document.write( "Question 100199This question is from textbook
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document.write( ": A chemist mixes distilled water with a 90% solution of sulfuric acid to produce a 50% solution. If 5 liters of distilled water is used, how much 50% solution is produced? \n" );
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Algebra.Com's Answer #72994 by ptaylor(2198)![]() ![]() You can put this solution on YOUR website! \n" ); document.write( "Let x=amount of 50% solution that is produced\r \n" ); document.write( "\n" ); document.write( "Then x-5=amount of 90% solution that was mixed with the distilled water\r \n" ); document.write( "\n" ); document.write( "(1) Pure sulfuric acid in the 90% solution=0.90(x-5)\r \n" ); document.write( "\n" ); document.write( "(2) Pure sulfuric acid in the distilled water=0\r \n" ); document.write( "\n" ); document.write( "(3) Pure sulfuric acid in the final mixture=0.50x\r \n" ); document.write( "\n" ); document.write( "Now we know that (1)+(2)=(3). So our equation to solve is:\r \n" ); document.write( "\n" ); document.write( "0.90(x-5)=0.50x get rid of parens\r \n" ); document.write( "\n" ); document.write( "0.90x-4.5=0.50x subtract 0.90x from both sides\r \n" ); document.write( "\n" ); document.write( "0.90x-0.90x-4.5=0.50x-0.90x collect like terms\r \n" ); document.write( "\n" ); document.write( "-4.5=-0.40x divide both sides by -0.40\r \n" ); document.write( "\n" ); document.write( "x=11.25 liters -------------amount of 50% solution produced\r \n" ); document.write( "\n" ); document.write( "CK\r \n" ); document.write( "\n" ); document.write( "0.90*(11.25-5)=0.50*(11.25) \n" ); document.write( "0.90(6.25)=5.625 \n" ); document.write( "5.625=5.625\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Hope this helps---ptaylor\r \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |