Download PX312-1718

PX312-1718
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1. What is the solubility product expression for Th(IO3)4?
A) Ksp = [Th4+][4IO3–]4
B) Ksp = [Th4+][IO3–]
C) Ksp = [Th][IO3]4
D) Ksp = [Th4+][IO3–]4
E) Ksp = [Th4+][IO3–]
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2. What is the solubility product expression for La2(CO3)3?
A) Ksp = [2La3+]2[3CO32–]3
B) Ksp = [La2+]2[CO32–]3
C) Ksp = [2La3+]2[CO32–]3
D) Ksp = [2La3+][3CO32–]
E) Ksp = [La3+]2[CO32–]3
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3. After mixing an excess PbCl2 with a fixed amount of water, it is found that the equilibrium
concentration of Pb2+ is 1.6  10–2 M. What is Ksp for PbCl2?
A) 4.0  10–6
B) 1.6  10–5
C) 2.5  10–4
D) 4.8  10–2
E) 1.0  10–6
____
4. The solubility of lead(II) sulfate is 4.0  10–2 g/L. What is the solubility product constant for
lead(II) sulfate?
A) 1.7  10–8
B) 1.3  10–4
C) 1.6  10–3
D) 4.6  10–15
E) 8.9  10–12
____
5. The silver-ion concentration in a saturated solution of silver(I) chromate is 1.3  10–4 M. What is
Ksp for silver(I) chromate?
A) 2.9  10–16
B) 4.2  10–9
C) 8.8  10–12
D) 1.1  10 12
E) 1.7  10–8
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6. Which of the following salts has the lowest molar solubility?
A) SrCO3 (Ksp = 9.3  10–10)
B) MnS (Ksp = 2.5  10–10)
C) BaF2 (Ksp = 1.0  10–6)
D) BaSO4 (Ksp = 1.1  10–10)
E) AgCl (Ksp = 1.8  10–10)
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7. Rank the following metal sulfides in order of increasing molar solubility in water.
Salt
CoS
CuS
FeS
HgS
MnS
A)
B)
C)
D)
E)
Ksp
4  10–21
6  10–36
6  10–18
1.6  10–52
2.5  10–10
MnS < FeS < CoS < CuS < HgS
FeS < HgS < CoS < CuS < MnS
HgS < CuS < CoS < FeS < MnS
CuS < CoS < FeS < MnS < HgS
CoS < CuS < FeS < HgS < MnS
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8. What is the molar solubility of calcium sulfate at 25°C? The solubility product constant for
calcium sulfate is 2.4  10–5 at 25°C.
A) 2.4  10–5 M
B) 3.5  10–2 M
C) 1.2  10–5 M
D) 1.8  10–2 M
E) 4.9  10–3 M
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9. What is the molar solubility of barium fluoride at 25°C? The solubility product constant for
barium fluoride is 1.0  10–6 at 25°C.
A) 6.3  10–3 M
B) 1.0  10–6 M
C) 5.0  10–7 M
D) 1.0  10–3 M
E) 1.6  10–2 M
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10. What is the pH of a saturated solution of Ni(OH)2? For Ni(OH)2, Ksp = 2.0  10–15.
A) 4.80
B) 8.90
C) 5.10
D) 9.20
E) 7.00
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11. The insoluble salts AV, B2W, C2X3, DY2, and EZ3, which were formed from the metal ions A+, B+,
C3+, D2+, and E3+ and the nonmetals V1–, W2–, X2–, Y1–, and Z1–, all have the same Ksp value.
Which salt has the highest molar solubility?
A) AV
B) EZ3
C) DY2
D) B2W
E) C2X3
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12. In which of these solutions would silver(I) carbonate have the lowest molar solubility? For
silver(I) carbonate, Ksp = 8.5  10–12.
A) 0.03 M H2CO3
B) 0.1 M AgNO3
C) 0.01 M AgNO3
D) 0.1 M Na2CO3
E) pure water
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13. What is the concentration of silver(I) ion in a saturated solution of silver(I) carbonate containing
0.0030 M Na2CO3? For Ag2CO3, Ksp = 8.6  10–12.
A) 6.0  10–4 M
B) 2.0  10–9 M
C) 8.0  10–9 M
D) 5.4  10–5 M
E) 8.0  10–4 M
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14. What is the molar solubility of MgF2 in a 0.40 M Mg(NO3)2 solution? For MgF2, Ksp = 8.4  10–8.
