AP exam2 _ MCQ_ Review1 and 2 KEY
Diagnostic Test 1
1. (A) 10. (C) 19. (A) 28. (A) 37. (D)
2. (D) 11. (B) 20. (B) 29. (D) 38. (A)
3. (A) 12. (D) 21. (B) 30. (C) 39. (C)
4. (C) 13. (B) 22. (B) 31. (B) 40. (D)
5. (A) 14. (D) 23. (A) 32. (D) 41. (C)
6. (B) 15. (A) 24. (C) 33. (D) 42. (B)
7. (B) 16. (A) 25. (A) 34. (B) 43. (D)
8. (C) 17. (C) 26. (B) 35. (C) 44. (C)
9. (C) 18. (D) 27. (D) 36. (C) 45. (C)
1. (A) The basic equation is A,v = c, which reads "The wavelength times the frequency is
equal to the speed of light, which is 3.00 x 108 m s-1."
= (3.00 x 108 m 0)/ (6.00 x 1014 s-1)
k= 5.0 x 10-7 m = 500 x 10-8 m = 500 nm
If we convert the units to meters (m), centimeters (cm), picometers (pm), we do not end
up with any of the other choices.
2. (D) Among HBrO, HCIO, and HIO, the more electronegative central atom indicates a
stronger oxo acid. Therefore HC1O is the strongest of these three. For the same reason,
we choose HC103 over HI03 as the strongest acid in this pair. Comparing HC1O and
HC103, we select HC103 as the strongest because it has the larger number of unshared
or double-bonded oxygen atoms that stabilize the anion, creating a stronger acid.
3. (A) By convention, liquids are not written as part of the equilibrium law. Liquids always
have the same number of molecules per liter, which is a constant, combined into the
value of the K The only nonsolid substances in this reaction are 02(g) and CO2(g). In
this reaction, CO2 to the third power appears in the numerator because it is a product,
and 02 to the fifth power is in the denominator.
4. (C) Only reactions that are known to be elementary processes (i.e., actual collisions of
molecules) can be used to write a rate law. Most other reactions proceed by multistep
mechanisms that have to be deduced from experimental evidence.
5. (A) With a half-life of 36 minutes, there will be 120/36 = 3.33 half-lives. Therefore there
are more than 3 and less than 4 half-lives. Three half-lives leaves 1/8 of the material,
and 4 half-lives leaves 1/16 of the material. 1/8 of the initial 10 grams is a little more
than 1 gram, while 1/16 of 10 grams is a little more than 1/2 gram. The only answer that
falls in that range is 1.00 gram.
Use the integrated rate equation: 1n(go /gt) = kt, also ln(2) = kti /2.
Therefore k = (0.693)/36 min = 0.0193 min-1.
Then
ln(go) - ln(gt) = (0.0193 min-1)(120 min)
ln(10) -1n(gt) = 2.31
2.30 - ln(gt) = 2.31
-1n(gt) = 0.01
ln(gt) = -0.01
gt= 1.00
6. (B) Molarity is the moles of solute in each liter of solution; molarity = mol solute/L
solution
mol Ni(NO3)2 = 30.0 g Ni(NO3)2
= 0.1640 mol Ni(NO3)2
molarity Ni(NO3)2 = (0.1640 mol Ni(NO3)2)/0.250 L
= 0.656 mol Ni(NO3)2 /L
7. (B) Network crystals are held together with covalent bonds. These bonds make the
crystal one large, rigid, molecule. As a result, the macroscopic substance is very hard.
8. (C) The reaction is 2 HI(g) H H2(g) + I2(g), and the equilibrium law is
K= [H2 ][12 =21
[HI] 2
?mol L—
HI - 15 gHI [1 molHI] - 9.77 X 10-3 mol L-1
12 L 128 gHI
When HI reacts, 2x moles of HI form x moles of H2 and x moles of 12. The equilibrium
law is
(x) (x)
2 - 21 Take the square root of both sides of the
(9.77 x10-3 -2x) equation.
(x)
= 4.58
9.77x10-3 -2x
x = 4.47 X 10-2 - 9.16x
10.16x = 4.47 X 10-2
x = 4.4 X 10^ -3 mol L-1
Because x= [12], that is the concentration of 12 vapor expected. We keep only two
significant
figures because lc had only two significant figures.
9. (C) Molarity is a ratio of units (moles per liter); therefore, the setup of the calculation
must start with a ratio, such as another molarity.
