Chemistry 12
Solutions Manual Part A
Unit 4 Chemical Systems b. 2.55 g of solid in enough water to make 75.0 mL of
solution
and Equilibrium c. 913.23 g of solid in enough water to make 1.75 L of
solution
Answers to Unit 4 Preparation Questions
6. a. 46.7 mL of acid mixed with 153.3 mL of water
(Student textbook pages 414-7)
b. 0.25 mL of acid mixed with 12.25 mL of water
1. Accept all reasonable answers. Sample answer: Strong
acids and strong bases are corrosive and dangerous to c. 18.3 mL of acid mixed with 1.08 L of water
transport. Pouring strong acids and bases down the 7. a. LiOH(aq) + HCl(aq) → LiCl(aq) + H2O(ℓ)
drain cause unsafe situations at water treatment plants b. 2KCl(aq) + Pb(NO3)2(aq) →
and cause unnecessary wear and tear on pipes, valves, PbCl2(s) + 2KNO3(aq)
and other parts in the sewer system. Disposing of c. Na2S(aq) + 2AgNO3(aq) → Ag2S(s) + 2NaNO3(aq)
unneutralized acids and bases is both unsafe and costly.
8. b
2. a. Always add concentrated acid to water. Once the
9. e
concentrated acid is measured, it should be added
to the 45.00 mL of water, so that the exothermic 10. a. NO3-(aq) and Na+
reaction can dissipate the heat throughout the 45 mL b. silver chromate, Ag2CrO4(s)
of water. c. chromate, CrO42-
b. Rinse any clothing thoroughly and use the eyewash d. 2Ag+(aq) + CrO42-(aq) →Ag2CrO4(s)
station to rinse your eyes immediately and 11. a. A solid formed.
continuously for at least 15 minutes, get your lab
partner to inform the teacher. b. 2Ag+(aq) + 2NO3-(aq) + 2Na+(aq) + CrO4-(aq)
→
c. Using excessive amounts of chemicals is costly
Ag2CrO4(s) + 2NO3-(aq) + 2Na+(aq)
because the initial cost is higher than necessary for
the extra chemicals, and the disposal is more costly 2Ag+(aq) + CrO4-(aq) →Ag2CrO4(s)
because you have to safely dispose of excessive
amounts of chemicals.
3. a. Because lead(II) chloride solid forms, follow all
instructions given by the teacher in terms of its
disposal. Some teachers might ask for the solution
to be filtered and the filter paper disposed of in
hazardous waste containers, while other teachers
might ask for all solutions to be poured into a 12. a. H+(aq) + Br-(aq) + K+(aq) + OH-(aq) →
hazardous waste container. K+(aq) + Br-(aq) + H2O(ℓ)
b. When you wash solid materials down the drain b. H+(aq) + OH-(aq) → H2O(ℓ)
you risk clogging the drain. It can be hazardous to c. neutralization reaction
unclog a drain that contains unknown or a variety of
13. n = cV
chemical solids in it.
mol 1L
4. a. A straw that another student has used is = 0.254 (25.00 mL)
contaminated with germs. Illnesses and diseases can L 1000 mL
be spread among students that share straws. = 0.00635 mol
b. Eyewear should always be worn in the laboratory. 14. n = cV
Also, when you blow too hard through the straw, mol 1L
spray from the mixture can get into your eyes. = 0.125 (38.42 mL)
L 1000 mL
5. a. 25.16 g of solid in enough water to make 225 mL of = 0.00480 mol
solution
Unit 4 Part A • MHR 1
,15. n = cV 20. A strong base completely dissociates in water, whereas
n a weak base does not dissociate completely.
V = c
21. A weak base can be concentrated, because
0.425 moL
= concentration is based on the amount of material in
mol
0.125 L solution, not the degree of ionization; therefore, the
= 3.40 L statement is true.
16. n = cV 22. 0.182 mol/L
n 23. The first proton is lost from the sulfuric molecule due
c = V
to the interactions of the polar water molecule. Once
0.385 mol 1000 mL this proton is lost, HSO4- remains and due to the
= ×
225 mL 1L negative charge, the second proton (which is positively
= 1.71 mol charged) is more difficult for water molecules to strip
L off the ion.
