FINAL EXAM REVIEW – AP EXAM FORMAT
SOLUTIONS TO SELECT PROBLEMS
We will meet to continue analysis of practice problems Wednesday night, 1/19/11. Right after 5 th period exam. See
you then!
1994 AP Examination (used for review and found at:
http://apcentral.collegeboard.com/apc/public/courses/213049.html )
SECTION I: MULTIPLE CHOICE
Questions 1-4: 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.
A. Heisenberg uncertainty principle
B. Pauli exclusion principle (In a given atom no two electrons can have the same set of four quantum numbers: n, l,
ml, ms). An orbital can only hold 2 electrons and they must have opposite spins. pp 311-312)
C. Hunds rule (principle of maximum multiplicity) (the lowest energy configuration for an atom is the one having
the maximum number of unpaired electrons allowed by the Pauli principle in a particular set of degenerate
orbitals. P 318).
D. Shielding effect
E. Wave nature of matter
1. Can be used to predict that a gaseous carbon atom in its ground state in paramagnetic. (paramagnetic means an
atom contains one or more unpaired electrons, p432): C
2. Explains the experimental phenomenon of electron diffraction (p 296) : E
3. Indicates that an atomic orbital can hold no more than two electrons (p 311-312) : B
4. Predicts that it is impossible to determine simultaneously the exact position and the exact velocity of an electron
(p 305) : A
16. Commercial vinegar was titrated with NaOH solution to determine the content of acetic acid, HC2H3O2. For 20.2
milliliters of the vinegar, 26.7 milliliters of 0.600 molar NaOH solution was required. What was the concentration of
acetic acid in the vinegar if no other acid was present?
(B) 0.800 M
ANALYSIS: Use MAVA = MBVB (NOTE: convert volumes to L)
MA (0.0202L) = (0.600mol/L) (0.0267L)
MA = 0.800 M
18.
+1 -2 +7 -2
+3 -2
+4 -2
+7 -2
-2 +1
2H2O + 4MnO4 + 3ClO2  MnO2 + 3ClO4- + 4 OHWhich species acts as an oxidizing agent in the reaction represented above?
(E) MnO4ANALYSIS: An oxidizing agent is something that oxidizes another species. Another way to look at it is: an oxidizing agent
is what gets reduced. Analyze the oxidation numbers you’ve assigned and you will see that Mn goes from a +7  +4
state. It has been reduced. It is the oxidizing agent.
19. In which of the following compounds is the mass ratio of chromium to oxygen closest to 1.62: 1.00?
(B) CrO2
ANALYSIS: If 1.62 : 1.00 is the mass ratio of the compound we must find, then identify the mass of Cr and O in each
compound and reduce each ratio until you find the one that gives the desired ratio (1.62 : 1.00).
Cr: O2 = 52g: 32g
Reduced: 1.62 : 1.00
20. When the equation above is balanced with lowest whole-number coefficients, the coefficient for OH- is
ANSWER: (D) 6
ANALYSIS: Ag+ + AsH3(g) + 6 OH  Ag(s) H3AsO3(aq) + 3H2O
27. Which of the following sets of quantum numbers (n, l ml, ms)best describes the valence electron of highest energy in
a ground-state gallium atom (atomic number 31)?
ANSWER: (C)
ANALYSIS: the highest energy electron in gallium is located in a 4p orbital
(A) 4, 0, 0 , ½
NO b/c l= 0 indicates s sublevel
(B) 4, 0, 1, ½
NO b/c l = 0 indicates s sublevel
(C) 4, 1,1, ½
YES b/c n=4 indicates 4th energy level, l = 1 indicates p sublevel, ml = 1 is an allowed value, and ½ is an
allowed spin for an electron.
(D) 4, 1, 2, ½
NO b/c ml = 2 is not an allowed value when l = 1.
(E) 4, 2, 0 , ½
NO b/c l = 2 indicates d sublevel
28. Given that a solution is 5 percent sucrose by mass, what additional information is necessary to calculate the molarity
of the solution?