A) 8.0  10–8 M
B) 2.3  10–4 M
C) 2.0  10–-8 M
D) 4.6  10–4 M
E) 3.2  10–3 M
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15. The solubility of La(IO3)3 in a 0.62 M KIO3 solution is 1.0 10–7 mol/L. Calculate Ksp for La(IO3)3.
A) 6.2  10–8
B) 2.4  10–22
C) 2.4  10–1
D) 2.4  10–8
E) none of these
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16. How many moles of CaF2 will dissolve in 3.0 L of 0.051 M NaF solution? (Ksp for
CaF2 = 4.0  10–11)
A) 2.6  10–10
B) 1.5  10–8
C) 4.6  10–8
D) 5.1  10–9
E) none of these
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17. Which of Figures I–IV represent(s) the result of mixing aqueous solutions of Na2S and NiCl2 in
which the ion product Qc > Ksp for the insoluble product? (C = cation, A = anion)
A)
B)
C)
D)
E)
both I and II
only I
only II
only III
only IV
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18. Which of the following will apply to a saturated solution of an ionic compound?
A) Qc < Ksp
B) Qc > Ksp
C) Qc = Ksp
D) Ksp = 1
E) Qc = 1
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19. To 1.0 L of water, 3.0  10–6 mol of Pb(NO3)2, 4.0  10–6 mol of K2CrO4, and 1.0 mol of NaCl are
added. What will happen?
Salt
PbCrO4
PbCl2
A)
B)
C)
D)
E)
Ksp
1.8  10–14
1.6  10–5
A precipitate of KCl will form.
A precipitate of PbCrO4 will form.
A precipitate of PbCl2 will form.
No precipitate will form.
Both a precipitate of PbCl2 and a precipitate of PbCrO4 will form.
____
20. What will happen if 0.1 mol of solid silver(I) nitrate is added to 1.0 L of a saturated solution of
silver(I) chromate? For Ag2CrO4, Ksp = 2.4  10–12.
A) The AgNO3 will settle to the bottom without dissolving.
B) The concentration of CrO42– will increase.
C) Some Ag2CrO4 will precipitate.
D) Nothing will happen.
E) The concentration of Ag+ in solution will not change.
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21. What is the minimum mass of Na2CO3 that must be added to 24.6 mL of a 9.5  10–4 M AgNO3
solution in order for precipitation to occur? For Ag2CO3, Ksp = 8.6  10–12 .
A) 2.5  10–3 g
B) 3.1  10–4 g
C) 1.2  10–3 g
D) 2.4  10–8 g
E) 2.5  10–5 g
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22. What is the maximum concentration of carbonate ions that will precipitate BaCO3 but not MgCO3
from a solution that is 2.7  10 3 M each in Mg2+ and Ba2+? For MgCO3, Ksp = 1.0  10–5 and for
BaCO3, Ksp = 2.6  10–9.
A) 3.7  10 3 M
B) 9.6  10 7 M
C) 2.7  10 8 M
D) 7.0  10 12 M
E) 2.6  10–14 M
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23. Which of the following solutions should be added to a solution containing both copper(II) ions and
silver(I) ions in order to precipitate only one of the ions?
A) HCl(aq)
B) H2S(aq)
C) HNO3(aq)
D) H2S(aq) + HCl(aq)
E) H2S(aq) + HNO3(aq)
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24. For which pair of cations would the addition of dilute hydrobromic acid precipitate one but not the
other?
A) Ag+ and Ca2+
B) Hg22+ and Ag+
C) Ba2+ and Na+
D) Ca2+ and Ba2+
E) Pb2+ and Ag+
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25. Sodium chloride is added slowly to a solution that is 0.010 M in Cu+, Ag+, and Au+. The Ksp values
for the chloride salts are 1.9  10–7, 1.6  10–10, and 2.0  10–13, respectively. Which compound
will precipitate first?
A) AuCl(s)
B) All will precipitate at the same time.
C) It cannot be determined.
D) AgCl(s)
E) CuCl(s)
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26. Solid KCN is added to a solution composed of 0.10 M Ag+ and 0.10 M Zn2+ just until a precipitate
forms. What is the composition of this initial precipitate? AgCN Ksp = 2.2  10-16 and Zn(CN)2
Ksp = 3  10-16.
