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		<title>ELEKTROLISIS</title>
		<link>http://suprimadya.wordpress.com/2009/10/16/elektrolisis/</link>
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		<pubDate>Fri, 16 Oct 2009 06:31:04 +0000</pubDate>
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				<category><![CDATA[sains]]></category>

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		<description><![CDATA[ELEKTROLISIS Katode(-) : tempat reduksi Anoda(+) : tempat oksidasi hukum Faraday w = e. F W= e. Q/96500 w = e.i.t/96500 mol O2 = 1/4 F               &#8230;&#8230;.   F = VO2/5,6 mol X2 = 1/2 F               &#8230;&#8230;. F  = VCl2/11,2 1 mol e = F = mol H+ = mol OH- = mol penetral 1.  Pada [...]<img alt="" border="0" src="http://stats.wordpress.com/b.gif?host=suprimadya.wordpress.com&amp;blog=9961523&amp;post=8&amp;subd=suprimadya&amp;ref=&amp;feed=1" width="1" height="1" />]]></description>
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<p>ELEKTROLISIS</p>
<p>Katode(-) : tempat reduksi</p>
<p>Anoda(+) : tempat oksidasi</p>
<p>hukum Faraday</p>
<p>w = e. F</p>
<p>W= <em>e. Q/96500</em></p>
<p>w = e.i.t/96500</p>
<p>mol O2 = 1/4 F               &#8230;&#8230;.   F = VO2/5,6</p>
<p>mol X2 = 1/2 F               &#8230;&#8230;. F  = VCl2/11,2</p>
<p>1 mol e = F = mol H+ = mol OH- = mol penetral</p>
<p><a href="http://suprimadya.files.wordpress.com/2009/10/elektrolisis3.jpg"><img class="alignleft size-thumbnail wp-image-74" title="elektrolisis" src="http://suprimadya.files.wordpress.com/2009/10/elektrolisis3.jpg?w=150&#038;h=127" alt="elektrolisis" width="150" height="127" /></a></p>
<p>1.  Pada elektrolisis larutan 2 liter CuSO4  digunakan arus 0,5 F</p>
<p>a.hitung masssa CU yang mengendap</p>
<p>b. hitung  Ph larutan yang terjadi.</p>
<p>2. Pada elektrolisis larutan NiSO4 ditambahkan200 ml Ca(OH)2 1.0M .          larutan hasil elektrolisis ternaya dapat dinetralkan 100 ml HCl 0,1 M. tentukan massa Ni Yang mengendap Ar NI = 59 )</p>
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