Chem 1310 I
Answers for Assignment VI
Due September 27, 2004
p. 233, #33
Increasing the volume by a factor of 3.25 lowers the pressure by a factor of
3.25. Doubling the absolute temperature doubles the pressure. Thus, the new
pressure is 3.38 atm (= 5.50 × (2.00/3.25)).
p. 233, #39
Substituting into PV = nRT gives a temperature of 262 K. Note that
n = 29.8 g × (1 mol/32.0 g).
p. 233, #45
One mol of SF6 weighs 146.05 g. Substitute P= 0.942 atm, n= one mol and
T=288 K into PV = nRT to find that V =25.09 L. Thus, the density is
5.82 g/L (= 146.05 g mol–1 /25.09 L mol–1).
p. 234, #49
The formula mass of sodium chloride is 58.44 g/mol. Thus, 2500 kg of NaCl
contains 8.898 × 104 mol NaCl. The theoretical yield of hydrogen chloride
gas is 8.898 × 104 mol NaCl. Under the conditions specified (550 ˚C and
742.3 torr), the gas would occupy 6.15 × 106 L. (The volume is obtained by
substituting P= 0.9767 atm, n= 8.898 × 104 mol and T=823 K into
PV = nRT).
p. 234, #53
Under the conditions specified (35.3 ˚C and 735.5 torr), 4.48 L of gas
contains 0.1713 mol (a reasonable number; one mol would occupy 22.4 L at
STP, not far from our conditions and 4.48/22.4 = 0.200; of course, I got the
more precise number by substituting into PV = nRT after converting to atm
and K).
Looking at the balanced equation reveals that 0.1713 mol Cl2 is made from
0.1713 mol MnO2 which represents a mass of 14.98 g (molar mass of
manganese(IV) oxide is 86.938 g mol–1).
p. 236, #88
The Chemistry in Your Life essay is on p. 194. The reaction that occurs when
the air bag inflates is: 2 NaN3 → 2 Na + 3 N2.
nnitrogen = (1 atm)(70.0 L)/(0.08206 L atm mol–1 K–1)(298 K) = 2.86 mol N2
nsodium azide = 2.86 mol N2 × (2 mol NaN3/3 mol N2) = 1.91 mol NaN3
gsodium azide = 1.91 mol NaN3 × (65.20 g NaN3/mol NaN3) = 125 g NaN3
p. 236, #90
Let 2y represent the number of particles originally present. After the change,
y particles have reacted and their total number has been changed to 0.666 y.
Thus, P = 0.740 atm × (1.666y mol/2y mol) = 0.616 atm
p. 237, #93
The balanced equation for the reaction is:
CS2 (g) + 3 O2 (g) → CO2 (g) + 2 SO2 (g)
Before the reaction begins, the pressure of the gas mixture is 3.00 atm,
corresponding to 0.9796 mol of gas (use PV = nRT with V = 10.0 L and
T = 373.2 K).
After the reaction is complete, the pressure of the gas mixture is 2.40 atm,
corresponding to 0.7837 mol of gas (use PV = nRT with V = 10.0 L and
T = 373.2 K).
We are told that the initial mixture contained carbon disulfide and excess
oxygen, so we can construct an “i-c-e” table.
Mols Mols Mols Mols
CS2
O2
CS2 SO2
initial
3x + y
0
0
X
change –x
–3x
+ x +2x
end
0
2x
y
x
p. 237, #93 (con’t)
Initially, 4x + y = 0.9796 mol
Finally, 3x + y = 0.7837 mol
Thus, x = 0.1959 mol CS2 which is also the value of y (=0.1959 mol of O2 in
excess).
The molar mass of carbon disulfide is 76.13.
0.1959 mol CS2 × (76.13 g CS2/mol CS2) = 14.9 g CS2
p. 237, #98
Use the Ideal Gas Law to find the chemical amount of oxygen gas produced.
Substitute P = 751.3 torr × (1 atm/760 torr), V = 34.6 mL and T = 291.8 K
into PV = nRT, with R = 0.8206 mL atm mmol–1 K–1 to find that 1.428 mmol
of oxygen gas were produced.
The balanced equation is 2 NaClO3 (s) → 2 NaCl (s) + 3 O2 (g), meaning
that 101.3 mg of sodium chlorate reacted.
Thus, the sample was 61.4% NaClO3.
p. 289, #3
Attractions between particles are always affected, to some degree, by van
der Waals’ forces (also called dispersion forces or induced dipole
interactions).
a. In potassium fluoride, ion-ion forces will be far more important than van
der Waals’ forces.
b. In hydrogen iodide, dipole-dipole forces and van der Waals’ forces are
probably of equal importance.
c. In CH3OH (methanol), hydrogen bonding is more important than van der
Waals’ forces. (Hydrogen bonding may be regarded as a special case of
dipole-dipole interactions. Methanol is “wood alcohol”, a poisonous liquid.
Ethane, C2H6, has a molar mass comparable to that of methanol but ethane is
non-polar and is a gas. (Statements about the state of the compound refer to
conditions of room temperature and atmospheric pressure).
p. 289, #3 (con’t)
d. In radon, only van der Waals’ forces occur between atoms.
e. In (di)nitrogen, only van der Waals’ forces occur between molecules.
p. 289, #5
A sodium ion will be more strongly attracted to a bromide ion than to a
hydrogen bromide molecule or a krypton atom.
p. 289, #15
Pacetylene = Pwet gas – Pwater vapor = 700.9 torr
Substitute in PV = nRT (P = 700.9 torr × (1 atm/760 torr), V = 1 L (exactly)
and T = 313.2 K). This yields a chemical amount of 0.0356 mol of acetylene
in the 1 L volume visualized.
The molar mass of acetylene, C2H2 is 26.038. Thus, the density of the wet
gas is 0.9343 g/L.
p. 289, #17
When the metal is in a high oxidation state, the bonding is more likely to be
covalent than ionic. Waxes are typical covalent solids; they have melting
points on the order of 100 ˚C -200 ˚C. Table salt (sodium chloride) is a
typical ionic solid; its mp is 800 ˚C.
a. lowest mp
Mn2O7 < MnO2 < Mn3O4
highest mp
Note: The oxidation state of manganese in Mn3O4 is apparently +8/3.
The compound is really an equimolar mixture of MnO and Mn2O3. Thus,
Mn3O4 contains two Mn(III) for every Mn(II) for an average oxidation
number of 2⅔.
b. Potassium permanganate is an ionic compound that contains potassium
ions and permanganate ions. The manganese is in the +7 state in the
permanganate ion, a polyatomic ion in which oxygens are covalently bound
to Mn(VII) ions.