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<div class="moz-cite-prefix">On 10/21/14 20:21, wrote:<br>
</div>
<blockquote
cite="mid:C40B2F181831EF44A88CD73525827803130CFB7716@MERCERMAIL.MercerU.local"
type="cite">
<div dir="ltr">Hello Dr. Pounds,
<div><br>
</div>
<div>I would like to ask for the answers to Ch 9: 50, 66, 82,
84, 86, 88; Ch 10: 46, 64; Ch 11: 32.</div>
<div><br>
</div>
<div>Thank you,</div>
<div><br>
</div>
</div>
</blockquote>
<br>
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...<br>
<br>
9.50 a) T = 424 K, b) T = 442 K<br>
<br>
9.66 a) density is 1.15 g/L b) density is 1.25 g/L<br>
<br>
9.82 The fraction of methane that is lost is 0.0075<br>
<br>
9.84 a) the diffusion constant is <img style="vertical-align:
middle" src="cid:part1.05090003.01020406@mercer.edu"
alt="$3.6\times 10^{-5}$"> m^2/s<br>
<br>
b) 1.5 years<br>
<br>
9.86 a) <img style="vertical-align: middle"
src="cid:part2.07050305.08040004@mercer.edu" alt="$1.4\times
10^{-19}$"> atm b) <img style="vertical-align: middle"
src="cid:part3.07020408.01000607@mercer.edu" alt="$1.57 \times
10^{3}$"> m/s<br>
<br>
9.88 a) $1.0 \time 10^{10}$ s<sup>-1</sup><br>
<br>
b) $1.0 \time 10^{3}$ s<sup>-1</sup><br>
<br>
<br>
10.46<br>
<img src="cid:part4.02080902.05090003@mercer.edu" alt=""
height="225" width="389"><br>
<br>
<br>
10.64 If the pressure inside the lighter is not to exceed 1 atm,
then the butane must be a gas.<br>
Pressurization is the only way to keep the butane a liquid at room
temperature, because room<br>
temperature exceeds its normal boiling point. Estimating the amount
of gaseous butane in<br>
a lighter with a storage volume of 10 mL requires substitution in
the ideal gas equation and<br>
solving for n. The conditions are: P = 1 atm, V = 0.01 L, and T =
298 K. Doing the arithmetic gives<br>
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).<br>
This is about 1/200 of the butane in a standard lighter.<br>
<br>
11.32 In acidic solution, H3O+ and H2O may be added either as
reactants or as products to achieve<br>
balance.<br>
a) 8 MnO4- (aq)+ 5 H2S(aq)+14 H3O+ (aq) --> 8 Mn2+ (aq)+ 5 SO42-
(aq)+ 26 H2O(l)<br>
b) 4Zn(s)+NO3- (aq)+10 H3O+ (aq) --> 4 Zn2+ (aq)+NH4+ (aq)+13
H2O(l)<br>
c) 5 H2O2 (aq)+ 2 MnO4- (aq)+ 6 H3O+ (aq) --> 5 O2 (g)+ 2 Mn2+
(aq)+14 H2O(l)<br>
d) 2 Sn(s)+ 2 NO3- (aq)+10 H3O+ (aq) --> 2 Sn4+ (aq)+N2O(g)+15
H2O(l)<br>
e) 3 UO22+ (aq)+ Te(s)+ 4 H3O+ (aq) -->3 U4+ (aq)+ TeO42- (g)+ 6
H2O(l)<br>
<br>
<pre class="moz-signature" cols="72">--
Andrew J. Pounds, Ph.D. (<a class="moz-txt-link-abbreviated" href="mailto:pounds_aj@mercer.edu">pounds_aj@mercer.edu</a>)
Professor of Chemistry and Computer Science
Mercer University, Macon, GA 31207 (478) 301-5627
<a class="moz-txt-link-freetext" href="http://faculty.mercer.edu/pounds_aj">http://faculty.mercer.edu/pounds_aj</a>
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