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 =
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 =
. In which orbital is this
____ 13. Which element has the ground state electron configuration [Ne] 3s23px13py1 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?
____ 16.
____ 17.
____ 18.
____ 19.
____ 20.
____ 21.
a. O2 and Cl
b. Ca2+ and Cl
c. F and N2
d. Li and Na
e. K and Kr
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
Which process produces the most energy?
a. cooking
b. chemical reaction
c. physical change
d. dissolution
e. nuclear reaction
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
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
In an exothermic reaction, Hrxn is . . .
a. always positive
b. always negative
c. negative or positive
d. zero
e. not possible to measure
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?
____ 30.
____ 31.
____ 32.
____ 33.
____ 34.
____ 35.
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.
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.
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)
d. 2Na(aq)  O2(aq)  Na2O(s)
e. all of the above
Use Le Châtelier’s principle to predict what will happen if silver acetate is added to the following equilibrium
system.
Co(H2O)62(aq)  4Cl (aq)  CoCl42(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.
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.
Co(H2O)62(aq)  4Cl(aq)  CoCl42(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.
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.
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
____ 36.
____ 37.
____ 38.
____ 39.
____ 40.
____ 41.
____ 42.
d. Qsp  Ksp
e. Qsp = Ksp
Using the solubility rules, which compound does not form a precipitate in water?
a. PbSO4
b. CaO
c. CaCO3
d. CaBr2
e. Hg3PO4
Which salt produces an acidic solution?
a. NaCl
b. NaCH3COO
c. K2SO4
d. NaNO3
e. NH4NO3
Which salt produces a basic solution?
a. KNO2
b. Na2SO4
c. NH4ClO4
d. Ca(NO3)2
e. NH4CH3COO
A weak acid (Ka = 4.5  105) is neutralized with a weak base (Kb = 5.6  108). What is the pH of the
resulting solution?
a. acidic
b. basic
c. neutral
d. pH = 3.2
e. pH = 7
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
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.
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
____ 45. What is the oxidation number of P in PO33?
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?
____ 51.
____ 52.
____ 53.
____ 54.
a. Na
b. Zn2
c. H
d. Cu2
e. Mg2
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
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.
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.
Consider the following half-reactions.
Pb2(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?
2Ag(s) + Pb2(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.
MnO4 + SO32  Mn+2 + SO42
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.
Ni2(aq)  6NH3(aq)  Ni(NH3)62(aq)
green
blue
Stress
addition of nickel(II) nitrate
removal of ammonia
addition of water
addition of inert gas at constant pressure
removal of Ni(NH3)62(aq)
Colour
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.
2
X (aq)
Y2(aq)
Z2(aq)
X(s)
No reaction
Crystals formed
Crystals formed
Y(s)
No reaction
No reaction
No reaction
Z(s)
No reaction
Crystals formed
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
ethanol
carbon dioxide
Formula
CH3CH2OH
CO2
Molecular mass
46.0 u
44.0 u
Boiling point
78.5°C
-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
1
2
3
Initial concentration of SO2Cl2(g) (mol/L)
0.100
0.200
0.300
Initial reaction rate [mol/(Ls)]
2.2  106
4.4  106
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)
80. The following equilibrium system has an equilibrium constant of 4.06 at 500oC. 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?
