document.write( "Question 1200052: What amount of pure acid must be added to 500 mL of a 25% acid solution to produce a 50% acid solution?
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Algebra.Com's Answer #834079 by josgarithmetic(39617)\"\" \"About 
You can put this solution on YOUR website!
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document.write( "MATERIALS       CONC.%     VOLUME       PURE\r\n" );
document.write( "Pure Acid       100         v           100v\r\n" );
document.write( "Lower Conc.      25         500         (25)(500)\r\n" );
document.write( "ResultingMix     50        v+500        (50)(v+500)\r\n" );
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\n" ); document.write( "The amount of pure acid in the parts to be mixed is the same as the pure acid of the resulting mixing.\r
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\n" ); document.write( "\n" ); document.write( "\"100v%2B%2825%29%28500%29=50%28v%2B500%29\"\r
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\n" ); document.write( "\n" ); document.write( "\"100v=50v%2B%2850%29%28500%29-%2825%29%28500%29\"\r
\n" ); document.write( "\n" ); document.write( "\"100v-50v=500%2A50-500%2A25\"\r
\n" ); document.write( "\n" ); document.write( "\"%28100-50%29v=500%2850-25%29\"\r
\n" ); document.write( "\n" ); document.write( "\"highlight_green%28v=500%28%2850-25%29%2F%28100-50%29%29%29\"----------simplify and compute. \n" ); document.write( "
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