document.write( "Question 102385: how much pure acid should be mixed with 7 gallons of a 40% acid solution in order to get a 70% acid solution? \n" ); document.write( "

Algebra.Com's Answer #74533 by ptaylor(2198) $\"\"$ $\"About$

You can put this solution on YOUR website!
Let x=amount of pure acid needed\r
\n" ); document.write( "\n" ); document.write( "Now we know that the amount of pure acid in the acid that's added (x) plus the amount of pure acid in the 40% solution ((7)*(0.40)) equals the amount of pure acid in the final mixture ((0.70)(7+x)). So our equation to solve is:\r
\n" ); document.write( "\n" ); document.write( "x+7*0.40=0.70(7+x) get rid of parens\r
\n" ); document.write( "\n" ); document.write( "x+2.8=4.9+0.70x subtract 0.70x and also 2.8 from both sides\r
\n" ); document.write( "\n" ); document.write( "x+2.8-2.8-0.70x=4.9-2.8+0.70x-0.70x collect like terms\r
\n" ); document.write( "\n" ); document.write( "0.30x= 2.1 divide both sides by 0.30\r
\n" ); document.write( "\n" ); document.write( "x=7 gallons---------------------------amount of pure acid needed\r
\n" ); document.write( "\n" ); document.write( "CK
\n" ); document.write( "7+7*0.40=0.70(14)\r
\n" ); document.write( "\n" ); document.write( "7+2.8=9.8
\n" ); document.write( "9.8=9.8\r
\n" ); document.write( "\n" ); document.write( "Hope this helps---ptaylor \n" ); document.write( "

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