82. A 35.0 mL sample of (monoprotic) lactic acid, C3H6O3, is titrated with 20.0 mL of a 4.0  104 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
ions, without a precipitate of BaF2 forming? Ksp for barium fluoride is 1.7  106.
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
orange dichromate, Cr2O72, to green Cr3+. 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
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DIF:
DIF:
DIF:
DIF:
DIF:
DIF:
DIF:
DIF:
DIF:
DIF:
DIF:
DIF:
DIF:
DIF:
DIF:
DIF:
DIF:
DIF:
DIF:
DIF:
easy
easy
easy
average
easy
difficult
average
easy
average
easy
easy
average
easy
average
average
easy
easy
average
easy
easy
average
average
easy
average
average
average
easy
easy
average
average
easy
difficult
average
difficult
easy
easy
average
difficult
average
easy
easy
REF:
REF:
REF:
REF:
REF:
REF:
REF:
REF:
REF:
REF:
REF:
REF:
REF:
REF:
REF:
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REF:
REF:
REF:
REF:
REF:
REF:
REF:
REF:
K/U
K/U
K/U
K/U
K/U
K/U
K/U, I
K/U
K/U
K/U
K/U
K/U
K/U
K/U
K/U
K/U
K/U
I
K/U
K/U
K/U
K/U
K/U, I
K/U
I
K/U
K/U
K/U
K/U
K/U
K/U
K/U
K/U
K/U
K/U
K/U
K/U
K/U
K/U
K/U
K/U
LOC:
LOC:
LOC:
LOC:
LOC:
LOC:
LOC:
LOC:
LOC:
LOC:
LOC:
LOC:
LOC:
LOC:
LOC:
LOC:
LOC:
LOC:
LOC:
LOC:
LOC:
LOC:
LOC:
LOC:
LOC:
LOC:
LOC:
LOC:
LOC:
LOC:
LOC:
LOC:
LOC:
LOC:
LOC:
LOC:
LOC:
LOC:
LOC:
LOC:
LOC:
OC 2.02
OC V.02
OC V.01, OC 1.01
OC 1.02
OC 1.02
OC V.01
OC V.01
OC 1.03, OC V.01
OC 1.03
SP 1.02, SP V.01
SP 1.02, SP V.01
SP 1.02, SP V.01
SP 1.02, SP 1.03, SP V.01
SP 1.02, SP V.01
SP 1.02, SP V.01
SP 2.03, SP V.02
EC 1.01
EC 2.03
EC 2.01
EC 2.01
EC V.01, EC 2.06
EC 1.04
EC 1.06
EC 1.04
EC 1.03
CS 1.01, CS V.01
CS 1.05
CS 1.05
CS 1.02, CS V.02
CS 1.03
CS 1.04
CS 1.03
CS 1.03
CS 1.03
CS 2.05
CS 1.06, CS 2.01
CS 2.07
CS 2.07
CS 2.07
CS 2.08
CS 1.05
42.
43.
44.
45.
46.
47.
48.
49.
50.
51.
52.
53.
54.
ANS:
ANS:
ANS:
ANS:
ANS:
ANS:
ANS:
ANS:
ANS:
ANS:
ANS:
ANS:
ANS:
C
D
C
C
C
C
A
C
A
B
E
E
A
DIF:
DIF:
DIF:
DIF:
DIF:
DIF:
DIF:
DIF:
DIF:
DIF:
DIF:
DIF:
DIF:
difficult
average
average
easy
average
average
average
average
easy
average
average
difficult
difficult
REF:
REF:
REF:
REF:
REF:
REF:
REF:
REF:
REF:
REF:
REF:
REF:
REF:
I
I
I
K/U
K/U
K/U, I
K/U, I
K/U
K/U
I
K/U
K/U
I
LOC:
LOC:
LOC:
LOC:
LOC:
LOC:
LOC:
LOC:
LOC:
LOC:
LOC:
LOC:
LOC:
CS 2.01
CS 2.04
CS 2.04
EL 1.01
EL 1.01, EL 2.03
EL 1.01
EL V.01
EL V.01
EL V.01
EL 1.02
EL 1.02
EL 1.02
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
60. ANS:
REF: K/U, C
LOC: EC 2.02
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:
The oxidizing agent is MnO4 because it gained electrons from SO32 to allow SO32 to be oxidized. The
reducing agent is SO32 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
substitution
Class of organic product
chlorinated hydrocarbon
elimination
alkene
neutralization
salt of an acid
hydrolysis
organic acid, amine
addition polymerization
polymer (polystyrene)
DIF: average
68. ANS:
REF: K/U, I
LOC: OC V.01, OC 1.04, OC 2.05
DIF: average
69. ANS:
REF: MC
LOC: OC 2.05
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
72. ANS:
REF: K/U
Stress
addition of nickel (II) nitrate
removal of ammonia
addition of water
addition of inert gas at constant pressure
removal of Ni(NH3)62(aq)
LOC: EC 1.05
Colour
increase in intensity of blue
increase in intensity of green
dilution and reduction in intensity of
colour
no effect
decrease in intensity of green
DIF: difficult
73. ANS:
Y2 > Z2 > X2
REF: I
LOC: CS 1.03, CS 2.02
DIF: difficult
REF: I
74. ANS:
a) Fe(s)Fe2(aq)Ag(aq)Ag(s)
b)
LOC: EL 2.02
c) Anode: Fe(s)  Fe2(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
76. ANS:
REF: I, C
Qlost by mercury
mcT
(26.6 g)(0.139 J/gC)(110.0°C  Tf)
406.7  3.70Tf
519.3Tf
Tf
LOC: SP 2.05, SP 1.04, SP 2.06






