DLRC -Land peculiar is that it defies the laws and theories behind the structure of atoms.

Scientists noticed that the elements in Earth are different compared to the elements in DLRC Land. Pauli’s Exclusion Principle, Aufbau’s Principle, Hund’s Rule, Heisenberg’s Uncertainty, et c. all apply to DLRC, except with the quantum numbers. By calculations they concluded that they have the quantum numbers d, lr, and c.   

d corresponds to quantum number n. lr corresponds to both quantum numbers l and ml. This means lr = 0, ±1, ±2, ±3…± (n -1). c corresponds to quantum number ms, The elements in DLRC-Land are named by their atomic number.

a) DRAW the first five periods of the DLRC-Land periodic table based on the information given above (You just need to draw blocks.) b) WRITE and DRAW the electronic configuration of DLRC-Land elements 28, 24, and 21. Give the quantum numbers of their last-entering electron. c) ARRANGE the following elements by their periodicity. Assume that like elements in Earth, DLRC-Land elements have their cationic and anionic forms. a. Atomic Radius: 7-, 162+ 62b. Ionization Energy: 2, 9, 13, 14 d) What rules in covalent bonding does DLRC- Land follow if our planet’s elements follow the octet rule and 18-electron rule? EXPLAIN. e) INDICATE which of the following hypothetical covalent compounds are possible to exist in DLRC-Land. If possible, DRAW their Lewis structures and indicate the electron-group geometry and moleculargroup geometry in each central atom. a. (6)(5)3 c. Hypothetical “carbon chain” (5) 3(1)5 b. Hypothetical “carbon chain” (5) 2(1)6 d. Hypothetical anion (13)(7) 4 f) GIVE the hybridized DLRC-Land orbital that corresponds to (a) tetrahedral, (b) trigonal bipyramidal, and (c) octahedral electron-group geometry.

Problem 4: “Qualitative” Analysis Dessa Eibaf Conor, a Chem 16 student in UP Diliman, was asked to get 17 reagents from the IC Stock Room for the qualitative analysis experiment. However, Dessa is a mischievous student, so she removed the labels of the bottles and replaced them with labels A to Q. Air Asserc, the topnotcher of the 1 st Chem 16 LE, tried to find out the compounds based on her vast knowledge of Chem 16. These are the results of her tests/observations:   

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Dessa accidentally placed one drop of L on a tissue paper. The tissue paper acquired a burnt mark. Based on her lab manual, all fresh (soluble) anionic and cationic samples have molarity of 1.0 M. When G is added with 5 drops A, a deep red solution formed. When it was added with 5 drops M, a blue precipitate formed. When 5 drops of C was added with 5 drops D, a blue precipitate formed, but when mossy zinc was immersed before adding D, a yellow precipitate formed.


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When 5 drops of E was added with 5 drops J, a brown precipitate formed. The precipitate turned colorless when added with 1 drop of L, but turned green when added with 3 drops of K. Solid F dissolves into a blue solution when added with L. When added with excess O, the solution turned deep blue. Only B and A released a basic gas (by litmus paper test) when added with 2 drops K. G, N, and P gave a brown ring when the test for nitrate was done. When 5 drops I was added with 2 drops 0.5 M P, a white precipitate formed, and it dissolved (with effervescence) in excess Q. When H was studied for the toluene elimination test, the toluene and aqueous layers turned yellowishbrown. When 10 mL I was added with 10 mL G, 0.972 g of brown-orange precipitate formed. When 5 mL N was added with 1 mL of 6 M K, 0.37 g of white precipitate formed. When 5 mL H was added with 5 mL of 1.0 M silver nitrate, 0.939 g of precipitate formed. When 10 mL B was added with 5 mL of 0.5 M P, 1.50 g of white precipitate formed, and it dissolved in excess organic acid Q. By records in the stock room, solution I was made by dissolving 73.89 g of it in 500 mL distilled water. By records in the stock room, solution N was made by dissolving 98.45 g of it in 600 mL distilled water.

After doing the said tests and observations, Air Asserc knew the corresponding compounds from A to Q. GIVE the corresponding compounds from A to Q. (NOTE: Some of the compounds have slight modifications from the compounds in the Chem 16 manual. Qualitative analysis alone is not enough to solve the problem.)

