[CHM 115] Solutions to Even Problems
Andrew J. Pounds
pounds_aj at mercer.edu
Tue Oct 21 21:16:57 EDT 2014
On 10/21/14 20:21, wrote:
> Hello Dr. Pounds,
>
> I would like to ask for the answers to Ch 9: 50, 66, 82, 84, 86, 88;
> Ch 10: 46, 64; Ch 11: 32.
>
> Thank you,
>
Hmm... seems kinda late to ask me to work all those with detailed
solutions -- but here are the solutions that I can look up quickly for
you...
9.50 a) T = 424 K, b) T = 442 K
9.66 a) density is 1.15 g/L b) density is 1.25 g/L
9.82 The fraction of methane that is lost is 0.0075
9.84 a) the diffusion constant is $3.6\times 10^{-5}$ m^2/s
b) 1.5 years
9.86 a) $1.4\times 10^{-19}$ atm b) $1.57 \times 10^{3}$ m/s
9.88 a) $1.0 \time 10^{10}$ s^-1
b) $1.0 \time 10^{3}$ s^-1
10.46
10.64 If the pressure inside the lighter is not to exceed 1 atm, then
the butane must be a gas.
Pressurization is the only way to keep the butane a liquid at room
temperature, because room
temperature exceeds its normal boiling point. Estimating the amount of
gaseous butane in
a lighter with a storage volume of 10 mL requires substitution in the
ideal gas equation and
solving for n. The conditions are: P = 1 atm, V = 0.01 L, and T = 298 K.
Doing the arithmetic gives
n = 4.1 × 10–4 mol of butane which amounts to 0.024 g of butane (the M
of butane is 58.1 g mol–1).
This is about 1/200 of the butane in a standard lighter.
11.32 In acidic solution, H3O+ and H2O may be added either as reactants
or as products to achieve
balance.
a) 8 MnO4- (aq)+ 5 H2S(aq)+14 H3O+ (aq) --> 8 Mn2+ (aq)+ 5 SO42- (aq)+
26 H2O(l)
b) 4Zn(s)+NO3- (aq)+10 H3O+ (aq) --> 4 Zn2+ (aq)+NH4+ (aq)+13 H2O(l)
c) 5 H2O2 (aq)+ 2 MnO4- (aq)+ 6 H3O+ (aq) --> 5 O2 (g)+ 2 Mn2+ (aq)+14
H2O(l)
d) 2 Sn(s)+ 2 NO3- (aq)+10 H3O+ (aq) --> 2 Sn4+ (aq)+N2O(g)+15 H2O(l)
e) 3 UO22+ (aq)+ Te(s)+ 4 H3O+ (aq) -->3 U4+ (aq)+ TeO42- (g)+ 6 H2O(l)
--
Andrew J. Pounds, Ph.D. (pounds_aj at mercer.edu)
Professor of Chemistry and Computer Science
Mercer University, Macon, GA 31207 (478) 301-5627
http://faculty.mercer.edu/pounds_aj
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