Name: __________ ______________
Version A
Period: ____________
AP* Chemistry: 1984 Released Multiple Choice Exam
NO CALCULATORS MAY BE USED
Note: For all questions, assume that the temperature is 298 K, the pressure is 1.00 atmosphere, and solutions are aqueous
unless otherwise specified.
Throughout the test the following symbols have the definitions specified unless otherwise noted.
Directions: Each set of lettered choices below refers to the numbered questions or statements immediately following it. Select the
one lettered choice that best answers each question or best fits each statement and then fill in the corresponding oval on the answer
sheet. A choice may be used once, more than once, or not at all in each set. Before turning in your answer sheet, count the number
of questions that you have skipped and place that number next to your name ON YOUR ANSWER SHEET and circle it.
Questions 1-3 refer to the following elements.
3. Forms oxides that are common air pollutants and
that yield acidic solution in water
(A) F
(B) S
(C) Mg
(D) Ar
(E) Mn
1. Forms monatomic ions with 2¯ charge in solutions
2. Forms a compound having the formula KXO4
(1) Test Questions are Copyright © 1984-2002 by College Entrance Examination Board, Princeton, NJ. All rights reserved. For face-to-face teaching purposes, classroom teachers are permitted to reproduce the questions.
Web or Mass distribution prohibited. (2) AP® is a registered trademark of the College Entrance Examination Board. The College Entrance Examination Board was not involved in the production of and does not endorse this
product. Permission is granted for individual classroom teachers to reproduce the activity sheets and illustrations for their own classroom use. Any other type of reproduction of these materials is strictly
prohibited.
19
Version A
Use the following answers for questions 4 - 7.
9. Silicon dioxide, SiO2
(A) Hydrofluoric acid
Questions 10-13 refer to the following laboratory
scenario.
(B) Carbon dioxide
(C) Aluminum hydroxide
Assume that you have an "unknown" consisting of
an aqueous solution of a salt that contains one of
the ions listed above. Which ions must be absent
on the basis of each of the following observations
of the "unknown"?
(D) Ammonia
(E) Hydrogen peroxide
4. Is a good oxidizing agent
A) CO32–
B) Cr2O72–
5. Is used to etch glass chemically
C) NH4+
D) Ba2+
6. Is used extensively for the production of fertilizers
E) Al3+
7. Has amphoteric properties
10. The solution is colorless.
Questions 8-9 refer to the following types of solids.
11. The solution gives no apparent reaction with dilute
hydrochloric acid.
(A) A network solid with covalent
bonding
(B) A molecular solid with zero
dipole moment
12. No odor can be detected when a sample of the
solution is added drop by drop to a warm solution
of sodium hydroxide.
(C) A molecular solid with
hydrogen bonding
13. No precipitate is formed when a dilute solution of
H2SO4 is added to a sample of the solution.
(D) An ionic solid
(E) A metallic solid
8. Solid ethyl alcohol, C 2H5OH
2
Version A
Questions 14-17 refer to the following electrochemical cell. The spontaneous reaction that occurs when the cell
above operates is:
2 Ag+ + Cd(s) → 2 Ag(s) + Cd2+
(A)
Voltage increases.
(B)
Voltage decreases.
(C)
Voltage becomes zero and remains at zero.
(D)
No change in voltage occurs.
(E)
Direction of voltage change cannot be predicted without additional information.
Which of the above occurs for each of the following circumstances?
14. A 50-milliliter sample of a 2-molar Cd(NO3)2
solution is added to the left beaker.
16. The salt bridge is replaced by a platinum wire.
17. Current is allowed to flow for 5 minutes.
15. The silver electrode is made larger.
3
Version A
Directions: Each of the questions or incomplete statements below is followed by five suggested answers or completions. Select the
one that is best in each case and then fill in the corresponding oval on the answer sheet.
18.
Hydrogen Halide
HF
HCl
HBr
HI
Normal Boiling Point, °C
+19
– 85
– 67
– 35
The liquefied hydrogen halides have the normal boiling points given above. The relatively high boiling point of HF
can be correctly explained by which of the following?
A)
B)
C)
D)
E)
HF gas is more ideal.
HF is the strongest acid.
HF molecules have a smaller dipole moment.
HF is much less soluble in water.
HF molecules tend to form hydrogen bonds.
19. Which of the following represents a pair of isotopes?
A)
B)
C)
D)
E)
I.
II.
I.
II.
I.
II.
I.
II.
I.
II.
Atomic
Number
6
7
6
14
6
14
7
7
8
16
Atomic
Mass Number
14
14
7
14
14
28
13
14
16
20
20. .....Mg(s) + .....NO3¯(aq) +.....H+(aq) → ......Mg2+(aq) + ....NH4+(aq) + ....H2O(l)
When the skeleton equation above is balanced and all coefficients reduced to their lowest whole-number terms,
what is the coefficient for H+?
A) 4
B) 6
C) 8
D) 9
E) 10
4
Version A
Questions 25-26
21. When a sample of oxygen gas in a closed container
of constant volume is heated until its absolute
temperature is doubled, which of the following is
also doubled?
A)
B)
C)
D)
E)
H3AsO4 + 3 I– + 2 H3O+ → H3AsO3 + I3– + H2O
The oxidation of iodide ions by arsenic acid in
acidic aqueous solution occurs according to the
stoichiometry shown above. The experimental rate
law of the reaction is:
The density of the gas
The pressure of the gas
The average velocity of the gas molecules
The number of molecules per cm3
The potential energy of the molecules
Rate = k [H3AsO4] [I–] [H3O+]
22.
25. What is the order of the reaction with respect to I –?
1s2 2s22p6 3s23p3
A)
B)
C)
D)
E)
Atoms of an element, X, have the electronic
configuration shown above. The compound most
likely formed with magnesium, Mg, is
A)
B)
C)
D)
E)
MgX
Mg2X
MgX2
MgX3
Mg3X2
26. According to the rate law for the reaction, an
increase in the concentration of hydronium ion has
what effect on this reaction?
A)
B)
C)
D)
The rate of reaction increases.
The rate of reaction decreases.
The value of the equilibrium constant increases.
The value of the equilibrium constant
decreases.
E) Neither the rate nor the value of the equilibrium
constant is changed.
23. The density of an unknown gas is 2.00 grams per
liter at 3.00 atmospheres pressure and 127°C. What
is the molecular weight of this gas?
A)
B)
C)
D)
E)
254/3 R
188 R
800/3 R
600 R
800 R
27. The critical temperature of a substance is the
A) temperature at which the vapor pressure of the
liquid is equal to the external pressure
B) temperature at which the vapor pressure of the
liquid is equal to 760 mm Hg
C) temperature at which the solid, liquid, and
vapor phases are all in equilibrium
D) Temperature at which liquid and vapor phases
are in equilibrium at 1atmosphere
E) lowest temperature above which a substance
cannot be liquefied at any applied pressure
24. The formula for potassium hexacyanoferrate(II) is
A)
B)
C)
D)
E)
1
2
3
5
6
K4[Fe(CN)6]
K3[Fe(CN)6]
K2[Pt(CN)4]
K2[Pt(CN)6]
KCN
5
Version A
28.
2 A(g) + B(g)
Po decays, the emission consists
30. When 214
84
consecutively of an alpha particle, then two beta
particles, and finally another alpha particle. The
resulting stable nucleus is
2 C(g)
When the concentration of substance B in the
reaction above is doubled, all other factors being
held constant, it is found that the rate of the
reaction remains unchanged. The most probable
explanation for this observation is that
A)
B)
C)
A) the order of the reaction with respect to
substance B is 1
B) substance B is not involved in any of the steps
in the mechanism of the reaction
C) substance B is not involved in the
rate-determining step of the mechanism, but is
involved in subsequent steps
D) substance B is probably a catalyst, and as such,
its effect on the rate of the reaction does not
depend on its concentration
E) the reactant with the smallest coefficient in the
balanced equation generally has little or no
effect on the rate of the reaction
D)
E)
206
83
210
83
206
82
208
82
210
81
Bi
Bi
Pb
Pb
Tl
31. A 0.1-molar solution of which of the following ions
is orange?
A)
B)
C)
D)
E)
29.
Cu(s) + 2 Ag+ → Cu2+ + 2 Ag(s)
If the equilibrium constant for the reaction above is
3.7 × 1015, which of the following correctly
describes the standard voltage, E°, and the standard
free energy change, ΔG°, for this reaction?
A) E° is positive and ΔG° is negative.
B) E° is negative and ΔG° is positive.
C) E° and ΔG° are both positive.
D) E° and ΔG° are both negative.
E) E° and ΔG° are both zero.
