Download ExamView - 1999 AP Chemistry Exam.tst

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Version A
Period: ____________
AP* Chemistry: 1999 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 1-4 refer to the following types of
energy.
2. The energy change that occurs in the conversion of
an ionic solid to widely separated gaseous ions
A) Activation energy
B) Free energy
C) Ionization energy
D) Kinetic energy
E) Lattice energy
3. The energy in a chemical or physical change that is
available to do useful work
4. The energy required to form the transition state in a
chemical reaction
1. The energy required to convert a ground-state atom
in the gas phase to a gaseous positive ion
(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.
1
Version A
Questions 5-8 refer to the following orbital notations.
A) 1s
2s
↑
B) 1s
↓↑ 2s
↓↑
C) 1s
↓↑ 2s
↓↑ 2p ↑
D) 1s
↓↑ 2s
↓↑ 2p
E) [Ar] 4s
↓↑ 3d
↓↑
↑
↓↑
↓↑
↓↑
↑
↑
↑
↑
5. Represents an atom that is chemically unreactive
Questions 13-16 refer to the following descriptions
of bonding in different types of solids.
6. Represents an atom in an excited state
A) Lattice of positive and negative ions
held together by electrostatic forces
B) Closely packed lattice with delocalized
electrons throughout
C) Strong single covalent bonds with weak
intermolecular forces
D) Strong multiple covalent bonds
(including π bonds) with weak intermolecular
forces
E) Macromolecules held together with
strong polar bonds
7. Represents an atom that has four valence electrons
8. Represents an atom of a transition metal
Question 9-12 refer to aqueous solutions containing
1:1 mole ratios of the following pairs of substances.
Assume all concentrations are 1 M.
A) NH3 and NH4Cl
B) H3PO4 and NaH2PO4
C) HCl and NaCl
D) NaOH and NH3
E) NH3 and HC2H3O2 (acetic acid)
13. Cesium chloride, CsCl(s)
14. Gold, Au(s)
15. Carbon dioxide, CO2(s)
9. The solution with the lowest pH
10. The most nearly neutral solution
16. Methane, CH4(s)
11. A buffer at a pH > 8
Questions 17-18 refer to the following elements.
A) Lithium
B) Nickel
C) Bromine
D) Uranium
E) Fluorine
12. A buffer at a pH < 6
17. Is a gas in its standard state at 298 K
18. Reacts with water to form a strong base
2
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.
19. Which of the following best describes the role of the spark from the spark plug in an automobile engine?
A)
B)
C)
D)
E)
The spark decreases the energy of activation for the slow step.
The spark increases the concentration of the volatile reactant.
The spark supplies some of the energy of activation for the combustion reaction.
The spark provides a more favorable activated complex for the combustion reaction.
The spark provides the heat of vaporization for the volatile hydrocarbon.
20. What mass of Au is produced when 0.0500 mol of Au 2S3 is reduced completely with excess H2?
A)
B)
C)
D)
E)
9.85 g
19.7 g
24.5 g
39.4 g
48.9 g
21. When a solution of sodium chloride is vaporized in a flame, the color of the flame is
A)
B)
C)
D)
E)
blue
yellow
green
violet
white
22. Of the following reaction, which involves the largest decrease in entropy?
A) CaCO3(s) → CaO(s) + CO2(g)
B) 2 CO(g) + O2(g) → 2 CO2(g)
C) Pb(NO3)3(s) + 2 KI(s) → PbI2(s) + 2 KNO3(s)
D) C3H8(g) + O2(g) → 3 CO2(g) + 4 H2O(g)
E) 4 La(s) + 3 O2(g) → 2 La2O3(s)
3
Version A
24. The safest and most effective emergency procedure
to treat an acid splash on skin is to do which of the
following immediately?
23.
A) Dry the affected area with paper towels.
B) Sprinkle the affected area with powdered
Na2SO4(s).
C) Flush the affected area with water and then
with a dilute NaOH solution.
D) Flush the affected area with water and then
with a dilute NaHCO3 solution.
E) Flush the affected area with water and then
with a dilute vinegar solution.
A hot-air balloon, shown above, rises. Which of the
following is the best explanation for this
observation?
A) The pressure on the walls of the balloon
increases with increasing temperature.
B) The difference in temperature between the air
inside and outside the balloon produces
convection currents.
C) The cooler air outside the balloon pushes in on
the walls of the balloon.
D) The rate of diffusion of cooler air is less than
that of warmer air.
E) The air density inside the balloon is less than
that of the surrounding air.
4
Version A
25.
The cooling curve for a pure substance as it changes from a liquid to a solid is shown above. The solid and the
liquid coexist at
A)
B)
C)
D)
E)
point Q only
point R only
all points on the curve between Q and S
all points on the curve between R and T
no point on the curve
26.
. . . C10H12O4S(s) + . . O2(g) → . . . CO2(g) + . . . SO2(g) + . . . H2O(g)
When the equation above is balanced and all coefficients are reduced to their lowest whole-number terms, the
coefficient for O2(g) is?
A) 6
B) 7
C) 12
D) 14
E) 28
27. Appropriate uses of a visible-light spectrophotometer include which of the following?
I. Determining the concentration of a solution of Cu(NO 3)2
II. Measuring the conductivity of a solution of KMnO 4
III. Determining which ions are present in a solution that may contain Na +, Mg2+, Al3+
A)
B)
C)
D)
E)
I only
II only
III only
I and II only
I and III only
5
Version A
28. The melting point of MgO is higher than that of NaF. Explanations for this observation include which of the
following?
I. Mg2+ is more positively charged than Na+
II. O2– is more negatively charged than F–
III. The O2– ion is smaller than the F– ion
A)
B)
C)
D)
E)
II only
I and II only
I and III only
II and III only
I, II, and III
29.
The organic compound represented above is an example of
A)
B)
C)
D)
E)
an organic acid
an alcohol
an ether
an aldehyde
a ketone
30.
H2Se(g) + 4 O2F2(g) → SeF6(g) + 2 HF(g) + 4 O2(g)
Which of the following is true regarding the reaction represented above?
A)
B)
C)
D)
E)
The oxidation number of O does not change.
The oxidation number of H changes from –1 to +1.
The oxidation number of F changes from +1 to –1.
The oxidation number of Se changes from –2 to +6.
It is a disproportionation reaction for F.
6
Version A
Questions 34-35 refer to an electrolytic cell that
involves the following half-reaction.
31. If the temperature of an aqueous solution of NaCl is
increased from 20°C to 90°C, which of the
following statements is true?
AlF63– + 3 e– → Al + 6 F–
A) The density of the solution remains unchanged.
B) The molarity of the solution remains
unchanged.
C) The molality of the solution remains
unchanged.
D) The mole fraction of solute decreases.
E) The mole fraction of solute increases.
34. Which of the following occurs in the reaction?
A) AlF 63– is reduced at the cathode.
B) Al is oxidized at the anode.
C) Aluminum is converted from the –3 oxidation
state to the 0 oxidation state.
D) F– acts as a reducing agent.
E) F– is reduced at the cathode.
32. Types of hybridization exhibited by the C atoms in
propene, CH3CHCH2, include which of the
following?
35. As steady current of 10 amperes in passed though
an aluminum-production cell for 15 minutes. Which
of the following is the correct expressions for
calculating the number of grams of aluminum
produced? (1 faraday = 96,500 coulombs)
I. sp
II. sp2
III. sp3
A)
B)
C)
D)
E)
I only
III only
I and II only
II and III only
I, II, and III
A)
33. A 1.0 L sample of an aqueous solution contains
0.10 mol of NaCl and 0.10 mol of CaCl 2. What is
the minimum number of moles of AgNO3 that must
be added to the solution in order to precipitate all of
the Cl– as AgCl(s) ? (Assume that AgCl is
insoluble.)
B)
C)
(10) (15) (60) (27)
g
ÁÊ 96,500 ˜ˆ (3)
Ë
¯
D)
0.10 mol
0.20 mol
0.30 mol
0.40 mol
0.60 mol
7
g
(27) (60)
(10) (15) (27)
g
(60) ÁÊË 96,500˜ˆ¯
E)
A)
B)
C)
D)
E)
(10) (15) ÊÁË 96,500 ˆ˜¯
ÊÁ 96,500 ˆ˜ (27)
Ë
¯
g
(10) (15) (60) (3)
(27) (3)
g
ÁÊ 96,500 ˜ˆ (10) (15) (60)
Ë
¯
Version A
36.
1
0.10
0.10
Initial Rate of
Formation of NO2
(mol L–1 s–1)
2.5 × 10–4
2
0.20
0.10
5.0 × 10–4
3
0.20
0.40
8.0 × 10–3
Experiment
Initial [NO] Initial [O2]
(mol L–1)
(mol L–1)
The initial-rate data in the table above were obtained for the reaction represented below.
2 NO + O2 → 2 NO2
What is the experimental rate law for the reaction?
A)
B)
C)
D)
E)
rate = k[NO] [O2]
rate = k[NO] [O2]2
rate = k[NO]2 [O2]
rate = k[NO]2 [O2]2
rate = k[NO] / [O2]
37.
Ionization Energies for element X (kJ mol-1)
First
Second
Third
Fourth
Five
580
1,815
2,740
11,600
14,800
The ionization energies for element X are listed in the table above. On the basis of the data, element X is most
likely to be
A)
B)
C)
D)
E)
Na
Mg
Al
Si
P
38. A molecule or an ion is classified as a Lewis acid if
it
A)
B)
C)
D)
E)
accepts a proton from water
accepts a pair of electrons to form a bond
donates a pair of electrons to form a bond
donates a proton to water
has resonance Lewis electron-dot structures
8
Version A
39.
40. Of the following molecules, which has the largest
dipole moment?
A)
B)
C)
D)
E)
CO
CO2
O2
HF
F2
41.
2 SO3 (g) p 2 SO2 (g) + O2 (g)
The phase diagram for a pure substance is shown
above. Which point on the diagram corresponds to
the equilibrium between the solid and liquid phases
at the normal melting point?
A)
B)
C)
D)
E)
After the equilibrium represented above is
established, some pure O2 (g) is injected into the
reaction vessel at constant temperature. After
equilibrium is reestablished, which of the following
has a lower value compared to its value at the
original equilibrium?
A
B
C
D
E
A)
B)
C)
D)
E)
Keq for the reaction.
The total pressure in the reaction vessel.
The amount of SO3 (g) in the reaction vessel.
The amount of O2 (g) in the reaction vessel.
The amount of SO2 (g) in the reaction vessel.
42.
. . . Li3N(s) + . . . H2O(l) → . . . Li+ (aq) + . . . OH–(aq) + . . . NH3(g)
When the equation above is balanced and all coefficients reduced to lowest whole number terms, the coefficient for
OH–(aq) is
A)
B)
C)
D)
E)
1
2
3
4
6
43. A sample of 61.8 g of H3BO3, a weak acid is dissolved in 1,000 g of water to make a 1.0-molal solution. Which of
the following would be the best procedure to determine to molarity of the solution? (Assume no additional
information is available.)
A)
B)
C)
D)
E)
Titration of the solution with standard acid
Measurement of the pH with a pH meter
Determination of the boiling point of the solution
Measurement of the total volume of the solution
Measurement of the specific heat of the solution
9
Version A
44. A rigid metal tank contains oxygen gas. Which of the following applies to the gas in the tank when additional
oxygen is added at constant temperature?
A)
B)
C)
D)
E)
The volume of the gas increase.
The pressure of the gas decreases.
The average speed of the gas molecules remains the same.
The total number of gas molecules remains the same.
The average distance between the gas molecules increases.
45. What is the H+(aq) concentration in 0.05 M HCN
(aq)? (The Ka for HCN is 5.0 × 10–10)
A)
B)
C)
D)
E)
48. If 87.5 percent of a sample of pure 131I decays in 24
days, what is the half-life of 131I?
A) 6 days
B) 8 days
C) 12 days
D) 14 days
E) 21 days
2.5 × 10–11
2.5 × 10–10
5.0 × 10–10
5.0 × 10–6
5.0 × 10–4
49. Which of the following techniques is most
appropriate for the recovery of solid KNO 3 from an
aqueous solution of KNO3?
46. Which of the following occurs when excess
concentrated NH3(aq) is mixed thoroughly with 0.1
M Cu(NO3)2(aq)?
A)
B)
C)
D)
E)
A) A dark red precipitate forms and settles out.
B) Separate layers of immiscible liquids form with
a blue layer on top.
C) The color of the solution turns from light blue
to dark blue.
D) Bubbles of ammonia gas form.
E) The pH of the solution decreases.
50. In the periodic table, as the atomic number
increases from 11 to 17, what happens to the atomic
radius?
47. When hafnium metal is heated in an atmosphere of
chlorine gas, the product of the reaction is found to
contain 62.2 percent Hf by mass and 37.4 percent
Cl by mass. What is the empirical formula for this
compound?
A)
B)
C)
D)
E)
Paper chromatography
Filtration
Titration
Electrolysis
Evaporation to dryness
A)
B)
C)
D)
E)
HfCl
HfCl2
HfCl3
HfCl4
Hf2Cl3
10
It remains constant.
