4U Chemistry Practice Exam

Multiple Choice Identify the letter of the choice that best completes the statement or answers the question.

____ 1. Name CH3CH(OH)CH3. a. iso-ethanol b. tertiary-propanol c. butanol d. propanol e. 2-propanol

____ 2. Which organic compound is unsaturated? a. ethylcyclopentane b. 2-methyl-3-ethylpentyne c. 1,1-dimethylhexane d. cyclohexane e. 1,3,5-trimethyloctane

____ 3. To which family of organic compounds does CH3COCH2CH2CH3 belong? a. alcohol b. aldehyde c. alkyne d. ketone e. carboxylic acid

____ 4. What is the difference between an amine and an amide? a. There is no carbon-oxygen bond in an amine, but there is in an amide. b. Amines are non-polar molecules. c. Amines always have a larger molecular weight than amides. d. Amines always have a nitrogen atom attached to two carbon atoms. e. Amines can be found in proteins, but amides can not.

____ 5. Why does the boiling point of an alkane increase as its chain length increases? a. There is more hydrogen bonding. b. There are more hydrogen atoms available for hydrogen bonds to form. c. The number of dipole-dipole interactions increases. d. The strength of the dispersion forces increases with increased molecular size. e. Heavier molecules cannot float on the surface of water as well.

____ 6. How many unshared electron pairs are found in a molecule of formaldehyde, HNCHO? Hint: Draw a Lewis structure to help you answer this question. a. none b. three c. two d. four e. one

____ 7. How many isomers have the molecular formula C5H10O? a. four b. five c. six d. seven e. three

____ 8. What type of reaction occurs when bromine reacts with pentyne to make dibromopentene? a. substitution b. isomerism c. oxidation d. reduction e. addition

____ 9. Which reaction might form cis-trans isomers? a. addition to an alkene b. addition to an alkyne c. elimination of benzene d. reduction of an alcohol e. substitution of an alkane

____ 10. Which situation must be true for two electrons to occupy the same orbital? a. The electrons must have the same principal quantum number, but the other quantum numbers must be different. b. The electrons must have the same spin. c. The electrons must have identical sets of quantum numbers. d. The electrons must have low energy. e. The electrons must have the opposite spin.

____ 11. An electron has the following set of quantum numbers: n = 3, l = 1, ml = 1, ms = . In which orbital is this electron found? a. 3s b. 3p c. 3d d. 3f e. 4p

____ 12. Which set of quantum numbers is not possible? a. n = 3, l = 0, ml = 0, ms = b. n = 5, l = 3, ml = 2, ms = c. n = 4, l = 3, ml = 1, ms = d. n = 5, l = 3, ml = 3, ms = e. n = 4, l = 4, ml = 2, ms =

2 1 1 ____ 13. Which element has the ground state electron configuration [Ne] 3s 3px 3py for its valence electrons? a. Mg b. Al c. Si d. P e. S

____ 14. What is the total number of electrons in the 2p orbitals of a sulfur atom at ground state? a. 8 b. 6 c. 4 d. 3 e. 2

____ 15. Which pair of atoms and/or ions is isoelectronic? a. O2 and Cl b. Ca2+ and Cl c. F and N2 d. Li and Na e. K and Kr

____ 16. What is the bond angle in a bent molecule, such as water? a. 90 b. 104.5 c. 107.3 d. 109.5 e. 120

____ 17. Which process produces the most energy? a. cooking b. chemical reaction c. physical change d. dissolution e. nuclear reaction

____ 18. 5.00 g of sodium hydroxide is dissolved in 150 mL of water. Using a coffee-cup calorimeter, the temperature change of the water is measured to be 2.2C. The specific heat capacity of water is 4.184 J/gC. What is the heat change for this dissolution? a. 46 J b. 1.4  103 J c. 5.8  103 J d. 1.7  105 J e. 58 J

____ 19. What is the definition of the temperature of a substance? a. the total heat content of a substance b. the speed of the fastest particles in the substance c. the speed of the slowest particles in the substance, subtracted from the speed of the fastest particles in the substance d. the heat capacity of the substance times its mass e. the average kinetic energy of a system ____ 20. In an exothermic reaction, Hrxn is . . . a. always positive b. always negative c. negative or positive d. zero e. not possible to measure

____ 21. In the following reaction, butane is consumed at the rate of 0.0333 mol/(Ls). Determine the rate at which

CO2 is produced.

C4H10(g)  O2(g)  4CO2(g)  5H2O(g) a. 0.008 25 mol/(Ls) b. 0.0165 mol/(Ls) c. 0.0333 mol/(Ls) d. 0.0667 mol/(Ls) e. 0.133 mol/(Ls)

____ 22. Which statement about the factors that affect reaction rates is false? a. Decreasing the concentrations of the reacting particles decreases the chance of collision. b. A collision with poor orientation requires a higher activation energy than a collision with optimum orientation. c. Increasing the pressure in a gaseous reaction increases the chance of collision. d. A reaction occurs every time particles of the reactants collide. e. Increasing the temperature increases the reaction rate.

