(^28) › STEP 2. Determine Your Test Readiness
❯ Answers and Explanations
Chapter 5
- D—The others (nonmetals) form anions.
- B—Decreasing radii is related to increasing
charges, or for going up a column (with equal
charges), or moving toward the right in a period
of the periodic table. This explanation will not be
sufficient for the free-response portion of the test,
where it is necessary to address such factors as the
effective nuclear charge. - C—The element that is farthest away from F on
the periodic table. - A—Hexaammine = (NH 3 ) 6 ; cobalt(III) = Co^3 +;
and nitrate = NO 3 -. - D—Hexaammine = (NH 3 ) 6 ; chromium(III) =
Cr^3 +; chloride = Cl- - D—Rutherford, and students, determined this
by bombarding gold foil with alpha particles
and detecting the deflection of some of the
particles.
Chapter 6
- A—The balanced chemical equation is:
3 Mn(OH) 2 (s) + 2 H 3 AsO 4 (aq) →
Mn 3 (AsO 4 ) 2 (s) + 6 H 2 O(l)
- D—Ca(OH) 2 , NaOH, and Na 2 CO 3 are strong
electrolytes (strong bases or soluble salts) and should
be separated. You should know all the strong bases
and that sodium compounds are soluble. Cancel all
spectator ions (Na+ and OH-). - C—The hydroxide is low because it combined
with some of the iron, so Fe^2 + will be low. There
is no other ion that the hydroxide ion could
combine with to form a precipitate. The nitrate is
double the potassium because there are two moles
of nitrate per mole of iron(II) nitrate instead of
one ion per mole, as in potassium hydroxide. - C—Copper is blue, not red, and carbonate and
aluminum are colorless. Iron slowly hydrolyzes
(reacts with water) to form solid Fe(OH) 3 (rust).
Chapter 7
- C—(0.1000 mol Cr 2 O 72 - /1,000 mL)(45.20 mL)
(6 mol Fe^2 +/1 mol Cr 2 O 72 - )(1/75.00 mL)
(1,000 mL/L) - C—Either calculate the percent Mn in each
oxide: (A) 77.4%; (B) 69.6%; (C) 49.5%;
(D) 63.2%, or determine the empirical formula
from the percent manganese and the percent
oxygen ( = 100.0 - 49.5). - A—H 2 C 2 O 4 is the limiting reagent as the amount is
less than the stoichiometric ratio indicates. The cal-
culation is
(2.5molHCO)
2molMnSO(^224) 5molHCO
4
224
.
- A—The coefficients in the balanced equation are 2,
2, and 3. Therefore, (1.0 mol KClO 3 ) (3 mol O 2 /
2 mol KClO 3 ) = 1.5 mol.
Chapter 8
- A—(0.400 mol Ba)(1 mol H 2 /1 mol Ba)
(22.4 L/mol). Note that the 22.4 L/mol only
works at STP. - D—This is an application of Charles’s law,
which relates volume to temperature. There is
a direct relationship between volume and the
absolute temperature. Doubling either volume or
temperature, with moles and pressure remaining
constant, doubles the other. - D—The average kinetic energy of a gas depends
upon the temperature. Since the temperature
of the two gases is the same, the average kinetic
energy of the gases is the same. The moles are the
same, so the number of particles and the volumes
must be the same. Density is mass over volume,
and since the balloons have the same volume, the
one with more mass will have the higher density. - A—These are the basic differences between ideal
and real gases. - D—It is necessary to first write the balanced
chemical equation:
2 Al(s) + 6 HCl(g) → 2 AlCl 3 (s) + 3 H 2 (g)
(13.5 g Al)(1 mol Al/27.0 g Al)(3 mol H 2 /
2 mol Al)(22.4 L/mol H 2 )