6
Fe(H2O)42+
Cu(NH3)42+
Zn(OH)42–
Zn(NH3)42+
Cr2O72–
Version A
32. The net ionic equation for the reaction between silver carbonate and hydrochloric acid is
A) Ag2CO3(s) + 2 H+ + 2 Cl– → 2 AgCl(s) + H2O + CO2(g)
B) 2 Ag+ + CO32– + 2 H+ + 2 Cl– → 2 AgCl(s) + H2O + CO2(g)
C) CO32– + 2 H+ → H2O + CO2(g)
D) Ag+ + Cl– → AgCl(s)
E) Ag2CO3(s) + 2 H+ → 2Ag+ + H2CO3
35. The addition of an oxidizing agent such as chlorine
water to a clear solution of an unknown compound
results in the appearance of a brown color. When
this solution is shaken with the organic solvent,
methylene dichloride, the organic solvent layer
turns purple. The unknown compound probably
contains
33. The pH of 0.1-molar ammonia is approximately
A)
B)
C)
D)
E)
1
4
7
11
14
A)
B)
C)
D)
E)
34.
...CrO2– + ...OH– → ... CrO42– + ... H2O + ... e–
When the equation for the half-reaction above is
balanced, what is the ratio of the coefficients OH – :
CrO2– ?
A)
B)
C)
D)
E)
K+
Br–
NO3–
I–
Co2+
1:1
2:1
3:1
4:1
5:1
36.
CuO(s) + H2(g)
Cu(s) + H2O(g)
ΔH = –2.0 kilojoules mol–1
When the substances in the equation above are at equilibrium at pressure P and temperature T, the equilibrium can
be shifted to favor the products by
A)
B)
C)
D)
E)
increasing the pressure by means of a moving piston at constant T
increasing the pressure by adding an inert gas such as nitrogen
decreasing the temperature
allowing some gases to escape at constant P and T
adding a catalyst
7
Version A
41. Which of the following molecules has the shortest
bond length?
37. The molality of the glucose in a 1.0-molar glucose
solution can be obtained by using which of the
following?
A)
B)
C)
D)
E)
A)
B)
C)
D)
E)
Volume of the solution
Temperature of the solution
Solubility of glucose in water
Degree of dissociation of glucose
Density of the solution
42. Metallic copper is heated strongly with
concentrated sulfuric acid. The products of this
reaction are
38. The radioactive decay of 146 C to 147 N occurs by the
process of
A)
B)
C)
D)
E)
A)
B)
C)
D)
E)
beta particle emission
alpha particle emission
positron emission
electron capture
neutron capture
39. Equal masses of three different ideal gases, X, Y,
and Z, are mixed in a sealed rigid container. If the
temperature of the system remains constant, which
of the following statements about the partial
pressure of gas X is correct?
CuSO4(s) and H2(g) only
Cu2+, SO2(g), and H2O
Cu2+, H2(g), and H2O
CuSO4(s), H2(g), and SO2(g)
Cu2+, SO3(g), and H2O
43. The elements in which of the following have most
nearly the same atomic radius?
A)
B)
C)
D)
E)
A) It is equal to 1/3 the total pressure.
B) It depends on the intermolecular forces of
attraction between molecules of X, Y, and Z.
C) It depends on the relative molecular masses of
X, Y, and Z.
D) It depends on the average distance traveled
between molecular collisions.
E) It can be calculated with knowledge only of the
volume of the container.
Be, B, C, N
Ne, Ar, Kr, Xe
Mg, Ca, Sr, Ba
C, P, Se, I
Cr, Mn, Fe, Co
44. What number of moles of O2 is needed to produce
142 grams of P4O10 from P? (Molecular weight
P4O10 = 284)
A)
B)
C)
D)
E)
40. The geometry of the SO3 molecule is best described
as
A)
B)
C)
D)
E)
N2
O2
Cl2
Br2
I2
trigonal planar
trigonal pyramidal
square pyramidal
bent
tetrahedral
8
0.500 mole
0.625 mole
1.25 mole
2.50 mole
5.00 mole
Version A
46. If 0.060 faraday is passed through an electrolytic
cell containing a solution of In 3+ ions, the
maximum number of moles of In that could be
deposited at the cathode is
45. The alkenes are compounds of carbon and
hydrogen with the general formula C nH2n. If 3.50
gram of any alkene is burned in excess oxygen,
what number of moles of H2O is formed?
A)
B)
C)
D)
E)
A)
B)
C)
D)
E)
0.250 mole
0.375 mole
0.500 mole
2.50 mole
5.00 mole
0.010 mole
0.020 mole
0.030 mole
0.060 mole
0.18 mole
47.
CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l)
ΔΗrxn = –889.1 kJ/mol
ΔHf° H2O(l) = –285.8 kJ /mol
ΔHf° CO2(g) = –393.3 kJ /mol
What is the standard heat of formation of methane, ΔHf° CH4(g), as calculated from the data above?
A) –210.0 kJ/mole
B) –107.5 kJ/mole
C) –75.8 kJ/mole
D)
75.8 kJ/mole
E) 210.0 kJ/mole
50. Two flexible containers for gases are at the same
temperature and pressure. One holds 0.50 gram of
hydrogen and the other holds 8.0 grams of oxygen.
Which of the following statements regarding these
gas samples is FALSE?
48. Which of the following ions is the strongest Lewis
acid?
A)
B)
C)
D)
E)
Na+
Cl–
CH3COO–
Mg2+
Al3+
A) The volume of the hydrogen container is the
same as the volume of the oxygen container.
B) The number of molecules in the hydrogen
container is the same as the number of
molecules in the oxygen container.
C) The density of the hydrogen sample is less than
that of the oxygen sample.
D) The average kinetic energy of the hydrogen
molecules is the same as the average kinetic
energy of the oxygen molecules.
E) The average speed of the hydrogen molecules
is the same as the average speed of the oxygen
molecules.
49. Each of the following can act as both a Brönsted
acid and a Brönsted base EXCEPT
A)
B)
C)
D)
E)
HCO3–
H2PO4–
NH4+
H2O
HS–
9
Version A
51. Pi bonding occurs in each of the following species
EXCEPT
A)
B)
C)
D)
E)
54. Which of the following statements is always true
about the phase diagram of any one-component
system?
CO2
C2H4
CN–
C6H6
CH4
A) The slope of the curve representing equilibrium
between the vapor and liquid phases is positive.
B) The slope of the curve representing equilibrium
between the liquid and solid phases is negative.
C) The slope of the curve representing equilibrium
between the liquid and solid phases is positive.
D) the temperature at the triple point is greater
than the normal freezing point.
E) The pressure at the triple point is greater than 1
atmosphere.
52.
3 Ag(s) + 4 HNO3
3 AgNO3 + NO(g) + 2 H2O
The reaction of silver metal and dilute nitric acid
proceeds according to the equation above. If 0.10
mole of powdered silver is added to 10. milliliters
of 6.0-molar nitric acid, the number of moles of NO
gas that can be formed is
A)
B)
C)
D)
E)
55. At 20. °C, the vapor pressure of toluene is 25
millimeters of mercury and that of benzene is 75
millimeters of mercury. An ideal solution,
equimolar in toluene and benzene, is prepared. At
20. °C, what is the mole fraction of benzene in the
vapor in equilibrium with this solution?
0.015 mole
0.020 mole
0.030 mole
0.045 mole
0.090 mole
A)
B)
C)
D)
E)
53. Which, if any, of the following species is in the
greatest concentration in a 0.100-molar solution of
H2SO4 in water?
A)
B)
C)
D)
E)
0.25
0.33
0.50
0.75
0.83
56. A cube of ice is added to some hot water in a rigid,
insulated container, which is then sealed. There is
no heat exchange with the surroundings. What has
happened to the total energy and the total entropy
when the system reaches equilibrium?
H2SO4 molecules
H3O+ ions
HSO4– ions
SO42– ions
All species are in equilibrium and therefore
have the same concentrations.
A)
B)
C)
D)
E)
10
Energy
Entropy
Remains constant
Remains constant
Remains constant
Decreases
Increases
Remains constant
Decreases
Increases
Increases
Decreases
Version A
57. For the reaction A(g)
61. When a solution of potassium dichromate is added
to an acidified solution of iron(II) sulfate, the
products of the reaction are
B(g) + C(g), the
equilibrium constant, Kp, is 2 × 10–4 at 25 °C. A
mixture of the three gases at 25 °C is placed in a
reaction flask and the initial pressures are PA = 2
atmosphere, PB = 0.5 atmosphere, and PC = 1
atmosphere. At the instant of mixing, which of the
following is true for the reaction as written?
A)
B)
C)
D)
E)
A) ΔG < 0
B) ΔG > 0
62. A student pipetted five 25.00-milliliter samples of
hydrochloric acid and transferred each sample to an
Erlenmeyer flask, diluted it with distilled water, and
added a few drops of phenolphthalein to each. Each
sample was then titrated with a sodium hydroxide
solution to the appearance of the first permanent
faint pink color. The following results were
obtained.