____ 23. Given the following reaction mechanism, what is the equation for the overall reaction? 2A  B  2C (slow) B  C  D  E (fast) C  D  E  F (fast) a. 2A  2E  F b. 2A  B  2C  D  2E  F c. 2A  2C  2E  F d. 2A  B  2C  D  B  2C  D  2E  F e. 2A  C  2E  F

____ 24. Which quantity does not increase when the temperature of a reaction system is raised? a. activation energy b. number of collisions c. number of effective collisions d. average kinetic energy of the particles e. all of the above increase

____ 25. The isotope has a half-life of 165 days. The isotope has a half-life of 330 days. Which statement about these two isotopes is false? a. One year represents slightly more than two half-lives of a sample of the calcium isotope. b. Almost all of 100 g of the calcium isotope will decay in one year’s time. c. 100 g of the vanadium isotope will decay to somewhat less than 50 g in a year. d. The calcium isotope decays at twice the rate of the vanadium isotope. e. All of these statements are true.

____ 26. Which condition is not a condition for an equilibrium system? a. equal rates for opposing changes b. a closed system c. reactants in the same phase d. constant temperature e. all of the above

____ 27. Which condition represents a spontaneous reaction? a. negative change in enthalpy and negative change in entropy b. negative change in enthalpy and positive change in entropy c. positive change in enthalpy and negative change in entropy d. positive change in enthalpy and positive change in entropy e. all of the above

____ 28. Which condition represents an unfavourable reaction, regardless of temperature? a. negative change in enthalpy and negative change in entropy b. negative change in enthalpy and positive change in entropy c. positive change in enthalpy and negative change in entropy d. positive change in enthalpy and positive change in entropy e. all of the above

____ 29. A reaction quotient is calculated to be 3.2  105. The equilibrium constant for the same reaction is 5.4  105. Which statement is correct? a. The system is at equilibrium. b. The concentrations of the products are greater than their concentrations at equilibrium. c. The system will attain equilibrium by moving to the right. d. The system will attain equilibrium by moving to the left. e. More information is needed to determine which statement is correct.

____ 30. Use Le Châtelier’s principle to predict what will happen to the following equilibrium if the pressure is decreased.

2NH3(g)  N2(g)  3H2(g) a. The system will not change. b. Equilibrium will shift toward the products. c. Equilibrium will shift toward the reactants.

d. The concentration of N2 will decrease. e. The concentration of NH3 will increase.

____ 31. Which reaction represents a favourable change in entropy?

a. I2(s)  I2(g)

b. H2O(l)  H2O(s)

c. N2(g)  3H2(g)  2NH3(g)  2 d. 2Na (aq)  O (aq)  Na2O(s) e. all of the above ____ 32. Use Le Châtelier’s principle to predict what will happen if silver acetate is added to the following equilibrium system. 2  2 Co(H2O)6 (aq)  4Cl (aq)  CoCl4 (aq)  6H2O(l) pink purple/blue a. The solution will become more pink. b. The solution will become more purple. c. There will be no change to the equilibrium system. d. The concentration of chloride will increase. e. The pH of the system will increase.

____ 33. Use Le Châtelier’s principle to predict what will happen if the temperature of the following equilibrium system is decreased. The reaction in the forward direction is endothermic. 2  2 Co(H2O)6 (aq)  4Cl (aq)  CoCl4 (aq)  6H2O(l) pink purple/blue a. The equilibrium will shift to the left. b. The equilibrium solution will become more purple. c. The equilibrium will shift to the right. d. The concentration of chloride ion will decrease. e. There will be no change.

____ 34. What will happen to the following equilibrium if an inert gas is added while the volume remains constant?

2IBr(g)  I2(g)  Br2(g) a. The concentration of IBr will increase.

b. The concentration of I2 will increase. c. The concentration of Br2 will decrease. d. There will be no change to the equilibrium system. e. The pressure will decrease.

