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  1. #1
    New to the CF scene
    Join Date
    Mar 2009
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    getElementById Problem.

    Hey guys!

    I've been getting a little problem with my getElementById script. It works when I click on one of my <span> links to reveal a different <div> id, but when you try to click on another one it doesn't work. I've checked the error console, it says it turns out null. I've ran out of ideas and don't know what to do.

    If anyone could help me, I'd appreciate it.

    The source:

    <html>

    <head>

    <title>ChemEASY</title>

    <link rel="stylesheet" type="text/css" href="MasterStyle.css">

    <script type="text/javascript">

    function changeNavigation(id)
    {document.getElementById('Content').innerHTML=document.getElementById(id).innerHTML}

    </script>

    </head>

    <body>

    <div id="Main">

    <div id="TopDesign">

    </div>

    <div id="Navigator">

    <span class="nav" onClick="changeNavigation('Intro')">Home</span>
    <span class="nav" onClick="changeNavigation('Stoich')">Stoichiometry</span>
    <span class="nav" onClick="changeNavigation('Probs')">Problems</span>
    <span class="nav" onClick="changeNavigation('Other')">Other Refrences</span>


    </div>

    <div id="Content">

    <div id="Intro">

    <h2>Welcome!</h2>

    <hr NOSHADE size="5" color="black">

    <p>Hello! Welcome to ChemEASY,chemistry made EASY! This site aims to help students with highschool chemistry,
    specifically stoichiometry.</p>

    <p>In this site, you can find a guide and introduction to stoichiometry, example problems to work on and other refrences to help you with other
    chemistry subjects!</p>

    <p>So get out there, get solving, and get learning!
    <br>-The WebMaster</p>

    </div>

    <div id="Stoich" style="display: none;">

    <h2>Stoichiometry</h2>


    <hr NOSHADE size="5" color="black">

    <p>In chemistry, calculations that relate quantities of substances are known as <b>stoichiometry</b>(stoi-kee-AHM-uh-tree) problems.
    <b>Stoichiometry is the study of the quantitative, or measurable, relationships that exist in chemical formulas and chemical reactions.
    </b> The term is derived from the Greek words <i>stoicheion</i>, meaning element, and <i>metron</i>, meaning measure. Converting moles, particles,
    mass, and volume of chemical formulas compromise one aspect of stoichiometry, which you should already be familiar with. Another aspect, which this site will taclke,
    is concerned with chemical reactions and involves the relationships between reactants and productsin a chemical reaction.</p>

    <p>Take note that the coefficients in balanced equations do not represent the actual number of grams or liters of the substance, instead, they represent the number of particles of the substance.
    The measurements needed by industry and laboratories are not explicitly stated in chemical equations. However they can be obtained through stoichiometry.</p>


    <h2>Mole-Mole Conversions</h2>

    <p>Lets take this chemical equation as an example:<br><br>

    AgI + Fe<sub>2</sub>(CO<sub>3</sub>)<sub>3</sub> -> FeI<sub>3</sub> + Ag<sub>2</sub>CO<sub>3</sub>


    <br><br>
    The balanced form of which would be:
    <br><br>

    6AgI + Fe<sub>2</sub>(CO<sub>3</sub>)<sub>3</sub> -> 2FeI<sub>3</sub> + 3Ag<sub>2</sub>CO<sub>3</sub>

    <br><br>
    That would mean that there are 6 moles of AgI reacting with 1 mole of Fe<sub>2</sub>(CO<sub>3</sub>)<sub>3</sub> to produce 2FeI<sub>3</sub> and 3Ag<sub>2</sub>CO<sub>3</sub>.
    </p>