17. Moles of sulfate:
24. a. x = -1.56 or x = 2.56
n = cV
b. x = -1.46 or x = 5.46
mol 1L
= 0.200 (50.0 mL) c. x = -1.85 or x = 0.18
L 1000 mL
25. a. x = 0.50
= 0.0100 mol
b. 4.20 × 10-4
Moles of lead: 26. a. 26
n = cV
b. a10
mol 1L
= 0.100 (80.0 mL) c. x4y4
L 1000 mL
d. 34
= 0.00800 mol
27. a. 3
The limiting ion is lead so only 0.00800 moles of lead b. 4
sulfate can form. c. 8
Molar mass of PbSO4(s): d. 12
MPbSO4 = MPb + Ms + 4MO e. -12
g g g
= 207.2 + 32.07 + 4 × 16.00 28. a. 2.87 × 102
mol mol mol
b. 5.95 × 10-3
g
= 303.27 c. 1.79 × 106
mol
Mass of 0.00800 mol of lead sulfate: d. 4.12 × 10-7
m 29. a. The function key LOG is used to take the logarithm
n = M
of a number. The function key 10x is used to take the
m = nM
antilogarithm or to find the number that you raise
g the base to in order to obtain a specific number.
= (0.00800 mol) 303.27
mol b. Sample answer: log 0.001 = x and 102.961 = x;
= 2.42616 g c. The pH scale for acids and bases is based on
= 2.43 g logarithms.
18. a. lead (II) chromate
Chapter 7 Chemical Equilibrium
b. 2K+(aq) + CrO42-(aq) + Pb2+(aq) + 2NO3-(aq)
→
PbCrO4(s) + 2K+(aq) + 2NO3-(aq) Answers to Learning Check Questions
c. 0.58 g (Student textbook page 422)
19. A concentrated acid has a large number of acid 1. A reversible reaction is a reaction that can proceed in
molecules in solution, while a dilute acid has a smaller both the forward and the reverse directions.
number of acid molecules in solution.
2 MHR • Chemistry 12 Solutions Manual 978-0-07-106042-4
, 2. The double arrows indicate that the reaction is and since the denominator contains information
reversible. regarding reactants, reactant formation is favoured.
3. The rope-pulling team as a whole (comparable to a
(Student textbook page 435)
chemical system) may not appear to be moving, but
the individual participants (comparable to particles 13. When an external influence (such as a change in
in a chemical system) are engaged in movement to temperature or pressure) on a system at equilibrium
maintain their position. causes a change, the system will eventually move to a
new equilibrium.
4. Equilibrium does not mean an equal balance in
concentrations of reactants and products but, rather, a 14. The forward reaction is favoured, as initially
balance in the reaction rates of the forward and reverse more reactant particle collisions will occur, which
processes. temporarily increases the rate of the forward reaction,
until equilibrium is restored.
5. Changing: the reaction is still occurring, so reactants
still form products while products are re-forming 15. You exhale CO2(g), which causes a decrease in the
reactants Balanced: these two processes are occurring carbon dioxide gas concentration in the system. As a
at the same rate result, the reaction will shift in the forward direction
to form more CO2(g) and this causes a decrease in the
6. CO(g) + 3H2(g) ⇋ CH4(g) + H2O(g)
carbonic acid level in the blood.
(Student textbook page 427) 16. The addition of the inert gas will increase the pressure
7. Only the liquid and the water vapour in the sealed jar inside the container, but it does not have an effect on
can reach equilibrium because it is the only system that the equilibrium, as this increase in pressure affects
is closed to the surroundings. the reactants and products equally, so both rates of
reaction remain the same.
8. Similarities: both involve a balance in forward and
reverse reaction rates, and both involve constant 17. With a pressure increase (due to a volume decrease),
macroscopic properties. the system will shift to favour the reaction that takes
up the smaller volume, which will be in the direction
Differences: different states of matter exist in the
of the lesser amount in moles of gas. With a pressure
heterogeneous equilibrium, whereas all substances in a
decrease (due to a volume increase), the system will
homogeneous equilibrium are in the same state.
shift to favour the reaction that takes up the larger
9. If the stopper were removed, the iodine gas would volume, which will be in the direction of the greater
escape, thus preventing the reverse reaction from amount in moles of gas.
occurring. As a result, the open system could not reach
18. Assuming the reaction vessel is kept at a constant
equilibrium.
temperature:
10. The law of mass action states that there is a constant
a. No change
ratio between the concentrations of products and
b. No change
the concentrations of reactants. This ratio (where
the balanced coefficients of the reaction are used as c. No change
exponents) is the equilibrium constant. d. No change
11. If a chemical system is not at equilibrium, then it is not e. No change
accurate to write an equilibrium constant expression or f. No change
a constant. A reaction quotient is written for chemical
(Student textbook page 438)
systems when they are not at equilibrium.