A) The precipitate is pure AgCN(s).
B) The precipitateis pure Zn(CN)2(s).
C) The precipitate is a mixture of AgCN(s) and Zn(CN)2(s).
D) The precipitate is a mixture of KCN(s) and AgCN(s).
E) The precipitate is a mixture of KCN(s) and Zn(CN)2(s).
____
27. Suppose hydrogen sulfide is added to a solution that is 0.0010 M in Fe2+, Cd2+, Co2+, and Mn2+
such that the concentration of H2S is 0.10 M. When the pH of the solution is adjusted to 3, a
precipitate forms. What is the composition of the precipitate?
H2S(aq) + 2H2O(l)
Salt
FeS
CdS
CoS
MnS
A)
B)
C)
D)
E)
____
2H3O+(aq) + S2–(aq); Kc = 1.1  10–20
Ksp
6.0  10–18
8.0  10–27
4.0  10–21
2.5  10–10
CdS only
CdS, CoS, FeS, and MnS
CdS, CoS, and FeS
CdS and FeS
CdS and CoS
28. Suppose sodium hydroxide is added to a 0.0016 M solution of zinc nitrate such that the pH of the
solution is 13.42. What is the equilibrium concentration of Zn2+?
Zn2+(aq) + 4OH–(aq)
A)
B)
C)
D)
E)
Zn(OH)42–(aq); Kf = 2.8  1015
2.2  10–18 M
1.6  10–3 M
6.6  10–2 M
1.2  10–16 M
2.2  10–18 M
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29. In the sulfide scheme for qualitative analysis, the cations of Analytical Group IV are precipitated
as phosphates or carbonates. Analytical Group IV consists of
A) alkaline earth elements.
B) the halogens.
C) alkali metals.
D) transition metals having +2 ions.
E) none of these
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30. Consider a solution containing the following cations: Na+, Hg2+, Mn2+, Al3+ and Ag+. Treatment of
the solution with dilute, HCl followed by saturation with H2S, results in formation of precipitate(s).
Which ions still remain in solution (did not precipitate)?
A) Na+, Hg2+, Al3+
B) Na+ only
C) Ag+ and Hg2+
D) Ag+ only
E) Na+, Al3+, and Mn2+
____
31. The following reaction represents a step in the separation of which analytical group of cations?
Cu2+(aq) + S2–(aq)  CuS(s)
A)
B)
C)
D)
E)
____
Analytical Group I
Analytical Group III
Analytical Group V
Analytical Group IV
Analytical Group II
32. Which of the following statements concerning entropy change is/are true?
1.
2.
3.
A)
B)
C)
D)
E)
For a spontaneous process,
For a spontaneous process,
.
.
For an equilibrium process (such as a phase change),
.
1 only
2 only
3 only
2 and 3
1, 2, and 3
____
33. The total entropy of a system and its surroundings always increases for a spontaneous process.
This is a statement of
A) the third law of thermodynamics.
B) the law of constant composition.
C) the second law of thermodynamics.
D) the law of conservation of matter.
E) the first law of thermodynamics.
____
34. At the normal boiling point of o-xylene, H°vap = 36.2 kJ/mol and S°vap = 86.7 J/(molK).
What is the normal boiling point of o-xylene?
A) 314 K
B) 373 K
C) 115 K
D) 867 K
E) 418 K
____
35. Which of the following is not a spontaneous process at 25°C and 1 atm pressure?
A) salt dissolving
B) ice melting
C) water boiling
D) iron rusting
E) steam condensing
____
36. For the process Cl2(g)  2Cl(g),
A) H is + and S is + for the reaction.
B) H is + and S = 0 for the reaction.
C) H is – and S is – for the reaction.
D) H is – and S is + for the reaction.
E) H is + and S is – for the reaction.
____
37. What is the change in entropy when 0.646 g of water decomposes to form hydrogen gas and
oxygen gas at 298 K?
