Take a Diagnostic Exam ‹ 29
- A—The mole fraction of CO times the total pres-
sure yields the partial pressure. The mole fraction
of CO is the moles of CO (4.0) divided by the
total moles (10.0). - A— n = PV/RT = (0.993 atm)(0.237 L)/
(0.0821 L atm/K mol)(373 K)
= 7.69 × 10 -^3 mol
molar mass = 0.548 g/7.69 × 10 -^3 mol
= 71.3 g/mol
This example illustrates the importance of
rounding in calculations where no calculator is
available. The answers are not close together;
therefore, a rough calculation will lead to the
correct answer. Also, you should notice the
answer D is impossible for any substance.
- A—For the rate to be one-half, the molar mass
of the other gas must be the square of the molar
mass of helium (4^2 = 16). - A—The average kinetic energy of the molecules
depends on the temperature. The correct answer
involves a temperature difference (333 K - 303 K).
Do not forget that ALL gas law calculations require
Kelvin temperatures.
Chapter 9
- D—This is the definition of the ionization energy.
- C—This is a basic postulate of kinetic molecular
theory. - A—This is one of the properties of free energy.
- B—This is the definition of the lattice energy.
- B—This is an application of Hess’s law.
+
→+[2ClF(g) O(g)
Cl O(g) OF (g)](^122)
22 12 (167.5kJ)
- →
[2 F(g) O(g)
2OF(g)]
(^1222)
2 12 ( 43.5− kJ)
→+
[ClO(g)3OF (g)
2ClF (l)2O(g)]
(^1222)
32 −^12 (394.1kJ)
ClF(g) + F 2 (g) → ClF 3 (l) -135.1 kJ
As always, rounding and estimating will save time.
- C—The one with the greatest increase in the
moles of gas.
30. D—Nonspontaneous means DG > 0. For a reac-
tion to become spontaneous at lower temperature
(DG < 0) means DH < 0 and DS < 0.
Chapter 10
- C—Atoms with completely filled shells or subshells
are not paramagnetic; they are diamagnetic. From
the choices in this problem, these are Be, Mg,
He, Kr, and Zn; therefore, any answer containing
one of these cannot be the correct choice. It is not
necessary to work through a possible solution until
encountering a diamagnetic species. Also, it might
be helpful to look on the periodic table. - C—Transition metal ions are, in general, s^0 and
p^0 or p^6 with the possibility of having one or
more electrons in the d orbitals. C could be Cr^3 +. - B—The noble gases, except helium, are ns^2 np^6.
In this case, n = 4, and the gas is krypton. - D—Halogens are ns^2 np^5. In this case, n = 2, and
the halogen is F. - A—The 1p orbital does not exist.
- D—This is a statement of the uncertainty principle.
- C—According to Hund’s rule, the nitrogen 2p
electrons enter the 2p orbitals individually (with
spins parallel). - A—The Pauli exclusion principle states this limi-
tation for all orbitals. - B—The d orbitals are less effectively shielded
than the s orbitals. Due to this difference, the s
orbitals have lower energy. - C—Mg becomes Mg^2 +. The element is N, which
can become N^3 -.
Chapter 11
- C—The iodine has five bonding pairs and one
lone pair. - B—This is the only one with only single bonds.
The other molecules have double or triple bonds.
All double and triple bonds are a combination of
s and p bonds. - C—Use VSEPR; only the tetrahedral CF 4 is non-
polar. The other materials form a square pyramidal
(IF 5 ), T-shaped (BrF 3 ), and irregular tetrahedral
(SF 4 ), and, therefore, are polar.