5 Steps to a 5 AP Chemistry

Take a Diagnostic Exam  25

73.

What is the equilibrium constant for the above

reaction? The successive acid dissociation con-

stants for H 2 S are 9.5 × 10 −^8 (Ka1) and 1 × 10 −^19

(Ka2). The Ksp, the solubility product constant,

for SnS equals 1.0 × 10 −^25.

(A) 9.5 × 10 −^8 /1.0 × 10 −^25

(B) 9.5 × 10 −^27 /1.0 × 10 −^25

(C) 1.0 × 10 −^25 /9.5 × 10 −^27

(D) 1 ×l0−^19 /1.0 × 10 −^25

(E) 1.0 × 10 −^25 /9.5 × 10 −^8

74. endothermic

An equilibrium mixture of the compounds is

placed in a sealed container at 150°C. The

amount of the product may be increased by which

of the following changes?

I. Adding 1 mole of Ar(g) to the container

II. Increasing the volume of the container

III. Raising the temperature of the container

(A) III only

(B) II and III

(C) II only

(D) II and III

(E) I only

75.The Kspfor LaF 3 is 2 × 10 −^9. What is the molar
solubility of this compound in water?

(A)

(B)

(C)

(D)

(E) 2 × 10 −^19

Chapter 16

76.I−(aq) +H+(aq) +MnO 4 −(aq)→ Mn^2 +(aq) +
H 2 O(l) +I 2 (s)
What is the coefficient of H+when the above
reaction is balanced?
(A) 12
(B) 32
(C) 16
(D) 8
(E) 2

77.How many moles of Au may be deposited on the

cathode when 0.60 Faradays of electricity is

passed through a 1.0 M solution of Au^3 +?

(A) 0.60 mol

(B) 0.30 mol

(C) 0.40 mol

(D) 0.20 mol

(E) 1.0 mol

78.Sn^2 ++2 Fe^3 +→ Sn^4 ++2 Fe^2 +

The reaction shown above was used in an

electrolytic cell. The voltage measured for the cell

was not equal to the calculated E°for the cell.

This discrepancy could be caused by which of the

following?

(A) Both of the solutions were at 25°C instead

of 0°C.

(B) The anode and cathode were different sizes.

(C) The anion in the anode compartment was

chloride instead of nitrate, as in the cathode

compartment.

(D) One or more of the ion concentrations was

not 1 M.

(E) The solution in the salt bridge was Na 2 SO 4

instead of KNO 3.

Questions 79–80 are concerned with the following

half-reaction in an electrolytic cell:

2 SO 42 −+10 H++8 e−→ S 2 O 32 −+5 H 2 O

79.Choose the correct statement from the following list.

(A) The sulfur is oxidized.

(B) This is the cathode reaction.

(C) The oxidation state of sulfur does not change.

(D) The H+serves as a catalyst.

(E) This is the anode reaction.

80.If a current of 0.60 amperes is passed through the

electrolytic cell for 0.75 hours, how should you

calculate the grams of S 2 O 32 −formed?

(A) (0.60)(0.75)(3600)(112) / (96500)(8)

(B) (0.60)(0.75)(3600)(112) / (96500)(10)

(C) (0.60)(0.75)(60)(32) / (96500)(8)

(D) (0.60)(0.75)(3600)(112) / (10)

(E) (0.60)(0.75)(32) / (96500)(8)

(^4) 210 27× − (^9) /
(^2210) × − 19
(^4210) × − 19

210 27× −^9 /

NO g 24 () 2 NO g 2 ()

SnS s()++ 2 H aq++()Sn^2 ()aq H S aq 2 ()