____ 35. What is the relationship between the ion product and the solubility product constant when a precipitate does not form?

a. Qsp > Ksp

b. Qsp  Ksp c. Qsp < Ksp

d. Qsp  Ksp e. Qsp = Ksp

____ 36. Using the solubility rules, which compound does not form a precipitate in water?

a. PbSO4 b. CaO

c. CaCO3 d. CaBr2 e. Hg3PO4

____ 37. Which salt produces an acidic solution? a. NaCl

b. NaCH3COO c. K2SO4 d. NaNO3 e. NH4NO3 ____ 38. Which salt produces a basic solution?

a. KNO2 b. Na2SO4 c. NH4ClO4 d. Ca(NO3)2 e. NH4CH3COO

5 8 ____ 39. A weak acid (Ka = 4.5  10 ) is neutralized with a weak base (Kb = 5.6  10 ). What is the pH of the resulting solution? a. acidic b. basic c. neutral d. pH = 3.2 e. pH = 7

____ 40. The equivalence point of a neutralization reaction is 8.5. What is the best indicator to use? a. crystal violet b. bromocresol green c. thymol blue d. methyl red e. alizatin

____ 41. Which statement about solubility is correct? a. Solubility is always favourable because dissolving ions results in an increase in entropy. b. Solubility is always unfavourable because energy is needed to separate the anions from the cations. c. Solubility is unfavourable if the energy that is needed to separate the ions is greater than the energy that is released to surround each ion in water. d. Solubility is unfavourable if the energy that is released when the ions are surrounded in water is greater than the energy that is needed to separate the ions. e. All ionic compounds are soluble.

____ 42. 250.0 mL of a 0.250 mol/L solution of sodium iodide is added to 1.00 L of a 0.100 mol/L solution of copper(II) iodide. What is the new concentration of iodide ions? a. 0.263 mol/L b. 0.250 mol/L c. 0.210 mol/L d. 0.162 mol/L e. 0.130 mol/L

____ 43. 9.54  104 mol/L is the solubility of lead(II) iodide in water at 0C. What is the solubility product constant for lead(II) iodide at the same temperature? a. 9.54  104 b. 1.82  106 c. 9.10  107 d. 3.48  109 e. 8.68  1010 ____ 44. The solubility of iron(II) carbonate is 6.7  103 g/100 mL at 25C. What is this solubility in mol/L? a. 6.7  103 mol/L b. 3.8  104 mol/L c. 5.9  104 mol/L d. 3.8  105 mol/L e. 5.9  105 mol/L

3 ____ 45. What is the oxidation number of P in PO3 ? a. 6 b. 3 c. +3 d. +6 e. +5

____ 46. How many electrons need to be added to balance the following half-reaction?  Br2(g)  2Br (aq) a. one electron to the right side b. one electron to the left side c. two electrons to the left side d. two electrons to the right side e. no electrons are needed for this equation

____ 47. Which statement about the following reaction is correct?

6NaOH + 3Br2  NaBrO3 + 5NaBr + 3H2O a. Bromine is reduced, and sodium is oxidized. b. Hydrogen is oxidized, and oxygen is reduced. c. Bromine is both oxidized and reduced. d. Oxygen is reduced, and sodium is oxidized. e. This is not a redox reaction.

____ 48. Which statement about the following equation is correct?

3Cu + 8HNO3  3Cu(NO3)2 + 2NO + 4H2O a. Nitrogen is reduced, and copper is oxidized. b. Nitrogen is oxidized, and copper is oxidized. c. Nitrogen is reduced, and oxygen is oxidized. d. Nitrogen is oxidized, and oxygen is oxidized. e. This is not a reduction-oxidation reaction.

____ 49. When a solid zinc strip is placed in a blue copper ion solution, the solution turns clear. What happened? a. The blue colour was bleached out. b. The blue ions bonded onto the zinc strip, so the strip became heavier. c. The blue copper ions were reduced to copper metal, and the zinc strip corroded into aqueous ions. d. Nothing happened. e. The solution would produce electricity until the beaker got too hot to touch.

____ 50. Which ion is the strongest reducing agent? a. Na b. Zn2 c. H d. Cu2 e. Mg2

____ 51. A voltaic cell has a solid lead electrode in a solution of 1.0 mol/L Pb(NO3)2 and a solid zinc electrode in a solution of 1.0 mol/L Zn(NO3)2. What reactant will be reduced? a. Zn2 b. Pb2 c. Pb d. Zn  e. NO3

____ 52. Which statement about electrochemical cells is not correct? a. A salt bridge is required for proper operation. b. Electrons always flow from the anode to the cathode. c. Negative ions within a cell migrate towards the anode. d. Oxidation always occurs at the anode. e. The value of E for an operating electrochemical cell is negative.

____ 53. An electrochemical cell consists of an electrode of solid lead immersed in a solution of 1.0 mol/L Pb(NO3)2 and an electrode of solid copper immersed in a solution of 1.0 mol/L Cu(NO3)2. The two half-cells are separated by a porous barrier. What happens in the cell when the electrodes are joined by a conducting wire? a. Electrons migrate from the cathode to the anode through the wire.  b. NO3 ions migrate from the anode to the cathode. c. Nitrogen dioxide gas is produced. d. Pb2 ions migrate to the Pb electrode. e. Cu2+ ions migrate to the Cu electrode.