    <p>Now, lets say we want to find how many moles of Fe<sub>2</sub>(CO<sub>3</sub>)<sub>3</sub> is needed to react with 4 moles of AgI to produce 2FeI<sub>3</sub> and 3Ag<sub>2</sub>CO<sub>3</sub>.
    In order to do that, we have to get the molar ratio between Fe<sub>2</sub>(CO<sub>3</sub>)<sub>3</sub> and AgI. Since the balanced equation indicates 6 moles of AgI and 1 mole of Fe<sub>2</sub>(CO<sub>3</sub>)<sub>3</sub>,
    the molar ratio is 6:1. Now, we multiply the 4 moles of AgI with the reciprocal of AgI and Fe<sub>2</sub>(CO<sub>3</sub>)<sub>3</sub> in order to cancel out the AgI.
    <br><br>


    Let n = moles,<br><br>

    n<sub>Fe<sub>2</sub>(CO<sub>3</sub>)<sub>3</sub></sub> = 4mol AgI x [1mol Fe<sub>2</sub>(CO<sub>3</sub>)<sub>3</sub> / 6mol AgI]

    <br><br>


    n<sub>Fe<sub>2</sub>(CO<sub>3</sub>)<sub>3</sub></sub> = 0.66666666666666, or about .067 moles.

    <br><br>What you have just seen is an example of mole-mole conversions, one of the simpler stoichiometry problems.
    </p>

    <h2>Mass-Mass Conversions</h2>

    <p>Now, let us move on to mass-mass conversions. Lets say I want to produce 20 grams of FeI<sub>3</sub>. How many grams of Ag<sub>2</sub>CO<sub>3</sub> would also be produced?
    To solve this, we need to convert the mass of FeI<sub>3</sub> into moles. So,<br><br>


    n<sub>FeI<sub>3</sub></sub> = 20 grams FeI<sub>3</sub> / [2 moles x 437g/mol]

    <br><br>

    n<sub>FeI<sub>3</sub></sub> = 0.022883295 moles or 0.0229 moles of FeI<sub>3</sub>


    <br><br>

    After doing so, we now get the number of moles of Ag<sub>2</sub>CO<sub>3</sub>.

    <br><br>

    n<sub>Ag<sub>2</sub>CO<sub>3</sub></sub> = 0.0229 moles FeI<sub>3</sub> x [1 mol Ag<sub>2</sub>CO<sub>3</sub> / 1 mol FeI<sub>3</sub>]

    <br><br>


    n<sub>Ag<sub>2</sub>CO<sub>3</sub></sub> = 0.0229 moles Ag<sub>2</sub>CO<sub>3</sub>

    <br><br>

    Finally, we convert the moles into grams.

    <br><br>


    g<sub>Ag<sub>2</sub>CO<sub>3</sub></sub> = 0.0229 moles Ag<sub>2</sub>CO<sub>3</sub> x 276g/mol Ag<sub>2</sub>CO<sub>3</sub>


    <br><br>

    g<sub>Ag<sub>2</sub>CO<sub>3</sub></sub> = 6.320grams of Ag<sub>2</sub>CO<sub>3</sub> would also be produced.

    </p>

    <h2>Mass-Volume Conversions</h2>


    <p>Mass-volume computations are very much alike to mass-mass conversions, except instead of converting the moles of the unknown into mass, you convert them to volume.
    </p>

    <p>Lets take this as an example: Suppose I have 42 grams of Fe<sub>2</sub>(CO<sub>3</sub>)<sub>3</sub>. How many liters of FeI<sub>3</sub> would be produced after reacting
    Fe<sub>2</sub>(CO<sub>3</sub>)<sub>3</sub> with AgI?</p>


    <p>First, we convert the 42 grams of Fe<sub>2</sub>(CO<sub>3</sub>)<sub>3</sub> into moles. So,
    <br><br>

    n<sub>Fe<sub>2</sub>(CO<sub>3</sub>)<sub>3</sub></sub> = 42 grams Fe<sub>2</sub>(CO<sub>3</sub>)<sub>3</sub> / [172 grams / mole Fe<sub>2</sub>(CO<sub>3</sub>)<sub>3</sub>]

    <br><br>

    n<sub>Fe<sub>2</sub>(CO<sub>3</sub>)<sub>3</sub></sub> = 0.244186046 moles Fe<sub>2</sub>(CO<sub>3</sub>)<sub>3</sub> or 0.244 moles of Fe<sub>2</sub>(CO<sub>3</sub>)<sub>3</sub>.