It increases only.
It increases, then decreases.
It decreases only.
It decreases, then increases.
Version A
51. Which of the following is a correct interpretation of
the results of Rutherford's experiments in which
gold atoms were bombarded with alpha particles?
53.
W(g) + X(g) → Y(g) + Z(g)
Gases W and X react in a closed, rigid vessel to
form gases Y and Z according to the equation
above. The initial pressure of W(g) is 1.20 atm and
that of X(g) is 1.60 atm. No Y(g) or Z(g) is initially
present. The experiment is carried out at constant
temperature. What is the partial pressure of Z(g)
when the partial pressure of W(g) has decreased to
1.0 atm?
A) Atoms have equal numbers of positive and
negative charges.
B) Electrons in atoms are arranged in shells.
C) Neutrons are at the center of an atom.
D) Neutrons and protons in atoms have nearly
equal mass.
E) The positive charge of an atom is concentrated
in a small region.
A)
B)
C)
D)
E)
52. Under which of the following sets of conditions
could the most O2(g) be dissolved in H2O(l)?
Pressure of O2(g)
Above H2O(l)
(atm)
A)
B)
C)
D)
E)
5.0
5.0
1.0
1.0
0.5
Temperature
of H2O(l)
(°C)
0.20 atm
0.40 atm
1.0 atm
1.2 atm
1.4 atm
80
20
80
20
20
54.
ΔH < 0
2 NO(g) + O2(g) p 2 NO2(g)
Which of the following changes alone would cause a decrease in the value of Keq for the reaction represented
above?
A)
B)
C)
D)
E)
Decreasing the temperature
Increasing the temperature
Decreasing the volume of the reaction vessel
Increasing the volume of the reaction vessel
Adding a catalyst
11
Version A
55.
10 HI + 2 KMnO4 + 3 H2SO4 → 5 I2 + 2 MnSO4 + K2SO4 + 8 H2O
According to the balanced equation above, how many moles of HI would be necessary to produce 2.5 mol of I 2,
starting with 4.0 mol of KMnO4 and 3.0 mol of H2SO4?
A) 20
B) 10
C) 8.0
D) 5.0
E) 2.5
56. A yellow precipitate forms when 0.5 M NaI(aq) is added to a 0.5 M solution of which of the following ions?
A)
B)
C)
D)
E)
57.
Pb2+(aq)
Zn2+(aq)
CrO42–(aq)
SO42–(aq)
OH–(aq)
M(s) + 3 Ag+(aq) → 3 Ag(s) + M3+(aq)
E° = +2.46 V
Ag+(aq) + e– → Ag(s)
E° = +0.80 V
According to the information above, what is the standard reduction potential for the half-reaction M 3+(aq) + 3 e- →
M(s)?
A) –1.66 V
B) –0.06 V
C) 0.06 V
D) 1.66 V
E) 3.26 V
12
Version A
60.
58. On a mountaintop, it is observed that water boils at
90°C, not at 100°C as at sea level. This
phenomenon occurs because on the mountaintop
the
NH4NO3(s) → N2O(g) + 2 H2O(g)
A 0.03 mol sample of NH4NO3(s) is placed in a 1 L
evacuated flask, which is then sealed and heated.
The NH4NO3(s) decomposes completely according
to the balanced equation above. The total pressure
in the flask measured at 400 K is closest to which
of the following? ( The value of the gas constant, R,
is 0.082 L atm mol–1 K–1)
A) equilibrium water vapor pressure is higher due
to the higher atmospheric pressure
B) equilibrium water vapor pressure is lower due
to the higher atmospheric pressure
C) equilibrium water vapor pressure equals the
atmospheric pressure at a lower temperature
D) water molecules have a higher average kinetic
energy due to the lower atmospheric pressure
E) water contains a greater concentration of
dissolved gases
A)
B)
C)
D)
E)
59. A 40.0 mL sample of 0.25 M KOH is added to 60.0
mL of 0.15 M Ba(OH)2. What is the molar
concentration of OH–(aq) in the resulting solution?
(Assume that the volumes are additive.)
A)
B)
C)
D)
E)
3 atm
1 atm
0.5 atm
0.1 atm
0.03 atm
61.
C2H4(g) + 3 O2(g) → 2 CO2(g) + 2 H2O(g)
0.10 M
0.19 M
0.28 M
0.40 M
0.55 M
For the reaction of ethylene represented above, ΔH
is –1,323 kJ mol–1. What is the value of ΔH if the
combustion produced liquid water H 2O(l), rather
than water vapor H2O(g)? (ΔH for the phase
change H2O(g) → H2O(l) is –44 kJ mol–1.)
A)
B)
C)
D)
E)
–1,235 kJ/mol
–1,279 kJ/mol
–1,323 kJ/mol
–1,367 kJ/mol
–1,411 kJ/mol
62.
HC2H3O2(aq) + CN–(aq) p HCN(aq) + C2H3O2–(aq)
The reaction represented above has an equilibrium constant equal to 3.7 × 10 4. Which of the following can be
concluded from this information?
A)
B)
C)
D)
E)
CN–(aq) is a stronger base than C2H3O2–(aq).
HCN(aq) is a stronger acid than HC2H3O2(aq).
The conjugate base of CN–(aq) is C2H3O2–(aq).
The equilibrium constant will increase with an increase in temperature.
The pH of a solution containing equimolar amounts of CN –(aq) and HC2H3O2(aq) is 7.0.
13
Version A
63.
The graph above shows the results of a study of the reaction of X with a large excess of Y to yield Z. The
concentrations of X and Y were measured over a period of time. According to the results, which of the following
can be concluded about the rate law for the reaction under the conditions studied?
A)
B)
C)
D)
E)
It is zero order in [X].
It is first order in [X].
It is second order in [X].
It is the first order in [Y].
The overall order of the reaction is 2.
64. Equal numbers of moles of He(g), Ar(g), and Ne(g) are placed in a glass vessel at room temperature. If the vessel
has a pinhole-sized leak, which of the following will be true regarding the relative values of the partial pressures of
the gases remaining in the vessel after some of the gas mixture has effused?
A)
B)
C)
D)
E)
PHe < PNe < PAr
PHe < PAr < PNe
PNe < PAr < PHe
PAr < PHe < PNe
PHe = PAr = PNe
65. Which of the following compounds is NOT
appreciably soluble in water but is soluble in dilute
hydrochloric acid?
A)
B)
C)
D)
E)
Mg(OH)2(s)
(NH4)2CO3(s)
CuSO4(s)
(NH4)2SO4(s)
Sr(NO3)2(s)
14
Version A
66. When solid ammonium chloride, NH4Cl(s) is added
to water at 25 °C, it dissolves and the temperature
of the solution decreases. Which of the following is
true for the values of ΔH and ΔS for the dissolving
process?
A)
B)
C)
D)
E)
ΔH
Positive
Positive
Positive
Negative
Negative
70. When 100 mL of 1.0 M Na3PO4 is mixed with 100
mL of 1.0 M AgNO3, a yellow precipitate forms
and [Ag+] becomes negligibly small. Which of the
following is a correct listing of the ions remaining
in solution in order of increasing concentration?
A)
B)
C)
D)
E)
ΔS
Positive
Negative
Equal to zero
Positive
Negative
71. In a qualitative analysis for the presence of Pb 2+,
Fe2+, and Cu2+ ions in a aqueous solution, which of
the following will allow the separation of Pb 2+ from
the other ions at room temperature?
67. What is the molar solubility in water of Ag 2CrO4?
(The Ksp for Ag2CrO4 is 8 × 10–12.)
A) 8 × 10–12 M
B) 2 × 10–12 M
A)
B)
C)
D)
E)
(4 × 10 −12 ) M
C)
D)
3
(4 × 10 −12 ) M
E)
3
(2 × 10 −12 ) M
Adding dilute Na2S(aq) solution
Adding dilute HCl(aq) solution
Adding dilute NaOH(aq) solution
Adding dilute NH3(aq) solution
Adding dilute HNO3(aq) solution
72. After completing an experiment to determine
gravimetrically the percentage of water in a
hydrate, a student reported a value of 38 percent.
The correct value for the percentage of water in the
hydrate is 51 percent. Which of the following is the
most likely explanation for this difference?
68. In which of the following processes are covalent
bonds broken?
A) I2(s) → I2(g)
B) CO2(s) → CO2(g)
A) Strong initial heating caused some of the
hydrate sample to spatter out of the crucible.
B) The dehydrated sample absorbed moisture after
heating.
C) The amount of the hydrate sample used was too
small.
D) The crucible was not heated to constant mass
before use.
E) Excess heating caused the dehydrated sample to
decompose.
C) NaCl(s) → NaCl(l)
D) C(diamond) → C(g)
E) Fe(s) → Fe(l)
69. What is the final concentration of barium ions,
[Ba2+], in solution when 100. mL of 0.10 M
BaCl2(aq) is mixed with 100. mL of 0.050 M
H2SO4(aq)?
A)
B)
C)
D)
E)
[PO43–] < [NO3–] < [Na+]
[PO43–] < [Na+] < [NO3–]
[NO3–] < [PO43–] < [Na+]
[Na+] < [NO3–] < [PO43–]
[Na+] < [PO43–] < [NO3–]
0.00 M
0.012 M
0.025 M
0.075 M
0.10 M
15
Version A
73. The volume of distilled water that should be added
to 10.0 mL of 6.00 M HCl(aq) in order to prepare a
0.500 M HCl(aq) solution is approximately
A) 50.0 mL
B) 60.0 mL
C) 100. mL
D) 110. mL
E) 120. mL
74. Which of the following gases deviates most from
ideal behavior?
A)
B)
C)
D)
E)
SO2
Ne
CH4
N2
H2
75. Which of the following pairs of liquids forms the
solution that is most ideal (most closely follows
Raoult's law)?
A)
B)
C)
D)
E)
C8H18(l) and H2O(l)
CH3CH2CH2OH(l) and H2O(l)
CH3CH2CH2OH(l) and C8H18(l)
C6H14(l) and C8H18(l)
H2SO4(l) and H2O(l)
16
AP Chemistry
1999 Free-Response Questions
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Copyright © 2001 by College Entrance Examination Board. All rights reserved. College Board, Advanced Placement Program, AP, and the acorn logo are registered
trademarks of the College Entrance Examination Board.
1999
The College Board
Advanced Placement Examination
CHEMISTRY
SECTION II
MATERIAL IN THE FOLLOWING TABLE MAY BE USEFUL IN ANSWERING THE QUESTIONS IN THIS
SECTION OF THE EXAMINATION.
Copyright © 1999 by College Entrance Examination Board and Educational Testing Service. All rights reserved.
1999
STANDARD REDUCTION POTENTIALS IN AQUEOUS SOLUTION AT 25°C
E °( V )
Half-reaction
Li + + e -
+ eK + eRb + + e Ba 2 + + 2 e Sr 2 + + 2 e Ca 2 + + 2 e Na + + e Mg 2 + + 2 e Be 2 + + 2 e Al 3 + + 3 e Mn 2 + + 2 e Zn 2 + + 2 e Cr 3 + + 3 e Fe 2 + + 2 e Cr 3 + + e Cd 2 + + 2 e Tl + + e Co 2 + + 2 e Ni 2 + + 2 e Sn 2 + + 2 e Pb 2 + + 2 e 2 H + + 2 eS( s) + 2 H + + 2 e Sn 4 + + 2 e Cu 2 + + e Cu 2 + + 2 e Cu + + e I 2 ( s) + 2 e Fe 3 + + e Hg 2 2 + + 2 e Ag + + e Hg 2 + + 2 e 2 Hg 2 + + 2 e Br 2 (l ) + 2 e -
“
“
“
“
“
“
“
“
“
“
“
“
“
“
“
“
“
“
“
“
“
“
“
“
“
“
“
“
“
“
“
“
“
“
“
Li( s)
Cs( s)
K ( s)
Rb( s)
Ba ( s)
Sr( s)
Ca ( s)
Na ( s)
Mg( s)
Be( s)
Al( s)
Mn( s)
Zn( s)
Cr ( s)
Fe( s)
Cr 2 +
Cd( s)
Tl( s)
Co( s)
Ni( s)
Sn( s)
Pb( s)
H 2 ( g)
H 2S ( g)
Sn 2 +
Cu +
Cu( s)
Cu( s)
2 IFe 2 +
2 Hg(l )
Ag( s)
Hg(l )
Hg 2 2 +
2 Br -
-3.05
-2.92
-2.92
-2.92
-2.90
-2.89
-2.87
-2.71
-2.37
-1.70
-1.66
-118
.
-0.76
-0.74
-0.44
-0.41
-0.40
-0.34
-0.28
-0.25
-0.14
-0.13
0.00
0.14
0.15
0.15
0.34
0.52
0.53
0.77
0.79
0.80
0.85
0.92
1.07
O 2 ( g) + 4 H + + 4 e Cl 2 ( g ) + 2 e Au 3 + + 3 e Co 3 + + e F2 ( g ) + 2 e -
“
“
“
“
“
2 H 2 O(l )
2 Cl Au( s)
Co 2 +
2 F-
1.23
1.36
1.50
1.82
2.87
Cs
+
+
1999
ADVANCED PLACEMENT CHEMISTRY EQUATIONS AND CONSTANTS
ATOMIC STRUCTURE
D E = hv
c = lv
=
=
=
=
u =
n =
m =
E
v
l
p
l = mhu
p = mu
En =
−2.178 × 10 −18
joule
n2
EQUILIBRIUM
Speed of light, c = 3.0 × 10 8 m s −1
Ka =
[H + ] [A − ]
[ HA ]
Kb =
[OH − ] [ HB + ]
[ B]
Planck’s constant, h = 6.63 × 10 −34 J s
K w = [OH − ] [ H + ] = 1.0 × 10 −14 @ 25o C
= Ka × K b
pH = − log [ H + ], pOH = − log [OH − ]
14 = pH + pOH
pH = pKa + log
[A − ]
[ HA ]
[HB + ]
pOH = pKb + log
[ B]
pKa = − log Ka , pKb = − log Kb
K p = Kc ( RT )
Dn
energy
frequency
wavelength
momentum
velocity
principal quantum number
mass
,
where D n = moles product gas − moles reactant gas
Boltzmann’s constant, k = 1.38 × 10 −23 J K −1
Avogadro’s number = 6.022 × 10 23 molecules mol −1
Electron charge, e = −1.602 × 10 −19 coulomb
1 electron volt per atom = 96.5 kJ mol −1
Equilibrium Constants
Ka
Kb
Kw
Kp
(weak acid)
(weak base)
(water)
(gas pressure)
Kc (molar concentrations)
S o = standard entropy
THERMOCHEMISTRY
∑ S o products −∑ S o reactants
= ∑ DHfo products − ∑ DH fo reactants
= ∑ DGfo products − ∑ DGfo reactants
DS o =
DH o
DG o
DG o = DH o − TD S o
= − RT ln K = −2.303 RT log K
= −n Ᏺ E o
DG = DG o + RT ln Q = DG o + 2.303 RT log Q
q = mcDT
DH
Cp =
DT
H o = standard enthalpy
G o = standard free energy
Eo
T
n
m
q
c
Cp
=
=
=
=
=
=
=
standard reduction potential
temperature
moles
mass
heat
specific heat capacity
molar heat capacity at constant pressure
1 faraday Ᏺ = 96,500 coulombs
1999
GASES, LIQUIDS, AND SOLUTIONS
PV = nRT
P + n a (V - nb) = nRT
V 2
2
PA = Ptotal ™ X A , where X A =
moles A
total moles
Ptotal = PA + PB + PC + . . .
m
n =
M
K = o C + 273
P1V1
PV
= 2 2
T1
T2
m
D =
V
3kT
3 RT
urms =
=
m
M
1
KE per molecule = mu 2
2
3
KE per mole =
RT
2
r1
M2
=
r2
M1
molarity, M = moles solute per liter solution
molality = moles solute per kilogram solvent
DT f = iK f ™ molality
DTb = iKb ™ molality
p = nRT i
V
=
=
=
=
=
=
u =
P
V
T
n
D
m
pressure
volume
temperature
number of moles
density
mass
velocity
urms
KE
r
M
p
i
Kf
=
=
=
=
=
=
=
root-mean-square speed
kinetic energy
rate of effusion
molar mass
osmotic pressure
van’t Hoff factor
molal freezing-point depression constant
Kb
Q
I
q
t
=
=
=
=
=
molal boiling-point elevation constant
reaction quotient
current (amperes)
charge (coulombs)
time (seconds)
E o = standard reduction potential
K = equilibrium constant
Gas constant, R = 8.31 J mol −1 K −1
= 0.0821 L atm mol −1 K −1
= 8.31 volt coulomb mol −1 K −1
OXIDATION-REDUCTION; ELECTROCHEMISTRY
Boltzmann’s constant, k = 1.38 × 10 −23 J K −1
K f for H 2 O = 1.86 K kg mol −1
Q =
[C ] [ D ]
c
d
a
b
, where a A + b B → c C + d D
[ A ] [ B]
q
I =
t
o − RT ln Q = E o − 0.0592 log Q @ 25o C
E cell = E cell
cell
nᏲ
n
log K =
nE o
0.0592
Kb for H 2 O = 0.512 K kg mol −1
STP = 0.000 o C and 1.000 atm
Faraday’s constant, Ᏺ = 96,500 coulombs per mole
of electrons
1999
CHEMISTRY
SECTION II
(Total time—90 minutes)
Part A
Time—40 minutes
YOU MAY USE YOUR CALCULATOR FOR PART A.
CLEARLY SHOW THE METHOD USED AND STEPS INVOLVED IN ARRIVING AT YOUR ANSWERS. It is
to your advantage to do this, because you may earn partial credit if you do and you will receive little or no credit if
you do not. Attention should be paid to significant figures. Be sure to write all your answers to the questions on the
lined pages following each question in this booklet.
Answer Question 1 below. The Section II score weighting for this question is 20 percent.
→ NH +(aq) + OH -(aq)
NH3(aq) + H2O(l) ←
4
1. In aqueous solution, ammonia reacts as represented above. In 0.0180 M NH3(aq) at 25°C, the hydroxide ion
concentration, [OH -] , is 5.60 ¥ 10-4 M. In answering the following, assume that temperature is constant at
25°C and that volumes are additive.
(a) Write the equilibrium-constant expression for the reaction represented above.
(b) Determine the pH of 0.0180 M NH3(aq).
(c) Determine the value of the base ionization constant, Kb , for NH3(aq).
(d) Determine the percent ionization of NH3 in 0.0180 M NH3(aq).
(e) In an experiment, a 20.0 mL sample of 0.0180 M NH3(aq) was placed in a flask and titrated to the
equivalence point and beyond using 0.0120 M HCl(aq).
(i) Determine the volume of 0.0120 M HCl(aq) that was added to reach the equivalence point.
(ii) Determine the pH of the solution in the flask after a total of 15.0 mL of 0.0120 M HCl(aq) was added.
(iii) Determine the pH of the solution in the flask after a total of 40.0 mL of 0.0120 M HCl(aq) was added.
1999
Answer EITHER Question 2 below OR Question 3 printed on the next page. Only one of these two questions will be
graded. If you start both questions, be sure to cross out the question you do not want graded. The Section II score
weighting for the question you choose is 20 percent.
2. Answer the following questions regarding light and its interactions with molecules, atoms, and ions.
(a) The longest wavelength of light with enough energy to break the Cl–Cl bond in Cl2(g) is 495 nm.
(i) Calculate the frequency, in s−1, of the light.
(ii) Calculate the energy, in J, of a photon of the light.
(iii) Calculate the minimum energy, in kJ mol−1, of the Cl–Cl bond.
(b) A certain line in the spectrum of atomic hydrogen is associated with the electronic transition in the H atom
from the sixth energy level (n = 6) to the second energy level (n = 2).
(i) Indicate whether the H atom emits energy or whether it absorbs energy during the transition. Justify
your answer.
(ii) Calculate the wavelength, in nm, of the radiation associated with the spectral line.
(iii) Account for the observation that the amount of energy associated with the same electronic transition
(n = 6 to n = 2) in the He+ ion is greater than that associated with the corresponding transition in
the H atom.
1999
2 NO(g) + Br2(g) → 2 NOBr(g)
3. A rate study of the reaction represented above was conducted at 25°C. The data that were obtained are shown in
the table below.
Initial Rate of
Appearance of
NOBr
(mol L-1 s-1)
Initial [Br2]
Experiment
Initial [NO]
(mol L−1)
1
0.0160
0.0120
3.24 ¥ 10- 4
2
0.0160
0.0240
6.38 ¥ 10- 4
3
0.0320
0.0060
6.42 ¥ 10- 4
(mol L−1)
(a) Calculate the initial rate of disappearance of Br2(g) in experiment 1.
(b) Determine the order of the reaction with respect to each reactant, Br2(g) and NO(g). In each case,
explain your reasoning.
(c) For the reaction,
(i) write the rate law that is consistent with the data, and
(ii) calculate the value of the specific rate constant, k, and specify units.
(d) The following mechanism was proposed for the reaction:
Br2(g) + NO(g) → NOBr2(g)
slow
NOBr2(g) + NO(g) → 2 NOBr(g)
fast
Is this mechanism consistent with the given experimental observations? Justify your answer.
STOP
DO NOT GO ON TO PART B UNTIL YOU ARE TOLD TO DO SO.
1999
CHEMISTRY
Part B
Time—50 minutes
NO CALCULATORS MAY BE USED WITH PART B.
Answer Question 4 below. The Section II score weighting for this question is 15 percent.
4. Write the formulas to show the reactants and the products for any FIVE of the laboratory situations described
below. Answers to more than five choices will not be graded. In all cases a reaction occurs. Assume that solutions are aqueous unless otherwise indicated. Represent substances in solution as ions if the substances are
extensively ionized. Omit formulas for any ions or molecules that are unchanged by the reaction. You need not
balance the equations.
Example: A strip of magnesium is added to a solution of silver nitrate.
(a) Calcium oxide powder is added to distilled water.
(b) Solid ammonium nitrate is heated to temperatures above 300°C.
(c) Liquid bromine is shaken with a 0.5 M sodium iodide solution.
(d) Solid lead(II) carbonate is added to a 0.5 M sulfuric acid solution.
(e) A mixture of powdered iron(III) oxide and powdered aluminum metal is heated strongly.
(f) Methylamine gas is bubbled into distilled water.
(g) Carbon dioxide gas is passed over hot, solid sodium oxide.
(h) A 0.2 M barium nitrate solution is added to an alkaline 0.2 M potassium chromate solution.
1999
Your responses to the rest of the questions in this part of the examination will be graded on the basis of the accuracy
and relevance of the information cited. Explanations should be clear and well organized. Examples and equations
may be included in your responses where appropriate. Specific answers are preferable to broad, diffuse responses.
Answer BOTH Question 5 below AND Question 6 printed on the next page. Both of these questions will be graded.
The Section II score weighting for these questions is 30 percent (15 percent each).
Collected Gas
Unknown Gas
Water
5. A student performs an experiment to determine the molar mass of an unknown gas. A small amount of the pure
gas is released from a pressurized container and collected in a graduated tube over water at room temperature, as
shown in the diagram above. The collection tube containing the gas is allowed to stand for several minutes, and
its depth is adjusted until the water levels inside and outside the tube are the same. Assume that:
• the gas is not appreciably soluble in water
• the gas collected in the graduated tube and the water are in thermal equilibrium
• a barometer, a thermometer, an analytical balance, and a table of the equilibrium vapor pressure of water
at various temperatures are also available.
(a) Write the equation(s) needed to calculate the molar mass of the gas.
(b) List the measurements that must be made in order to calculate the molar mass of the gas.
(c) Explain the purpose of equalizing the water levels inside and outside the gas collection tube.
(d) The student determines the molar mass of the gas to be 64 g mol -1. Write the expression (set-up)
for calculating the percent error in the experimental value, assuming that the unknown gas is butane
(molar mass 58 g mol−1). Calculations are not required.
(e) If the student fails to use information from the table of the equilibrium vapor pressures of water in the
calculation, the calculated value for the molar mass of the unknown gas will be smaller than the actual value.
Explain.
1999
6. Answer the following questions in terms of thermodynamic principles and concepts of kinetic molecular theory.
(a) Consider the reaction represented below, which is spontaneous at 298 K.
CO2(g) + 2 NH3(g) → CO(NH2)2(s) + H2O(l)
D
D H298
= -134 kJ
D
, is positive, or negative, or zero.
(i) For the reaction, indicate whether the standard entropy change, DS298
Justify your answer.
D
D
(ii) Which factor, the change in enthalpy, D H298
, or the change in entropy, DS298
, provides the principal
driving force for the reaction at 298 K? Explain.
(iii) For the reaction, how is the value of the standard free energy change, DG D , affected by an increase in
temperature? Explain.
(b) Some reactions that are predicted by their sign of DG D to be spontaneous at room temperature do not
proceed at a measurable rate at room temperature.
(i) Account for this apparent contradiction.
(ii) A suitable catalyst increases the rate of such a reaction. What effect does the catalyst have on DG D
for the reaction? Explain.
1999
Answer EITHER Question 7 below OR Question 8 below. Only one of these two questions will be graded. If you
start both questions, be sure to cross out the question you do not want graded. The Section II score weighting for the
question you choose is 15 percent.
0.10 M
NaF
0.10 M
MgCl2
0.10 M
C2H5OH
0.10 M
CH3COOH
7. Answer the following questions, which refer to the 100 mL samples of aqueous solutions at 25°C in the
stoppered flasks shown above.
(a) Which solution has the lowest electrical conductivity? Explain.
(b) Which solution has the lowest freezing point? Explain.
(c) Above which solution is the pressure of water vapor greatest? Explain.
(d) Which solution has the highest pH? Explain.
8. Answer the following questions using principles of chemical bonding and molecular structure.
(a) Consider the carbon dioxide molecule, CO2 , and the carbonate ion, CO32−.