____ 54. Consider the following half-reactions. 2  Pb (aq)  2e  Pb(s) E = –0.1262 V   Ag (aq) + e  Ag(s) E = 0.800 V What is the standard cell potential for the reaction below? 2  2Ag(s) + Pb (aq)  Pb(s)  2Ag (aq) a. 0.926 V b. 0.674 V c. 0.674 V d. 0.926 V e. 0.463 V

Short Answer

For the following questions, write the most appropriate answer in the space provided.

55. Write the condensed structure of the ester that is produced by the reaction of methanol and formic acid. Name the ester. 56. State three ways that proteins differ from each other. 57. What contributions did Planck and Einstein make to the current model of the atom? 58. Outline the steps you would take to determine the molecular shape of a compound. 59. The neutralization of nitric acid with potassium hydroxide has an enthalpy change of 53.4 kJ/mol. Write a thermochemical equation for this reaction. 60. Methane burns in oxygen to form carbon dioxide and water. This process releases 882 kJ/mol of methane. a) Write the thermochemical equation for this reaction. b) If 15.0 g of methane is burned, how much heat is released? 61. What are the two major requirements for a reaction to occur? 62. The rate constant for the following reaction is 6.0  104 s1. What is the half-life of this reaction?

N2O5(g)  2NO2(g)  O2(g)

63. The equilibrium constant for the following reaction is 4.8  103 at 25ºC.

N2O4(g)  2NO2(g) a) Which direction does the equilibrium normally favour? b) Which species has the greater concentration? c) What is the equilibrium constant for the reverse reaction? 64. Based on your knowledge of strong and weak acids and bases, is the solution of each salt acidic, basic, or neutral? Explain your choice. a) KCN

b) Na2SO4 c) CuClO3 65. Distinguish between the end-point and the equivalence point in a titration. 66. Identify the oxidizing agent and the reducing agent in the following redox reaction. Explain your answers.  2 +2 2 MnO4 + SO3  Mn + SO4

Problem

Graphics For the following questions, use the graphics provided to review terms or skills. Add any missing labels, draw any missing parts, or use the graphics to help you answer a question.

67. Complete the following table.

Reaction Type of reaction Class of organic product 68. Draw the chemical reaction that links an alcohol to each organic compound. Identify the type of chemical reaction. a) aldehyde b) ketone c) carboxylic acid d) ester 69. Draw the three-dimensional representation of each molecule.

a) CCl4 (tetrahedral)  b) BrCl4 (square planar) c) NF3 (trigonal pyramidal) 70. a) Given equations (1) and (2), calculate the enthalpy change for equation (3).

(1) Pb(s)  PbO2(s)  2SO3(g)  2PbSO4(s) H  775 kJ

(2) SO3(g)  H2O(l)  H2SO4(aq) H  133 kJ

(3) Pb(s)  PbO2(s)  2H2SO4(aq)  2PbSO4(s)  2H2O(l) b) Draw an enthalpy diagram to represent equation (3). 71. Use the following diagram to answer the questions below.

a) Is the reaction exothermic or endothermic? Explain. b) What letter represents the activation energy of the forward reaction? c) What letter represents the heat of reaction? d) What letter represents the activation energy of the reverse reaction? 72. Complete the following table, based on the following equilibrium system. 2 2 Ni (aq)  6NH3(aq)  Ni(NH3)6 (aq) green blue

Stress Colour addition of nickel(II) nitrate removal of ammonia addition of water addition of inert gas at constant pressure 2 removal of Ni(NH3)6 (aq)

73. In an experiment, strips of three metalsX, Y, and Zwere placed in beakers that contained solutions of the aqueous ions X2, Y2, and Z2and were allowed to react. The following data were obtained. Based on these data, what is the order of the metal ions, in decreasing ability to be reduced.

X(s) Y(s) Z(s) 2 X (aq) No reaction No reaction No reaction 2 Y (aq) Crystals formed No reaction Crystals formed 2 Z (aq) Crystals formed No reaction No reaction

74. An electrochemical cell consists of two half-cells connected by a salt bridge. One half-cell contains an iron electrode in 1.0 mol/L iron(II) nitrate solution. The other half-cell contains a silver electrode in 1.0 mol/L silver nitrate solution. The salt bridge contains sodium sulfate. a) Write the shorthand representation for the electrochemical cell. b) Sketch the electrochemical cell. Mark the anode, the cathode, and the charge of each electrode. Indicate the direction of electron and ion movement. c) Write the half-reactions that occur at the anode and the cathode.

Critical Thinking For the following questions, write the answer in the space provided. Use complete sentences in your answer. If the question requires mathematical calculations, show all of your work. Write a final statement that gives your solution.

75. The two substances in the following table have similar molecular masses but very different boiling points. Explain how this is possible.

