AP Chemistry Practice Exam 1 ❮ 323❯ Answers and Explanations for Exam 1, Section 1
(Multiple Choice)- A—The reaction is an oxidation-reduction reaction
since the sulfide ion undergoes oxidation to elemen-
tal sulfur and the azide ion undergoes reduction. - B—It is necessary to remove the ammonia
from the gas; otherwise the part of the volume
generated would be due to the ammonia. Since
ammonia is a base, it will react with the acid. - D—The pressure inside the flask is the sum of
the partial pressures. Therefore, the pressure of
hydrogen gas is the total pressure (748.2 torr)
minus the vapor pressure of water (27.0 torr).
The leveling of the water in the beaker and flask
adjusted the pressure in the flask to the external
(barometric) pressure. - B—When the two liquid levels are the same then
the pressures must be equal. In this case, both
pressures are equal to the barometric pressure. - D—The ideal gas equation (PV = nRT) gives the
moles of hydrogen gas formed.
Moles H 2 = n =PV
RT
= (^) −−
(721.2torr/760 torr)(0.315L)
(0.0821Latm molK^11 )(300.2K)
= 0.0121 moles H 2
It is easier to calculate the answer by simple
rounding as:
= (^) −−
(1 atm)(0.3 L)
(0.08Latm molK^11 )(300 K)
≈
(1)
(0.1)
0.3
300
=
0.3
30.0
= 0.01 moles- C—The moles of nitrogen gas formed equal the
moles of sodium azide reacting (see the balanced
chemical equation). If the sample were pure
sodium azide, the mass of sodium azide would
be (176.604 - 175.245) g = 1.359 g sample
(= NaN 3 ). The moles of sodium azide are the
mass of sodium azide divided by its molar mass
(65.0 g mol-^1 ).Moles N 2 = moles NaN 3 = (^) −
1.359g
65.0gmol^1
≈
1.359
65
≈
1.3
5(13)
≈
0.13
5
≈ 0.026
Due to rounding, the actual answer must be a
little smaller (0.0209 moles).- A—Sulfur dioxide gas is soluble in water and,
while less soluble in dilute acid, some would still
dissolve to give a smaller volume. - A—This represents an ion-dipole force, which is
stronger than a hydrogen bond (B), a dipole-dipole
force (C), or a London dispersion force (D). - C—After one half-life, 50% would remain.
Another half-life would reduce this by one-half
to 25% and a third half-life would reduce the
remaining material by 12.5%. Thus, three half-
lives = (50 + 25 + 12.5)% = 87.5% decayed.
The total amount decayed is 88%. For this
reason, 24 days years must be three half-lives of
about 8 days each. - C—The best choice to prepare the buffer is the
one where the pKa is closest to the desired pH. It
is possible to estimate the pKa for an acid (with-
out a calculator) by taking the negative of the
exponent for the different Ka values. This gives
1 for HIO 3 , 2 for HIO 4 , 4 for HNO 2 , and 10
for HCN. The HNO 2 has the value closest to
the desired pH. It would not be a good choice
since the pKa is so far from the pH; however, it
is the best choice.
20-Moore_PE01_p307-340.indd 323 31/05/18 1:58 pm