Setup:
?MNaOH = 0.216 Mlic1
Expand molarities to mol/L and insert conversion factors:
mol NaOH 0.108 mol HCl 1 mol NaOH 0.0426 L HCl x x
L NaOH L HCI 1 mol HCl 0.0400 L NaOH
= 0.115 M NaOH
10. (C) ONLY temperature changes will result in changes in the numerical value of the
equilibrium constant.
11. (B) Alkenes have a double bond. The double bond consists of one sigma and one pi
bond connecting a pair of carbon atoms. The simplest alkene, ethane, has one pi bond
and five sigma bonds. Geometrically, they are triangular planar around the carbon
atoms (sp2 hybrid). Because of that geometry, the simplest alkene molecule does not
have isomers or optical activity
12. (D) Graham's law of effusion is written as
rate' = mass2
rate2 mass'
Defining helium as compound 1 in the equation above, we can substitute
5.33 —
mass2
4
Square both sides:
mass2
28.4 =
4
mass2 = 4 x 28.4 = 114 g/ mol
The molecular masses of the possible compounds are
(A) CO2 = 44, (B) CH4 = 16, (C) C51112 = 72, (D) C81-118 = 114, is obviously the sample
in
this experiment.
13. (B) The Pauli exclusion principle states that no two electrons in an element can have
the same set of four quantum numbers. This is another way of saying that no two electrons
can occupy the same orbital with the same spin. Response (A) is false—electrons
often have the same energy. Response (C) is wrong because although electrons can
occupy separate orbitals, this.has nothing to do with the question. Response (D) is a
statement of Hund's rule.
14. (D) Hydrogen bonds can form when a compound contains a fluorine, an oxygen, or a
nitrogen atom with a hydrogen bonded to it. In this question, this criterion is fulfilled
only by CH3NH2.
15. (A) The neutron, with no charge, has tremendous penetrating ability. Many can pass
through the earth without stopping.
16. (A) The Lewis structures of all molecules except PC14F are symmetric. The fact that
two different atoms are attached to the central P atom immediately suggests that this
molecule is not symmetric. We test it first by drawing its Lewis structure. We find that
structure to be a trigonal bipyramid that is not symmetrical, and we predict it will be polar.
17. (C) All of the ions in this question are isoelectronic with the noble gas krypton except
arsenic. Arsenic would have to be a 3- ion to be isoelectronic with krypton. It is written
as a 5+ ion.
18. (D) Generally, with a pair of elements, the one closest to fluorine in the periodic table
is negative (largest electronegativity), and the atom farthest from fluorine is positive
(lowest electronegativity). The only one of the five pairs where the second element is
closest to fluorine is the P—Cl pair.
19. (A) To determine the molar mass, we need to calculate the moles of the acid and divide
it into the mass of the acid used in this experiment. Since each acid molecule has one
proton (it is monoprotic), determining the moles of protons gives us the moles of the
acid.
moles of protons = (0.0263 L) (0.122 M KOH) = 3.21 X 10-3 mol
molar mass = (0.682 gram)/ ( 3.21 X 10-3 mol) = 212 g mol-1
20. (B) Hydrogen bonding is the strongest intermolecular attractive force and is listed first.
Induced dipoles are the weakest attractive forces and last for a short period of time.
21. (B) An intermediate is defined as a substance that is neither a reactant nor a product.
Intermediates are often difficult to detect. Catalysts fit the above description, but they
are substances that are added to the reaction mixture and can be isolated afterward.
22.' (B) Mathematically, we start with ? mol C3H8 = 6.2 g C3H8. We then use the
conversion
factor 1 mol C3H8 = 44 g C3H8 to convert g to mol.
1 mol C3H8
44 g C3C8
= 1.4 X 10-1 mol C3H8
23. (A) Work can be determined from PV data. The equation is
work ( w) = -P V.
We can use P11/1 = P2 V2 to calculate the final volume in the problem above:
(2.4 atm) (2.0 L) = (0.80 atm) (x liters)
x liters = 6.0 L
work = -(0.80 atm) (6.0 L - 2.0 L) = -3.2 L atm
24. (C) This is the best answer because all group 1A, 2A, and 3A metals, as well as some
transition metals, have ions that are isoelectronic with a noble gas.
? mol C3H8 = 6.2 g C3H8
25. (A) From the integrated rate law we can derive that In(2) = kt1/2, where t112 is the
half-life.
Converting 34 minutes to seconds yields 34 X 60 = 2040 seconds. The rate constant is
calculated as
k = ln(2) /2040 s
= 0.693/2040 s
= 3.4 X 10-4 s-1
We round the final answer to two significant figures because 34 minutes has only two
significant figures.