Name Formula Molecular mass Boiling point

ethanol CH3CH2OH 46.0 u 78.5°C

carbon dioxide CO2 44.0 u -78.5°C

76. A 26.6 g sample of mercury is heated to 110.0C and then placed in 125 g of water in a coffee-cup calorimeter. The initial temperature of the water is 23.00C. The specific heat capacity of water is 4.184 J/gC, and the specific heat capacity of mercury is 0.139 J/gC. What is the final temperature of the water and the mercury?

77. The experimental data in the table below were collected for the following decomposition of SO2Cl2(g). What is the rate law for this reaction?

SO2Cl2(g)  SO2(g)  Cl2(g)

Trial Initial concentration of SO2Cl2(g) (mol/L) Initial reaction rate [mol/(Ls)] 1 0.100 2.2  106 2 0.200 4.4  106 3 0.300 6.6  106 78. 0.150 mol of SO3 and 0.150 mol of NO are placed in a 1 L container and allowed to react as follows. At equilibrium, the concentration of both SO2 and NO2 is 0.0621 mol/L. What is the equilibrium constant?

SO3(g)  NO(g)  SO2(g)  NO2(g)

79. The equilibrium constant for the following reaction is 14.3. If the equilibrium concentration of O2 is 0.360 mol/L, how much NO2 was introduced into the container?

2NO2(g)  2NO(g)  O2(g)

o 80. The following equilibrium system has an equilibrium constant of 4.06 at 500 C. 0.500 mol of CO(g) and 0.500 mol of H2O(g) are added to a 2.5 L container at this temperature. Determine the equilibrium concentrations.

CO(g)  H2O(g)  CO2(g)  H2(g)

81. 30.0 mL of a solution of a diprotic acid, oxalic acid (C2H2O4), is titrated with 56 mL of a 0.050 mol/L solution of potassium hydroxide. What was the concentration of the oxalic acid?

4 82. A 35.0 mL sample of (monoprotic) lactic acid, C3H6O3, is titrated with 20.0 mL of a 4.0  10 mol/L sodium hydroxide solution. What is the pH of the resulting solution at the equivalence point, if Ka for lactic acid is 1.4  104? 83. What is the largest concentration of barium ions that can be added to a 2.3  102 mol/L solution of fluoride 6 ions, without a precipitate of BaF2 forming? Ksp for barium fluoride is 1.7  10 . 84. A buffer solution is made by mixing 400.0 mL of 0.15 mol/L acetic acid with 325 mL of 0.20 mol/L sodium acetate. What is the pH of the buffer solution? 85. When there is sufficient ethanol in a person’s breath, the chromium ions in a police breathalyzer change from 2 3+ orange dichromate, Cr2O7 , to green Cr . The carbon is oxidized from 2 in the ethanol molecule to 0 in the C2H4O2 product molecule. Use the oxidation number method or the half-reaction method to write and balance the equation that occurs in acidic conditions. 4U Chemistry Practice Exam Answer Section