19. Similarities: both involve a difference in thermal energy
12. If Keq > 1, the denominator of the equilibrium constant
during the reaction.
is less than the numerator, and since the denominator
contains information regarding reactants, product Differences: exothermic reactions give off thermal
formation is favoured. If Keq ≈ 1, the numerator energy, whereas endothermic reactions absorb thermal
is approximately equal to the denominator, and energy.
approximately equal concentrations of reactants and 20. a. When heated, the endothermic reaction (that is,
products exist. If Keq < 1, the denominator of the forward reaction) is favoured to absorb the heat
equilibrium constant is greater than the numerator, added, and more heat is available to enable the
reaction to occur.
Unit 4 Part A • MHR 3
, b. The product concentrations will increase and the 33. When the expression involving the x is not a perfect
reactant concentrations will decrease. square or cannot be factored, the quadratic formula
21. The time needed to reach equilibrium is reduced when must be used to solve the problem.
a catalyst is added, but the equilibrium is not shifted in 34. Any negative roots (as a negative concentration
either direction. has no meaning) and any root that exceeds initial
22. There is no change in the Keq value when a catalyst is concentrations must be eliminated, leaving the other
added, because there is no change in the amount of root as the correct answer.
reactant or product; rather, the rate of the reaction 35. The approximation method can be used to solve
changes. equilibrium calculations when the Keq value is
23. a. The endothermic reaction (that is, the reverse extremely small.
reaction) is favoured to absorb the heat added. 36. In general, if the initial concentrations are at least 1000
b. The product concentration will decrease and the times greater than the value of Keq, the approximation
reactant concentration will increase. method can be used.
24. In question 20, the forward reaction is an 37. If Keq = 4.7 × 10-9 and the initial concentrations of
endothermic reaction. When the system is heated, reactants are approximately 1.0 mol/L.
the product concentrations increase and the reactant
concentrations decrease. Because the equation for Keq (Student textbook page 466)
involves product concentrations divided by reactant 38. Nitrogen narcosis occurs with an elevated
concentrations, Keq increases. concentration of nitrogen in the blood, which can
In question 23, the forward reaction is exothermic.
cause a diver to experience symptoms of intoxication.
When the system is heated, the reverse reaction is 39. To avoid the bends, a diver must surface slowly,
favoured and the reactant concentrations increase allowing the extra dissolved gases in the blood to leave
and the product concentrations decrease. Keq solution slowly.
decreases because the numerator is decreasing and the 40. If inhaled, carbon monoxide binds the hemoglobin
denominator is increasing. protein (which would normally bind to the oxygen that
25. Pressure can be used to monitor this reaction, or cells need). The carbon monoxide binds more readily
colour intensity can be used (dinitrogen tetroxide is than does the oxygen and thus does not let go as easily.
colourless, whereas nitrogen dioxide is a dark-brown As a result, there is a decreased capacity for the blood
colour). to carry oxygen to the cells in need of the oxygen. This
can lead to death.
(Student textbook page 452)
41. Since the carbon monoxide binds more readily to the
26. Keq is calculated using the equilibrium concentrations hemoglobin, the equilibrium constant for the reaction
of the products and reactants at a given temperature. with CO(g) would be larger than that for the reaction
27. Within experimental error, the investigation reveals with oxygen.
that Keq is a constant for a reaction, even when 42. The increased pressure in the chamber, coupled with
concentrations vary. an increased oxygen level in the chamber, causes a
28. Kp represents the equilibrium constant calculated using shift in the equilibrium process and the toxic gases are
partial pressures of the gases in the system. removed from the blood.
29. Kp uses partial pressures of all gases at a constant 43. Increased amounts of carbon dioxide in the
temperature, unlike Keq that uses concentrations. atmosphere cause less calcium carbonate to be
30. The units of partial pressure and the units for pressure available because of a shift in the equilibrium reaction
in R must be the same. of calcium carbonate, calcium ions, and carbonate ions,
and as a result, there is less calcium carbonate available
31. When initial concentrations of reactants and Keq
to coral polyps for making shells. Thus, the corals grow
are given, an ICE table should be used to calculate
more slowly.
equilibrium concentrations.
(Student textbook page 469)
(Student textbook page 458)
44. The supply of sodium nitrate from Chile to Germany
32. Any equation of the form ax2 + bx + c = 0 is a
could easily have been interrupted or even cut off in
quadratic equation.