    <br><br>


    Now, we use the moles of Fe<sub>2</sub>(CO<sub>3</sub>)<sub>3</sub> to find the moles of FeI<sub>3</sub>.

    <br><br>

    n<sub>FeI<sub>3</sub></sub> = 0.244 moles Fe<sub>2</sub>(CO<sub>3</sub>)<sub>3</sub> [ 2 moles FeI<sub>3</sub> / 1 mole Fe<sub>2</sub>(CO<sub>3</sub>)<sub>3</sub>]

    <br><br>

    n<sub>FeI<sub>3</sub></sub> = 0.488 moles FeI<sub>3</sub>

    <br><br>

    Now, recall that the constant of the volume of elements is 22.4 liters / mole. We simply multiply the moles of FeI<sub>3</sub> with the constant to get the volume.

    <br><br>


    V<sub>FeI<sub>3</sub></sub> = 10.9312 L of FeI<sub>3</sub> or 10.9 L of FeI<sub>3</sub>

    </p>

    <h2>Volume-Volume Conversions</h2>

    <p>Recall that 1 mole of any gas occupies the same volume as 1 mole of any other gas. Thus if the coefficients in a chemical equation represnet the ratio of moles,
    they also represent the ratio of the volumes of gases involved in the reaction. Volume-volume problems are very much like mole-mole problems.</p>

    <p>For example: If I have 10.5 liters of Ag<sub>2</sub>CO<sub>3</sub>, how many liters of AgI reacted with Fe<sub>2</sub>(CO<sub>3</sub>)<sub>3</sub> to produce it?
    <br><br>


    V<sub>AgI</sub> = 10.5 liters Ag<sub>2</sub>CO<sub>3</sub> x [6 liters AgI / 3 liters Ag<sub>2</sub>CO<sub>3</sub>]

    <br><br>

    V<sub>AgI</sub> = 21 liters of AgI.</p>

    <h2>Others</h2>

    <p>Not all the problems you will encounter will be exactly like those this page has covered. Some may be variations. You will still be able to solve them, however, if you have
    a good understanding of the general steps. For example, consider a problem in which you are given an amount of a product in moles and are asked to determine the mass of one of the
    reactants. Although this is not exactly a mass-mass problem, you can still solve it. In fact, it is a little easier since the first step has already been done for you!</p>

    <hr NOSHADE size="5" color="black">

    <p>If you find all of this confusing, fear not! All you need is practice, practice, practice!</P>

    </div>


    <div id="Probs" style="display: none;">

    <script language="javascript" src="MasterScript.js">
    </script>

    <h2>Review Problems</h2>

    <hr NOSHADE size="5" color="black">

    <p>An equation is given. First, balance the equation. Then, follow the instructions that appear after.</p>

    <p>1.) NaNO<sub>3</sub> + PbO -> Pb(NO3)2 + Na<sub>2</sub>O

    <form><input type="button" value="Ready to answer!" onClick="BalanceOne()"></form></p>

    <p>2.) C<sub>2</sub>H<sub>4</sub>O<sub>2</sub> + O<sub>2</sub> -> CO<sub>2</sub> + H<sub>2</sub>O

    <form><input type="button" value="Ready to answer!" onClick="BalanceTwo()"></form></p>


    <p>3.) AgBr + GaPO<sub>4</sub> -> Ag<sub>3</sub>PO<sub>4</sub> + GaBr<sub>3</sub>

    <form><input type="button" value="Ready to answer!" onClick="BalanceThree()"></form></p>

    <p>4.) Fe + AgNO<sub>3</sub> -> Fe(NO<sub>3</sub>)<sub>2</sub> + Ag

    <form><input type="button" value="Ready to answer!" onClick="BalanceFour()"></form></p>


    <p>5.) C<sub>2</sub>H<sub>6</sub> + O<sub>2</sub> -> CO<sub>2</sub> + H<sub>2</sub>O