2H2O(l)  2H2(g) + O2(g); S° = 326.3 J/K at 298 K
A)
B)
C)
D)
E)
____
38. Determine G° for the following reaction:
CH4(g) + 2O2(g)
CO2(g) + 2H2O(l)
Substance
Gf° (kJ/mol)
CH4(g)
–50.72
O2(g)
0
CO2(g)
–394.4
H2O(l)
–237.4
A)
B)
C)
D)
E)
____
5.85 J/K
0.0785 J/K
23.4 J/K
11.7 J/K
211 J/K
–581.1 kJ
–818.5 kJ
131.1 kJ
–682.5 kJ
–919.9 kJ
39. Based on the following data, what is the standard Gibbs free energy of formation of the sulfate ion
at 298 K? (R = 8.31 J/(Kmol))
CaSO4(s)
Substance
Ca2+(aq)
CaSO4(s)
A)
B)
C)
D)
E)
____
Ca2+(aq) + SO42–(aq); Ksp = 2.4  10–5
G°f (kJ/mol) at 298 K
–553.5
–1322.0
–768.5 kJ/mol
–1849.2 kJ/mol
–1875.5 kJ/mol
–742.2 kJ/mol
794.8 kJ/mol
40. A certain reaction is found to be product favored. Which of the following is a correct description
of the reaction?
A) G° < 0, K > 1
B) G° > 0, K < 1
C) G° > 0, K > 1
D) G° < 0, K < 1
E) G° = 0, K > 1
____
41. The standard free energy of formation of nitric oxide, NO, at 1000. K (roughly the temperature in
an automobile engine during ignition) is 78.0 kJ/mol. Calculate the equilibrium constant for the
reaction
2NO(g)
at 1000. K. (R = 8.31 J/(Kmol))
A) 0.95
B) 7.0  10–9
C) 1.6  105
D) –15
E) 8.4  10–5
____
42. Consider the following hypothetical reaction (at 316.8 K). Standard free energies, in kJ/mol, are
given in parentheses.
A
(–32.2)
B
(207.8)
+
C
(–237.0)
G° = ?
What is the value of the equilibrium constant for the reaction at 316.8 K?
A) 0.42
B) 1.0
C) 273
D) 6.5  104
E) 0.32
____
43. A 0.0997 M solution of a particular monoprotic weak acid, HA, has a pH of 6.00 at 298 K. What
is G° for the following equilibrium?
HA(aq) + H2O(l)
A)
B)
C)
D)
E)
H3O+(aq) + A–(aq)
28.5 kJ
34.2 kJ
620 kJ
5.71 kJ
62.7 kJ
____
44. The reaction CaO(s) + SO3(g)  CaSO4(s) is nonspontaneous at 2200 K, whereas it is
spontaneous at room temperature. Which of the following statements is false?
A) The change in enthalpy is the main driving force of the reaction.
B) Both H and S are negative for the reaction.
C) G is negative at room temperature.
D) The change in entropy is the main driving force of the reaction.
E) G becomes zero at a temperature between 300 and 2200 K.
____
45. For the reaction CaCO3(s)  CaO(s) + O2(g) at 1 atm pressure, the values of H and S are both
positive, and the process is spontaneous at high temperatures. Which of the following statements
about this reaction is true?
A) The change in entropy is the driving force for the reaction.
B) The process is exothermic at high temperatures and endothermic at room
temperature.
C) The reverse reaction is endothermic.
D) The reverse reaction is nonspontaneous at room temperature.
E) G at room temperature is negative.
____
46. Which of the following statements is true concerning the reaction below?
CH4(g) + N2(g)  HCN(g) + NH3(g); H° = 164.1 kJ; G° = 159.1 kJ at 298 K
A)
B)
C)
D)
E)
____
It is nonspontaneous at all temperatures.
It is spontaneous at relatively low temperatures only.
It is spontaneous at all temperatures.
It is at equilibrium at 298 K.
It is spontaneous at relatively high temperatures only.
47. Which of the following statements is true for the following reaction?
NH4HS(s)  NH3(g) + H2S(g); H° = 93 kJ
A)
B)
C)
D)
E)
The reaction is not spontaneous at any temperature.
The reaction is spontaneous only at relatively high temperatures.
The reaction is at equilibrium under standard-state conditions.
The reaction is spontaneous at all temperatures.
The reaction is spontaneous only at relatively low temperatures.
____
48. Sublimation is an example of a process for which
A) H is positive and S is negative at all temperatures.
B) H is negative and S is positive at all temperatures.
C) H, S, and G are negative at all temperatures.
D) H and S are positive at all temperatures.
E) H, S, and G are positive at all temperatures.
____
49. For the reaction
reaction at 700. K?