MULTIPLE CHOICE

1. ANS: E DIF: easy REF: K/U LOC: OC 2.02 2. ANS: B DIF: easy REF: K/U LOC: OC V.02 3. ANS: D DIF: easy REF: K/U LOC: OC V.01, OC 1.01 4. ANS: A DIF: average REF: K/U LOC: OC 1.02 5. ANS: D DIF: easy REF: K/U LOC: OC 1.02 6. ANS: B DIF: difficult REF: K/U LOC: OC V.01 7. ANS: D DIF: average REF: K/U, I LOC: OC V.01 8. ANS: E DIF: easy REF: K/U LOC: OC 1.03, OC V.01 9. ANS: B DIF: average REF: K/U LOC: OC 1.03 10. ANS: E DIF: easy REF: K/U LOC: SP 1.02, SP V.01 11. ANS: B DIF: easy REF: K/U LOC: SP 1.02, SP V.01 12. ANS: E DIF: average REF: K/U LOC: SP 1.02, SP V.01 13. ANS: C DIF: easy REF: K/U LOC: SP 1.02, SP 1.03, SP V.01 14. ANS: C DIF: average REF: K/U LOC: SP 1.02, SP V.01 15. ANS: B DIF: average REF: K/U LOC: SP 1.02, SP V.01 16. ANS: B DIF: easy REF: K/U LOC: SP 2.03, SP V.02 17. ANS: E DIF: easy REF: K/U LOC: EC 1.01 18. ANS: B DIF: average REF: I LOC: EC 2.03 19. ANS: E DIF: easy REF: K/U LOC: EC 2.01 20. ANS: B DIF: easy REF: K/U LOC: EC 2.01 21. ANS: E DIF: average REF: K/U LOC: EC V.01, EC 2.06 22. ANS: D DIF: average REF: K/U LOC: EC 1.04 23. ANS: A DIF: easy REF: K/U, I LOC: EC 1.06 24. ANS: A DIF: average REF: K/U LOC: EC 1.04 25. ANS: B DIF: average REF: I LOC: EC 1.03 26. ANS: C DIF: average REF: K/U LOC: CS 1.01, CS V.01 27. ANS: B DIF: easy REF: K/U LOC: CS 1.05 28. ANS: C DIF: easy REF: K/U LOC: CS 1.05 29. ANS: C DIF: average REF: K/U LOC: CS 1.02, CS V.02 30. ANS: B DIF: average REF: K/U LOC: CS 1.03 31. ANS: A DIF: easy REF: K/U LOC: CS 1.04 32. ANS: A DIF: difficult REF: K/U LOC: CS 1.03 33. ANS: A DIF: average REF: K/U LOC: CS 1.03 34. ANS: D DIF: difficult REF: K/U LOC: CS 1.03 35. ANS: C DIF: easy REF: K/U LOC: CS 2.05 36. ANS: D DIF: easy REF: K/U LOC: CS 1.06, CS 2.01 37. ANS: E DIF: average REF: K/U LOC: CS 2.07 38. ANS: A DIF: difficult REF: K/U LOC: CS 2.07 39. ANS: A DIF: average REF: K/U LOC: CS 2.07 40. ANS: C DIF: easy REF: K/U LOC: CS 2.08 41. ANS: C DIF: easy REF: K/U LOC: CS 1.05 42. ANS: C DIF: difficult REF: I LOC: CS 2.01 43. ANS: D DIF: average REF: I LOC: CS 2.04 44. ANS: C DIF: average REF: I LOC: CS 2.04 45. ANS: C DIF: easy REF: K/U LOC: EL 1.01 46. ANS: C DIF: average REF: K/U LOC: EL 1.01, EL 2.03 47. ANS: C DIF: average REF: K/U, I LOC: EL 1.01 48. ANS: A DIF: average REF: K/U, I LOC: EL V.01 49. ANS: C DIF: average REF: K/U LOC: EL V.01 50. ANS: A DIF: easy REF: K/U LOC: EL V.01 51. ANS: B DIF: average REF: I LOC: EL 1.02 52. ANS: E DIF: average REF: K/U LOC: EL 1.02 53. ANS: E DIF: difficult REF: K/U LOC: EL 1.02 54. ANS: A DIF: difficult REF: I LOC: EL 2.05

SHORT ANSWER

55. ANS:

CHO–O–CH3, methyl formate

DIF: average REF: I LOC: OC 1.02 56. ANS: 1) the order of their amino acids 2) the way their chains are coiled, folded, or twisted 3) the type of bonding that holds each particular protein in its specific shape

DIF: average REF: K/U, I, C LOC: OC 1.05 57. ANS: Planck proposed that matter at the atomic level can absorb or emit only discrete quantities of energy. In other words, Planck said that the energy of the atom is quantized. Einstein proposed that all forms of electromagnetic energy travel as photons of energy.

DIF: easy REF: K/U, C LOC: SP 1.02, SP V.01 58. ANS: 1) Draw a Lewis structure of the molecule. 2) Determine the number of pairs of valence electrons around the central atom 3) Determine which one of the five geometric arrangements can accommodate this total number of electron groups. 4) Determine the molecular shape from the positions occupied by the bonding pairs and lone pairs.

DIF: easy REF: K/U, C LOC: SP 2.03 59. ANS: Since the enthalpy change is negative, the reaction is exothermic. Therefore, the energy term is written on the product side of the reaction equation.

HNO3(aq)  KOH(aq)  KNO3(aq)  H2(l)  53.4 kJ

DIF: easy REF: K/U, C LOC: EC 2.02 60. ANS: a) CH4(g)  O2(g)  CO2(g)  H2O(g)  882 kJ

b)

Burning 15.0 g of methane releases 825 kJ of heat.

DIF: average REF: I LOC: EC 2.02 61. ANS: The following two requirements must be met for a reaction to occur: - A collision must occur between reactant particles that have the correct orientation. - Reactants must have sufficient collision energy to be able to overcome the activation energy barrier.

DIF: easy REF: K/U LOC: EC 1.04 62. ANS:

The equation has only N2O5(g) as a reactant, and the molar coefficient is one. Therefore, this is a first-order reaction. For a first-order reaction,  0.693  k  0.693  (6.0  104 s1)  1.16  103 s  19.3 min The half-life is 19.3 min.

DIF: difficult REF: I LOC: EC 1.03 63. ANS: a) Since the equilibrium constant is less than 1, the equilibrium favours the reactants.

b) N2O4(g) has the greater concentration. c) The equilibrium constant for the reverse reaction is the inverse of the equilibrium constant for the forward reaction.

DIF: average REF: K/U, I LOC: CS 1.02, CS 1.07 64. ANS: a) The solution is basic, since KCN is the salt of a strong base and a weak acid.

b) The solution is neutral, since Na2SO4 is the salt of a strong base and a strong acid. c) The solution is acidic, since CuClO3 is the salt of a weak base and a strong acid.

