AP Chemistry Practice Exam 2 339
- D—Hydrogen bonding may occur when hydro-
gen is attached directly to N, O, or F.
- B—B has three equivalent sp^2 orbitals.
- A—C and D have one unshared pair, whereas E
has none.
- A—All three carbons have four single bonds.
- D—Hydrogen bonding is possible when hydro-
gen is attached to N, O, and F.
- D—A consequence of metallic bonding.
- C—The two molecules are hydrogen bonded
together.
- B—The presence of the −OH group makes
hydrogen bonding possible.
- B—This is a dimensional analysis problem.
- C—A is a soluble ionic compound; all the others
are acids or bases. C is a neutral molecular
compound.
- D—(0.50 mol/L)(4.0 L)(122.6 g/mol) =245.2 g
and dilute to volume.
- E—The substance producing the largest concen-
tration of ions will have the highest boiling point.
- D—The substance producing the largest concen-
tration of ions will have the lowest freezing point.
- E—Placing a larger number into the rate law will
give a larger rate. The rate constant, k, is constant
(unless the temperature changes or a catalyst is
added).
- D—Add the equations together and cancel any
species that appear on both sides.
- D—The reaction is first-order in H 2 and second-
order in NO.
- A—Strong acids and weak bases have pH =7 at
the equivalence point. The presence of a weak
base lowers this value.
58. B—The carbonate ion is the conjugate base of a
weak acid, and will produce a basic solution. The
potassium ion comes from a strong base and will
not undergo hydrolysis or change the pH.
59. E—[OH–] =(0.10 × 9 × 10 –9)1/2= 3 × 10 –5M
60. B—K= Ksp/Ka1Ka2=5.0 × 10 –18/(9.5 × 10 –8)
(1 × 10 –19)
61. E—I will yield no change, and III will decrease
the amount of product.
62. B—Ksp = [Cr^3 +][OH–]^3 = (x)(3x)^3 = 27 x^4 =
1.6 × 10 –30. Solve for x.
63. C—This will lower the concentration of Zn^2 +,
causing a shift to the right.
64. A—The size of the electrode is irrelevant.
65. B—A source of cations and anions is needed.
66. D—As the cell begins to run, the voltage
decreases.
67. E—(0.60 F)(l mol Cr/3F)
68. B—log K=nE°/0.0592 =
[2 ×(0.76 – 0.28)]/0.0592 =16, K= 1016
69. B—Bi^3 +is the oxidizing agent.
70. D—Reduction reactions like this one always
occur at the cathode.
71. D—Dimensional analysis:
(7.50 coul/s)(0.45 h)(3600 s/h)(253.8 g I 2 /mol I 2 )/
(96500 coul/F)(10 F)
- B—Mass difference =234 – 230 =4, and atomic
number difference =92 – 90 =2. These corre-
spond to an αparticle.
- C—Mass difference =236 – 4(1) – 136 =96.
Atomic number difference =92 – 4(0) – 53 =39.
- D—Based on classification of organic compounds.
- E—Definition
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Multiply the number of wrong answers by 0.25 and subtract this value from the number of correct answers.
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