4 MHR • Chemistry 12 Solutions Manual 978-0-07-106042-4
Solutions Manual Part A
Unit 4 Chemical Systems b. 2.55 g of solid in enough water to make 75.0 mL of
solution
and Equilibrium c. 913.23 g of solid in enough water to make 1.75 L of
solution
Answers to Unit 4 Preparation Questions
6. a. 46.7 mL of acid mixed with 153.3 mL of water
(Student textbook pages 414-7)
b. 0.25 mL of acid mixed with 12.25 mL of water
1. Accept all reasonable answers. Sample answer: Strong
acids and strong bases are corrosive and dangerous to c. 18.3 mL of acid mixed with 1.08 L of water
transport. Pouring strong acids and bases down the 7. a. LiOH(aq) + HCl(aq) → LiCl(aq) + H2O(ℓ)
drain cause unsafe situations at water treatment plants b. 2KCl(aq) + Pb(NO3)2(aq) →
and cause unnecessary wear and tear on pipes, valves, PbCl2(s) + 2KNO3(aq)
and other parts in the sewer system. Disposing of c. Na2S(aq) + 2AgNO3(aq) → Ag2S(s) + 2NaNO3(aq)
unneutralized acids and bases is both unsafe and costly.
8. b
2. a. Always add concentrated acid to water. Once the
9. e
concentrated acid is measured, it should be added
to the 45.00 mL of water, so that the exothermic 10. a. NO3-(aq) and Na+
reaction can dissipate the heat throughout the 45 mL b. silver chromate, Ag2CrO4(s)
of water. c. chromate, CrO42-
b. Rinse any clothing thoroughly and use the eyewash d. 2Ag+(aq) + CrO42-(aq) →Ag2CrO4(s)
station to rinse your eyes immediately and 11. a. A solid formed.
continuously for at least 15 minutes, get your lab
partner to inform the teacher. b. 2Ag+(aq) + 2NO3-(aq) + 2Na+(aq) + CrO4-(aq)
→
c. Using excessive amounts of chemicals is costly
Ag2CrO4(s) + 2NO3-(aq) + 2Na+(aq)
because the initial cost is higher than necessary for
the extra chemicals, and the disposal is more costly 2Ag+(aq) + CrO4-(aq) →Ag2CrO4(s)
because you have to safely dispose of excessive
amounts of chemicals.
3. a. Because lead(II) chloride solid forms, follow all
instructions given by the teacher in terms of its
disposal. Some teachers might ask for the solution
to be filtered and the filter paper disposed of in
hazardous waste containers, while other teachers
might ask for all solutions to be poured into a 12. a. H+(aq) + Br-(aq) + K+(aq) + OH-(aq) →
hazardous waste container. K+(aq) + Br-(aq) + H2O(ℓ)
b. When you wash solid materials down the drain b. H+(aq) + OH-(aq) → H2O(ℓ)
you risk clogging the drain. It can be hazardous to c. neutralization reaction
unclog a drain that contains unknown or a variety of
13. n = cV
chemical solids in it.
mol 1L
4. a. A straw that another student has used is = 0.254 (25.00 mL)
contaminated with germs. Illnesses and diseases can L 1000 mL
be spread among students that share straws. = 0.00635 mol
b. Eyewear should always be worn in the laboratory. 14. n = cV
Also, when you blow too hard through the straw, mol 1L
spray from the mixture can get into your eyes. = 0.125 (38.42 mL)
L 1000 mL
5. a. 25.16 g of solid in enough water to make 225 mL of = 0.00480 mol
solution
Unit 4 Part A • MHR 1
,15. n = cV 20. A strong base completely dissociates in water, whereas
n a weak base does not dissociate completely.
V = c
21. A weak base can be concentrated, because
0.425 moL
= concentration is based on the amount of material in
mol
0.125 L solution, not the degree of ionization; therefore, the
= 3.40 L statement is true.
16. n = cV 22. 0.182 mol/L
n 23. The first proton is lost from the sulfuric molecule due
c = V
to the interactions of the polar water molecule. Once
0.385 mol 1000 mL this proton is lost, HSO4- remains and due to the
= ×
225 mL 1L negative charge, the second proton (which is positively
= 1.71 mol charged) is more difficult for water molecules to strip
L off the ion.