DIF: average REF: I, K/U LOC: CS 2.07 65. ANS: The end-point is the point at which the indicator changes colour. The equivalence point is the point at which just enough acid and base have been mixed for a complete reaction to occur, with no excess of either reactant.

DIF: easy REF: K/U LOC: CS 2.01 66. ANS:  2 2 The oxidizing agent is MnO4 because it gained electrons from SO3 to allow SO3 to be oxidized. The 2   reducing agent is SO3 because it lost electrons to MnO4 to allow MnO4 to be reduced.

DIF: average REF: K/U, C LOC: EL 2.01, EL V.01, EL 2.03

PROBLEM

67. ANS: Reaction Type of reaction Class of organic product substitution chlorinated hydrocarbon

elimination alkene

neutralization salt of an acid

hydrolysis organic acid, amine

addition polymerization polymer (polystyrene)

DIF: average REF: K/U, I LOC: OC V.01, OC 1.04, OC 2.05 68. ANS:

DIF: average REF: MC LOC: OC 2.05 69. ANS: DIF: easy REF: K/U LOC: SP 1.05 70. ANS:

a) Required equation: Pb(s) + PbO2(s) + 2H2SO4(aq)  2PbSO4(s) + 2H2O(l)

Pb(s)  PbO2(s)  2SO3(g)  2PbSO4(s) H  775 kJ

2H2SO4(aq)  2SO3(g)  2H2O(l) H  +266 kJ

Pb(s)  PbO2(s)  2H2SO4(aq)  2PbSO4(s) + 2H2O(l) H  509 kJ The enthalpy change for equation (3) is 509 kJ/mol. b)

DIF: difficult REF: I LOC: EC 2.04, EC 1.05 71. ANS: a) The reaction is endothermic. The enthalpy of the products is higher than the enthalpy of the reactants. This means that energy is absorbed as the reaction proceeds. b) A c) C d) B

DIF: easy REF: K/U LOC: EC 1.05 72. ANS:

Stress Colour addition of nickel (II) nitrate increase in intensity of blue removal of ammonia increase in intensity of green addition of water dilution and reduction in intensity of colour addition of inert gas at constant pressure no effect 2 removal of Ni(NH3)6 (aq) decrease in intensity of green DIF: difficult REF: I LOC: CS 1.03, CS 2.02 73. ANS: Y2 > Z2 > X2

DIF: difficult REF: I LOC: EL 2.02 74. ANS: 2  a) Fe(s)Fe (aq)Ag (aq)Ag(s) b)

2  c) Anode: Fe(s)  Fe (aq) + 2e   Cathode: 2Ag (aq)  2e  2Ag

DIF: average REF: I LOC: EL 2.04, EL 2.01 75. ANS:

The CO2 molecule has a linear shape, with the oxygen atoms located on either side of the carbon atom. This results in a non-polar molecule. The only forces that are holding the CO2 molecules together are dispersion forces. CH3CH2OH has oxygen atoms bonded to hydrogen atoms, resulting in hydrogen bonds between molecules. These hydrogen bonds are much stronger than the dispersion forces between CO2 molecules. Therefore, CH3CH2OH has a much higher boiling point.

DIF: average REF: I, C LOC: SP 2.05, SP 1.04, SP 2.06 76. ANS:

Qlost by mercury  Qgained by water mcT  mcT

(26.6 g)(0.139 J/gC)(110.0°C  Tf)  (125 g)(4.184 J/gC)(Tf  23.00C)

406.7  3.70Tf  523Tf  12 029 519.3Tf  124 357

Tf  23.9C

The final temperature of both the mercury and the water is 23.9C.

DIF: difficult REF: I LOC: EC 2.04 77. ANS:

Compare trials 1 and 2: Doubling the concentration of SO2Cl2(g) causes the reaction rate to double. Therefore, the reaction rate is first order with respect to the concentration of SO2Cl2(g). Comparing trials 1 and 3 verifies this conclusion. The tripling of the concentration of SO2Cl2(g) causes the reaction rate to triple. 1 Reaction rate  k[SO2Cl2 (g)]

DIF: average REF: I LOC: EC 1.03 78. ANS:

Concentration (mol/L) SO3(g)  NO(g)  SO2(g)  NO2(g) Initial 0.150 0.150 0.0 0.0 Change 0.0621 0.0621 0.0621 0.0621 Final 0.0879 0.0879 0.0621 0.0621

Keq 



 4.99  101 The equilibrium constant is 4.99  101.