26. (B) In this molten salt the only possible products are magnesium and bromine. The
question is at which electrode are these products found. Remembering that oxidation
always occurs at the anode, we see that 2 Br- —> Br2 + 2 e is the oxidation process.
Therefore bromine is produced at the anode, and magnesium at the cathode.
27. (D) To calculate the oxidation number of an element in a polyatomic ion, the charges
of the atoms must add up to the charge on the ion.
Charge of ion = (ox. no. CI) + 4(ox. no. 0)
-1 = (ox. no. Cl) + 4(-2)
1+ 8 = ox. no. Cl
+ 7 = ox. no. Cl
28. (A) The Gibbs free energy equation is AG° = AH° - T AS°. To determine AG° at a
temperature
other than 298 K we need AH° and AS° so that the above equation can be
solved at a temperature different from 298.
AG° = -43.2 kJ mol-1 - (1073 K) (22.0 J mol-1 K-1)
= -43.2 kJ mot-1 - (23.6 x 103 J. mol-1)
= -43.2 kJ mo1-1 - (23.6 kJ mot-1)
= -66.8 kJ mol-1
29. (D) The symbol represents an isotope of chromium. Since there is no charge, it is not
an ion that may have gained or lost electrons. Therefore the numbers of electrons and
protons are equal (24 of each), which eliminates response (C). Response (B) does not
have 24 protons, and response (D) is the remaining one where the protons and neutrons
add to 52.
30. (C) This response is true because of the large distance between gas particles. The other
four responses are not necessarily true about any given gas sample.
31. (B) This pair differs by only one H+ between their two formulas.
32. (D) This electron configuration has 20 electrons in the lowest possible energy levels
according to the Aufbau ordering. Element 20 is calcium.
33. (D) The vapor pressure of a mixture is found to be
'total = PAXA + PBXB
= 345 torr (0.285) + 213 torr (0.715)
= 98.3 torr + 152.3 torr
= 250.6 torr = 251 torr (when rounded correctly)
34. (B) The most important factor in entropy change is whether or not there is a change
in volume as denoted by the change in the number of gas molecules in the chemical
equation (Ang). For the reactions given, the Ang is: (A) 0, (B) -2, (C) 0, (D) -1. Reaction
(B) has the greatest decrease in the moles of gas and should have the largest decrease
in entropy.
35. (C) The conjugate acid of a base is obtained by adding one hydrogen atom and one
positive charge to the base. Therefore, H2PO4 + H+ -3 H3PO4.
36. (C) Numbering the chain from left to right results in the lowest possible numbers for
the chloro and bromo substituents.
37. (D) The nitrate ion has two oxygen atoms bonded to the nitrogen with single bonds,
and one oxygen is bonded to the nitrogen with a double bond. This adds up to three
sigma and one pi bond.
38. (A) The definition of molality is moles of solute per kilogram of solvent. In dilute
solutions
the mass of solute is negligible and the mass of one liter of water is one kilogram.
Therefore, molarity and molality are essentially equal. We will calculate molarity as
(g I M)I(L of solution)
molality = molarity = (25 X 10-3 g sucrose/342 g mol-1)/1.00 L
= 7.31 X 10-5 molar = 7.3 X 10-5 molal (when correctly rounded)
39. (C) Silicon is a metalloid and is better known as a semiconductor material used in
transistors.
40. (D) Mathematically, the fraction left is 1 microgram/10 kilograms, which is a ratio of
(10-6)/ (10 X 103) or 1/10-1°. For half-lives, the fraction remaining is (1/2)n, where n is
the number of half-lives. By equating (1/2)n = (1/10-10), we can take the logarithm of
both sides of the equation to get n log (1/2) = —10. We can invert the fraction within the
log term to get n log (2) = 10. (Note that this operation changes the sign.) Finally, n =
10/ (log 2) = 33.2. This needs to be rounded to the next highest half-life, 34. This may be
solved without a calculator by recalling that log (2) = 0.3.
41. (C) Half-reactions 1 and 2 are reductions because the charge of the copper ions
decreases in the process. The iron half-reaction is an oxidation because the oxidation
number increases from +2 to +3.
42. (B) The difference in electronegativity is 1.9. The others are (a) 1.1, (c) 0.5, and (d) 0.6.
This answer can be estimated since the P and F are more widely separated than the
other pairs that are adjacent atoms in the periodic table.
43. (D) Because chromium has more than one isotope and this question does not specify
which isotope, it is impossible to state the number of neutrons. The atomic mass in the
periodic table is NOT an isotope mass and cannot be used for that purpose.