Problem 5: Polynitrogen Chemistry In 1999, Karl O. Christe and his team of researchers from the Air Force Research Laboratory synthesized a polynitrogen species other than N 2, and azide (N 3-). Pentazenium (N5+), an open chained “polynitride” like the other compounds, was synthesized for a less toxic and more environmental -friendly alternative to hydrazine, a gas used in rocket fuel. a) DRAW the Lewis structures (and resonance structures if possible) for azide anion. b) DRAW the three most dominant resonance structures of pentazenium. INDICATE the formal charges of each atom. GIVE the hybridization overlaps between the five nitrogen atoms for each resonance structure. Karl Christe proposed a reaction between gaseous hydroazoic acid and N 2F+ gas to give pentazenium and hydrogen fluoride as the byproduct. The bond energies of the atoms are given below: H – H 432 kJ/mol F – F 159 kJ/mol N = N 418 kJ/mol H – F 565 kJ/mol F – N 283 kJ/mol N ≡ N 945 kJ/mol H – N 391 kJ/mol N – N 160 kJ/mol c) WRITE the balanced equation of the proposed mechanism, using the Lewis structures of the compounds/ions (instead of writing with chemical formulas). d) CALCULATE the change in bond enthalpy of the proposed mechanism. However, a new polynitrogen species is being studied. Pentazole anion, or N 5-, is an anion from aromatic pentazole. Due to its all-nitrogen cyclical structure, chemists have been studying this anion as 8|DLRC Chem 16 LE2 Samplex

another alternative for propellants in space travel, but so far pentazole is known to be unstable and it does not last longer than a few seconds. e) DRAW the three most dominant resonance structures of pentazole anion. INDICATE the formal charges of each atom. GIVE the hybridization overlaps between the five nitrogen atoms for each resonance structure.

Problem 6: Molecular Orbital Theory Using Molecular Orbital Theory, PROVE that: a) Binary (homonuclear) molecules from diamagnetic atoms cannot exist. b) Charged binary (homonuclear) molecules from diamagnetic atoms can exist. c) Using your answer in (a), can binary noble gases (Ne 2, He2, etc.) exist? Consider binary fluorine species F 2+, F2, and F2-, and binary boron species B 2+, B2, and B2-. ARRANGE the groups of species (separately) in increasing order by: d) Bond Energy e) Bond Length f) CALCULATE the bond order of all the species. INDICATE their magnetic properties.

Problem 7: Activity of Lanthanum Lanthanum metal forms cations with a charge of 3+. Consider the following observations about the chemistry of lanthanum: When lanthanum metal is exposed to air, a white solid A is formed that contains lanthanum and one other element. When lanthanum metal is added to water, gas bubbles are observed and a different white solid B is formed. Both A and B dissolve in hydrochloric acid to give a clear solution. When either of these solutions is evaporated, a soluble white solid C remains. If compound C is dissolved in water and sulfuric acid is added, a white precipitate D forms. 

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a) PROPOSE identities for the substances A, B, C, and D. b) WRITE net ionic equations for all the reactions described. c) Is Lanthanum, in the activity series, above hydrogen or below hydrogen? EXPLAIN. What you will see below are free energy changes of redox reactions (regardless of spontaneity) between lanthanum and certain metals: La(s) + Al +(aq) → La +(aq) + Al(s) -208 kJ/mol 2Al(s) + 3Yb +(aq) → 2Al +(aq) + 3Yb(s) 212 kJ/mol 3Rb+(aq) + La(s) → 3Rb(s) + La3+(aq) 174 kJ/mol 2Rb(s) + Yb2+(aq) → 2Rb+(aq) + Yb(s) -42 kJ/mol 3Yb(s) + 2La +(aq) → 3Yb +(aq) + 2La(s) -221 kJ/mol Ti(s) + Yb +(aq) →Ti +(aq) + Yb(s) 218 kJ/mol 3+ 2+ + 2+ 3Ti(s) + 2La (aq) → 3Ti (aq) + 2La(s) 434 kJ/mol Ti(s) + 2Rb (aq) → Ti (aq) + 2Rb(s) 261 kJ/mol 3+ + 3Rb(s)+ Al (aq) → 3Rb (aq) + Al(s) -382 kJ/mol 3Ti(s) + 2Al3+(aq) → 3Ti2+ + 2Al(s) 19 kJ/mol d) ARRANGE the five elements (in the table) in decreasing reducing property. e) INDICATE which elements can act as a sacrificial anode for lanthanum. 9|DLRC Chem 16 LE2 Samplex

f)

INDICATE which elements will be protected by lanthanum from corrosion.