DIF: average REF: I LOC: CS 2.06 79. ANS:

Concentration (mol/L) 2NO2(g)  2NO(g) + O2(g) Initial y 0.0 0.0 Change 0.720 2(0.360) = 0.720 0.360 Final y  0.720 0.720 0.360

Keq 

 14.3

(y  0.720)2 

(y  0.720)2  1.31  10–2 Take the square root of both sides. y  0.720  0.114 y 0.834

Therefore, 0.834 mol of NO2 was introduced for each litre of volume.

DIF: average REF: I LOC: CS 2.06 80. ANS:

Concentration (mol/L) CO(g)  H2O(g)  CO2(g)  H2(g) Initial 0.500 0.500 0.0 0.0 Change x x x x Final 0.500  x 0.500  x x x Keq 



 4.06 Take the square root of both sides.

 2.015

x  2.015(0.500  x) x  1.008  2.015x 2.015x  1.008 x  0.334 mol/L

The equilibrium concentrations of CO2(g) and H2(g) are both 0.334 mol/L. The equilibrium concentrations of

CO(g) and H2O(g) are both 0.500  0.334  0.166 mol/L.

DIF: average REF: I LOC: CS 2.06 81. ANS:

2KOH(aq) + C2H2O4(aq)  K2C2O4(aq) + 2H2O(l) Moles KOH = C  V = 0.056 L  0.050 mol/L = 2.8  103 mol

3 Moles C2H2O4 = 2.8  10 mol KOH 

= 1.4  103 mol

= 1.4 

= 4.7  102 mol/L The concentration of the oxalic acid was 4.7  102 mol/L.

DIF: average REF: I LOC: CS 2.08 82. ANS:

C3H6O3(aq) + NaOH(aq)  NaC3H5O3(aq) + H2O(l) n  c  V Moles NaOH  0.020 L  (4.0  104 mol/L)  8.0  106 mol 6 From the one to one ratio, there are 8.0  10 mol of NaC3H5O3(aq). Total volume  35.0 mL + 20.0 mL  55.0 mL 6 [NaC3H5O3(aq)]  8.0  10 mol/0.055 L  1.45  104 mol/L +  + The salt forms K (aq) and C3H5O3 (aq) in solution. K (aq) is the conjugate acid of a strong base, so it does not react  with water. C3H5O3 (aq) is the conjugate base of a weak acid, so it does react with water. The pH is determined by the extent of the following reaction.   C3H5O3 (aq) + H2O(l)  C3H6O3(aq) + OH (aq)

Kb 

 7.14  1011  Since x is very small, the change in [C3H5O3 (aq)] can be ignored.   Concentration C3H5O3 (aq) + H2O(l)  C3H6O3(aq) + OH (aq) (mol/L) Initial 1.45  104 0 0 Change x x x Equilibrium (1.45  104)  x 1.45  104 x x

7.14  1011 =

x2 = 1.0353  1014 x = 1.0175  107 pOH = log x = log(1.0175  107) = 6.99 pH = 14  6.99 = 7.01 The pH at the equivalence point is 7.01.

DIF: difficult REF: I LOC: CS 2.08 83. ANS: 2+  BaF2(s)  Ba (aq) + 2F (aq) This problem involves the common ion effect. 2+ - 2 Ksp = [Ba (aq)][F (aq)]

2+  Concentration (mol/L) BaF2(s)  Ba (aq) + 2F (aq) Initial ---- 0 2.3  102 Change ---- x x Equilibrium ---- x (2.3  102) + x

Since x is small, use an approximation. (2.3  102) + x  2.3  102 2+  2 Ksp = [Ba (aq)][F (aq)] 2 2 Ksp = (x)(2.3  10 ) 9.0  109 = (x)(2.3  102)2 x = 1.70  105 mol/L The largest concentration of barium ions that can be added to a 2.3  102 mol/L solution of fluoride ions, 5 without a precipitate of BaF2 forming, is 1.70  10 mol/L. DIF: difficult REF: I LOC: CS 2.06 84. ANS:

[HC2H3O2(aq)] = 0.15 mol/L 

= 0.15 mol/L  = 0.008 mol/L

 [C2H3O2 (aq)] = 0.20 mol/L 

= 0.20 mol/L  = 0.090 mol/L

 + Concentration (mol/L) HC2H3O2(aq) + H2O(l) C2H3O2 (aq) + H3O (aq) Initial 0.008 0.090  0 Change x +x +x Equilibrium 0.008 – x 0.090 + x x

Ka =

 [C2H3O2 (aq)]  0.090 and [HC2H3O2(aq)]  0.008 mol/L

1.8  105 =

x = 1.6  106 mol/L pH = log(1.6  106) = 5.79

DIF: difficult REF: I LOC: CS 1.09, CS 2.06 85. ANS: + 2 3+ 16H + 2Cr2O7 + 3C2H5OH  4Cr + 3C2H4O2 + 11H2O

DIF: difficult REF: I, MC LOC: EL 2.02, EL 2.03