C) ΔS = 0
D) ΔG° = 0
E) ΔG° < 0
58. Which of the following represents the ground state
electron configuration for the Mn 3+ ion? (Atomic
number Mn = 25)
A)
B)
C)
D)
E)
Volumes of NaOH Solution
First Sample..................35.22 mL
Second Sample..............36.14 mL
Third Sample.................36.13 mL
Fourth Sample...............36.15 mL
Fifth Sample.................36.12 mL
1s2 2s2 2p6 3s2 3p6 3d4
1s2 2s2 2p6 3s2 3p6 3d5 4s2
1s2 2s2 2p6 3s2 3p6 3d2 4s2
1s2 2s2 2p6 3s2 3p6 3d8 4s2
1s2 2s2 2p6 3s2 3p6 3d3 4s1
Which of the following is the most probable
explanation for the variation in the student's
results?
59. When 70. milliliters of 3.0-molar Na2CO3 is added
to 30. milliliters of 1.0-molar NaHCO3 the resulting
concentration of Na+ is
A)
B)
C)
D)
E)
A) The burette was not rinsed with NaOH solution.
B) The student misread a 5 for a 6 on the burette
when the first sample was titrated.
C) A different amount of water was added to the
first sample.
D) The pipette was not rinsed with the HCl
solution.
E) The student added too little indicator to the first
sample.
2.0 M
2.4 M
4.0 M
4.5 M
7.0 M
60. Which of the following has a zero dipole moment?
A)
B)
C)
D)
E)
FeCr2O7(s) and H2O
FeCrO4(s) and H2O
Fe3+, CrO42–, and H2O
Fe3+, Cr3+, and H2O
Fe2(SO4)3(s), Cr3+ and H2O
HCN
NH3
SO2
NO2
PF5
11
Version A
63.
Acid
Acid Dissociation
Constant, Ka
H3PO4
7 × 10–3
H2PO4–
8 × 10–8
HPO42–
5 × 10–13
On the basis of the information above, a buffer with a pH = 9 can best be made by using
A)
B)
C)
D)
E)
pure NaH2PO4
H3PO4 + H2PO4–
H2PO4– + PO43–
H2PO4– + HPO42–
HPO42– + PO43–
64. The net ionic equation for the reaction that occurs
during the titration of nitrous acid with sodium
hydroxide is
A) HNO2 +
Na+
+
OH–
66.
Ca, V, Co, Zn, As
Gaseous atoms of which of the elements above are
paramagnetic?
→ NaNO2 + H2O
B) HNO2 + NaOH → Na+ + NO2– + H2O
A)
B)
C)
D)
E)
C) H+ + OH– → H2O
D) HNO2 + H2O → NO2– + H3O+
E) HNO2 + OH– → NO2¯ + H2O
67. A student wishes to prepare 2.00 liters of
0.100-molar KIO3 (molecular weight 214). The
proper procedure is to weigh out
65. Which of the following species CANNOT function
as an oxidizing agent?
A)
B)
C)
D)
E)
Ca and As only
Zn and As only
Ca, V, and Co only
V, Co, and As only
V, Co, and Zn only
Cr2O72–
MnO4–
NO3–
S
I–
A) 42.8 grams of KIO3 and add 2.00 kilograms of
H2O
B) 42.8 grams of KIO3 and add H2O until the final
homogeneous solution has a volume of 2.00
liters
C) 21.4 grams of KIO3 and add H2O until the final
homogeneous solution has a volume of 2.00
liters
D) 42.8 grams of KIO3 and add 2.00 liters of H2O
E) 21.4 grams of KIO3 and add 2.00 liters of H2O
12
Version A
68. A 20.0-milliliter sample of 0.200-molar K2CO3
solution is added to 30.0 milliliters of 0.400-molar
Ba(NO3)2 solution. Barium carbonate precipitates.
The concentration of barium ion, Ba 2+, in solution
after reaction is
A)
B)
C)
D)
E)
72. A compound is heated to produce a gas whose
molecular weight is to be determined. The gas is
collected by displacing water in a water-filled flask
inverted in a trough of water. Which of the
following is necessary to calculate the molecular
weight of the gas, but does NOT need to be
measured during the experiment?
0.150 M
0.160 M
0.200 M
0.240 M
0.267 M
A)
B)
C)
D)
E)
69. What is the approximate mole fraction of ethanol,
C2H5OH, in an aqueous solution in which the
ethanol concentration is 11.0 molal?
A)
B)
C)
D)
E)
73. A 27.0-gram sample of an unknown hydrocarbon
was burned in excess oxygen to form 88.0 grams of
carbon dioxide and 27.0 grams of water. What is a
possible molecular formula of the hydrocarbon?
0.0011
0.011
0.170
0.200
0.600
A)
B)
C)
D)
E)
70. One of the outermost electrons in a strontium atom
in the ground state can be described by which of the
following sets of four quantum numbers?
A)
B)
C)
D)
E)
Mass of the compound used in the experiment
Temperature of the water in the trough
Vapor pressure of the water
Barometric pressure
Volume of water displaced from the flask
CH4
C2H2
C4H3
C4H6
C4H10
74. How many moles of NaF must be dissolved in 1.00
liter of a saturated solution of PbF 2 at 25 °C to
5, 2, 0, ½
5, 1, 1, ½
5, 1, 0, ½
5, 0, 1, ½
5, 0, 0, ½
reduce the [Pb2+] to 1 × 10–6 molar? (Ksp of PbF2 at
25 °C = 4.0 × 10–8)
A)
B)
C)
D)
E)
71. Which of the following reactions does NOT
proceed significantly to the right in aqueous
solutions?
A) H3O+ + OH– → 2 H2O
B) HCN + OH– → H2O + CN–
C) Cu(H2O)42+ + 4 NH3 → Cu(NH3)42+ + 4H2O
D) H2SO4 + H2O → H3O+ + HSO4–
E) H2O + HSO4– → H2SO4 + OH–
13
0.020 mole
0.040 mole
0.10 mole
0.20 mole
0.40 mole
Version A
75. If the acid dissociation constant, Ka, for an acid HA is 8 × 10–4 at 25 °C, what percent of the acid is dissociated in a
0.50-molar solution of HA at 25 °C?
A)
B)
C)
D)
E)
0.08%
0.2%
1%
2%
4%
76.
HgO(s) + 4 I– + H2O
HgI42– + 2 OH–
ΔH < 0
Consider the equilibrium above. Which of the following changes will increase the concentration of HgI 42–?
A)
B)
C)
D)
E)
Increasing the concentration of OH–
Adding 6 M HNO3
Increasing the mass of HgO present
Increasing the temperature
Adding a catalyst
14
Version A
77. Which of the following compounds exhibits optical
isomerism?
78. When the actual gas volume is greater than the
volume predicted by the ideal gas law, the
explanation lies in the fact that the ideal gas law
does NOT include a factor for molecular
A)
A)
B)
C)
D)
E)
volume
mass
velocity
attractions
shape
B)
79.
5 Fe2+ + MnO4– + 8 H+
5 Fe3+ + Mn2+ + 4 H2O
In a titration experiment based on the equation
above, 100.0 milliliters of an acidified Fe 2+ solution
requires 14.0 milliliters of standard 0.050-molar
MnO4– solution to reach the equivalence point. The
concentration of Fe2+ in the original solution is
C)
A)
B)
C)
D)
E)
D)
0.0035 M
0.0070 M
0.035 M
0.070 M
0.14 M
80. For which of the following molecules are resonance
structures necessary to describe the bonding
satisfactorily?
E)
A)
B)
C)
D)
E)
15
H2S
SO2
CO2
OF2
PF3
Version A
81. What is the net ionic equation for the reaction that
occurs when aqueous copper(II) sulfate is added to
excess 6-molar ammonia?
A) Cu2+ + SO42– + 2 NH4+ + 2 OH– → (NH4)2SO4
+ Cu(OH)2
B) Cu2+ + 4 NH3 + 4 H2O → Cu(OH)42– + 4 NH4+
C) Cu2+ + 2 NH3 + 2 H2O → Cu(OH)2 + 2 NH4+
D) Cu2+ + 4 NH3 → Cu(NH3)42+
E) Cu2+ + 2 NH3 + H2O → CuO + 2 NH4+
82.
N2HO2– + H+
Step 1
N2H2O2
Step 2
N2HO2– → N2O + OH–
(slow)
Step 3
H+ + OH– → H2O
(fast)
(fast equilibrium)
Nitramide, N2H2O2, decomposes slowly in aqueous solution. This decomposition is believed to occur according to
the reaction mechanism above. The rate law for the decomposition of nitramide that is consistent with this
mechanism is given by which of the following?
A)
B)
C)
D)
E)
Rate = k [N2H2O2]
Rate = k [N2H2O2] [H+]
Rate = (k [N2H2O2]) / [H+]
Rate = (k [N2H2O2]) / [N2HO2–]
Rate = k [N2H2O2] [OH–]
83.
5
O (g)
H2NCH2COOH(s) + 3 H2O(l)
2 2
At constant temperature, ΔH, the change in enthalpy for the reaction above is approximately equal to
NH3(g) + 2 CH4(g) +
11
RT
2
7
ΔE – RT
2
ΔE + RT
7
ΔE + RT
2
˜ˆ
ÁÊÁ 11
ΔE ÁÁÁ RT ˜˜˜˜
¯
Ë 2
A) ΔE –
B)
C)
D)
E)
16
Version A
84. Which of the following aqueous solutions has the
highest boiling point?