    <form><input type="button" value="Ready to answer!" onClick="BalanceFive()"></form></p>

    </div>


    <div id="Other" style="display: none;">

    <h2>Help Sites</h2>

    <hr NOSHADE size="5" color="black">

    <table cellpadding="10px" cellspacing="10px">

    <tr>

    <td align="center"><span class="header">Balancing Chemical Equations</span></td>
    <td align="center"><span class="header">Mole Conversions</span></td>

    </tr>

    <tr>

    <td>
    <ul>
    <li><a href="http://www.geocities.com/jigers19">Leonora Sabandal's Help Site</a>
    <li><a href="http://www.geocities.com/">Robert Valencia's Help Site</a>
    <li><a href="http://www.geocities.com/">Link Here</a>

    <li><a href="http://www.geocities.com/">Link Here</a>
    </ul>
    </td>

    <td>
    <ul>
    <li><a href="http://www.geocities.com/notsobad93">Jesha Saladaga's Help Site</a><br>
    <li><a href="http://www.geocities.com/">Nathan Maybuena's Help Site</a>

    <li><a href="http://www.geocities.com/">Link Here</a>
    <li><a href="http://www.geocities.com/">Link Here</a>
    </td>

    </tr>
    </table>

    </div>

    </div>


    </div>

    </div>

    </body>

    </html>
    The link:

    http://www.geocities.com/madsimster44

    Thanks! =]

  • #2
    Regular Coder
    Join Date
    Dec 2008
    Location
    Tannhäuser Gate
    Posts
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    Thanks
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    Thanked 58 Times in 57 Posts
    There's no problem with your Javascript, only with your HTML. Your Content DIV includes all the other DIVs with text you want to display on clicking the link. So, after you do it once, whole Content is being overwritten with the text of one of the DIVs [Probs, Stoich etc.] - and none of those include others. Below adjusted code:
    Code:
    <html>
    <head>
    <title>ChemEASY</title>
    <link rel="stylesheet" type="text/css" href="MasterStyle.css">
    <script type="text/javascript">
    function changeNavigation(id){document.getElementById('Content').innerHTML=document.getElementById(id).innerHTML;}
    </script>
    </head>
    <body>
    <div id="Main">
    <div id="TopDesign"></div>
    
    <div id="Navigator">
    <span class="nav" onClick="changeNavigation('Intro')">Home |</span>
    <span class="nav" onClick="changeNavigation('Stoich')">Stoichiometry |</span>
    <span class="nav" onClick="changeNavigation('Probs')">Problems |</span>
    <span class="nav" onClick="changeNavigation('Other')">Other Refrences </span>
    </div>
    
    <div id="Content"><!--start Content DIV that is going to be changed on clicking the link-->
    <!--initial content, it is going to be destroyed, so can't place here anything you want to refer later to-->
    <h2>Welcome!</h2>
    <hr NOSHADE size="5" color="black">
    <p>Hello! Welcome to ChemEASY,chemistry made EASY! This site aims to help students with highschool chemistry,
    specifically stoichiometry.</p>
    <p>In this site, you can find a guide and introduction to stoichiometry, example problems to work on and other refrences to help you with other
    chemistry subjects!</p>
    <p>So get out there, get solving, and get learning!
    <br>-The WebMaster</p>
    
    </div> <!--end of your Content DIV here------>
    
    <div id="Intro" style="display: none;"><!-- moved out of Content, otherwise your link to HOME won't work-->
    <h2>Welcome!</h2>
    <hr NOSHADE size="5" color="black">
    <p>Hello! Welcome to ChemEASY,chemistry made EASY! This site aims to help students with highschool chemistry,
    specifically stoichiometry.</p>
    <p>In this site, you can find a guide and introduction to stoichiometry, example problems to work on and other refrences to help you with other
    chemistry subjects!</p>
    <p>So get out there, get solving, and get learning!
    <br>-The WebMaster</p>
    </div><!--end of your Intro DIV here------>
    
