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<div class="moz-cite-prefix">On 10/10/2017 01:27 PM,Jones wrote:<br>
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<blockquote type="cite"
cite="mid:7564493b70914e44b0759c63e2810290@spiderman.MercerU.local">
<pre wrap="">Dr. Pounds,
Hello again. In your powerpoint of chapter 4 about oxidation reduction reactions. H2 + F2——>2HF. H2 has a charge of zero on reaction side. And 1 on product side. How? Same goes for every compound/ element ever with these problems.
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<p><font face="serif">H<sub>2</sub> and F<sub>2</sub> on the
reactant side are ELEMENTS in their STANDARD STATE, and
therefore have oxidation numbers of zero. On the product side
in the compound HF H has a +1 oxidation state to balance out the
F with an oxidation state of -1.</font></p>
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<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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