A)
B)
C)
D)
E)
0.10 M potassium sulfate, K2SO4
0.10 M hydrochloric acid, HCl
0.10 M ammonium nitrate, NH4NO3
0.10 M magnesium sulfate, MgSO4
0.20 M sucrose, C12H22O11
85. A sample of 9.00 grams of aluminum metal is
added to an excess of hydrochloric acid. The
volume of hydrogen gas produced at standard
temperature and pressure is
A)
B)
C)
D)
E)
22.4 liters
11.2 liters
7.46 liters
5.60 liters
3.74 liters
17
INFORMATION IN THE TABLE BELOW AND IN THE TABLES ON PAGES 23-25 MAY BE USEFUL IN
ANSWERING THE QUESTIONS IN THIS SECTION OF THE EXAMINATION.
GO ON TO THE NEXT PAGE.
-22-
ID: A
AP* Chemistry: 1984 Released Multiple Choice Exam
Answer Section
OTHER
1. ANS:
B
Elements in the oxygen family form monatomic ions with 2¯ charges (oxidation numbers). Halogens form
monatomic ions with 1¯ charges. II A metals form monatomic ions with 2+ charges. Noble gases do not form ions.
Manganese is a metal and has several oxidation states all of which are positive.
DIF: Easy
2. ANS:
TOP: Periodicity
MSC: 1984 #1
NOT: 78% answered correctly
E
Think “polyatomic ion with a –1 charge”. MnO 4 has a negative one charge. S would form SO 42–.
DIF: Easy
3. ANS:
TOP: Periodicity
MSC: 1984 #2
NOT: 68% answered correctly
B
In general, nonmetal oxides form acids in aqueous solution and rain counts as “aqueous”. SO x and NOx
compounds (pronounced “socks and knocks”) are particularly notorious for causing acid rain. The acid formed
retains the oxidation number of the non metal.
DIF: Easy
4. ANS:
TOP: Periodicity
MSC: 1984 #3
NOT: 74% answered correctly
E
Remember “ is a good oxidizing agent” is code for “is itself easily reduced”. These would be substances with
freakish oxidation states like peroxide O 22–, or substances with metals with really high positive oxidation states
such as MnO4–, and Cr2O72–.
DIF: Hard
TOP: Chemical Reactions
NOT: 34% answered correctly
5. ANS:
MSC: 1984 #4
A
Although HF is a weak acid that does not make it a harmless acid! Aside from etching glass it will also remove
calcium ions directly from human bone, so it should always be handled with care.
DIF:
Easy
TOP: Acid-Base
MSC: 1984 #5
1
NOT: 62% answered correctly
ID: A
6. ANS:
D
Ammonia contributes significantly to the nutritional needs of terrestrial organisms by serving as a
precursor to foodstuffs and fertilizers. Ammonia, either directly or indirectly, also is a building block for
the synthesis of many pharmaceuticals. Although in wide use, ammonia is both caustic and hazardous.
DIF: Medium
7. ANS:
TOP: Descriptive
MSC: 1984 #6
NOT: 55% answered correctly
C
Used to describe a compound, such as H2O or Al(OH)3, that can act as either an acid or a base.
DIF: Hard
8. ANS:
TOP: Acid-Base
MSC: 1984 #7
NOT: 29% answered correctly
C
The structure of ethanol is found below (note the 2 unshared pairs of electrons surrounding oxygen are not shown):
The –OH group makes the molecule polar and that fact that a hydrogen is bound to “a highly electronegative”
element (such as F, O, or N) coupled with the presence of 2 lone pairs of electrons (not shown on the O) makes
ethanol capable of hydrogen bonding.
DIF: Easy
TOP: Bonding & Molecular Structure
NOT: 66% answered correctly
9. ANS:
MSC: 1984 #8
A
Sand on the beach, quartz and the glass from which we make bottles all contain silicon dioxide. Silicon
and carbon compounds (diamond) are capable of network covalent bonding.
DIF: Medium
TOP: Bonding & Molecular Structure
NOT: 56% answered correctly
10. ANS:
MSC: 1984 #9
B
IF dichromate ion were present, the solution would be colored bright orange. The only solutions that exhibit color
have ions with electron configurations that possess unpaired d-electrons. Dichromate ion contains the only
transition metal listed in the answer choices.
DIF:
Medium
TOP: Lab
MSC: 1984 #10
2
NOT: 58% answered correctly
ID: A
11. ANS:
A
IF the carbonate ion (or bicarbonate ion) were present it would react with the hydrogen ion of HCl to produce
carbon dioxide bubbles.
DIF: Hard
12. ANS:
TOP: Lab
MSC: 1984 #11
NOT: 23% answered correctly
C
Ammonium ion is the conjugate acid of ammonia. If it were dropped into a solution of sodium hydroxide
(depending on temperature and concentration), an acid-base neutralization reaction would occur and ammonia gas
would be produced.
DIF: Medium
13. ANS:
TOP: Lab
NOT: 53% answered correctly
D
If barium ion were in solution, very insoluble barium sulfate would be produced upon the addition of any soluble
sulfate solution.
DIF: Medium
14. ANS:
TOP: Lab
MSC: 1984 #13
NOT: 46% answered correctly
B
ÍÈÍ 2 + ˙˘˙
ÍÍ Cd ˙˙
˚
Î
.
According to the balanced reaction given, 2 Ag + + Cd(s) → 2 Ag(s) + Cd2+, Q = È
ÍÍ + ˘˙˙ 2
ÍÍ Ag ˙˙
Î
˚
If the concentration of cadmium(II) ion increases, then numerator of the Q expression increases above standard
conditions of 1.0 and Q is greater than one. Thus the log of Q is positive, so the –0.059/2 log Q term is indeed
subtracted reducing the voltage of the cell.
DIF: Hard
TOP: Electrochemistry
NOT: 24% answered correctly
15. ANS:
MSC: 1984 #14
D
The size or mass of an electrode is not included in the Nernst equation, so it has no effect on the voltage.
DIF: Medium
TOP: Electrochemistry
NOT: 52% answered correctly
MSC: 1984 #15
3
ID: A
16. ANS:
C
Replacing the salt bridge with a platinum wire stops the flow of ions that balance the charge, so the voltage will
drop to zero and stay there until the salt bridge is reinstated.
DIF: Medium
TOP: Electrochemistry
NOT: 50% answered correctly
17. ANS:
MSC: 1984 #16
B
As current flows, voltage drops. Once equilibrium is established the voltage becomes zero. Thus, batteries don’t
“die”, they just reach equilibrium!
DIF: Medium
TOP: Electrochemistry
NOT: 45% answered correctly
MSC: 1984 #17
MULTIPLE CHOICE
18. ANS: E
Always be on the lookout for physical properties (boiling, melting, vaporizing, etc.) tied to hydrogen bonding.
Hydrogen bonding occurs when a H is bound to a “highly electronegative atom” (F, N or O) AND said hydrogen is
attracted to an unshared electron pair or another highly electronegative atom on a neighboring molecule. The
graph below is a classic! Think about the Lewis structures of HF, NH 3, and H2O.
DIF: Medium
TOP: IMFs
MSC: 1984 #18
NOT: 65% answered correctly
19. ANS: D
You’re looking for elements with the same atomic number (# of protons) but a different number of neutrons, thus a
different mass number. This was supposed to be a really easy question!
DIF: Easy
TOP: Atomic Structure
NOT: 87% answered correctly
MSC: 1984 #19
4
ID: A
20. ANS: E
Red: NO3¯(aq) + 10 H+(aq) + 8 e– → NH4+(aq) + 3 H2O(l)
Ox: 4 (Mg(s) → Mg2+(aq) + 2e–)
[you can stop once you see that there is no multiplier for the reduction reaction]
SUM: NO3¯(aq) + 10 H+(aq) + 4 Mg(s) → 4 Mg2+ + NH4+(aq) + 3 H2O(l)
DIF: Easy
TOP: Chemical Reactions
NOT: 82% answered correctly
21. ANS: B
When volume is held constant, apply Boyle’s Law: P1T2 = P2T1
MSC: 1984 #20
So, if T1 is doubled, then the left side of the equation becomes P1(2T1) and the right side of the equation must then
= (2)P2T1 so that the law is obeyed.
DIF: Easy
TOP: Gas Laws
MSC: 1984 #21
NOT: 75% answered correctly
22. ANS: E
Element X has a valence configuration of 3s23p3 , which indicates five valence electrons, thus it belongs to Group
VA (or 15 or the nitrogen family). That means it is a nonmetal with a common oxidation state of –3.
Meanwhile...Mg has a common oxidation state of +2 since it is a member of Group IIA (or 2 or the alkaline earth
metals). So, we have a “two-three trick” where we need 3 of the plus two’s and 2 of the minus threes to create a
neutral compound. Therefore, Mg3X2 is the best answer choice.