A) 1.00
B) 1.54
C) 10.1
D) 2.31
E) none of these
____
50. Consider the following reaction:
G°700K = –13.457 kJ. What is Kp for this
2C(s) + 3H2(g)  C2H6(g); H° = –84.68 kJ; S° = –173.8 J/K at 298 K
What is the equilibrium constant at 400.0 K for this reaction?
A) 1.0  10–2
B) 2.0  104
C) 1.0
D) 9.5  101
E) 8.6  10–10
PX312-1718
Answer Section
1. ANS:
OBJ:
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10. ANS:
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11. ANS:
OBJ:
D
PTS: 1
DIF: easy
REF: 17.1
Write solubility product expressions. (Example 17.1)
solubility | solubility equilibria
KEY: solubility product constant
general chemistry
E
PTS: 1
DIF: easy
REF: 17.1
Write solubility product expressions. (Example 17.1)
solubility | solubility equilibria
KEY: solubility product constant
general chemistry
B
PTS: 1
DIF: moderate
REF: 17.1
Calculate Ksp from the solubility (more complicated example).(Example 17.3)
solubility | solubility equilibria
KEY: solubility product constant
general chemistry
A
PTS: 1
DIF: moderate
REF: 17.1
Calculate Ksp from the solubility (more complicated example).(Example 17.3)
solubility | solubility equilibria
KEY: solubility product constant
general chemistry
D
PTS: 1
DIF: difficult
REF: 17.1
Calculate Ksp from the solubility (more complicated example).(Example 17.3)
solubility | solubility equilibria
KEY: solubility product constant
general chemistry
D
PTS: 1
DIF: easy
REF: 17.1
Calculate the solubility from Ksp. (Example 17.4)
solubility | solubility equilibria
solubility product constant | relative solubilities
MSC: general chemistry
C
PTS: 1
DIF: easy
REF: 17.1
Calculate the solubility from Ksp. (Example 17.4)
solubility | solubility equilibria
solubility product constant | relative solubilities
MSC: general chemistry
E
PTS: 1
DIF: easy
REF: 17.1
Calculate the solubility from Ksp. (Example 17.4)
solubility | solubility equilibria
KEY: solubility product constant
general chemistry
A
PTS: 1
DIF: easy
REF: 17.1
Calculate the solubility from Ksp. (Example 17.4)
solubility | solubility equilibria
KEY: solubility product constant
general chemistry
D
PTS: 1
DIF: moderate
REF: 17.1
Calculate the solubility from Ksp. (Example 17.4)
solubility | solubility equilibria
KEY: solubility product constant
general chemistry
E
PTS: 1
DIF: difficult
REF: 17.1
Calculate the solubility from Ksp. (Example 17.4)
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
TOP: solubility | solubility equilibria
KEY: solubility product constant | relative solubilities
MSC: general chemistry
ANS: B
PTS: 1
DIF: easy
REF: 17.2
OBJ: Explain how the solubility of a salt is affected by another salt that has the same cation or
anion (common ion).
TOP: solubility | solubility equilibria
KEY: solubility and the common-ion effect
MSC: general
chemistry
ANS: D
PTS: 1
DIF: moderate
REF: 17.2
OBJ: Calculate the solubility of a slightly soluble salt in a solution of a common ion. (Example
17.5) TOP:
solubility | solubility equilibria
KEY: solubility and the common-ion effect
MSC: general
chemistry
ANS: B
PTS: 1
DIF: moderate
REF: 17.2
OBJ: Calculate the solubility of a slightly soluble salt in a solution of a common ion. (Example
17.5) TOP:
solubility | solubility equilibria
KEY: solubility and the common-ion effect
MSC: general
chemistry
ANS: D
PTS: 1
DIF: moderate
REF: 17.2
OBJ: Calculate the solubility of a slightly soluble salt in a solution of a common ion. (Example
17.5) TOP:
solubility | solubility equilibria
KEY: solubility and the common-ion effect
MSC: general
chemistry
ANS: C
PTS: 1
DIF: moderate
REF: 17.2
OBJ: Calculate the solubility of a slightly soluble salt in a solution of a common ion. (Example
17.5) TOP:
solubility | solubility equilibria
KEY: solubility and the common-ion effect
MSC: general
chemistry
ANS: D
PTS: 1
DIF: moderate
REF: 17.3
OBJ: State the criterion for precipitation.