Problem 8: Permanganometry Permanganometry is a special redox titration that involves the use of permanganates, and it is an alternative to acid-base titration because permanganates can transform to different manganese species, depending on the pH of the solution. Standardization of KMnO 4 is done by oxalic acid as the primary standard (You should know the formula of oxalic acid, based on inorganic nomenclature.) Oxalic acid was prepared by dissolving 0.63 g of oxalic acid crystals in 100 mL water. 20 mL of 0.1 M sulfuric acid is added to the oxalic acid solution to acidify the medium. The results of the standardization are as follows: Trial 1 2 3

Starting Point 0.0 mL 3.7 mL 7.5 mL

Endpoint 3.7 mL 7.5 mL 11.0 mL

Organic acids (Acids containing C, H, and O) turn into carbon dioxide gas and water during titration. a) WRITE the balanced redox equation for standardization. b) CALCULATE the molarity of KMnO 4 in each trial, and the average molarity of KMnO 4. A Chem 16 student wants to find out the amount of Ascorbic Acid (MF = C 6H8O6) in a sample of orange juice. A 25 mL orange juice sample was added with 25 mL distilled water, and was titrated with the KMnO 4 solution. While titrating, the analyte slowly turned into a solution with brown precipitate. Trial Starting Point Endpoint 1 0.0 mL 5.6 mL 2 5.6 mL 11.6 mL 3 11.6 mL 17.0 mL c) WRITE the balanced redox equation for titration of ascorbic acid. d) CALCULATE the molarity of ascorbic acid in each trial, and the average molarity of ascorbic acid.

Problem 9: Olympicene Olympicene (MW = 240.286 g/mol) is a special aromatic compound with five carbon rings stacked together like a benzene ring. It was recently synthesized by Anish Mistry and David Fox of the University of Warwick only for the purpose of celebrating the London 2012 Olympics. 1.00 g of olympicene (only containing C and H) was burned in excess oxygen and the combustion yielded 3.48 g carbon dioxide. a) CALCULATE the molecular formula of olympicene.

10 | D L R C C h e m 1 6 L E 2 S a m p l e x

The picture above is the carbon skeletal structure of olympicine. Based on experiments, only Ring 2 cannot undergo pi-bond delocalization. b) Can Ring 2 have all of its carbon atoms sp 2 hybridized? EXPLAIN. c) GIVE three possible resonance structures of olympicene by assigning the C and H atoms, and double bonds. d) GIVE all possible hybridization overlaps in olympicene.

Problem 10: Solvay Process The Solvay Process, invented by Ernest Solvay in the 1860s, is an industrial process used to make sodium carbonate, and this chemicals are needed in glass making and oxidizing bleaches. The following reactions are steps for Solvay Process: NaCl(aq) + CO2(g) + NH3(g) + H2O(l) → NaHCO3(s) + NH4Cl(s) CaCO3(s) → CaO(s) + CO2(g) CaO(s) + H2O(l) → Ca(OH)2(s) Ca(OH)2(s) + 2NH4Cl(s) → 2NH3(g) + CaCl2(aq) + 2H2O(l) 2 NaHCO3(s) → Na2CO3(s) + H2O(l) + CO2(g)

∆Hrxn = -129.48 kJ/mol ∆Hrxn = 178.3 kJ/mol ∆Hrxn = -65.17 kJ/mol ∆Hrxn = 73.75 kJ/mol ∆Hrxn = -85.17 kJ/mol

∆Srxn = -394.81 J/mol*K ∆Srxn = 160.47 J/mol*K ∆Srxn =-26.26 J/mol*K ∆Srxn =356.82 J/mol*K ∆Srxn =215.31 J/mol*K

The overall reaction of the spontaneous process is: 2 NaCl (aq) + CaCO3(s) → Na2CO3(s) + CaCl2(aq). a) CALCULATE the overall ∆Hrxn of the Solvay process. b) CALCULATE the overall free energy change of the Solvay process in 25 °C c) If 1324.2 kJ of free energy were released in the process (in excess calcium carbonate): a. CALCULATE the grams of sodium carbonate synthesized in the process. b. CALCULATE the grams of sodium chloride used in the process.

11 | D L R C C h e m 1 6 L E 2 S a m p l e x

DLRC Chemistry 16 Comprehensive Samplex for Long Exam 2

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