    <div id="Stoich" style="display: none;">
    <h2>Stoichiometry</h2>
    <hr NOSHADE size="5" color="black">
    <p>In chemistry, calculations that relate quantities of substances are known as <b>stoichiometry</b>(stoi-kee-AHM-uh-tree) problems.
    <b>Stoichiometry is the study of the quantitative, or measurable, relationships that exist in chemical formulas and chemical reactions.
    </b> The term is derived from the Greek words <i>stoicheion</i>, meaning element, and <i>metron</i>, meaning measure. Converting moles, particles,
    mass, and volume of chemical formulas compromise one aspect of stoichiometry, which you should already be familiar with. Another aspect, which this site will taclke,
    is concerned with chemical reactions and involves the relationships between reactants and productsin a chemical reaction.</p>
    <p>Take note that the coefficients in balanced equations do not represent the actual number of grams or liters of the substance, instead, they represent the number of particles of the substance.
    The measurements needed by industry and laboratories are not explicitly stated in chemical equations. However they can be obtained through stoichiometry.</p>
    <h2>Mole-Mole Conversions</h2>
    <p>Lets take this chemical equation as an example:<br><br>
    AgI + Fe<sub>2</sub>(CO<sub>3</sub>)<sub>3</sub> -> FeI<sub>3</sub> + Ag<sub>2</sub>CO<sub>3</sub>
    <br><br>
    The balanced form of which would be:
    <br><br>
    6AgI + Fe<sub>2</sub>(CO<sub>3</sub>)<sub>3</sub> -> 2FeI<sub>3</sub> + 3Ag<sub>2</sub>CO<sub>3</sub>
    <br><br>
    That would mean that there are 6 moles of AgI reacting with 1 mole of Fe<sub>2</sub>(CO<sub>3</sub>)<sub>3</sub> to produce 2FeI<sub>3</sub> and 3Ag<sub>2</sub>CO<sub>3</sub>.
    </p>
    <p>Now, lets say we want to find how many moles of Fe<sub>2</sub>(CO<sub>3</sub>)<sub>3</sub> is needed to react with 4 moles of AgI to produce 2FeI<sub>3</sub> and 3Ag<sub>2</sub>CO<sub>3</sub>.
    In order to do that, we have to get the molar ratio between Fe<sub>2</sub>(CO<sub>3</sub>)<sub>3</sub> and AgI. Since the balanced equation indicates 6 moles of AgI and 1 mole of Fe<sub>2</sub>(CO<sub>3</sub>)<sub>3</sub>,
    the molar ratio is 6:1. Now, we multiply the 4 moles of AgI with the reciprocal of AgI and Fe<sub>2</sub>(CO<sub>3</sub>)<sub>3</sub> in order to cancel out the AgI.
    <br><br>
    Let n = moles,<br><br>
    n<sub>Fe<sub>2</sub>(CO<sub>3</sub>)<sub>3</sub></sub> = 4mol AgI x [1mol Fe<sub>2</sub>(CO<sub>3</sub>)<sub>3</sub> / 6mol AgI]
    <br><br>
    n<sub>Fe<sub>2</sub>(CO<sub>3</sub>)<sub>3</sub></sub> = 0.66666666666666, or about .067 moles.
    <br><br>What you have just seen is an example of mole-mole conversions, one of the simpler stoichiometry problems.
    </p>
    <h2>Mass-Mass Conversions</h2>
    <p>Now, let us move on to mass-mass conversions. Lets say I want to produce 20 grams of FeI<sub>3</sub>. How many grams of Ag<sub>2</sub>CO<sub>3</sub> would also be produced?
    To solve this, we need to convert the mass of FeI<sub>3</sub> into moles. So,<br><br>
    n<sub>FeI<sub>3</sub></sub> = 20 grams FeI<sub>3</sub> / [2 moles x 437g/mol]
    <br><br>
    n<sub>FeI<sub>3</sub></sub> = 0.022883295 moles or 0.0229 moles of FeI<sub>3</sub>
    <br><br>
    After doing so, we now get the number of moles of Ag<sub>2</sub>CO<sub>3</sub>.
    <br><br>
    n<sub>Ag<sub>2</sub>CO<sub>3</sub></sub> = 0.0229 moles FeI<sub>3</sub> x [1 mol Ag<sub>2</sub>CO<sub>3</sub> / 1 mol FeI<sub>3</sub>]
    <br><br>
    n<sub>Ag<sub>2</sub>CO<sub>3</sub></sub> = 0.0229 moles Ag<sub>2</sub>CO<sub>3</sub>
    <br><br>
    Finally, we convert the moles into grams.
    <br><br>
    g<sub>Ag<sub>2</sub>CO<sub>3</sub></sub> = 0.0229 moles Ag<sub>2</sub>CO<sub>3</sub> x 276g/mol Ag<sub>2</sub>CO<sub>3</sub>
    <br><br>
    g<sub>Ag<sub>2</sub>CO<sub>3</sub></sub> = 6.320grams of Ag<sub>2</sub>CO<sub>3</sub> would also be produced.
    </p>
    <h2>Mass-Volume Conversions</h2>
    <p>Mass-volume computations are very much alike to mass-mass conversions, except instead of converting the moles of the unknown into mass, you convert them to volume.
    </p>
    <p>Lets take this as an example: Suppose I have 42 grams of Fe<sub>2</sub>(CO<sub>3</sub>)<sub>3</sub>. How many liters of FeI<sub>3</sub> would be produced after reacting
    Fe<sub>2</sub>(CO<sub>3</sub>)<sub>3</sub> with AgI?</p>
    <p>First, we convert the 42 grams of Fe<sub>2</sub>(CO<sub>3</sub>)<sub>3</sub> into moles. So,
    <br><br>
    n<sub>Fe<sub>2</sub>(CO<sub>3</sub>)<sub>3</sub></sub> = 42 grams Fe<sub>2</sub>(CO<sub>3</sub>)<sub>3</sub> / [172 grams / mole Fe<sub>2</sub>(CO<sub>3</sub>)<sub>3</sub>]
    <br><br>
    n<sub>Fe<sub>2</sub>(CO<sub>3</sub>)<sub>3</sub></sub> = 0.244186046 moles Fe<sub>2</sub>(CO<sub>3</sub>)<sub>3</sub> or 0.244 moles of Fe<sub>2</sub>(CO<sub>3</sub>)<sub>3</sub>.
    <br><br>
    Now, we use the moles of Fe<sub>2</sub>(CO<sub>3</sub>)<sub>3</sub> to find the moles of FeI<sub>3</sub>.
    <br><br>
    n<sub>FeI<sub>3</sub></sub> = 0.244 moles Fe<sub>2</sub>(CO<sub>3</sub>)<sub>3</sub> [ 2 moles FeI<sub>3</sub> / 1 mole Fe<sub>2</sub>(CO<sub>3</sub>)<sub>3</sub>]
    <br><br>
    n<sub>FeI<sub>3</sub></sub> = 0.488 moles FeI<sub>3</sub>
    <br><br>
    Now, recall that the constant of the volume of elements is 22.4 liters / mole. We simply multiply the moles of FeI<sub>3</sub> with the constant to get the volume.
    <br><br>
    V<sub>FeI<sub>3</sub></sub> = 10.9312 L of FeI<sub>3</sub> or 10.9 L of FeI<sub>3</sub>
    </p>
    <h2>Volume-Volume Conversions</h2>
    <p>Recall that 1 mole of any gas occupies the same volume as 1 mole of any other gas. Thus if the coefficients in a chemical equation represnet the ratio of moles,
    they also represent the ratio of the volumes of gases involved in the reaction. Volume-volume problems are very much like mole-mole problems.</p>
    <p>For example: If I have 10.5 liters of Ag<sub>2</sub>CO<sub>3</sub>, how many liters of AgI reacted with Fe<sub>2</sub>(CO<sub>3</sub>)<sub>3</sub> to produce it?
    <br><br>
    V<sub>AgI</sub> = 10.5 liters Ag<sub>2</sub>CO<sub>3</sub> x [6 liters AgI / 3 liters Ag<sub>2</sub>CO<sub>3</sub>]
    <br><br>
    V<sub>AgI</sub> = 21 liters of AgI.</p>
    <h2>Others</h2>
    <p>Not all the problems you will encounter will be exactly like those this page has covered. Some may be variations. You will still be able to solve them, however, if you have
    a good understanding of the general steps. For example, consider a problem in which you are given an amount of a product in moles and are asked to determine the mass of one of the
    reactants. Although this is not exactly a mass-mass problem, you can still solve it. In fact, it is a little easier since the first step has already been done for you!</p>
    <hr NOSHADE size="5" color="black">
    <p>If you find all of this confusing, fear not! All you need is practice, practice, practice!</P>
    </div><!--end of your Stoich DIV here------>
    