Qgained by water
mcT
(125 g)(4.184 J/gC)(Tf  23.00C)
523Tf  12 029
124 357
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.
Reaction rate  k[SO2Cl2 (g)]1
DIF: average
REF: I
78. ANS:
Concentration (mol/L)
SO3(g)
Initial
0.150
Change
0.0621
Final
0.0879
Keq 
LOC: EC 1.03

NO(g)
0.150
0.0621
0.0879

SO2(g)
0.0
0.0621
0.0621

NO2(g)
0.0
0.0621
0.0621



 4.99  101
The equilibrium constant is 4.99  101.
DIF: average
REF:
79. ANS:
Concentration (mol/L)
Initial
Change
Final
I
LOC: CS 2.06
2NO2(g)
y
0.720
y  0.720

2NO(g)
0.0
2(0.360) = 0.720
0.720
+
O2(g)
0.0
0.360
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
80. ANS:
Concentration (mol/L) CO(g)
Initial
0.500
Change
x
Final
0.500  x
Keq 
LOC: CS 2.06
 H2O(g)
0.500
x
0.500  x

CO2(g)
0.0
x
x

H2(g)
0.0
x
x

 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
Moles C2H2O4 = 2.8  103 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
From the one to one ratio, there are 8.0  106 mol of NaC3H5O3(aq).
Total volume  35.0 mL + 20.0 mL
 55.0 mL
[NaC3H5O3(aq)]  8.0  106 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
+ H2O(l)
C3H5O3(aq)
(mol/L)
Initial
1.45  104
Change
x
Equilibrium
(1.45  104)  x 1.45  104
 C3H6O3(aq)
+ OH(aq)
0
x
x
0
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:
BaF2(s)  Ba2+(aq) + 2F(aq)
This problem involves the common ion effect.
Ksp = [Ba2+(aq)][F-(aq)]2
Concentration (mol/L)
Initial
Change
Equilibrium
BaF2(s) 
----------
Ba2+(aq) +
0
x
x
2F(aq)
2.3  102
x
(2.3  102) + x
Since x is small, use an approximation.
(2.3  102) + x  2.3  102
Ksp = [Ba2+(aq)][F(aq)]2
Ksp = (x)(2.3  102)2
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,
without a precipitate of BaF2 forming, is 1.70  105 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)
Initial
Change
Equilibrium
HC2H3O2(aq) + H2O(l)
0.008
x
0.008 – x
C2H3O2(aq) +
0.090
+x
0.090 + 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:
16H+ + 2Cr2O72 + 3C2H5OH  4Cr3+ + 3C2H4O2 + 11H2O
DIF: difficult
REF: I, MC
LOC: EL 2.02, EL 2.03
H3O+(aq)
0
+x
x