<?xml-stylesheet type="text/xsl" href="http://www.college-cram.com/study/chemistry/rss/rssstyles.xsl"?>
<rss version='2.0'   xmlns:dc='http://purl.org/dc/elements/1.1/'>
    <channel xml:base='http://www.college-cram.com/study/chemistry/'>
        <title><![CDATA[Chemistry : Activity]]></title>
        <description><![CDATA[Activity for Chemistry, hosted on College-Cram.]]></description>
        <generator>Elgg</generator>
        <link>http://www.college-cram.com/study/chemistry/</link>        
        <item>
            <title><![CDATA[Periodic Table Cramlet Review]]></title>
            <link>http://www.college-cram.com/study/chemistry/weblog/periodic-table-cramlet-review</link>
            <guid isPermaLink="true">http://www.college-cram.com/study/chemistry/weblog/periodic-table-cramlet-review</guid>
            <pubDate>Thu, 09 Aug 2007 21:24:59 GMT</pubDate>
		<dc:subject><![CDATA[Periodic Table]]></dc:subject>
		<dc:subject><![CDATA[Period]]></dc:subject>
		<dc:subject><![CDATA[Name]]></dc:subject>
		<dc:subject><![CDATA[Meyer]]></dc:subject>
		<dc:subject><![CDATA[Mendeleev]]></dc:subject>
		<dc:subject><![CDATA[Group]]></dc:subject>
		<dc:subject><![CDATA[Elements]]></dc:subject>
		<dc:subject><![CDATA[Electronegativity]]></dc:subject>
		<dc:subject><![CDATA[Electron configuration]]></dc:subject>
		<dc:subject><![CDATA[Chemistry]]></dc:subject>
		<dc:subject><![CDATA[Chemical symbol]]></dc:subject>
		<dc:subject><![CDATA[Biology]]></dc:subject>
		<dc:subject><![CDATA[Atomic number]]></dc:subject>
		<dc:subject><![CDATA[Periodic Table Cramlet Review]]></dc:subject>
		<dc:subject><![CDATA[Science]]></dc:subject>
		<dc:subject><![CDATA[Type]]></dc:subject>
		<dc:subject><![CDATA[Atomic mass]]></dc:subject>
            <description><![CDATA[<p>(Click here to use our printable <span class="titleCram"><a href="http://www.college-cram.com/study/chemistry/presentations/504"  title="Periodic table">Periodic Table</a></span> or our <a href="http://www.college-cram.com/study/chemistry/presentations/508"  title="Chemical symbol lookup table">Chemical Symbol Lookup Table</a>.)</p><p>The modern periodic table has been at the heart of chemistry, biology, and other sciences since the pioneering work of Meyer and Mendeleev in the mid-1800s. A periodic table has graced the inside front cover of countless science texts for as long as I can remember, but textbooks don&#39;t always do such a good job of explaining what it is and how to use it.</p><p>Unlike these static textbook tables, the <strong>Periodic Table</strong> Cramlet&trade; has several features that make it effective where the book is not:</p><div class="icon"><img src="http://www.college-cram.com/images/gif/screen3.gif"  border="0"  alt="Study the Periodic Table"  width="281"  height="210" /></div><div class="bullet"><ul><li><strong>Color coding</strong> helps to visually distinguish between the different element groups and types.</li><li><strong>Printouts</strong> are just a mouse click away, giving a color printout with an additional page showing the symbols and names in order of their atomic number. Print it out and laminate it for a great carry-around reference!</li><li><strong>How to Use</strong> section explains in detail how to use the table and what each of the terms means. (This is especially useful for us non-scientific types.) Does your textbook do that? I don&#39;t think so!</li><li><strong>Key information</strong> about any element is displayed below the table when you roll over that element with your mouse:</li></ul><ol><li>Name</li><li>Chemical symbol</li><li>Type</li><li>Group</li><li>Electronegativity</li><li>Electron configuration</li><li>Atomic mass</li><li>Atomic number</li><li>Period</li></ol></div><p>The Periodic Table is one of our most popular Cramlets&trade;. Of the features listed here, the best one is that, unlike a textbook, it is available <strong>for free</strong> to every student who signs up for a <a href="http://www.college-cram.com/study/_invite/register.php"  title="Free Registration">free registration</a>.</p><div id="signature"><p>We are star stuff,</p><p>Professor Cram</p></div>]]></description>