TOP: solubility | solubility equilibria
KEY: precipitation calculations
MSC: general chemistry
ANS: C
PTS: 1
DIF: easy
REF: 17.3
OBJ: State the criterion for precipitation.
TOP: solubility | solubility equilibria
KEY: precipitation calculations | criterion for precipitation
MSC: general chemistry
ANS: B
PTS: 1
DIF: easy
REF: 17.3
OBJ: Predict whether precipitation will occur (given ion concentrations). (Example 17.6)
TOP: solubility | solubility equilibria
KEY: precipitation calculations | criterion for precipitation
MSC: general chemistry
ANS: C
PTS: 1
DIF: easy
REF: 17.3
OBJ: Predict whether precipitation will occur (given ion concentrations). (Example 17.6)
TOP: solubility | solubility equilibria
KEY: precipitation calculations | criterion for precipitation
MSC: general chemistry
ANS: E
PTS: 1
DIF: moderate
REF: 17.3
OBJ: Predict whether precipitation will occur (given solution volumes and concentrations).
(Example 17.7)
TOP: solubility | solubility equilibria
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KEY:
ANS:
OBJ:
TOP:
KEY:
ANS:
OBJ:
ANS:
OBJ:
TOP:
precipitation calculations | criterion for precipitation
MSC: general chemistry
A
PTS: 1
DIF: moderate
REF: 17.3
Explain how two ions can be separated using fractional precipitation.
solubility | solubility equilibria
precipitation calculations | fractional precipitation
MSC: general chemistry
A
PTS: 1
DIF: easy
REF: 17.3
Explain how two ions can be separated using fractional precipitation.
solubility | solubility equilibria
precipitation calculations | fractional precipitation
MSC: general chemistry
A
PTS: 1
DIF: easy
REF: 17.3
Explain how two ions can be separated using fractional precipitation.
solubility | solubility equilibria
precipitation calculations | fractional precipitation
MSC: general chemistry
A
PTS: 1
DIF: moderate
REF: 17.3
Explain how two ions can be separated using fractional precipitation.
solubility | solubility equilibria
precipitation calculations | fractional precipitation
MSC: general chemistry
A
PTS: 1
DIF: moderate
REF: 17.3
Explain how two ions can be separated using fractional precipitation.
solubility | solubility equilibria
E
PTS: 1
DIF: difficult
REF: 17.4
Explain the basis for the sulfide scheme to separate a mixture of metal ions.
solubility | solubility equilibria
KEY: effect of pH on solubility
general chemistry
D
PTS: 1
DIF: moderate
REF: 17.5
Calculate the concentration of a metal ion in equilibrium with a complex ion. (Example
TOP:
solubility | complex ion equilibria
complex ion formation | equilibrium calculations with Kf MSC: general chemistry
A
PTS: 1
DIF: moderate
REF: 17.7
Describe the main outline of the sulfide scheme for qualitative analysis.
solubility | applications of solubility equilibria
qualitative analysis of metal ions
MSC: general chemistry
E
PTS: 1
DIF: moderate
REF: 17.7
Describe the main outline of the sulfide scheme for qualitative analysis.
solubility | applications of solubility equilibria
qualitative analysis of metal ions
MSC: general chemistry
E
PTS: 1
DIF: easy
REF: 17.7
Describe the main outline of the sulfide scheme for qualitative analysis.
solubility | applications of solubility equilibria
qualitative analysis of metal ions
MSC: general chemistry
E
PTS: 1
DIF: easy
REF: 18.2
Entropy.
TOP: thermochemistry | thermodynamics
C
PTS: 1
DIF: easy
REF: 18.2
State the second law of thermodynamics in terms of system plus surroundings..
thermochemistry | thermodynamics
34.
35.
36.
37.
38.
39.
40.
41.
42.
43.
44.
KEY: second law of thermodynamics | entropy
MSC: general chemistry
ANS: E
PTS: 1
DIF: easy
REF: 18.2
OBJ: Calculate the entropy change for a phase transition. (Example 18.1)
TOP: thermochemistry | thermodynamics
KEY: second law of thermodynamics | entropy change for a phase transition
MSC: general chemistry
ANS: C
PTS: 1
DIF: easy
REF: 18.2
OBJ: Describe how deltaH – TdeltaS functions as a criterion of a spontaneous reaction.