(i) Draw the complete Lewis electron-dot structure for each species.
(ii) Account for the fact that the carbon-oxygen bond length in CO32- is greater than the carbon-oxygen
bond length in CO2 .
(b) Consider the molecules CF4 and SF4 .
(i) Draw the complete Lewis electron-dot structure for each molecule.
(ii) In terms of molecular geometry, account for the fact that the CF4 molecule is nonpolar, whereas the
SF4 molecule is polar.
END OF EXAMINATION
ID: A
AP* Chemistry: 1999 Released Multiple Choice Exam
Answer Section
OTHER
1. ANS:
C
This is the definition for ionization energy. Be sure and state “gas phase” in the free response if you are asked
about ionization energy.
DIF: Easy
2. ANS:
E
TOP: Periodicity
MSC: 1999 #1
NOT: 74% answered correctly
This energy overcomes the electrostatic attractions between closely spaced oppositely charged ions. You can also
link this to melting points, boiling points, etc.
DIF: Medium
3. ANS:
B
TOP: Periodicity
MSC: 1999 #2
NOT: 48% answered correctly
The classic definition for Gibbs’ free energy, ΔG.
DIF: Medium
TOP: Thermodynamics
NOT: 42% answered correctly
4. ANS:
A
MSC: 1999 #3
The classic definition for activation energy. Remember that it can be decreased by the addition of a catalyst.
Why? The catalyst provides a surface for more effective collisions.
DIF: Easy
TOP: Thermodynamics
NOT: 71% answered correctly
5. ANS:
D
MSC: 1999 #4
The classic ns2np6 noble gas configuration.
DIF: Easy
TOP: Atomic Structure
NOT: 75% answered correctly
MSC: 1999 #5
1
ID: A
6. ANS:
A
If only one electron is present, the ground state orbital notation should correspond to 1 s1. This atom is excited since
its electron resides in a higher energy level.
DIF: Easy
TOP: Atomic Structure
NOT: 78% answered correctly
7. ANS:
C
MSC: 1999 #6
Remember valence sublevels have the same (and largest) value of the principle quantum number, n. So, the orbital
notation showing 4 electrons with n = 2 is:
2s ↓↑ 2p ↑
↑
DIF: Medium
TOP: Atomic Structure
NOT: 55% answered correctly
8. ANS:
E
MSC: 1999 #7
By definition, the ground state electron configuration of a transition metal contains d-electrons. Note that this
element will also be paramagnetic since it contains some unpaired d-electrons and it will form colored aqueous
solutions.
DIF: Easy
TOP: Atomic Structure
MSC: 1999 #8
NOT: 83% answered correctly
9. ANS:
C
Lowest pH means most acidic. This is the only pair that contains a strong acid.
NH3 and NH4Cl → weak base + soluble salt containing conjugate acid ∴ a basic buffer is formed
H3PO4 and NaH2PO4 → weak acid + soluble salt containing conjugate base ∴ an acidic buffer is formed
HCl and NaCl → strong acid + neutral soluble salt ∴ no buffer is formed, but the result is very acidic (low pH)
NaOH and NH3→ strong base + weak base ∴ no buffer is formed, but the result is very basic (high pH)
NH3 and HC2H3O2 (acetic acid) → weak base + weak acid ∴ a buffer is formed since the salts of both the base and
the acid are in solution
DIF:
Medium
TOP: Acid-Base
MSC: 1999 #9
2
NOT: 52% answered correctly
ID: A
10. ANS:
E
Since the Ka and Kb of acetic acid and ammonia are both 1.8 × 10–5, this solution is closest to neutral.
NH3 and NH4Cl → weak base + soluble salt containing conjugate acid ∴ a basic buffer is formed
H3PO4 and NaH2PO4 → weak acid + soluble salt containing conjugate base ∴ an acidic buffer is formed
HCl and NaCl → strong acid + neutral soluble salt ∴ no buffer is formed, but the result is very acidic (low pH)
NaOH and NH3→ strong base + weak base ∴ no buffer is formed, but the result is very basic (high pH)
NH3 and HC2H3O2 (acetic acid) → weak base + weak acid ∴ a buffer is formed since the salts of both the base and
the acid are in solution
DIF: Medium
11. ANS:
A
MSC: 1999 #10
NOT: 41% answered correctly
Looking for a basic buffer, so look for a solution containing a weak base and its salt.
NH3 and NH4Cl → weak base + soluble salt containing conjugate acid ∴ a basic buffer is formed
H3PO4 and NaH2PO4 → weak acid + soluble salt containing conjugate base ∴ an acidic buffer is formed
HCl and NaCl → strong acid + neutral soluble salt ∴ no buffer is formed, but the result is very acidic (low pH)
NaOH and NH3→ strong base + weak base ∴ no buffer is formed, but the result is very basic (high pH)
NH3 and HC2H3O2 (acetic acid) → weak base + weak acid ∴ a buffer is formed since the salts of both the base and
the acid are in solution
DIF:
Hard
MSC: 1999 #11
NOT: 33% answered correctly
3
ID: A
12. ANS:
B
Looking for an acidic buffer, so look for a solution containing a weak acid and its salt.
NH3 and NH4Cl → weak base + soluble salt containing conjugate acid ∴ a basic buffer is formed
H3PO4 and NaH2PO4 → weak acid + soluble salt containing conjugate base ∴ an acidic buffer is formed
HCl and NaCl → strong acid + neutral soluble salt ∴ no buffer is formed, but the result is very acidic (low pH)
NaOH and NH3→ strong base + weak base ∴ no buffer is formed, but the result is very basic (high pH)
NH3 and HC2H3O2 (acetic acid) → weak base + weak acid ∴ a buffer is formed since the salts of both the base and
the acid are in solution
DIF: Hard
13. ANS:
A
MSC: 1999 #12
NOT: 35% answered correctly
A) Implies a solid of ionic character, so look for combinations of metal and nonmetal (from opposite sides of the
periodic table--high melting points)
B) Implies a solid metal since “delocalized electrons” is a major hammer hitting you over the head--and the
hammer is made of METAL
C) Implies a solid of covalent character, so look for combinations of nonmetal with nonmetal (near each other on
the periodic table--low melting points)
D) Implies a solid combination from the elements CNOPS (pronounce “seenops”, rhymes with cyclops) since
those are the only elements capable of forming multiple covalent bonds (including π bonds).
E) Implies a huge molecule like a polymer, protein, starch, etc.
DIF: Medium
TOP: Bonding & Molecular Structure
NOT: 60% answered correctly
14. ANS:
B
MSC: 1999 #13
A) Implies a solid of ionic character, so look for combinations of metal and nonmetal (from opposite sides of the
periodic table--high melting points)
B) Implies a solid metal since “delocalized electrons” is a major hammer hitting you over the head--and the
hammer is made of METAL
C) Implies a solid of covalent character, so look for combinations of nonmetal with nonmetal (near each other on
the periodic table--low melting points)
D) Implies a solid combination from the elements CNOPS (pronounce “seenops”, rhymes with cyclops) since
those are the only elements capable of forming multiple covalent bonds (including π bonds).
E) Implies a huge molecule like a polymer, protein, starch, etc.
DIF: Medium
TOP: Bonding & Molecular Structure
NOT: 64% answered correctly
4
MSC: 1999 #14
ID: A
15. ANS:
D
A) Implies a solid of ionic character, so look for combinations of metal and nonmetal (from opposite sides of the
periodic table--high melting points)
B) Implies a solid metal since “delocalized electrons” is a major hammer hitting you over the head--and the
hammer is made of METAL
C) Implies a solid of covalent character, so look for combinations of nonmetal with nonmetal (near each other on
the periodic table--low melting points)
D) Implies a solid combination from the elements CNOPS (pronounce “seenops”, rhymes with cyclops) since
those are the only elements capable of forming multiple covalent bonds (including π bonds).
E) Implies a huge molecule like a polymer, protein, starch, etc.
DIF: Medium
TOP: Bonding & Molecular Structure
NOT: 65% answered correctly
16. ANS:
C
MSC: 1999 #15
A) Implies a solid of ionic character, so look for combinations of metal and nonmetal (from opposite sides of the
periodic table--high melting points)
B) Implies a solid metal since “delocalized electrons” is a major hammer hitting you over the head--and the
hammer is made of METAL
C) Implies a solid of covalent character, so look for combinations of nonmetal with nonmetal (near each other on
the periodic table--low melting points)
D) Implies a solid combination from the elements CNOPS (pronounce “seenops”, rhymes with cyclops) since
those are the only elements capable of forming multiple covalent bonds (including π bonds).
E) Implies a huge molecule like a polymer, protein, starch, etc.
DIF: Medium
TOP: Bonding & Molecular Structure
NOT: 61% answered correctly
17. ANS:
E
MSC: 1999 #16
Fluorine gas is a yellow irritating toxic flammable gas and a powerful oxidizing agent.
Fluorine is the smallest of the halogens, therefore it has the least number of electrons and is least polarizable. That
means it has the weakest IMFs among its family which is why fluorine and chlorine (also small) are gases, bromine
is a liquid and iodine is a gas at room temperature.
DIF: Easy
18. ANS:
A
TOP: Periodicity
MSC: 1999 #17
NOT: 70% answered correctly
The IA and IIA (with some solubility issues--Be and Mg are not considered very soluble) metal hydroxides and
oxides form strong bases in solution.
DIF:
Easy
TOP: Periodicity
MSC: 1999 #18
5
NOT: 67% answered correctly
ID: A
MULTIPLE CHOICE
19. ANS: C
The combustion of octane [primary component of gasoline] is thermodynamically spontaneous. The spark supplies
energy that contributes to the acquisition of the activation energy for the reaction. Temperature also plays a role
since it factors into molecular motion. The spark is an electric spark.
The spark is not a catalyst and cannot decrease the activation energy.
The spark does not vaporize, thus cannot increase the concentration of the volatile reactant.
The spark is not a reactant and thus cannot combine with other reactants to form an activated complex.
The spark does not provide the heat of vaporization for the volatile hydrocarbon. Heat of vaporization relates to
evaporation, not combustion.
DIF: Easy
TOP: Kinetics
MSC: 1999 #19
NOT: 73% answered correctly
20. ANS: B
You were only given one starting amount (and it says excess hydrogen), so you have not entered the “land of
limiting reactant”. Whew! You were given the number of moles of reactant, but must calculate the mass of the
product which will require that you supply a balanced equation.
Au2S3 + H2 → Au + HxSy
If 0.0500 moles of Au2S3 completely reacts (not a limiting reactant), then 0.100 moles of Au was formed. The MM
of Au is 197, so 1/10 mole × 197 g/mol = 19.7 g
DIF: Medium
TOP: Stoichiometry
MSC: 1999 #20
NOT: 55% answered correctly
21. ANS: B
It is the sodium that is making the color in the flame. When the sodium electrons in the are excited, they jump to
higher energy levels. As the electrons fall back down, and leave the excited state, energy is re-emitted, the
wavelength of which refers to the discrete lines of the emission spectrum. The only way to see the entire spectrum
is to view the flame test through a diffraction grating. Otherwise, you get a general, qualitative result as shown
below.
Go to: http://en.wikipedia.org/wiki/Flame_test for a chart of which metals produce which colors.
DIF:
Hard
TOP: Lab
MSC: 1999 #21
6
NOT: 33% answered correctly
ID: A
22. ANS: E
In other words, which reaction system becomes more ordered as the reaction proceeds?
Look for formation of the more condensed states (preferably solids). The reaction in (C) forms three moles of solid
from three moles of solid while the reaction in E forms only two moles of solid from four moles of solid and three
moles of gas, bringing much more order to the system.
DIF: Easy
TOP: Thermodynamics
MSC: 1999 #22
NOT: 68% answered correctly
23. ANS: E
A helium balloon floats because it contains a gas that is lighter than the surrounding air. An air-filled balloon
floats because the hot air inside the balloon is less dense than the surrounding air. As the air is warmed, the
distances between gas particles increase causing the gas to be less dense.
Remember, heat does not rise! Heat travels from hot to cold, not “up”. Saying “heat” rises is a misnomer. What
we should say is that warmed air expands and is less dense [again, due to increased distance between particles],
thus it floats and carries thermal energy with it.
DIF: Easy
TOP: Gas Laws
MSC: 1999 #23
NOT: 67% answered correctly
24. ANS: D
Obviously, flushing the area is the first step. Since acid is the splash culprit, it would be best to follow with a weak
base wash as well to neutralize any residual acid on the skin.
DIF: Hard
TOP: Lab
MSC: 1999 #24
NOT: 20% answered correctly
25. ANS: C
P to Q shows the liquid cooling to a supercooled state. Most likely the liquid was not stirred during the cooling.
Once it is stirred, or a seed crystal is added, the temperature rises as the heat of crystallization is released (points Q
to R) and the freezing point is reached. Note that the temperature (KEave) remains constant throughout the freezing
process from points R to S.
So, solid must be present from points Q to S.
DIF: Medium
TOP: States of Matter
NOT: 57% answered correctly
26. ANS: C
The correctly balanced equation is:
MSC: 1999 #25
10 C10H12O4S(s) + 12 O2(g) → 10 CO2(g) + SO2(g) + 6 H2O(g)
Note that when compounds containing nonmetals are combusted, that oxides of the nonmetals form.
DIF: Easy
TOP: Chemical Reactions
MSC: 1999 #26
NOT: 74% answered correctly
27. ANS: A
A visible-light spectrophotometer is used for measuring absorbance or transmittance of light in a colored solution
at a certain wavelength. I. Solutions containing ions of transition metals with unpaired d-electrons are colored.
Those containing Cu2+ are blue. II. We use a conductivity apparatus or conductivity probe to measure the
conductivity of a solution. III. The ions listed are not transition metals with unpaired d-electrons, thus solutions
containing these ions are colorless.
DIF:
Hard
TOP: Lab
MSC: 1999 #27
7
NOT: 16% answered correctly
ID: A
28. ANS: B
When asked about melting points, (or FP, BP, vapor pressure, etc.) your thought process should run something like
this: Is this compound covalent (nonmetal-nonmetal) or ionic (metal-nonmetal)? If it is covalent, think IMFs and
consider polarizability of the molecule due to how many overall electrons are present. The more electrons present,
the larger the electron cloud, the more polarizable it is.
If it is ionic, think in terms of Coulomb’s Law and do the math!
(+2) (−2)
(+1) (−1)
and for NaF, F =
2
d
d
d2
The “players” in these molecules are neighbors, so their ionic radii d is about the same. It’s the difference in
charges that greatly increase the attractive Coulombic-force between the ions in the crystal. You can clearly see
that the Coulombic-force attraction ions in the MgO crystal is about 4 times greater than those attracting ions in the
NaF crystal, so MgO will have a much higher melting point.
F=
q1q2
2
; For MgO, F =
DIF: Medium
TOP: Bonding & Molecular Structure
MSC: 1999 #28
NOT: 53% answered correctly
29. ANS: E
Occasionally, you get to celebrate your memorization skills. All of these answer choices are basic examples of
organic functional groups. The “R” can be simple like an H atom or a more complex group. If two R groups
differ, we use R and R´ to designate that they are different. The compound given contains a ketone functional
group and two different R groups, the one on the left is a methyl and the one on the right is an ethyl.
Here are the others:
an organic acid
DIF:
Hard
an alcohol
TOP: Descriptive
an ether
MSC: 1999 #29
8
an aldehyde
NOT: 26% answered correctly
ID: A
30. ANS: D
The oxidation number for each atom is given above its symbol in the equation below:
+1,–2
+1 ,–1
+6,–1
+1,–1
0
H2Se(g) + 4 O2F2(g) → SeF6(g) + 2 HF(g) + 4 O2(g)
The oxidation number of O does change from +1 [very weird for oxygen] to 0.
The oxidation number of hydrogen does NOT change since neither a metal hydride nor H 2 is present.
The oxidation number of F does not change, it is –1 in all compounds given.
A disproportionation reaction is one in which the same element is BOTH oxidized and reduced which is not the
case here. Se is oxidized while O is reduced.
DIF: Easy
TOP: Chemical Reactions
MSC: 1999 #30
NOT: 75% answered correctly
31. ANS: C
An increase in temperature causes an expansion of solution volume. So, any term that contains V as a variable is
going to change. So, the only terms that are temperature independent are molality and mole fraction since they
depend on mass alone rather than mass and volume.
m
# mol
# mol
moles A
D=
M=
m=
χ=
V
V sol' n
kg solvent
moles total
DIF: Hard
TOP: Solutions
MSC: 1999 #31
NOT: 21% answered correctly
32. ANS: D
Draw the dang Lewis structure! Count the number of “sites” around each carbon. The carbon pictured left has 4
sites, so it is sp3 hybridized (4 sites, 4 letters, think: sppp), the two remaining carbons each have 3 sites of electron
density, so they are both sp2 hybridized (3 sites, 3 letters, think: spp). NEVER use the “letter” simplification ,
spp¸when answering a free-response--condense it to sp2!
DIF: Hard
TOP: Bonding & Molecular Structure
NOT: 38% answered correctly
33. ANS: C
Expect easy math! How much chloride ion is already present?
MSC: 1999 #32
0.10 mol of NaCl yields 0.10 mol Cl –
0.10 mol of CaCl2 yields 0.20 mol Cl–
For a total of 0.30 moles of Cl–, so you need the same amount of Ag + since there is one silver ion for each chloride
ion.
DIF: Medium
TOP: Stoichiometry
NOT: 54% answered correctly
MSC: 1999 #33
9
ID: A
34. ANS: A
Aluminum is being reduced from a +3 state to the elemental state where its oxidation state is zero. Reduction does
occur at the cathode (RED CAT). Oxidation occurs at the anode (AN OX), but aluminum is being reduced. F – is
being neither oxidized nor reduced, so it is neither an oxidizing agent nor a reducing agent.
DIF: Medium
TOP: Electrochemistry
MSC: 1999 #34
NOT: 42% answered correctly
35. ANS: C
Begin by calculating the number of coulombs produced: #coulombs = It; where time is in seconds. Also remember
an ampere is equal to 1c/s.
ÊÁ coulombs ˆ˜ ÊÁ
ˆ
˜˜ ÁÁ 15min× 60s ˜˜˜
#coulombs = It = ÁÁÁ 10
Á
˜
˜Á
ÁË
s
min ˜˜¯
¯Ë
Then, # coulombs can be converted to moles of electrons, moles of Al and finally grams of Al using the following:
ˆ˜ ÁÊ
ÁÊÁ 10 × 15 × 60
˜ˆ˜ ÊÁÁÁ
˜˜ ÁÁ 1 mol Al ˜ˆ˜˜ ÊÁÁÁ 27 g ˆ˜˜˜
1 mol e −
ÁÁ
˜˜ × ÁÁ
coulombs
Á
˜ ÁÁ 96,500 coulombs ˜˜˜˜ ÁÁ 3 mol e − ˜˜ ÁÁÁ 1 mol Al ˜˜˜
1
Ë
¯Ë
¯ Ë
¯
¯Ë
DIF: Medium
TOP: Electrochemistry
MSC: 1999 #35
NOT: 43% answered correctly
36. ANS: B
Experiments 1 and 2 hold [O2] constant and double [NO], the rate doubles, therefore the reaction is 1st order in
NO. Experiments 2 and 3 hold [NO] constant and quadruple [O 2], the rate increases by a factor of 16 which is
[quad or 4]2, therefore the reaction is 2nd order in O 2.
The rate law expression is rate = k[NO] [O2]2
DIF: Medium
TOP: Kinetics
MSC: 1999 #36
NOT: 52% answered correctly
37. ANS: C
There is a HUGE jump in ionization energy from the 3rd to the 4th. Such an increase indicates that the core has
been disrupted meaning the entire valence was removed at the 3rd ionization, so this element most likely makes a
+3 ion and is Al.
DIF: Hard
TOP: Periodicity
MSC: 1999 #37
NOT: 35% answered correctly
38. ANS: B
Lewis acids accept (appreciate the alliteration “acids accept”) an electron pair and form a coordinate covalent
bond. The classic example is ammonia reacting with boron trifluoride.
DIF:
Hard
TOP: Acid-Base
MSC: 1999 #38
10
NOT: 38% answered correctly
ID: A
39. ANS: C
The “normal” melting point (or boiling point or any other phase change) occurs at 1 atm or 760 mm Hg. Also
remember that each of the lines represents an equilibrium line where both phases exist. The usual phase labels
were left off to test if you knew them!
DIF: Easy
TOP: States of Matter
MSC: 1999 #39
NOT: 71% answered correctly
40. ANS: D
You should know that oxygen and fluorine, the diatomic molecules have zero dipole moment since they are linear
and cancel each other out. Draw the Lewis structures of the others and contemplate their dipole moments:
has a dipole moment as shown dipole moments cancel
has the greatest dipole moment since F
is much more electroneg ative
DIF: Medium
TOP: Bonding & Molecular Structure
MSC: 1999 #40
NOT: 54% answered correctly
41. ANS: E
Adding a reactant or a product causes a shift away from the addition. Adding O2 to the vessel causes more reactant
to form (shift left, shift toward reactant). Keq for the reaction would not change since it is the ratio of products to
reactants and is a constant at constant temperature. The total pressure will not change since 3 moles of gas is
shifted to 2 moles of gas.
DIF: Medium
TOP: Equilibrium MSC: 1999 #41
NOT: 65% answered correctly
42. ANS: C
Not a redox equation, although it looks like it could be. Nothing is neither oxidized nor reduced. The balanced
equation is:
Li3N(s) + 3 H2O(l) → Li+ (aq) + 3 OH–(aq) + NH3(g)
DIF: Easy
TOP: Chemical Reactions
NOT: 71% answered correctly
MSC: 1999 #42
11
ID: A
43. ANS: D
# mol
M=
V sol' n
m=
# mol
kg solvent
the only variable you are missing is the volume of the solution.
DIF: Medium
TOP: Solutions
MSC: 1999 #43
NOT: 60% answered correctly
44. ANS: C
Rigid is code for constant volume. The pressure would increase. Since the temperature is held constant the average
KE of the molecules remains the same and since their mass remains constant (only oxygen in the tank), their
velocity also remains constant. If you added oxygen to the tank you increased the number of molecules within the
tank, so the average distance between the molecules decreases (more crowded).
DIF: Medium
TOP: Gas Laws
45. ANS: D
Expect easy math!
HCN p H+ + CN–
Ka =
MSC: 1999 #44
NOT: 54% answered correctly
[H + ][CN − ]
x2
=
= 5.0 × 10 −10
[HCN]
0.05
∴ x 2 = 0.25 × 10 −10 ∴ x = [H + ] =
0.25 × 10−10 = 0.5 × 10 −5 = 5 × 10 −6
DIF: Medium
TOP: Acid-Base
MSC: 1999 #45
NOT: 40% answered correctly
46. ANS: C
If you have not performed this reaction or seen it as a demonstration, you are at a disadvantage for this question.
Here’s the reaction:
Cu2+(aq) + 4 NH3(aq) → Cu(NH3)42+(aq)
The tetraamminecopper(II) complex ion is a deep blue in solution and is shown on the right compared to
copper(II) nitrate on the left in the picture below.
DIF:
Hard
TOP: Lab
MSC: 1999 #46
12
NOT: 22% answered correctly
ID: A
47. ANS: C
Expect easy math and estimate!
62.2g
≈ 3 moles Hf
g
178.5
mol
and
37.4 g
≈ 1 mole Cl, so HfCl3
g
35.5
mol
DIF: Medium
TOP: Stoichiometry
MSC: 1999 #47
NOT: 50% answered correctly
48. ANS: B
Start with 100% and realize that if 87.5% decays, then 12.5% remains!
1
2
3
100 ⎯⎯
→ 50 ⎯⎯
→ 25 ⎯⎯
→ 12.5, so it takes 3 half-lives to decay to 12.5%, so each half-life is 24/3 or 8 days.
DIF: Medium
TOP: Nuclear
MSC: 1999 #48
NOT: 42% answered correctly
49. ANS: E
Paper chromatography works nicely to separate dyes or pigments in solution. Filtration is used to separate a solid
precipitate from a solution. Titration is used to determine the concentration of an known quantity unknown acid or
base with a known quantity of acid or base of known concentration. Electrolysis is using an electric current to
separate metals from molten ores or related compounds, convert water into hydrogen and oxygen gas, or plate
metals. Since KNO3 is such a soluble salt, evaporating away the water is the most appropriate.
DIF: Medium
TOP: Lab
MSC: 1999 #49
NOT: 53% answered correctly
50. ANS: D
The trend is that atomic radii generally decrease as you go across the periodic table since the effective nuclear
charge increases. When answering a question like this in the free-response section be sure to talk about why.
Discuss effective nuclear charge and attractive forces, just stating the trend will earn you no points since it is an
observation rather than an explanation!
DIF:
Medium
TOP: Periodicity
MSC: 1999 #50
13
NOT: 62% answered correctly
ID: A
51. ANS: E
Hans Geiger (as in the counter) and Earnest Marsden actually performed the experiments since they were graduate
students of Earnest Rutherford at the time. (Not a typo--Earnest was a popular name.) He made them repeat it
several times since he couldn’t believe the results. Rutherford said of the alpha particles that made a direct hit and
bounced straight back, “It was almost as incredible as if you fired a 15-inch shell at a piece of tissue paper
and it came back and hit you.”
Also, here’s a great animation with explanation:
http://www.mhhe.com/physsci/chemistry/essentialchemistry/flash/ruther14.swf
DIF: Medium
TOP: Atomic Structure
MSC: 1999 #51
NOT: 59% answered correctly
52. ANS: B
Gases, especially oxygen, are more soluble in water at high pressures and low temperatures. Think about
aquariums, if they get too warm, the fish suffocate and die. The highest pressure listed is 5.0 atm and the lowest
temperature is 20°C.
DIF: Hard
TOP: Solutions
MSC: 1999 #52
NOT: 31% answered correctly
53. ANS: A
A quick RICE table will help (even though this is not shown as an equilibrium):
R
I
C
E
W(g) + X(g) → Y(g)
1.20 atm 1.60 atm
0
–x
–x
+x
1.00 atm
+
Z(g)
0
+x
So, –x must equal –0.20 atm, thus +x must equal +0.20 atm
DIF: Medium
TOP: Stoichiometry
NOT: 46% answered correctly
MSC: 1999 #53
14
ID: A
54. ANS: B
ΔH < 0; thus it is a negative value, therefore the reaction is exothermic, so think of the reaction as:
2 NO(g) + O2(g) p 2 NO2(g) + heat
Thus, adding heat (increasing the temperature) shifts the reaction to the left (reactants favored) decreasing the
value of Keq.
DIF: Hard
TOP: Equilibrium MSC: 1999 #54
NOT: 35% answered correctly
55. ANS: D
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 three reactants. Determine the limiting reagent. Start by using I 2 since it is the
smallest starting amount coupled with the highest coefficient.
If 5 = 2.5 mol, then 10 = 5.0 mol. Calculate subsequent moles ( if 10 = 5 mol, then 1 = 1/2 mol, so 2 = 1.0 mol
and 3 = 1.5 mol) from that limiting amount of moles using the mole:mole.
10 HI
2 KMnO4
3 H2SO4
? mol
5.0 mol
4.0 mol
1.0 mol, so
plenty is
available
3.0 mol
1.5 mol,
so plenty
is
available
→
5 I2
2 MnSO4
K2SO4
8 H2 O
2.5 mol
limit!
DIF: Easy
TOP: Stoichiometry
MSC: 1999 #55
NOT: 69% answered correctly
56. ANS: A
A solubility rule moment. Salts containing Group I elements are soluble (Li +, Na+, K+, Cs+, Rb+), so the I– ion is
the precipitation culprit! That realization eliminates the negative ions from the possible answer choices, leaving
PbI2 or ZnI2 as the only possible choices. Salts containing Cl –, Br–, I– are generally soluble. Important exceptions
to this rule are halide salts of Ag+, Pb2+, and (Hg2)2+. PbI2 is indeed yellow!
DIF:
Hard
TOP: Lab
MSC: 1999 #56
15
NOT: 32% answered correctly
ID: A
57. ANS: A
The balanced redox reaction is given first, that means the cell potential is E°cell = +2.46 V. Also, it is clear that
silver is reduced and its reduction potential is also given. M is therefore oxidized, so solve for the oxidation
half-reaction AND DON’T FORGET TO CHANGE ITS SIGN since you are asked for M’s reduction potential.
E°cell = E°reduction + E°oxidation
The values are given, but you had to figure out the situation! E°cell = +2.46 V and E°reduction = + 0.80 V
∴ E°oxidation = E°cell – E°reduction = (+2.46 V) – (+0.80 V) = +1.66 V so M’s reduction potential is –1.66 V
DIF: Medium
TOP: Electrochemistry
MSC: 1999 #57
NOT: 44% answered correctly
58. ANS: C
On mountaintops, the atmospheric pressure is less. The “pull” of gravity originates at the center of earth and acts
on the gases in the atmosphere. The further away from the source of our gravitational pull, the less tightly the
atmospheric gases are held, the lower the density of air, thus the lower the atmospheric pressure. Water boils when
its vapor pressure is equal to atmospheric pressure. If the atmospheric pressure is less, less thermal energy must be
added to a sample of water to raise its VP to a (lesser) atmospheric pressure. Since the water is actually boiling at
a lower temperature, food must be boiled longer to compensate!
DIF: Hard
TOP: States of Matter
MSC: 1999 #58
NOT: 39% answered correctly
59. ANS: C
Expect easy math! Calculate the total number of moles of OH– and simply divide by the total volume.
moles
M=
∴ moles = M × L
V liters
(0.04 L)(0.25 M) = 0.01 moles of OH– from KOH
(0.06 L)(0.15 M)(2 OH– ions per mole Ba(OH)2 = 0.018 moles of OH– from Ba(OH)2
Total volume = 100 mL or 0.10 L
moles (0.028 mol)
∴M=
=
= 0.28 M
V liters
0.10 L
DIF: Hard
TOP: Solutions
MSC: 1999 #59
NOT: 28% answered correctly
60. ANS: A
Expect easy math! Use 8/100 for R and since the answer choices are so far apart...estimate!
NH4NO3(s) → N2O(g)
+
2 H2O(g)
0.03 mol solid ∴ → (0.03 mol
PV = nRT ∴ P =
DIF:
Hard
nRT
=
V
(0.09)(
+
0.06 mol) gas, so there is a total of 0.09 mol of gas in the 1 L flask
8
)(400)
100
1
= (0.09)(32) ≈
(32) ≈ 3atm
1
10
TOP: Gas Laws
MSC: 1999 #60
16
NOT: 22% answered correctly
ID: A
61. ANS: E
Realize this is a combustion reaction and heat energy is lost. Vaporization is an endothermic process and requires
energy. That means some of the energy lost by the combustion system is subsequently reused by the system to
vaporize water if H2O(g) is formed. This lowers the total energy yield.
As a result, more total energy is lost from the combustion reaction that produces liquid water compared to gaseous
water. So, expect that this reaction would yield 2(–44 kJ mol –1) more energy if liquid water is formed rather than
water vapor. Both answers D and E yield more energy, but D didn’t factor in that two moles of water are involved.
DIF: Hard
TOP: Thermochemistry
MSC: 1999 #61
NOT: 31% answered correctly
62. ANS: A
Write out the equilibrium expression and think about that tiny value for Ka.
È
˘
[HCN] ÍÍÎ C 2 H 3 O 2 − ˙˙˚
Ka = È
= 3.7 × 10 −4 which is much less than one, so the denominator is a larger term, therefore
ÍÍ HC H O ˘˙˙ ÍÈÍÍ CN − ˙˘˙˙
2 3
2˚Í
Î
˙˚
Î
reactants are favored, meaning acetic acid is a stronger acid than hydrocyanic acid AND cyanide ion is a stronger
base than acetate ion. Answer (E) is only true if the acid base mixture is of equimolar solutions of a strong acid
and strong base are mixed.
DIF: Hard
TOP: Acid-Base
MSC: 1999 #62
NOT: 31% answered correctly
63. ANS: B
It’s a concentration vs. time graph which means you must deal with integrated rate law. Ah, note that one of the
reactants Y, has a massive concentration compared to X. That is called “swamping” and is a technique used to
determine the order of the nonswamped reactant. IF the graph for reactant X were a straight line with a negative
slope, then we’d know it was zero order. Alas, the X line is a curve, so it is not zero order, it is either first or
second order, but at least answer (A) is eliminated.
What if the data were in a table?
[X]
from graph
0.20
0.10
0.05
[Y]
from graph
4.0
4.0
4.0
We know the rate is changing since the X
line is a curve, so the rate is dependent upon
decreasing [X] and independent of the
concentration of Y¸thus the order of Y is
zero--too bad that’s not an answer choice!
But at least answer (D) is eliminated.
Recall that rate = k[X]m[Y]n, and we just determined Y is zero order, so its exponent is zero and its value in the rate
law expression is 1! So, our rate law expression simplifies to rate = k[X]m. Thus, if (C) were correct, then (E)
would also be correct and that can’t be, so these choices are eliminated. Answer (B) is all that is left.
DIF:
Hard
TOP: Kinetics
MSC: 1999 #63
17
NOT: 34% answered correctly
ID: A
64. ANS: A
Light gas molecules move faster than heavy ones. So, He moved out quickest, with Ne in second place and Ar
dragging up the rear. That left more Ar in the container to exert a greater pressure (more molecules colliding with
the container) with Ne being in the middle and He having the fewest molecules remaining, thus exerting the least
pressure. So, PHe is less than PNe which is less than PAr.
DIF: Medium
TOP: Gas Laws
MSC: 1999 #64
NOT: 65% answered correctly
65. ANS: A
Do the not soluble in water part first: scratch the ammonium compounds as well as the nitrate since all of both are
always soluble in water. That leaves magnesium hydroxide and copper sulfate. Most sulfate salts are soluble.
Important exceptions to this rule include BaSO 4, PbSO4, Ag2SO4 and SrSO4, so that leaves only magnesium
hydroxide insoluble in water. No need to stress over the “soluble in dilute acid” part. It was a nice clue, that may
have led you straight to the only hydroxide.
It is important to note that hydroxides of IA and IIA metals are generally strong bases, but the “little IIA guys”, Be
and Mg are not very soluble, so they are often classified as weak and written undissociated on the net ionic
reaction question.
DIF: Hard
TOP: Chemical Reactions
MSC: 1999 #65
NOT: 35% answered correctly
66. ANS: A
If a solid dissolves, chaos abounds, thus disorder increases (+ΔS). If the temperature of the system decreases, heat
from the surroundings is being added to the system (+ΔH).
DIF: Hard
TOP: Thermodynamics
NOT: 31% answered correctly
67. ANS: E
Ag2CrO4 R 2 Ag+ + CrO42–
MSC: 1999 #66
The molar solubility is the concentration in moles/L of the salt that dissolves so it is equal to x and equal to
[CrO42–].
Therefore, Ksp = [Ag+]2[CrO42– ] = [2x]2[x] = 4x3 = 8 × 10–12
∴ x3 =
∴x=
3
8 × 10 −12
= 2 × 10−12
4
2 × 10 −12 M
= 4 × 10–12
DIF:
Hard
TOP: Equilibrium
MSC: 1999 #67
18
NOT: 25% answered correctly
ID: A
68. ANS: D
I2(s) → I2(g)--phase change; energy added to overcome dispersion IMFs in molecular solid
CO2(s) → CO2(g)--phase change; solid carbon dioxide is not a network covalent solid nor an ionic crystal, energy
added to overcome dispersion IMFs in a molecular solid
NaCl(s) → NaCl(l)--phase change; energy added to overcome lattice energy and melt ionic crystal
C(diamond) → C(g)--phase change; BUT diamond and graphite are network covalent solids, meaning four
covalent bonds exist between the carbons of diamond, that’s why it is so hard!
Fe(s) → Fe(l)--phase change; energy added to overcome attractions between charged iron ions and delocalized
electrons
DIF: Medium
TOP: States of Matter
MSC: 1999 #68
NOT: 54% answered correctly
69. ANS: C
Expect easy math!
Net ionic reaction: Ba2+ + SO42– → BaSO4
Calculate the total number of moles of Ba2+ and simply divide by the total volume.
moles
M=
∴ moles = M × L
V liters
(0.10 L)(0.10 M) = 0.01 moles of Ba2+ from BaCl2
(0.10 L)(0.05 M) = 0.005 moles of SO42– from H2SO4 which limits the amount of precipitate formed!
So, there is 0.005 unreacted Ba2+ in solution.
Total volume = 200 mL or 0.20 L ∴ M =
moles (0.005mol)
=
= 0.025M
V liters
0.200L
DIF: Hard
TOP: Solutions
MSC: 1999 #69
NOT: 31% answered correctly
70. ANS: A
It’s all about understanding your solubility rules! Who does Ag precipitate with? Phosphate, so it’s concentration
will be the smallest listed...only answers A & B are reasonable. Equal volumes and equal molarities mean equal
number of moles (moles = M × Vliters), so Na outnumbers nitrate by 3 to 1, it should be in greatest concentration
and listed last. The correct increasing concentration order is phosphate ion, nitrate ion, and sodium ion.
DIF: Hard
TOP: Chemical Reactions
NOT: 34% answered correctly
MSC: 1999 #70
19
ID: A
71. ANS: B
There is value in memorizing and applying your solubility rules as well as understanding the question. Separation
will be achieved by precipitate formation accompanied by subsequent filtering. In other words, you are trying to
cull lead ion from the herd of ions given. Pb 2+ ion is not soluble with halides (while halides of other metals are
soluble) so, go for the halide.
You could make a ppt. of PbS2, but that wouldn’t separate it from the others since iron(II) and copper(II) sulfides
would also form.
You could make a ppt. of Pb(OH)2, but that wouldn’t separate it from the others either since iron(II) and copper(II)
hydroxides would also form.
Lead(II) ion doesn’t even precipitate with ammonia.
Lead(II) ion would form lead(II) nitrate with nitric acid (the oxidizing acid), which is soluble.
DIF: Hard
TOP: Lab
MSC: 1989 #71
NOT: 32% answered correctly
72. ANS: B
Your initial thoughts should be:
1) Either the sample was still “wet”, not enough water was driven out OR
2) The mass of the hydrate was recorded as too large to begin with OR
3) There was a lot of humidity in the laboratory and the solid reabsorbed water.
Initially spattering would carry away the hydrate implying its waters of hydration are still intact, so it isn’t just
water that splattered out. The amount of the hydrate sample used is irrelevant. Not heating the crucible to constant
mass before use leaves water in the pores of the porcelain. The water trapped would be counted as waters of
hydration and make the percent reported too high, rather than too low. It is not likely that excess heating would
cause the dehydrated sample to decompose since the anhydrous compound is most likely ionic with an extremely
high melting point and even higher amounts of heat would be needed for decomposition to occur.
DIF: Hard
TOP: Lab
MSC: 1999 #72
NOT: 25% answered correctly
73. ANS: D
Two molarities, both HCl...dilution! M1V1 = M2V2.
(6.00 M ) (10.0mL)
1
V2 =
= 60 ÷ (or 60 × 2)
0.500 M
2
which is 120 mL of TOTAL volume, so add 50.0 mL since 10.0 mL was already in the container. You had to
suspect something when “added” was underlined!
DIF: Hard
TOP: Solutions
MSC: 1989 #73
NOT: 33% answered correctly
74. ANS: A
Deviations occur due to molecular volume (larger molecules have more mass as well) and attractive forces. The
more electrons present, the more polarizable a molecule, thus the greater the London dispersion (induced
dipole-induced dipole) attractive forces become. SO 2 has a higher molecular mass, more electrons and is more
polarizable than the other answer choices.
DIF:
Medium
TOP: Gas Laws
MSC: 1999 #74
20
NOT: 48% answered correctly
ID: A
75. ANS: D
It’s all about IMFs. Deviations from ideal behavior occur when IMFs enter the scenario. So, you are looking for
nonpolar-nonpolar combinations.
A) C8H18(l) and H2O(l)--polar solvent means lots of attractions even with octane
B) CH3CH2CH2OH(l) and H2O(l)--even more attractions with a polar alcohol
C) CH3CH2CH2OH(l) and C8H18(l)--better, while the alcohol is polar, octane isn’t
D) C6H14(l) and C8H18(l)--best, both very nonpolar
E) H2SO4(l) and H2O(l)--polar solvent with a very strong acid--not ideal at all!
DIF:
Hard
TOP: Solutions
MSC: 1999 #75
21
NOT: 33% answered correctly
AP® Chemistry
1999 Scoring Guidelines
The materials included in these files are intended for non-commercial use by AP
teachers for course and exam preparation; permission for any other use must be
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whole or in part, in limited quantities, for face-to-face teaching purposes but may
not mass distribute the materials, electronically or otherwise. These materials and
any copies made of them may not be resold, and the copyright notices must be
retained as they appear here. This permission does not apply to any third-party
copyrights contained herein.
These materials were produced by Educational Testing Service (ETS), which develops and administers the examinations of the Advanced Placement
Program for the College Board. The College Board and Educational Testing Service (ETS) are dedicated to the principle of equal opportunity, and their
programs, services, and employment policies are guided by that principle.
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are registered trademarks of the College Entrance Examination Board.
AP® CHEMISTRY
1999 SCORING GUIDELINES
Question 1
9 Points
One point deduction for mathematical error (maximum once per question)
One point deduction for error in significant figures* (maximum once per
question)
*number of significant figures must be correct within +/- one digit.
(except for pH: +/- two digits)
+
[ NH 4 ][OH − ]
[NH 3 ]
(a)
K=
(b)
[OH−]
(c)
Kb =
= 5.60 ×
10−4
1 pt
Þ {
(5.60 × 10 −4 ) 2
0.0180 − 5.60 × 10
−4
pOH = 3.252
or
} Þ pH = 10.748
[H+] = 1.79 × 10−11
= 1.74 × 10 −5 (or 1.80 × 10 −5 )
1 pt
2 pts
Note: 1st point for [NH4+] = [OH−] = 5.60 × 10−4; 2nd point for correct answer
5.60 × 10 −4
× 100 % = 3.11% (or 0.0311)
0.0180
(d)
% ionization =
(e)
NH3 + H+ → NH4+
(i)
mol NH3 = 0.0180 M × 0.0200 L = 3.60 × 10−4 mol = mol H+ needed
3.60 × 10 −4 mol
= 0.0300 L = 30.0 mL
vol HCl solution =
0.0120 M
(ii)
1 pt
1 pt
mol H+ added = mol 0.0120 M × 0.0150 L = 1.80 × 10−4 mol H+ added
= 1.80 × 10−4 mol NH4+ produced
[NH4+] =
1.80 × 10 −4 mol
= 0.00514 M = [NH3]
0.0350 L
1 pt
Note: Point earned for 1.80 × 10−4 mol, or 0.00514 M [NH3] or [NH4+],
or statement “halfway to equivalence point”.
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are registered trademarks of the College Entrance Examination Board.
2
AP® CHEMISTRY
1999 SCORING GUIDELINES
Question 1 (cont.)
Kb = 1.80 ×
10−5
+
[ NH 4 ][OH − ]
=
= [OH − ] Þ pOH = 4.745 Þ pH = 9.255
[ NH 3 ]
(= 1.74 × 10−5)
(iii)
(= 4.759)
1 pt
(= 9.241)
10.0 mL past equivalence point
0.0100 L × 0.0120 M = 1.20 × 10−4 mol H+ in 60.0 mL
[H + ] =
0.000120 mol
= 0.00200 M
0.0600 L
pH = − log (2.00 × 10−3) = 2.700
1 pt
______________________________________________________________________________________
One point deduction for mathematical error (maximum once per question)
One point deduction for error in significant figures* (maximum once per question)
*number of significant figures must be correct within +/− one digit (except for pH: +/− two digits)
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are registered trademarks of the College Entrance Examination Board.
3
AP® CHEMISTRY
1999 SCORING GUIDELINES
Question 2
9 points
One point deduction for mathematical error (maximum once per question)
One point deduction for error in significant figures* (maximum once per
question)
*number of significant figures must be correct within +/- one digit.
(a) (i) ν =
c 3.00 × 1017 nm/sec
3.00 × 10 8 m/sec
=
( or , =
) = 6.06 × 1014 sec −1
−
9
495 nm
λ
495 × 10 m
1 pt
(ii)
E = hν = (6.626 × 10−34 J sec)(6.06 × 1014 sec−1) = 4.02 × 10−19 J
1 pt
(iii)
(4.02 × 10−19 J)(6.022 × 1023 mol− 1)(0.00100 kJ/J) = 242 kJ/mol
1 pt
Note: No units required if answers are numerically same as above. No penalty
if answers are correct with different units and units are explicitly indicated
(e.g., for part (ii), 4.02 × 10−22 kJ is acceptable)
(b) (i) Energy is emitted.
1 pt
The n = 6 state is at a higher energy than the n = 2 state. Going from a high energy
state to a low energy state means that energy must be emitted.
1 pt
Note: The key idea is that the energy of the n = 6 state is higher (more excited) than
the lower (less excited) n = 2 state. The argument that “e− is farther away” at
the n = 6 level is not accepted.
(ii) E2 =
− 2.178 × 10−18 J
2
= −5.45 × 10−19 J ,
E6 =
− 2.178 × 10−18 J
2
= −6.05 × 10− 20 J
6
2
−20
−19
∆E = E6 − E2 = − 6.05 × 10 J − (−5.45 × 10 J) = 4.84 × 10−19 J
OR,
∆E = 2.178 × 10−18 (
1 pt
1
1
J = 2.178 × 10−18 (0.2222) J = 4.84 × 10−19 J
2 − 2 )
2
6
Note: Point earned for determining the energy of transition.
Negative energies acceptable.
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are registered trademarks of the College Entrance Examination Board.
4
AP® CHEMISTRY
1999 SCORING GUIDELINES
Question 2 (cont.)
E =
hc
λ
Þ
λ =
hc
E
OR,
1 pt
ν =
−19
4.84 × 10
J
E
=
= 7.30 × 1014 sec −1
− 34
h 6.626 × 10
J sec
Note: Point earned for writing or using E =
λ=
hc
, or for calculating the frequency, ν
λ
c 3.00 × 10 8 m sec −1 1 nm
×
= 411 nm
=
ν
7.30 × 1014 sec −1 10 −9 m
OR,
1 pt
λ=
(6.626 × 10
−34
17
J sec)(3.00 × 10
4.84 × 10
−19
−1
nm sec )
J
= 411 nm
Note: Point earned for correct wavelength; λ = 4.11 x 10−7 m accepted.
Negative wavelength not accepted.
(iii) The positive charge holding the electron is greater for He+, which has a 2+ nucleus,
than for H with its 1+ nucleus. The stronger attraction means that it requires more energy1 pt
for the electron to move to higher energy levels. Therefore, transitions from high energy
states to lower states will be more energetic for He+ than for H.
Note: Other arguments accepted, such as, “E is proportional to Z 2. Since Z = 2 for He+
and Z = 1 for H, all energy levels in He + are raised (by a factor of 4).” Other accepted
answers must refer to the increased charge on the He+ nucleus, and NOT the mass.
______________________________________________________________________________________
One point deduction for mathematical error (maximum once per question)
One point deduction for error in significant figures* (maximum once per question)
*number of significant figures must be correct within +/− one digit
Copyright © 2001 by College Entrance Examination Board. All rights reserved. College Board, Advanced Placement Program, AP, and the acorn logo
are registered trademarks of the College Entrance Examination Board.
5
AP® CHEMISTRY
1999 SCORING GUIDELINES
Question 3
9 points:
One point deduction for mathematical error (maximum once per question)
One point deduction for error in significant figures* (maximum once per
question)
*number of significant figures must be correct within +/- one digit.
(a)
Rate of Br2(g) loss occurs at
½
the rate of NOBr(g) formation.
−4
3.24 × 10 M
= 1.62 × 10−4 M sec−1 (or mol L−1 sec−1) 1 pt
2 sec
Note: No penalty for missing units; ignore + or − signs
Rate of Br2(g) loss =
(b)
Comparing experiments 1 and 2, [NO] remains constant, [Br2] doubles, and rate doubles;
therefore, rate ∝ [Br2]1 Þ reaction is first-order with respect to [Br2].
1 pt
6.38 × 10 −4
6.42 × 10 − 4
1 æ1ö
=ç ÷
4 è 2ø
k[ NO] x [Br2 ]
k[0.0160] x [0.0240]
x
æ1ö
≅1 =
=
= ç ÷ 4 =1
x
x
è2ø
k[ NO] [Br2 ]
k[0.0320] [0.0060]
Þ
x
Þ
x = 2 Þ reaction is second-order with respect to [NO]
2 pts.
Note: One point earned for a proper set-up, comparing experiments 2 and 3 (as is shown
here)
or experiments 1 and 3. Second point earned for solving the ratios correctly and
determining
that the exponent = 2. Also, credit can be earned for a non-mathematical approach (e.g.,
one point for describing the change in [Br2] and subsequent effect on rate, another point for
describing the change in [NO] and subsequent effect on rate).
(c)
(i)
Rate = k[NO]2[Br2]
1 pt
Note: Point earned for an expression that is not inconsistent with the answer in part
(b)
(ii) k =
Rate
2
[ NO] [ Br2 ]
=
3.24 × 10 −4 M sec −1
2
(0.0160) (0.0120) M
(Using rate of Br2(g) loss = 1.62 ×
correct.)
10−4
3
M
= 105 M −2 sec−1 (or 105 L2 mol−2 sec−1)
sec−1
Þ k = 52.7 M
−2
sec−1
2 pts.
is also
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6
AP® CHEMISTRY
1999 SCORING GUIDELINES
Question 3 (cont.)
Note: One point for solving for the value of the rate constant consistent with the
rate-law expression found in part (b) or stated in part (c); one point for the correct
units consistent with the rate-law expression found in part (b) or stated in part
(c).
(d)
No, it is not consistent with the given experimental observations.
1 pt
This mechanism gives a reaction that is first-order in [NO], and first-order in [Br2], as
those are the two reactants in the rate-determining step. Kinetic data show the reaction
is second-order in [NO] (and first-order in [Br2]), so this cannot be the mechanism.
1 pt
Note: One point earned for “No” [or for “Yes” if rate = k[NO][Br2] in part (b)].
One point earned for justifying why this mechanism is inconsistent with
the observed rate-law [or consistent with rate law stated earlier in response].
______________________________________________________________________________
One point deduction for mathematical error (maximum once per question)
One point deduction for error in significant figures* (maximum once per question)
*number of significant figures must be correct within +/− one digit
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are registered trademarks of the College Entrance Examination Board.
7
AP® CHEMISTRY
1999 SCORING GUIDELINES
Question 4
15 points:
Students choose five of the eight reactions. Only the answers in the boxes are graded
(unless clearly marked otherwise). Each correct answer earns 3 points, 1 point for
reactants and 2 points for products. All products must be correct to earn both product
points. Equations do not need to be balanced and phases need not be indicated. Any
spectator ions on the reactant side nullify the 1 possible reactant point, but if they appear
again on the product side, there is no product-point penalty. A fully molecular equation
(when it should be ionic) earns a maximum of one point. Ion charges must be correct.
(a) CaO + H2O → Ca(OH)2
•
(b)
No penalty for the set of products { Ca2+, OH−, and Ca(OH)2 }
NH4NO3 → N2 + O2 + H2O
OR
NH4NO3 → N2O + H2O
•
•
Two points earned for NH4NO2 → N2 + H2O
No penalty for other oxides of nitrogen (e.g., NO, NO2, N2O3, N2O4 – but not N2O5)
(c)
Br2 + I− → Br− + I2
(d)
PbCO3 + H+ + HSO4− (or SO42−) → PbSO4 + CO2 + H2O (or HCO3− )
•
•
(e)
•
No reactant point earned for H2SO4
No product point earned for H2CO3
Fe2O3 + Al → Al2O3 + Fe
No penalty for the set of products { FeO, Fe, and Al2O3 }
(f) CH3NH2 + H2O → CH3NH3+ + OH−
•
Two points earned for MeNH2 + H2O → MeNH3+ + OH−
(g)
CO2 + Na2O → Na2CO3
(h)
Ba2+ + CrO42− → BaCrO4
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are registered trademarks of the College Entrance Examination Board.
8
AP® CHEMISTRY
1999 SCORING GUIDELINES
Question 5
8 Points
(a)
PV = nRT AND n =
m
,
M
OR molar mass =
mRT
,
PV
OR
M=
DRT
P
(b) temperature, atmospheric pressure, volume of the gas, and mass of gas (mass of
pressurized container before and after releasing the gas)
1 pt
3 pts
Note: 1 point earned for any two of the above, 2 points earned for any three of them.
“The mass of the gas” is acceptable as a “measurement” for the 1st or 2nd point.
Extraneous measurements (e.g., density, volume of liquid, etc.) are ignored. To
earn 3rd point, “mass of pressurized container before and after releasing the gas”,
or “change in mass of container” must be indicated.
(c)
to equalize internal pressure with room pressure (atmospheric pressure), or the pressure(s) 1 pt
will be the same.
(d)
% error =
(64 − 58) g
× 100 %
58 g
(or
6
6
× 100%, or
)
58
58
Note: No points earned for generic response (e.g.,
1 pt
| (expt. − theor.) |
× 100 ),
theor.
6
× 100 % . No penalty if “× 100%” is absent or if value (10%)
64
is not calculated.
or for
(e)
Pressure will be larger, therefore number of moles will be larger
mass
, therefore calculated molar mass will be smaller
molar mass =
moles
DRT
mRT
(or M =
), and the denominator, PV, will be too large.
PV
P
DRT
mRT
Therefore, the value of the molar mass ( =
or
) will be too small.
PV
P
OR, M =
1 pt
1 pt
2 pts
OR, The pressure is larger, or the number of moles is larger, or since Ptotal = (Punknown − Pwater) 1 pt
we know that Ptotal > Punknown .
m
is missing in part (a) but present in part (e), 1 point is earned for part (a).
Note: If n =
M
8 points:
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9
AP® CHEMISTRY
1999 SCORING GUIDELINES
Question 65 (cont.)
(a)
(i)
∆S° is negative (−) OR ∆S° < 0 OR entropy is decreasing.
1 pt
1 pt
3 moles of gaseous particles are converted to 2 moles of solid/liquid.
•
•
•
One point earned for correct identification of (−) sign of ∆S°
One point earned for correct explanation (mention of phases is crucial for point)
No point earned if incorrect ∆S° sign is obtained from the presumed value of ∆G°
(ii)
∆H° drives the reaction.
1 pt
The decrease in entropy (∆S° < 0) cannot drive the reaction, so the decrease in enthalpy
(∆H° < 0) MUST drive the reaction.
OR
1 pt
∆G° = ∆H° − T∆S°; for a spontaneous reaction ∆G° < 0, and a negative value of ∆S°
positive ∆G°.
•
•
•
causes a
One point earned for identifying ∆H° as the principal driving force for the reaction
One point earned for correct justification
Justification point earned by mentioning the effects of changes in entropy and enthalpy on the
spontaneity of the reaction OR by a mathematical argument using the Gibbs-Helmholtz equation and
some implication about the comparison between the effects of ∆S° and ∆H°
(iii) Given that ∆G° = ∆H° − T∆S° and ∆S° < 0, an increase in temperature causes an increase in
the value of ∆G° (∆G° becomes less negative).
•
•
1 pt
One point earned for the description of the effect of an increase in temperature on ∆S° and
consequently on ∆G°
No point earned for an argument based on Le Châtelier’s principle
(b) (i)The reaction rate depends on the reaction kinetics, which is determined by the value of the activation
energy, Eact. If the activation energy is large, a reaction that is thermodynamically spontaneous may
1 pt
proceed very slowly (if at all).
•
One point earned for linking the rate of the reaction to the activation energy, which may be
explained verbally or shown on a reaction profile diagram
(ii)
The catalyst has no effect on the value of ∆G°.
The catalyst reduces the value of Eact, increasing the rate of reaction, but has no effect on
∆H° and ∆S°, so it cannot affect the thermodynamics of the reaction.
•
•
1 pt
the values of
One point earned for indicating no change in the value of ∆G°
One point earned for indicating (verbally, or with a reaction-profile diagram) that the
catalyst affects the activation energy
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are registered trademarks of the College Entrance Examination Board.
10
1 pt
AP® CHEMISTRY
1999 SCORING GUIDELINES
Question 7
8 points:
a. 1 point
C2H5OH (Flask #3)
1 point
Ethanol, a nonelectrolyte, does not break up or dissociate in
solution.
•
One point earned for identifying C2H5OH
•
One point earned for the correct explanation.
•
Explanation point earned for a description of a
nonelectrolyte (e.g., something that does not break up or
does not dissociate.)
•
No point earned for describing C2H5OH as organic, or as
the compound that contains the most hydrogens.
b. 1 point
MgCl2 (Flask #2)
1 point
The freezing-point depression is proportional to the concentration
of solute particles. All solutes are at the same concentration, but
the van't Hoff factor (i) is largest for MgCl2.
•
One point earned for identifying MgCl2 .
•
One point earned for the correct explanation.
c. 1 point
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are registered trademarks of the College Entrance Examination Board.
11
AP® CHEMISTRY
1999 SCORING GUIDELINES
Question 7 (cont.)
C2H5OH (Flask #3)
1 point
The lowering of vapor pressure of water is directly proportional to
the concentration of solute particles in solution. C2H5OH is the
only nonelectrolyte, so it will have the fewest solute particles in
solution.
•
One point earned for identifying C2H5OH .
•
One point earned for the correct explanation.
d. 1 point
NaF (Flask #1)
1 point
The F – ion, generated upon dissolution of NaF, is a weak base. It is
the only solution with pH > 7.
•
One point earned for identifying NaF.
•
One point earned for the correct explanation.
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are registered trademarks of the College Entrance Examination Board.
12
AP® CHEMISTRY
1999 SCORING GUIDELINES
Question 8
8 points
(a) (i)
2 pts
2-
O
O
C
O
C
O
O
•
•
•
One point earned for each Lewis electron-dot structure
Indication of lone pairs of electrons are required on each structure
Resonance forms of CO32− are not required
(ii) In CO2, the C−O interactions are double bonds, OR, in CO32− the C−O interactions
1 pt
are resonance forms (or figures below.)
O
O
C
O
C
O
O
O
O
C
O
O
1 pt
The carbon-oxygen bond length is greater in the resonance forms than in the double bonds.
1st point earned for indicating double bonds are present in CO2 OR resonance occurs in CO32−
•
2nd point earned for BOTH of the above AND indicating the relative lengths of the bond types
•
(b)
(i)
2 pts
F
F
C
F
•
•
F
F
F
..
S
F
One point earned for each Lewis electron-dot structure
Lone pairs of electrons are required on each structure
Copyright © 2001 by College Entrance Examination Board. All rights reserved. College Board, Advanced Placement Program, AP, and the acorn logo
are registered trademarks of the College Entrance Examination Board.
13
F
AP® CHEMISTRY
1999 SCORING GUIDELINES
Question 8 (cont.)
(ii) CF4 has a tetrahedral geometry, so the bond dipoles cancel, leading to a nonpolar molecule.
1 pt
With five pairs of electrons around the central S atom, SF4 exhibits a trigonal bipyramidal
electronic geometry, with the lone pair of electrons. In this configuration, the bond dipoles
do not cancel, and the molecule is polar.
1 pt
F
C
F
F
•
F
F
.. S
F
F
F
One point earned for each molecule for proper geometry and explanation
Copyright © 2001 by College Entrance Examination Board. All rights reserved. College Board, Advanced Placement Program, AP, and the acorn logo
are registered trademarks of the College Entrance Examination Board.
14
AP Chemistry Free­Response Question Means for Practice Exams 1999 AP Chemistry Q1 3.41 (all students answer) Q2 3.34 (students choose Q2 or Q3—the mean for Q2 was lower) Q3 4.32 Q4 6.33 (all students answer, remember 15 point Q) Q5 3.31 (all students answer) Q6 3.18 (all students answer) Q7 2.69 (students choose between Q7 and Q8—mean is much higher for Q8) Q8 4.37 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. Multiple Choice Diagnostics Guide for the 1999 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 (56.25 points in this case).
Atomic Theory: Percent of MC Exam Questions = 6.7%
Question #
5
6
7
8
51
Correct or Incorrect
% of Students
Answering Correctly
75
78
55
83
59
Periodic Properties: Percent of MC Exam Questions = 8.0%
Question #
1
2
17
18
37
Correct or Incorrect
% of Students
Answering Correctly
74
48
70
67
35
50
62
Bonding/Intermolecular Forces: Percent of MC Exam Questions = 13.3%
Question #
13
14
15
16
25
28
32
40
58
68
38
54
39
54
Solutions/Phase Diagrams: Percent of MC Exam Questions = 10.7%
Question #
31
39
43
52
59
69
73
75
Correct or Incorrect
% of Students
Answering Correctly
60
64
65
61
57
53
Nuclear: Percent of MC Exam Questions = 1.3%
Question #
48
Correct or Incorrect
% of Students
Answering Correctly
42
Gas Laws/Kinetic Theory: Percent of MC Exam Questions = 6.7%
Question #
23
44
60
64
74
Correct or Incorrect
% of Students
Answering Correctly
Correct or Incorrect
% of Students
Answering Correctly
67
21
54
71
22
60
65
31
48
28
31
33
33
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. Stoichiometry/Mole relationships: Percent of MC Exam Questions = 6.7%
Question #
20
33
47
53
55
Correct or Incorrect
% of Students
Answering Correctly
55
54
50
46
69
Kinetics: Percent of MC Exam Questions = 4.0%
Question #
19
36
63
Correct or Incorrect
% of Students
Answering Correctly
73
52
34
Equilibrium: Percent of MC Exam Questions = 4.0%
41
54
67
Question #
Correct or Incorrect
% of Students
Answering Correctly
65
35
25
Acid-Base Equilibrium & Buffer: Percent of MC Exam Questions = 9.3%
Question #
9
10
11
12
38
45
62
Correct or Incorrect
% of Students
Answering Correctly
52
41
33
35
38
40
31
Oxidation/Reduction/Electrochemistry: Percent of MC Exam Questions = 4.0%
Question #
34
35
57
Correct or Incorrect
% of Students
Answering Correctly
42
43
44
Thermodynamics: Percent of MC Exam Questions = 6.7%
Question #
3
4
22
61
66
Correct or Incorrect
% of Students
Answering Correctly
31
31
Reactions: Percent of MC Exam Questions = 6.7%
Question #
26
30
42
65
70
Correct or Incorrect
% of Students
Answering Correctly
42
74
71
75
68
71
35
34
2 Organic: Percent of MC Exam Questions = 1.3%
Question #
29
Correct or Incorrect
% of Students
Answering Correctly
26
Laboratory: Percent of MC Exam Questions = 10.7%
Question #
21
24
27
46
49
Correct or Incorrect
% of Students
Answering Correctly
33
20
16
22
53
56
71
72
32
32
25
3 AP* Chemistry 1999 Practice Exam Scoring Worksheet Section I: Multiple Choice ⎡
⎤
⎢
⎥
⎛
⎞⎥
− ⎜ 0.25 ×
=
⎢
⎟ × 0.9600 =
⎢ Number Correct
Number Wrong ⎠ ⎥
Multiple-Choice
Weighted
⎝
⎢
⎥
Section I Score
Score
⎣ ( out of 75 )
⎦
Section II: Free Response Question 1
(out of 9)
Question 2 or 3
Question 4
(out of 9)
(out of 15)
Question 5
(out of 8)
Question 6
(out of 8)
Question 7 or 8
(out of 8)
× 1.9556 =
× 1.9556 =
× 0.8800 =
× 1.6500 =
× 1.6500 =
× 1.6500 =
AP Grade Conversion Chart (Do not round)
(Do not round)
(Do not round)
(Do not round)
(Do not round)
Composite AP Grade Score Range* 100‐160 5 82‐99 4 57‐81 3 35‐56 2 0‐34 1 *The candidates’ scores are weighted as indicated on the exam sections. The Chief Faculty Consultant is responsible for converting composite scores to the 5‐point AP scale. (Do not round)
Total Section II Free Response SUM =
Weighted
Section II Score
Final Composite Score Weighted
Section I Score
+
Weighted
Section II Score
=
Composite Score
(round to the nearest integer)
Copyright © 2000 by College Entrance Examination Board. All rights reserved.