AP Chemistry Practice Exam 2 ❮ 355❯ Answers and Explanations for Exam 2, Section 1
(Multiple Choice)- B—This is a dipole-dipole force, which is stron-
ger than a dipole-induced dipole (A and C) or a
London dispersion force (D). - C—This will increase the concentration of Ag+,
causing a shift to the right, which will lead to an
increase in the cell voltage. - A—The zinc ion concentration remains the same
(1 M); therefore, the voltage will remain the same.
The identity of the anion associated with the
zinc is not important unless the compound is not
soluble. Zinc chloride must be soluble; otherwise
the student could not make a 1 M solution. - B—The cell voltage (1.56 V) is the sum of the
standard reduction potential for silver (0.80 V)
and the reverse of the standard reduction potential
for zinc. Therefore, (1.56 - 0.80) V = 0.76 V =
the reverse of the standard reduction potential
for zinc. Reversing the value gives -0.76 V as the
standard reduction potential for zinc. - B—As the cell begins to run, the voltage decreases
until the system reaches equilibrium where the
voltage is zero. - D—The mass of water is the difference in the
masses of the hydrated salt and the anhydrous
salt. In this case, (mass of crucible plus hydrated
salt) - (mass of crucible plus anhydrous salt) =
58.677 g - 57.857 g = 0.820 g. This assumes the
weight of the crucible is constant. - A—The percent water is 100% times the grams
of water divided by the mass of the hydrated salt.
The mass of water is:
(mass of crucible plus hydrated salt) - (mass of cru-
cible plus anhydrous salt) = (58.677 - 57.857) g =
0.820 g water.
The mass of the hydrated salt is:
(mass of crucible plus hydrated salt) - (mass of
crucible) = (58.677 - 53.120) g = 5.557 g hydrated
salt.
Finally, percent water =
0.820gHO
5.557g(^2) (100%) =
14.8%. The simplified calculation would be:
Percent water ≈≈
0.8 gHO
5.6 g
(100%)
1
7
(^2) (100%)
≈ 14%
- C—Leaving the sample overnight in the lab
drawer would cause the sample to be no longer
anhydrous. The mass of the “anhydrous” salt
would now be higher indicating a smaller amount
of water loss, which would lead to a lower per-
centage of water in the sample. - B—It is necessary to determine the empirical
formula of the compound. If the sample is 63%
water, then it is 37% anhydrous salt. Assuming
100 grams of compound, the masses of water and
anhydrous salt are 63 g and 37 g, respectively.
Converting each of these to moles gives:
Moles anhydrous salt = (37ganhydroussalt) ×
1moleanhydrous salt
106 ganhydrous salt= 0.35 mole anhydrous saltMoles water =
(63gHO)
1moleHO(^2) 18.0gHO
2
2
=
3.5 mole H 2 O
Since there are ten times as many moles of water
as moles of the anhydrous salt, the formula must
be AaXx•10H 2 O.- C—No calculations are necessary as this is the
only answer with the correct units. - A—Comparing experiments 1 and 3 shows that
changing the hydrogen ion concentration has no
effect upon the rate; therefore the reaction is zero
order in hydrogen ion. Comparing experiments
2 and 3 shows that doubling the urea concentra-
tion doubles the rate; therefore, the reaction is
first order with respect to urea.
21-Moore_PE02_p341-370.indd 355 31/05/18 1:54 pm