DIF: Easy
TOP: Atomic Structure
MSC: 1984 #22
NOT: 80% answered correctly
23. ANS: C
Expect easy math!
When density (or density data) is given and molar mass (weight) is required, think “molar mass kitty cat”. You
know, “every good cat puts “dirt” over its “pee”:
MM =
dRT
=
P
DIF:
Easy
ÊÁ 2g ˆ˜
ÁÁ
˜
ÁÁ 1L ˜˜˜ R (127 + 273K)
Ë
¯
3atm
TOP: Gas Laws
=
(2) (400)R 800
=
R
3
3
MSC: 1984 #23
5
NOT: 75% answered correctly
ID: A
24. ANS: A
Dissect the name starting at the back:
(II) indicates there is a +2 charge on the metal
ferrate indicates two things: the complex ion is negative and the metal is iron, so think Fe 2+
cyano indicates the ligand is CN and you know that CN has a charge of –1
hexa is a prefix meaning “6”, so there are 6 CN– ions present in the complex
Recap with mathematics included: So far, we have [Fe(CN)6], and its collective charge is a +2 for the iron with a
total of –6 for the six ligands present having a grand total of –4 for the complex ion [hence the –ate suffix
(negative complex) and switch to Latin root for the metal], which gives [Fe(CN)6]4–.
Potassium is just plain ole K+, so we need FOUR of them to cancel out the –4 charge on the complex and make a
neutral compound.
DIF: Medium
TOP: Descriptive MSC: 1984 #24
NOT: 68% answered correctly
25. ANS: A
The order of the reaction is the sum of the orders on each reactant. Recall that the order is simply the exponent on
each term, each reactant in this experimental rate law has an exponent of 1, there are 3 terms, therefore the overall
order of the reaction is 1 + 1 +1 or 3.
DIF: Easy
TOP: Kinetics
MSC: 1984 #25
NOT: 61% answered correctly
26. ANS: A
The rate law expression: Rate = k [H3AsO4] [I–] [H3O+], increasing the concentration of any of the reactants will
increase the calculated value of the rate, yet have no effect on the rate law constant, k. Imagine, the rate constant is
constant at constant temperature!
DIF: Easy
TOP: Kinetics
MSC: 1984 #26
NOT: 75% answered correctly
27. ANS: E
Critical temperature: A temperature beyond which a gas cannot be turned into a liquid no matter how much
pressure is applied. The process of liquefaction cannot occur above the critical temperature. Also called the critical
point. See the phase diagram below:
DIF: Easy
TOP: States of Matter
NOT: 67% answered correctly
MSC: 1984 #27
6
ID: A
28. ANS: C
Your first inclination is to say that the reaction is zero order in B, but that is not an answer choice.
Substance B is not involved in any step prior to the rate determining step nor in the rate determining step, but is
involved in subsequent steps. It does have to be involved in the mechanism as a reactant in a step or the
mechanism is invalid since all the steps of the mechanism must combine with the correct stoichiometry.
Substance B cannot be a catalyst. A catalyst is neither a reactant nor a product, substance B is clearly a reactant.
DIF: Easy
TOP: Kinetics
MSC: 1984 #28
NOT: 64% answered correctly
29. ANS: A
Spontaneous electrochemical cells have a positive voltage. Spontaneous reactions have a negative value of ΔG°.
Spontaneous reactions also have a K value of 1 or greater by definition (products are favored).
DIF: Medium
TOP: Electrochemistry
MSC: 1984 #29
NOT: 51% answered correctly
30. ANS: C
This is a transmutation reaction. It is essential to know that an alpha particle is akin to a helium nucleus,
4
2
He, and
that abeta particle is akin to an electron shot out of the nucleus, e or β (recall that a neutron consists of a
proton plus an electron and a bit of binding energy--when the electron is emitted, there is one less neutron but one
more proton). If you know that, the rest is simple math since the law of conservation of mass must be obeyed.
0
−1
214
84
Po →42 He +
0
−1
0
−1
β + −10 β +42 He + ?? X
(214 − 8)
X , so the element X has a mass of 206 and an atomic number of 82 which
So, ?? X must be equal to (84 − 4 − (−2) X = 206
82
is Pb.
DIF: Medium
TOP: Nuclear
MSC: 1984 #30
NOT: 46% answered correctly
31. ANS: E
Solutions containing transition metal ions are colored IF there are unpaired d-electrons present. So, iron, copper
and chromium are all candidates. Iron(III) is yellow in solutions, copper(II) is blue in solutions, zinc is colorless
and dichromate is a bright orange.
DIF: Medium
TOP: Descriptive MSC: 1984 #31
NOT: 46% answered correctly
32. ANS: A
Carbonates are generally insoluble, silver carbonate is really insoluble, so it is written “together” or undissociated
in an aqueous solution. Hydrochloric acid is a strong acid, so it completely dissociates in aqueous solution.
Adding acid to a carbonate makes bubbles [classic “volcano reaction”] so CO2 is formed, silver chloride is the
precipitate and water is the final product of the neutralization.
DIF: Hard
TOP: Chemical Reactions
NOT: 25% answered correctly
MSC: 1984 #32
7
ID: A
33. ANS: D
Expect easy math!
Realize that ammonia is a BASE! So it’s easiest to calculate pOH and then subtract from 14.
pOH = –log [OH–] = –log [10–1], so the log of [10–1] is simply –1, so the negative of –1 is plain old 1.0 [reported to
1 sig. fig.*]. Therefore the pH = 14 – 1 = 13.0 [still reported to 1 sig. fig.]
* Recall that sig. fig. rules for pH are different since it is a logarithmic scale. The number in front of the decimal
(the characteristic) is just a place holder, so the only significant figures are those after the decimal (the mantissa).
It’s explained in the appendix of your text if you don’t believe me!
DIF: Medium
TOP: Acid-Base
MSC: 1984 #33
NOT: 44% answered correctly
34. ANS: D
It is easier to take the OH– and water out of the equation. Next balance the half-reaction in acid media, then add
hydroxide to each side to neutralize any excess H + ions. Finally cancel waters from one side to “clean up” the
final balanced half-reaction.
The balanced half-reaction is:
→ CrO42– (skeleton without water or OH–)
CrO2–
CrO2– + 2 H2O → CrO42– + 4 H+ + 3 e–
+ 4 OH–
+ 4 OH – (makes 4 waters on the right, cancels 2 waters on the left)
SUM: CrO2– + 4 OH– → CrO42– + 2 H2O + 3 e–
DIF: Easy
TOP: Chemical Reactions
MSC: 1984 #34
NOT: 65% answered correctly
35. ANS: D
You either did a solvent extraction, saw a demonstration of a solvent extraction, or are at a severe disadvantage on
this lab question! Adding the chlorine water oxidizes the colorless I – ion and forms I2 which is very nonpolar!
This, it is not very soluble in water but, what does dissolve turns the solution amber to brown depending on the
concentration. Shaking the aqueous solution with an organic solvent will extract the iodine into the organic
solvent layer (likes dissolve likes) where it is a glorious purple.
Br– would also have been oxidized to Br 2 and then extracted and been a deep red.
Co2+ would have been pink, but would not have extracted into the organic solvent. Potassium nor nitrate ions
would have extracted, and they would remain colorless.
DIF:
Hard
TOP: Lab
MSC: 1984 #35
8
NOT: 40% answered correctly
ID: A
36. ANS: C
A negative ΔH value indicates an exothermic reaction. So, it may help to think of the equation this way:
CuO(s) + H2(g)
Cu(s) + H2O(g) + heat
If you remove heat, the system shifts to replace it. Since heat is essentially a product, that would shift the
equilibrium to favor the products. Since there are equal numbers of moles of gas on each side of the equation,
changing the pressure has no effect on equilibrium. A catalyst simply allows the system to reach equilibrium faster,
it never shifts an equilibrium.
DIF: Easy
TOP: Equilibrium MSC: 1984 #36
NOT: 62% answered correctly
37. ANS: E
moles of solute
moles of solute
Molality =
and Molarity =
kg of solution
liters of solution
You need the density of the solution to convert liters (a volume) into kilograms (a mass).
It is also important to note that molality is temperature independent since it is a ratio of masses which do not
expand or contract with temperature changes they way volumes do, which is why we use molality to calculate FP
depression and BP elevation.
DIF: Easy
TOP: Solutions
MSC: 1984 #37
NOT: 67% answered correctly
38. ANS: A
This is a transmutation reaction. It is essential to know that an alpha particle is akin to a helium nucleus,
4
2
He, and
that an alpha particle is akin to an electron shot out of the nucleus, e or β (recall that a neutron consists of a
proton plus an electron and a bit of binding energy--when the electron is emitted, there is one less neutron but one
more proton). If you know that the rest is simple math since the law of conservation of mass must be obeyed.