    <div id="Probs" style="display: none;">
    <script language="javascript" src="MasterScript.js"></script>
    <h2>Review Problems</h2>
    <hr NOSHADE size="5" color="black">
    <p>An equation is given. First, balance the equation. Then, follow the instructions that appear after.</p>
    <p>1.) NaNO<sub>3</sub> + PbO -> Pb(NO3)2 + Na<sub>2</sub>O
    <form><input type="button" value="Ready to answer!" onClick="BalanceOne()"></form></p>
    <p>2.) C<sub>2</sub>H<sub>4</sub>O<sub>2</sub> + O<sub>2</sub> -> CO<sub>2</sub> + H<sub>2</sub>O
    <form><input type="button" value="Ready to answer!" onClick="BalanceTwo()"></form></p>
    <p>3.) AgBr + GaPO<sub>4</sub> -> Ag<sub>3</sub>PO<sub>4</sub> + GaBr<sub>3</sub>
    <form><input type="button" value="Ready to answer!" onClick="BalanceThree()"></form></p>
    <p>4.) Fe + AgNO<sub>3</sub> -> Fe(NO<sub>3</sub>)<sub>2</sub> + Ag
    <form><input type="button" value="Ready to answer!" onClick="BalanceFour()"></form></p>
    <p>5.) C<sub>2</sub>H<sub>6</sub> + O<sub>2</sub> -> CO<sub>2</sub> + H<sub>2</sub>O
    <form><input type="button" value="Ready to answer!" onClick="BalanceFive()"></form></p>
    </div><!--end of your Probs DIV here------>
    