I. The density of water
II. The density of the solution
III. The molar mass of sucrose
ANSWER: (E) II and III
ANALYSIS
GIVEN
CONV FACTOR
UNKNOWN
(II)density of soln (I)molar mass
(molarity of soln)
↓
↓
↓
g sucrose x
1 mol sucrose
= mol sucrose
L sol’n
g sucrose
L sol’n
33. A hydrocarbon gas with an empirical formula CH2 has a density of 1.88 grams per liter at 0⁰ C and 1.00 atmosphere.
A possible formula for the hydrocarbon is
ANSWER: (C) C3H6
ANALYSIS: Quickly identify that the temperature and pressure are STP. Therefore, we know the volume of this gas is
22.4L/mol. Each of the possible answers has its own molar mass. If we can calculate the molar mass of the compound
by using the density, we can use this to identify the compound
1.88g cpd
1 L cpd
x
22.4 L cpd
1 mol cpd
=
42.112 g
mol cpd
THIS IS THE MOLAR MASS OF THE UNKNOWN CPD!
42.112 g/14 g = 3
DIVIDING THE MOLAR MASS BY THE EMPIRICAL MOLAR MASS, WE FIND
THAT THE COMPOUND IS 3x HEAVIER THAN THE EMPIRICAL FORMULA.
3(CH2)
C3H6
42. Mass of empty container
Mass of the container plus
the solid sample
Volume of the solid sample
THE ACTUAL FORMULA IS THEREFORE 3x the EMPIRICAL FORMULA
3.0 grams
25.0 grams
11.0 cubic centimeters
ANSWER: (D) 2.00 g/cm3
ANALYSIS: 25.0 g – 3.0 g = 22.0 g/11.0 cm3
= 2.00 g/cm3 (3 sig figs)
46. Which of the following solids dissolves in water to form a colorless solution?
ANSWER: (E) ZnCl2
ANALYSIS: From lab experience, we know that iron compounds are reddish-brown & copper compounds are bluish-green.
In addition, you may know that cobalt compounds tend to be blue. Finally, chromium compounds tend to have very vivid
colors (ranging from intense yellow to orange, black, red, and green) This leaves E as our only choice. Zinc, with its one
oxidation state, behaves much like main group metals, including forming colorless compounds.
50. Which of the following acids can be oxidized to form a stronger acid?
ANSWER: (E) H2SO3
ANALYSIS: H3PO4 and HNO3 are already very strong acids and have no stronger (more highly oxidized) form (ex: H3PO5)
H2SO3 looks a lot like its cousin, H2SO4. If we add another oxygen atom, we get H2SO4, a very strong acid.
56. It is suggested that SO2 (molar mass 64 grams), which contributes to acid rain, could be removed from a stream of
waste gases by bubbling the gases through 0.25 molar KOH, thereby producing K2SO3. What is the maximum mass of
SO2 that could be removed by 1,000 liters of the KOH solution?
ANSWER: (B) 8.0 kg
ANALYSIS:
SO2 (g) + 2KOH (aq)  K2SO3 (s) + H2O(l)
1000 L KOH x 0.25 mol KOH
1 L KOH soln
x 1 mol SO2
2 mol KOH
(often H2 gas or water is formed in rxns)
x
64 g SO2
1 mol SO2
=
8000 g SO2
Recognize that this is equal to 8.0 kg or convert with one more conversion factor!
59. When a 1.00 –gram sample of limestone was dissolved in acid, 0.38 gram of of CO2 was generated. If the rock
contained no carbonate other than CaCO3, what was the percent of CaCO3 by mass in the limestone?
ANSWER: (D) 86%.
ANALYSIS:
CaCO3(s) + 2H+(aq)  CO2(g) + H2O(l) + Ca(s)
This rxn is a little tricky. If you understand H2 or H2O is often formed in rxns involving acids or bases, you can likely figure
it out. In reality, H2CO3 is formed first, then it breaks down into H2O and CO2. (Note: H2CO3 = H2O + CO2)
Now, determine how much CaCO3 was present if 0.38 g of CO2 is formed.
0.38 g CO2 x 1 mol CO2
44 g CO2
0.86 g CaCO3 = 86%
1.00 g limestone
x 1 mol CaCO3
1 mol CO2
x 100g CaCO3 = 0.86 g CaCO3
1 mol CaCO3