TOP: thermochemistry | thermodynamics
KEY: second law of thermodynamics | entropy and molecular disorder
MSC: general chemistry
ANS: A
PTS: 1
DIF: moderate
REF: 18.3
OBJ: Predict the sign of the entropy change of a reaction. (Example 18.2)
TOP: thermochemistry | thermodynamics
ANS: A
PTS: 1
DIF: easy
REF: 18.3
OBJ: Calculate deltaS for a reaction. (Example 18.3)
TOP: thermochemistry | thermodynamics
KEY: third law of thermodynamics | entropy change for a reaction
MSC: general chemistry
ANS: B
PTS: 1
DIF: easy
REF: 18.4
OBJ: Calculate deltaG from standard free energies of formation. (Example 18.5)
TOP: thermochemistry | thermodynamics
KEY: free energy | standard free-energies of formation
MSC: general chemistry
ANS: D
PTS: 1
DIF: difficult
REF: 18.6
OBJ: Give the relation between free energy change of a reaction and the reaction quotient.
TOP: thermochemistry | thermodynamics
KEY: free energy
MSC: general chemistry
ANS: A
PTS: 1
DIF: easy
REF: 18.6
OBJ: Relate the standard free-energy change to the thermodynamic equilibrium constant.
TOP: thermochemistry | thermodynamics
ANS: B
PTS: 1
DIF: easy
REF: 18.6
OBJ: Calculate K from the standard free-energy change (molecular equation). (Example 18.8)
TOP: thermochemistry | thermodynamics
KEY: thermodynamic equilibrium constant (K)
MSC: general chemistry
ANS: E
PTS: 1
DIF: moderate
REF: 18.6
OBJ: Calculate K from the standard free-energy change (molecular equation). (Example 18.8)
TOP: thermochemistry | thermodynamics
KEY: thermodynamic equilibrium constant (K)
MSC: general chemistry
ANS: E
PTS: 1
DIF: difficult
REF: 18.6
OBJ: Calculate K from the standard free-energy change (molecular equation). (Example 18.8)
TOP: thermochemistry | thermodynamics
KEY: thermodynamic equilibrium constant (K)
MSC: general chemistry
ANS: D
PTS: 1
DIF: easy
REF: 18.7
OBJ: Describe the how spontaneity or nonspontaneity of a reaction is related to each of the four
possible choices of signs of deltaH and deltaS.
TOP: thermochemistry | thermodynamics
45.
46.
47.
48.
49.
50.
KEY: temperature dependence of free energy | spontaneity and temperature change
MSC: general chemistry
ANS: A
PTS: 1
DIF: easy
REF: 18.7
OBJ: Describe the how spontaneity or nonspontaneity of a reaction is related to each of the four
possible choices of signs of deltaH and deltaS.
TOP: thermochemistry | thermodynamics
KEY: temperature dependence of free energy | spontaneity and temperature change
MSC: general chemistry
ANS: E
PTS: 1
DIF: moderate
REF: 18.7
OBJ: Describe the how spontaneity or nonspontaneity of a reaction is related to each of the four
possible choices of signs of deltaH and deltaS.
TOP: thermochemistry | thermodynamics
KEY: temperature dependence of free energy | spontaneity and temperature change
MSC: general chemistry
ANS: B
PTS: 1
DIF: moderate
REF: 18.7
OBJ: Describe the how spontaneity or nonspontaneity of a reaction is related to each of the four
possible choices of signs of deltaH and deltaS.
TOP: thermochemistry | thermodynamics
KEY: temperature dependence of free energy | spontaneity and temperature change
MSC: general chemistry
ANS: D
PTS: 1
DIF: moderate
REF: 18.7
OBJ: Describe the how spontaneity or nonspontaneity of a reaction is related to each of the four
possible choices of signs of deltaH and deltaS.
TOP: thermochemistry | thermodynamics
KEY: temperature dependence of free energy | spontaneity and temperature change
MSC: general chemistry
ANS: C
PTS: 1
DIF: easy
REF: 18.7
OBJ: Calculate deltaG and K at various temperatures. (Example 18.10)
TOP: thermochemistry | thermodynamics
KEY: temperature dependence of free energy
MSC: general chemistry
ANS: D
PTS: 1
DIF: difficult
REF: 18.7
OBJ: Calculate deltaG and K at various temperatures. (Example 18.10)
TOP: thermochemistry | thermodynamics
KEY: temperature dependence of free energy
MSC: general chemistry