17. Moles of sulfate:
24. a. x = -1.56 or x = 2.56
n = cV
b. x = -1.46 or x = 5.46
mol 1L
= 0.200 (50.0 mL) c. x = -1.85 or x = 0.18
L 1000 mL
25. a. x = 0.50
= 0.0100 mol
b. 4.20 × 10-4
Moles of lead: 26. a. 26
n = cV
b. a10
mol 1L
= 0.100 (80.0 mL) c. x4y4
L 1000 mL
d. 34
= 0.00800 mol
27. a. 3
The limiting ion is lead so only 0.00800 moles of lead b. 4
sulfate can form. c. 8
Molar mass of PbSO4(s): d. 12
MPbSO4 = MPb + Ms + 4MO e. -12
g g g
= 207.2 + 32.07 + 4 × 16.00 28. a. 2.87 × 102
mol mol mol
b. 5.95 × 10-3
g
= 303.27 c. 1.79 × 106
mol
Mass of 0.00800 mol of lead sulfate: d. 4.12 × 10-7
m 29. a. The function key LOG is used to take the logarithm
n = M
of a number. The function key 10x is used to take the
m = nM
antilogarithm or to find the number that you raise
g the base to in order to obtain a specific number.
= (0.00800 mol) 303.27
mol b. Sample answer: log 0.001 = x and 102.961 = x;
= 2.42616 g c. The pH scale for acids and bases is based on
= 2.43 g logarithms.
18. a. lead (II) chromate
Chapter 7 Chemical Equilibrium
b. 2K+(aq) + CrO42-(aq) + Pb2+(aq) + 2NO3-(aq)
→
PbCrO4(s) + 2K+(aq) + 2NO3-(aq) Answers to Learning Check Questions
c. 0.58 g (Student textbook page 422)
19. A concentrated acid has a large number of acid 1. A reversible reaction is a reaction that can proceed in
molecules in solution, while a dilute acid has a smaller both the forward and the reverse directions.
number of acid molecules in solution.
2 MHR • Chemistry 12 Solutions Manual 978-0-07-106042-4
, 2. The double arrows indicate that the reaction is and since the denominator contains information
reversible. regarding reactants, reactant formation is favoured.
3. The rope-pulling team as a whole (comparable to a
(Student textbook page 435)
chemical system) may not appear to be moving, but
the individual participants (comparable to particles 13. When an external influence (such as a change in
in a chemical system) are engaged in movement to temperature or pressure) on a system at equilibrium
maintain their position. causes a change, the system will eventually move to a
new equilibrium.
4. Equilibrium does not mean an equal balance in
concentrations of reactants and products but, rather, a 14. The forward reaction is favoured, as initially
balance in the reaction rates of the forward and reverse more reactant particle collisions will occur, which
processes. temporarily increases the rate of the forward reaction,
until equilibrium is restored.
5. Changing: the reaction is still occurring, so reactants
still form products while products are re-forming 15. You exhale CO2(g), which causes a decrease in the
reactants Balanced: these two processes are occurring carbon dioxide gas concentration in the system. As a
at the same rate result, the reaction will shift in the forward direction
to form more CO2(g) and this causes a decrease in the
6. CO(g) + 3H2(g) ⇋ CH4(g) + H2O(g)
carbonic acid level in the blood.
(Student textbook page 427) 16. The addition of the inert gas will increase the pressure
7. Only the liquid and the water vapour in the sealed jar inside the container, but it does not have an effect on
can reach equilibrium because it is the only system that the equilibrium, as this increase in pressure affects
is closed to the surroundings. the reactants and products equally, so both rates of
reaction remain the same.
8. Similarities: both involve a balance in forward and
reverse reaction rates, and both involve constant 17. With a pressure increase (due to a volume decrease),
macroscopic properties. the system will shift to favour the reaction that takes
up the smaller volume, which will be in the direction
Differences: different states of matter exist in the
of the lesser amount in moles of gas. With a pressure
heterogeneous equilibrium, whereas all substances in a
decrease (due to a volume increase), the system will
homogeneous equilibrium are in the same state.
shift to favour the reaction that takes up the larger
9. If the stopper were removed, the iodine gas would volume, which will be in the direction of the greater
escape, thus preventing the reverse reaction from amount in moles of gas.
occurring. As a result, the open system could not reach
18. Assuming the reaction vessel is kept at a constant
equilibrium.
temperature:
10. The law of mass action states that there is a constant
a. No change
ratio between the concentrations of products and
b. No change
the concentrations of reactants. This ratio (where
the balanced coefficients of the reaction are used as c. No change
exponents) is the equilibrium constant. d. No change
11. If a chemical system is not at equilibrium, then it is not e. No change
accurate to write an equilibrium constant expression or f. No change
a constant. A reaction quotient is written for chemical
(Student textbook page 438)
systems when they are not at equilibrium.