0
−1
14
6
0
−1
C →147 N + ?? X
So, ?? X must be equal to −10 β
DIF: Medium
TOP: Nuclear
MSC: 1984 #38
NOT: 42% answered correctly
39. ANS: C
PV = nRT, but we have 3 constants: volume (rigid container), temperature, and R. So, this simplifies to P ∝ n, so
the pressure for each gas depends on the number of moles of each gas AND equal masses of each gas were placed
in the container, so P really depends on the molar mass of each gas. Also note the gases are “ideal”. Thus
collisions are elastic and IMFs are neglected.
DIF:
Medium
TOP: Gas Laws
MSC: 1984 #39
9
NOT: 54% answered correctly
ID: A
40. ANS: A
The structure of SO3 appears below (note it exhibits resonance):
There are three “sites” of electron density surrounding the central S atom, thus the molecular geometry is trigonal
planar with bond angles of 120°.
DIF: Medium
TOP: Bonding & Molecular Structure
MSC: 1984 #40
NOT: 54% answered correctly
41. ANS: A
The more bonds present, the shorter the bond length and the stronger the bond. The increased electron density
between 2 nuclei bring the two nuclei closer together. All of the answer choices are diatomic. Halogens are singly
bonded, oxygen is doubly bonded and nitrogen has a triple bond which is the shortest and the strongest of those
listed.
DIF: Medium
TOP: Bonding & Molecular Structure
MSC: 1984 #41
NOT: 49% answered correctly
42. ANS: B
Copper is going to dissolve or ionize in concentrated sulfuric acid! That’s an oxidation.
Bubbling is also going to happen. Copper has a more positive reduction potential than hydrogen ion from the acid,
so hydrogen is not reduced to hydrogen gas. So, the gas formed is a SO x compound.
Which SOx? Well, if copper is oxidized, and H + from the acid is not reduced, then the S atom in sulfuric acid (+6)
is reduced, so SO2 is formed rather than SO3.
DIF: Hard
TOP: Redox
MSC: 1984 #42
NOT: 13% answered correctly
43. ANS: E
Remember, a trend is NOT an explanation...but knowing the trends serves you well in the multiple choice!
Moving across the periodic table within a period, the atomic radius decreases due to and in crease in effective
nuclear charge (Zeff). So, answer (A) is not a good choice. Moving down the periodic table within a family, the
atomic radius increases due to adding an entire principal energy level. So, answers (B) and (C) are not good
choices. Answer (D) is all over the table, so it is not a good choice, either. But WHY is answer (E) the best
choice? Because they are all transition metals within the same period. Moving across a given period within the
transition metals adds 3d electrons, BUT the 4s electrons are also present, therefore the size of the radius stays
relatively constant.
DIF: Medium
TOP: Periodicity
MSC: 1984 #43
NOT: 48% answered correctly
44. ANS: D
Expect easy math!
4P + 5O2 → P4O10
142 grams of P4O10 is ½ a mole. So, you need ½ of 5 moles of O 2 or 2.5 moles.
DIF: Easy
TOP: Stoichiometry
NOT: 67% answered correctly
MSC: 1984 #44
10
ID: A
45. ANS: A
Expect easy math!
If you think of C nH2n as a “unit”, then the molar mass of the “unit” is (12 + 2) = 14 g/mol. Look for a simple
mathematical relationship: 3.50 g is ¼ of a mole and each unit has 2 H per mole, AND each water molecules
requires 2 H as well, so you have ¼ of a mole of water formed.
TOP: Stoichiometry
MSC: 1984 #45
NOT: 59% answered correctly
46. ANS: B
Expect easy math!
Recall that a faraday is simply a mole of electrons. The reduction reaction that would form In is: In 3+ + 3e– →
In°(s), so you need 3 moles of electrons to form 1 mole of product. You have 0.060 moles of electrons, so you can
form 0.020 moles of product or In.
DIF: Medium
TOP: Electrochem MSC: 1984 #46
47. ANS: C
Expect easy math!
The numerical values in the answers are far apart, so estimate!
NOT: 48% answered correctly
–900 = [–400 + 2(–300)] – x
–900 + x = –1000
x = –1000 – (–900) = –1000 + 900 = less than –100 kJ/mol (since all of the actual values were rounded up!)
DIF: Hard
TOP: Thermochemistry
MSC: 1984 #47
NOT: 38% answered correctly
48. ANS: E
Recall that Lewis acids “accept an electron pair”. It stands to reason that the answer choice with the highest
positive charge would attract electron pairs most strongly.
DIF: Hard
TOP: Acid-Base
MSC: 1984 #48
NOT: 34% answered correctly
49. ANS: C
Recall that Brönsted acids “donate a proton” and bases “donate accept a proton”. Since all of the answer choices
have a hydrogen ion or proton they can donate, all fit the definition for acids. Choices (A), (B), and (E) are
negative polyatomic ions that can readily accept a hydrogen ion, so they are bases. Answer (D) is not a negative
ion, but can readily accept a proton to form hydronium ion. Ammonium ion, however, cannot accept a proton,
therefore it cannon act as a base.
DIF: Easy
TOP: Acid-Base
MSC: 1984 #49
NOT: 63% answered correctly
50. ANS: E
Flexible is code for “volume not constant”. T and P are constant. Since both flexible containers hold ½ mole of
gas, the number of molecules in each container is the same, therefore, Avogadro’s Law holds true and their
volumes are also equal. The density of the hydrogen gas is less than the density of the oxygen gas since the molar
mass of hydrogen is less than oxygen’s yet they are in containers of equal volume. Since both gases are at the
same temperature, they have identical average kinetic energies. The hydrogen molecules move more rapidly since
their mass is smaller. (KEave = ½mv2)
DIF:
Hard
TOP: Gas Laws
MSC: 1984 #50
11
NOT: 39% answered correctly
ID: A
51. ANS: E
In any multiple bond, a sigma bond is present. A pi bond is the 2nd bond of a double bond as well and both the
2nd and 3rd bonds of a triple bond. Draw the Lewis structures! Only methane has no multiple bonds, therefore all
of its bonds are sigma.
DIF: Medium
TOP: Bonding & Molecular Structure
MSC: 1984 #51
NOT: 56% answered correctly
52. ANS: A
The “trick” to getting this one correct is to recognize that you have entered the “land of limiting reagent”! You
were given the number of moles of silver, but must calculate the moles of nitric acid, it’s two starting amounts
either way! Remember that molarity × liters = moles. Determine the limiting reagent and calculate subsequent
moles from that limiting amount of moles using the mole:mole.
3 Ag
mole:mole 3
# moles
0.10
divide by 3 = 0.033
+ 4 HNO3
4
=(0.010 liter)(6.0mol/L)
= 0.060 mol
divide by 4 = 0.015,
compare to 0.033
LIMITING! work from
this now...
3
AgNO3
3
+ NO
1
If 4 = 0.060,
what’s “1”
equal?
0.015 moles
NO formed
+ 2 H2 O
2
DIF: Medium
TOP: Stoichiometry
MSC: 1984 #52
NOT: 63% answered correctly
53. ANS: B
Sulfuric acid is a diprotic strong acid, therefore 2 H + ions are released from each molecule in water. The hydrogen
ions “ride piggy back” on a water molecule to form hydronium ions.
DIF:
Easy
TOP: Acid-Base
MSC: 1984 #53
12
NOT: 55% answered correctly
ID: A
54. ANS: A
Hopefully, this graph popped into your brain:
Each solid line represents an equilibrium. While the left diagram has a positive slope for the equilibrium line
between liquid and solid (most substances behave this way), that is not always the case as you can see (water is a
freak--its solid is less dense than its liquid phase).
DIF: Hard
TOP: States of Matter
MSC: 1984 #54
NOT: 30% answered correctly
55. ANS: D
Reread the question, this question scored low, but was supposed to be easy. It focuses on the VAPOR above the
solution and the mole fraction of ONE of the constituents. Dalton’s Law applies, there is a total pressure of 100
mmHg, where 75 mm Hg is due to benzene and 25 mm Hg is due to toluene. Therefore, the mole fraction of
benzene is 75/100 or 0.75
DIF: Hard
TOP: States of Matter
MSC: 1984 #55
NOT: 39% answered correctly
56. ANS: C
If melting occured, chaos abounds so entropy increases! No heat exchange means the total energy remains
constant.
The entropy increases as the ice melts. Only answers C and D were possible. The total energy would have
decreased had there been any heat exchange allowed...sort of tricky.
DIF: Hard
TOP: Thermodynamics
MSC: 1984 #56
NOT: 36% answered correctly
57. ANS: B
“At the instant of mixing” means equilibrium has not yet been established. Calculate Q to determine how the
reaction will proceed.
(0.5) (1)
= 0.25 which is greater than Kp, so the reaction shifts left initially, which is nonspontaneous
2
(reactants favored) and ΔG > 0.
Q=
DIF: Hard
TOP: Thermodynamics
NOT: 32% answered correctly
MSC: 1984 #57
13
ID: A
58. ANS: A
Mn’s ground state atomic configuration is 1s2 2s2 2p6 3s2 3p6 3d5 4s2, so the Mn3+ ion would have lost 3 electrons.
Those electrons come first from the 4s sublevel, then the 3d sublevel, leaving 1s2 2s2 2p6 3s2 3p6 3d4
DIF: Hard
TOP: Atomic Structure
MSC: 1984 #58
NOT: 33% answered correctly
59. ANS: D
Expect easy math! The total volume is 70 + 30 = 100 mL. Notice that 2 Na + ions are released from sodium
carbonate!
2(0.070 L × 3.0 M) + (0.030 L × 1.0 M) = 0.42 + 0.03 = 0.45 moles of ions in 0.100 L, so 4.5 M in Na+ ions.
DIF: Hard
TOP: Solutions
MSC: 1984 #59
NOT: 38% answered correctly
60. ANS: E
Draw the Lewis structures! It is evident that all the dipole moments cancel out in PF 5.
DIF: Hard
TOP: Bonding & Molecular Structure
MSC: 1984 #60
NOT: 34% answered correctly
61. ANS: D
It is a classic reaction that dichromate will be reduced to Cr3+ in acid solution. That leaves iron to be oxidized to
from Fe2+ to Fe3+ and if H+ is present on one side of a redox reaction, water is present on the other!
This was not supposed to be that hard of a question! Only 10% of the population answered correctly...ouch!
DIF: Hard
TOP: Redox
MSC: 1984 #61
NOT: 10% answered correctly
62. ANS: D
The first sample’s volume value is much lower than the others. If the pipet had been wet with water, then the first
sample was diluted and thus did not require as much NaOH to reach the equivalence point. This is a classic error!
All volumetric glassware should be rinsed with the “stuff” that is to be measured prior to conducting the
measurement.
If the buret had not been rinsed with NaOH, then the NaOH would have been diluted and more would have been
required for initial trials. The amount of water added to the flask does not affect the number of moles of acid in the
flask. The amount of indicator [within reason] added to the flask has no effect on the equivalence point.
DIF: Hard
TOP: Lab
MSC: 1984 #62
NOT: 27% answered correctly
63. ANS: D
If you want a buffer with a pH of 9, then choose a weak acid with a Ka near 10–9. H2PO4– is the best choice. The
tricky part is realizing that since it is weak, it will only lose one hydrogen ion, so the salt (conjugate base) it shou ld
be paired with is HPO42–.
DIF:
Hard
TOP: Acid-Base
MSC: 1984 #63
14
NOT: 32% answered correctly
ID: A
64. ANS: E
Nitrous acid is weak, so it does not dissociate and is written molecularly. Sodium hydroxide is a strong base and
completely dissociates, and sodium is a spectator...so you have HNO 2 + OH–, so far which leads you to answer (E).
To finish it off, you know acid + base yields salt and water, so the salt is sodium nitrite, but again...sodium ion is a
spectator and does not appear in a net ionic reaction.
DIF: Hard
TOP: Chemical Reactions
MSC: 1984 #64
NOT: 27% answered correctly
65. ANS: E
An oxidizing agent is itself reduced. Therefore, you are really being asked, “Which species cannot be reduced?”
I– cannot be reduced, all of its other oxidation states are either zero or positive which requires oxidation.
DIF: Hard
TOP: Electrochem MSC: 1984 #65
NOT: 38% answered correctly
66. ANS: D
Paramagnetic means that not all the electrons are paired. Paramagnetic substances are slightly magnetic or can be
induced to become slightly magnetic. Write the electron configurations [at least the valence part]. Think in terms
of orbital notation--sketch them if you must. It’s not about even or odd numbers of electrons!
Ca
V
Co
Zn
As
4s2 (all paired)
4s2 3d3 (3 unpaired)
4s2 3d7 (3 unpaired)
4s2 3d10 (all paired)
4s2 4p3 (3 unpaired)
DIF: Medium
TOP: Atomic Structure
MSC: 1984 #66
NOT: 42% answered correctly
67. ANS: B
Expect easy math! When given a concentration and volume, use Molarity × V to calculate the number of moles
required. (0.100 M × 2.00 L) = 0.200 moles which is 1/5th of a mole which is about 40+ grams. To make the
solution you’d mass the exact number of grams, transfer it to a volumetric flask, add DI water to dissolve and then
top off with DI water to the exact mark.
DIF: Easy
TOP: Solutions
MSC: 1984 #67
NOT: 64% answered correctly
68. ANS: B
Two starting amounts...your limiting reactant alarms should be sounding.
Expect easy math! When given a concentration and volume, use Molarity × V to calculate the number of moles
present. Don’t forget to track the total volume (20 + 30 = 50 mL).
For Ba2+ : (0.200 M × 20.0 mL) + (0.400 M × 30.0 mL) = 4.00 mmol + 12.0 mmol
So, 4.00 mmol of ppt. forms, and 8.00 mmol of excess, unreacted barium ion remains in 50 mL of solution. 8
mmol/50 mL (the milli’s cancel) which is equivalent to 16/100 and you get an answer of 0.16 M.
DIF: Medium
TOP: Stoichiometry
NOT: 48% answered correctly
MSC: 1984 #68
15
ID: A
69. ANS: C
Expect easy math!
11.0 molal translates to 11 moles of ethanol per kg of water. A kg of water is 1,000 grams, so round water’s molar
mass to 20g/mol [after all, they did say approximate!] and that’s about 50 moles. So the mole fraction of ethanol is
11
11
11
≈ . Resist the urge to scratch away at the exact answer...peruse the answer choices.
is close to
(11 + 50) 61
61
1/6th which is less than 1/5th (0.200 is 1/5), so go with answer (C).
The real answer with no approximations is 0.165.
DIF: Hard
TOP: Solutions
MSC: 1984 #69
NOT: 24% answered correctly
70. ANS: E
Sr’s valence electron configuration is 5s2. So, n = 5, = 0, thus m = 0 and ms can be either a +½ or a –½.
DIF: Medium
TOP: Atomic Structure
MSC: 1984 #70
NOT: 41% answered correctly
71. ANS: E
Sulfuric acid is a strong acid meaning it dissociates completely in water. This reaction represents the opposite of
that and is not at all likely!
DIF: Medium
TOP: Acid-Base
MSC: 1984 #71
NOT: 45% answered correctly
72. ANS: C
If you collect a gas under water, you must subtract the pressure due to the water vapor present in the “wet” gas. It
is not something you calculate or measure, you simply look it up on a table. It is temperature dependent, so you
need to measure the reaction temperature.
DIF: Medium
TOP: Lab
MSC: 1984 #72
NOT: 50% answered correctly
73. ANS: D
A classic “hydrocarbon burned...what’s the empirical formula? what’s the molecular formula?” type of problem.
Expect easy math!
moles CO2 = 88/44 = 2 moles CO2, therefore 2 moles C
moles H2O = 27/18 = 1.5 moles water, therefore 3.0 moles hydrogen
EF = C2H3 which is not an answer choice, so double, triple, quadruple, etc.
C4 H6 has the same ratio as our empirical formula.
DIF: Medium
TOP: Stoichiometry
MSC: 1984 #73
NOT: 44% answered correctly
74. ANS: D
Recognize that this question will take a lot of time which makes it a candidate for skipping! Calculate what you
need the [F–] concentration to become when [Pb2+] is equal to 1 × 10–6 molar.
Ksp = [Pb2+] [F–]2 = 4.0 × 10–8 = [1 × 10–6] [F–]2 therefore, [F–]2 = 4 × 10–2 = 0.04, take the square root and you
get [F–] = 0.20 molar which requires 0.2 moles of NaF in a 1.00 L of solution. The amount of F – already present in
the solution is negligible since the Ksp is so small to begin with.
DIF:
Hard
TOP: Equilibrium
MSC: 1984 #74
16
NOT: 20% answered correctly
ID: A
75. ANS: E
First, we need to know the [H+] at equilibrium...bring on the RICE table!
R
HA
I
0.50
C
–x
E 0.50 – x
Ka = 8 × 10 −4 =
H + + A–
0
0
+x
+x
x
x
[H + ][A − ]
x2
=
therefore, x =
0.50
[HA]
So, % dissociation =
4 × 10−4 = 2 × 10 −2
[H + ]
2 × 10/ −2
2
× 100/ =
= 4%
× 100 =
0.50
0.50
[HA]
DIF: Hard
TOP: Acid-Base
MSC: 1984 #75
NOT: 38% answered correctly
76. ANS: B
It’s an endothermic reaction, so think of heat as a reactant. If you want to increase the concentration of a product,
then lower the temperature, add a reactant that isn’t a solid, or remove a product to push the equilibrium to replace
it. A catalyst only allows equilibrium to be established faster, it has no effect on the equilibrium concentrations.
DIF: Hard
TOP: Equilibrium MSC: 1984 #76
NOT: 19% answered correctly
77. ANS: D
Optical isomerism is exhibited by molecules that have nonsuperimposable mirror images. Your hands are
nonsuperimposable mirror images. Such molecules are also described as chiral. The best way to internalize this
concept is to use a model kit to build pairs of each molecule and see if you can superimpose them or not. If not,
such as the molecule pictured in (D), then the molecule exhibits optical isomerism.
This is sort of an obscure question, I must admit! That’s why only 17% of the population answered correctly...less
than a random guess.
DIF: Hard
TOP: Bonding & Molecular Structure
MSC: 1984 #77
NOT: 17% answered correctly
78. ANS: A
An ideal gas has no molecular volume and no attractive forces between molecules. It’s collisions are perfectly
elastic [no energy lost].
DIF: Hard
TOP: Gas Laws
MSC: 1984 #78
79. ANS: C
5 Fe3+ + Mn2+ + 4 H2O
5 Fe2+ + MnO4– + 8 H+
NOT: 29% answered correctly
When in doubt, calculate the number of moles! (.05 mol/L × 14 mL) = 0.7 mmol of MnO 4–. So, 5 times that
amount is the required mmol of Fe2+. That’s 3.5 mmol, so the molarity is 3.5 mmol/100 mL (the “milli” cancels) =
0.035 M.
DIF: Hard
TOP: Stoichiometry
NOT: 29% answered correctly
MSC: 1984 #79
17
ID: A
80. ANS: B
Draw the Lewis structures! We use resonance to describe structures that have a combination of single-multiple
bonds among the same atoms. The bond order for SO 2 is 1.5.
DIF: Medium
TOP: Bonding & Molecular Structure
MSC: 1984 #80
NOT: 45% answered correctly
81. ANS: D
“Excess” stated in a net ionic equation is code for complexation! Copper sulfate is soluble, so the sulfate ion is a
spectator [since none of its “excepts” are present in this reaction]. The complex formed will have the ammonium
ion as a ligand, the oxidation number of copper is doubled to determine how many ligands [note: this doesn’t
always form the most common complex, but it does form a valid complex], so we need 4 ammonia ligands which
are neutral and the complex is [Cu(NH3)4]2+.
DIF: Hard
TOP: Chemical Reactions
MSC: 1984 #81
NOT: 29% answered correctly
82. ANS: C
In all proposed mechanisms, cross off catalysts and intermediates, write the molecularity of each step, slam on the
brakes at the slow step [so everything prior to and including the slow step is included in the rate law]. That gives
the following:
N2HO2– + H+ (fast equilibrium) so, RATE = k[N2H2O2]
Step 1
N2H2O2
Step 2
N2HO2– → N2O + OH–
(slow) so, the reactant is gone and there’s no molecularity!
Step 3
H+ + OH– → H2O (fast)
You think you’d get an overall rate law of RATE = k[N2H2O2], but you don’t! BECAUSE the first step is a fast
step AND prior to the slow step AND an equilibrium. That equilibrium forces a term in the denominator of the
[N 2 H 2 O 2 ]
mechanism (since the reaction is reversible). So, you get RATE = k
, answer C.
[H + ]
DIF:
Hard
TOP: Kinetics
MSC: 1984 #82
18
NOT: 11% answered correctly
ID: A
83. ANS: A
Peruse the answers...they all contain ΔE, so you are forced to contemplate
w = –PΔV and
ΔE = q + w = ΔH + w , solve for ΔH
ΔH = ΔE – w
The PΔV term equates to ΔnRT since the delta can only be “attached” to the number of moles...R is a constant and
the question states that temperature is held constant. Now we have:
˜ˆ
ÁÊ 11
11
RT
ΔH = ΔE – (–PΔV ) = ΔE +PΔV =ΔE +ΔnRT, so ΔE + ÁÁÁÁ − RT ˜˜˜˜ =ΔE –
2
2
¯
Ë
Note: The change in the number of moles of gas bears a negative sign since all of the moles of gas
ÊÁ 2 4 5 ˆ˜ 11
ÁÁ + + ˜˜ =
ÁÁ 2 2 2 ˜˜ 2 left the system, hence the negative sign.
¯
Ë
DIF: Hard
TOP: Thermodynamics
MSC: 1984 #83
NOT: 16% answered correctly
84. ANS: A
The substance with the highest concentration of charged particles will have the highest boiling point since the
presence of charged particles lowers water’s vapor pressure. How? Water molecules are attracted to each other,
but are even more attracted to charged particles, so more energy must be added to the system to overcome those
enhanced attractions so that water molecules vaporize.
Contemplate the van’t Hoff factor (i) for each substance since all the choices are soluble:
A)
B)
C)
D)
E)
0.10 M potassium sulfate, K2SO4; i = 3 thus the one that releases the most particles
0.10 M hydrochloric acid, HCl; i = 2
0.10 M ammonium nitrate, NH4NO3; i = 2
0.10 M magnesium sulfate, MgSO4; i = 2
0.20 M sucrose, C12H22O11; i = 1, but twice the molarity so it puts this choice level with the i = 2’s.
DIF: Hard
TOP: Solutions
MSC: 1984 #84
85. ANS: B
Expect easy math! 9 grams of Al is 1/3 mole.
NOT: 27% answered correctly
2 Al + 6 HCl → 2 AlCl3 +
3 H2
1/3 mol
therefore divide 1/3 by 2 and multiply by 3 which equal 0.5 mol or 11.2 liters at STP
DIF: Hard
TOP: Stoichiometry
NOT: 30% answered correctly
MSC: 1984 #85
19
Multiple Choice Diagnostics Guide for the 1984 AP* Chemistry Exam Place a 9in the box below the question number if you answered the question correctly.
Place a 8 in the box below the question number if you answered the question incorrectly.
If you skipped the question, simply leave the box below the question number blank.
Compare your answers to the “% of Students Answering Correctly”. If 50+% of the testing population
answered correctly, you should as well! If your skips pile up in any one section, then you need to study
that topic. Performing this analysis will help you target your areas of weakness and better structure your
remaining study time. Remember your goal is to get at least 75% of the points (63.75 points in this case).
Atomic Theory: Percent of MC Exam Questions = 5.9%
Question #
19
22
58
66
70
orrect or Incorrect
% of Students
Answering Correctly
87
80
33
42
41
Periodic Properties: Percent of MC Exam Questions = 4.7%
Question #
1
2
3
43
Correct or Incorrect
% of Students
Answering Correctly
78
68
74
48
Bonding/Intermolecular Forces: Percent of MC Exam Questions = 9.4%
Question #
8
9
18
40
41
51
60
Correct or Incorrect
% of Students
Answering Correctly
66
56
65
54
49
56
34
80
45
Nuclear: Percent of MC Exam Questions = 2.4%
Question #
30
38
Correct or Incorrect
% of Students
Answering Correctly
46
42
Gas Laws/Kinetic Theory: Percent of MC Exam Questions = 5.9%
Question #
21
23
39
50
78
Correct or Incorrect
% of Students
Answering Correctly
75
75
54
39
29
AP® is a registered trademark of the College Board. The College Board was not involved in the production of and does not endorse this product.
Copyright © 2008 by René McCormick. All rights reserved. Solutions/Phase Diagrams: Percent of MC Exam Questions = 9.4%
Question #
27
37
54
55
59
67
Correct or Incorrect
% of Students
Answering Correctly
67
54
30
39
38
64
69
84
24
27
Stoichiometry/Mole relationships: Percent of MC Exam Questions = 5.9%
Question #
44
45
52
73
85
Correct or Incorrect
% of Students
Answering Correctly
67
59
63
44
30
Kinetics: Percent of MC Exam Questions = 3.5%
Question #
25
26
82
Correct or Incorrect
% of Students
Answering Correctly
61
75
11
Equilibrium: Percent of MC Exam Questions = 4.7%
6
28
36
76
Question #
Correct or Incorrect
% of Students
Answering Correctly
55
64
62
19
Acid-Base Equilibrium & Buffer: Percent of MC Exam Questions = 9.4%
Question #
33
48
49
53
57
63
64
Correct or Incorrect
% of Students
Answering Correctly
44
34
63
55
32
32
27
75
38
Oxidation/Reduction/Electrochemistry: Percent of MC Exam Questions = 11.8%
Question #
4
14
15
16
17
20
34
46
Correct or Incorrect
% of Students
Answering Correctly
34
24
52
50
45
82
65
48
65
79
38
29
Thermodynamics: Percent of MC Exam Questions = 5.9%
Question #
29
47
56
57
83
Correct or Incorrect
% of Students
Answering Correctly
51
38
36
32
16
2 Reactions: Percent of MC Exam Questions = 8.2%
Question #
32
42
61
68
71
74
81
45
20
29
Laboratory: Percent of MC Exam Questions = 10.6%
Question #
10
11
12
13
24
31
35
62
72
46
40
27
50
Correct or Incorrect
% of Students
Answering Correctly
25
13
10
48
Organic: Percent of MC Exam Questions = 1.2%
Question #
77
Correct or Incorrect
% of Students
Answering Correctly
Correct or Incorrect
% of Students
Answering Correctly
17
58
23
53
46
68
3