<?xml version="1.0" encoding="UTF-8"?><rss version="2.0"
	xmlns:content="http://purl.org/rss/1.0/modules/content/"
	xmlns:dc="http://purl.org/dc/elements/1.1/"
	xmlns:atom="http://www.w3.org/2005/Atom"
	xmlns:sy="http://purl.org/rss/1.0/modules/syndication/"
	>
<channel>
	<title>Comments on: Ideal Gas: Reaction Stoichiometry And Mass Of Reagent To Volume  Confused?</title>
	<atom:link href="http://www.chemistrytutorsite.com/2009/05/12/ideal-gas-reaction-stoichiometry-and-mass-of-reagent-to-volume-confused/feed/" rel="self" type="application/rss+xml" />
	<link>http://www.chemistrytutorsite.com/2009/05/12/ideal-gas-reaction-stoichiometry-and-mass-of-reagent-to-volume-confused/</link>
	<description>Chemistry Tutor Site--Your Free Chemistry Tutor Site</description>
	<pubDate>Fri, 30 Jul 2010 08:26:43 +0000</pubDate>
	<generator>http://wordpress.org/?v=2.7</generator>
	<sy:updatePeriod>hourly</sy:updatePeriod>
	<sy:updateFrequency>1</sy:updateFrequency>
		<item>
		<title>By: Dr.A</title>
		<link>http://www.chemistrytutorsite.com/2009/05/12/ideal-gas-reaction-stoichiometry-and-mass-of-reagent-to-volume-confused/comment-page-1/#comment-1447</link>
		<dc:creator>Dr.A</dc:creator>
		<pubDate>Tue, 12 May 2009 23:00:33 +0000</pubDate>
		<guid isPermaLink="false">http://www.chemistrytutorsite.com/?p=664#comment-1447</guid>
		<description>Moles NaN3 = 44.8 g /  65.01 g/mol =0.689
the ratio between NaN3 and N2 is 2 : 3
moles N2 produced = 0.689 x 3 / 2 =1.03
T = 20 + 273 = 293 K
V = nRT / p = 1.03 x 0.0821 x 293 / 1.05 =23.7 L</description>
		<content:encoded><![CDATA[<p>Moles NaN3 = 44.8 g /  65.01 g/mol =0.689<br />
the ratio between NaN3 and N2 is 2 : 3<br />
moles N2 produced = 0.689 x 3 / 2 =1.03<br />
T = 20 + 273 = 293 K<br />
V = nRT / p = 1.03 x 0.0821 x 293 / 1.05 =23.7 L</p>
]]></content:encoded>
	</item>
</channel>
</rss>