19. Similarities: both involve a difference in thermal energy
12. If Keq > 1, the denominator of the equilibrium constant
during the reaction.
is less than the numerator, and since the denominator
contains information regarding reactants, product Differences: exothermic reactions give off thermal
formation is favoured. If Keq ≈ 1, the numerator energy, whereas endothermic reactions absorb thermal
is approximately equal to the denominator, and energy.
approximately equal concentrations of reactants and 20. a. When heated, the endothermic reaction (that is,
products exist. If Keq < 1, the denominator of the forward reaction) is favoured to absorb the heat
equilibrium constant is greater than the numerator, added, and more heat is available to enable the
reaction to occur.
Unit 4 Part A • MHR 3
, b. The product concentrations will increase and the 33. When the expression involving the x is not a perfect
reactant concentrations will decrease. square or cannot be factored, the quadratic formula
21. The time needed to reach equilibrium is reduced when must be used to solve the problem.
a catalyst is added, but the equilibrium is not shifted in 34. Any negative roots (as a negative concentration
either direction. has no meaning) and any root that exceeds initial
22. There is no change in the Keq value when a catalyst is concentrations must be eliminated, leaving the other
added, because there is no change in the amount of root as the correct answer.
reactant or product; rather, the rate of the reaction 35. The approximation method can be used to solve
changes. equilibrium calculations when the Keq value is
23. a. The endothermic reaction (that is, the reverse extremely small.
reaction) is favoured to absorb the heat added. 36. In general, if the initial concentrations are at least 1000
b. The product concentration will decrease and the times greater than the value of Keq, the approximation
reactant concentration will increase. method can be used.
24. In question 20, the forward reaction is an 37. If Keq = 4.7 × 10-9 and the initial concentrations of
endothermic reaction. When the system is heated, reactants are approximately 1.0 mol/L.
the product concentrations increase and the reactant
concentrations decrease. Because the equation for Keq (Student textbook page 466)
involves product concentrations divided by reactant 38. Nitrogen narcosis occurs with an elevated
concentrations, Keq increases. concentration of nitrogen in the blood, which can
In question 23, the forward reaction is exothermic.
cause a diver to experience symptoms of intoxication.
When the system is heated, the reverse reaction is 39. To avoid the bends, a diver must surface slowly,
favoured and the reactant concentrations increase allowing the extra dissolved gases in the blood to leave
and the product concentrations decrease. Keq solution slowly.
decreases because the numerator is decreasing and the 40. If inhaled, carbon monoxide binds the hemoglobin
denominator is increasing. protein (which would normally bind to the oxygen that
25. Pressure can be used to monitor this reaction, or cells need). The carbon monoxide binds more readily
colour intensity can be used (dinitrogen tetroxide is than does the oxygen and thus does not let go as easily.
colourless, whereas nitrogen dioxide is a dark-brown As a result, there is a decreased capacity for the blood
colour). to carry oxygen to the cells in need of the oxygen. This
can lead to death.
(Student textbook page 452)
41. Since the carbon monoxide binds more readily to the
26. Keq is calculated using the equilibrium concentrations hemoglobin, the equilibrium constant for the reaction
of the products and reactants at a given temperature. with CO(g) would be larger than that for the reaction
27. Within experimental error, the investigation reveals with oxygen.
that Keq is a constant for a reaction, even when 42. The increased pressure in the chamber, coupled with
concentrations vary. an increased oxygen level in the chamber, causes a
28. Kp represents the equilibrium constant calculated using shift in the equilibrium process and the toxic gases are
partial pressures of the gases in the system. removed from the blood.
29. Kp uses partial pressures of all gases at a constant 43. Increased amounts of carbon dioxide in the
temperature, unlike Keq that uses concentrations. atmosphere cause less calcium carbonate to be
30. The units of partial pressure and the units for pressure available because of a shift in the equilibrium reaction
in R must be the same. of calcium carbonate, calcium ions, and carbonate ions,
and as a result, there is less calcium carbonate available
31. When initial concentrations of reactants and Keq
to coral polyps for making shells. Thus, the corals grow
are given, an ICE table should be used to calculate
more slowly.
equilibrium concentrations.
(Student textbook page 469)
(Student textbook page 458)
44. The supply of sodium nitrate from Chile to Germany
32. Any equation of the form ax2 + bx + c = 0 is a
could easily have been interrupted or even cut off in
quadratic equation.
4 MHR • Chemistry 12 Solutions Manual 978-0-07-106042-4