        </item>
                
        <item>
            <title><![CDATA[Calculating Molar Solutions]]></title>
            <link>http://www.college-cram.com/study/chemistry/weblog/calculating-molar-solutions</link>
            <guid isPermaLink="true">http://www.college-cram.com/study/chemistry/weblog/calculating-molar-solutions</guid>
            <pubDate>Thu, 09 Aug 2007 18:19:10 GMT</pubDate>
		<dc:subject><![CDATA[Solutions]]></dc:subject>
		<dc:subject><![CDATA[Molar Solutions]]></dc:subject>
		<dc:subject><![CDATA[Molar]]></dc:subject>
		<dc:subject><![CDATA[Chemistry]]></dc:subject>
		<dc:subject><![CDATA[Calculating Molar Solutions]]></dc:subject>
		<dc:subject><![CDATA[mole conversion]]></dc:subject>
		<dc:subject><![CDATA[mole conversions]]></dc:subject>
		<dc:subject><![CDATA[science]]></dc:subject>
		<dc:subject><![CDATA[Bottomless Worksheet of Mole Conversions]]></dc:subject>
		<dc:subject><![CDATA[Calculating]]></dc:subject>
            <description><![CDATA[<div id="question"><p>Dear Professor Cram:</p><p>How many grams of NaCl would you dissolve in water to make a 0.5 M NaCl solution with 500 mL final volume? What are the steps in finding the answer?</p><p>Anneka L.</p></div><p>Thank you for using College-Cram.com and for your chemistry question.</p><p>This requires a multi-step process to determine the answer. First, we need to note that the &quot;M&quot; in a 0.5 M NaCl solution means there are 0.5 moles of NaCl per 1 liter of solution. Since there is in fact only 500mL of solution (i.e., half a liter), you really have only 0.25 moles of NaCl (half of 0.5).</p><p>Next, you can use our <a href="http://www.college-cram.com/study/chemistry/presentations/575"  title="Mole Conversions"><strong>Mole Conversions</strong></a> Cramlet&trade; to find how many grams it takes to produce this solution. Using your <a href="http://www.college-cram.com/study/chemistry/presentations/504"  title="Periodic Table"><strong>Periodic Table</strong></a>, you can determine the molar mass of one mole of NaCl by adding the atomic masses of the two elements:</p><ul><li>22.98977 + 35.453 = 58.44277 grams/mole</li></ul><p>Now multiply this by the number of moles of NaCl you have (0.25, as we found above) to find the number of grams of NaCl we needed to start with:</p><ul><li>(58.44277 grams/mole)(.25 moles) = 14.611 grams</li></ul><p>I hope this helps. Let us know if you need anything else.</p><div id="signature"><p>Good studying,</p><p>Professor Cram</p><p>P.S. Need more practice on this? Try our <a href="http://www.college-cram.com/study/chemistry/presentations/579"  title="bottomless worksheet of mole conversions">Bottomless Worksheet of Mole Conversions</a>!&nbsp;</p></div>]]></description>
        </item>
        
        <item>
            <title><![CDATA[Periodic Table]]></title>
            <link>http://www.college-cram.com/study/chemistry/files/17/114/MPeriodic2_1.swf</link>
            <enclosure url="http://www.college-cram.com/study/chemistry/files/17/114/MPeriodic2_1.swf" length="85873" type="application/x-shockwave-flash" />
            <pubDate>Thu, 17 Jan 2008 17:58:33 GMT</pubDate>
            <description><![CDATA[This is a periodic table for your Cell Phone or PDA written in FlashLite]]></description>
        </item>
    </channel>
</rss>