    <div id="Other" style="display: none;">
    <h2>Help Sites</h2>
    <hr NOSHADE size="5" color="black">
    <table cellpadding="10px" cellspacing="10px">
    <tr>
    <td align="center"><span class="header">Balancing Chemical Equations</span></td>
    <td align="center"><span class="header">Mole Conversions</span></td>
    </tr>
    <tr>
    <td>
    <ul>
    <li><a href="http://www.geocities.com/jigers19">Leonora Sabandal's Help Site</a>
    <li><a href="http://www.geocities.com/">Robert Valencia's Help Site</a>
    <li><a href="http://www.geocities.com/">Link Here</a>
    <li><a href="http://www.geocities.com/">Link Here</a>
    </ul>
    </td>
    <td>
    <ul>
    <li><a href="http://www.geocities.com/notsobad93">Jesha Saladaga's Help Site</a><br>
    <li><a href="http://www.geocities.com/">Nathan Maybuena's Help Site</a>
    <li><a href="http://www.geocities.com/">Link Here</a>
    <li><a href="http://www.geocities.com/">Link Here</a>
    </td>
    </tr>
    </table>
    </div><!--end of your Other DIV here------>
    </div><!--end of your Main DIV here------>
    </body>
    </html>

  • #3
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    Hey there freedom_razor,

    Thanks a whole bunch! Now I don't need to worry about my IT project.

    Glad to see this site has helpful people like you!

    Peace out!


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