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<div class="moz-cite-prefix">On 07/07/13 11:34, wrote:<br>
</div>
<blockquote
cite="mid:C40B2F181831EF44A88CD735258278030261C41B3E@MERCERMAIL.MercerU.local"
type="cite">
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<div style="direction: ltr;font-family: Tahoma;color:
#000000;font-size: 10pt;"><span class="Apple-style-span"
style="font-family: 'Segoe UI', Helvetica, Arial, sans-serif;
font-size: medium; "><span style="background-color: white; ">
<div dir="ltr"><font color="black" face="Tahoma" size="2"><span
dir="ltr" style="font-size: 10pt; ">In chapter 14
problem 118, I don't understand how you got each
number; I understand why you had to convert each time
but what numbers did you use to keep converting?</span></font></div>
</span></span></div>
</blockquote>
<br>
So in the first conversion I went from cm<sup>3</sup>/molecules s to
molecules s/cm<sup>3</sup> by taking the inverse of the number.<br>
<br>
In the second step I converted the cm<sup>3</sup> in the denominator
to L (which is 1000 cm<sup>3</sup>) by multiplying by 1000.<br>
<br>
To convert to moles in the third step I divided by Avogadro's
number.<br>
<br>
In the last step I took the inverse of this number to put me back it
the required units.<br>
<br>
<br>
<br>
By the way, I looked back at my solution and apparently forgot to
carry the seconds in to the numerator when I went to the second
step. Once it got into the numerator it would have been there for
the third and fourth steps. It reappeared in the denominator of the
last step when I took the inverse.<br>
<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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