document.write( "Question 566568:   my question:
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document.write( "A chemist mixes 12l of a solution tht is 45% acid wiht 8l of a sotution that is 70 % acid, what is the % of acid of the mixture? \n" );
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| Algebra.Com's Answer #366269 by Maths68(1474)     You can put this solution on YOUR website! Solution A \n" ); document.write( "Amount =121 \n" ); document.write( "Concentration =45% = 0.45\r \n" ); document.write( "\n" ); document.write( "Solution B \n" ); document.write( "Amount = 81 \n" ); document.write( "Concentration =70% = 0.70\r \n" ); document.write( "\n" ); document.write( "Mixture \n" ); document.write( "Amount =(81+121)=202 \n" ); document.write( "Concentration =x\r \n" ); document.write( "\n" ); document.write( "[Amount Solution A * Concentration A] + [Amount Solution B * Concentration of B] = Amount of Mixture * Concentration of Mixture\r \n" ); document.write( "\n" ); document.write( "(121)(0.45)+(81)(0.70)=(202)(x) \n" ); document.write( "54.45+56.70=202x \n" ); document.write( "111.15=202x \n" ); document.write( "111.15/202=202x/202 \n" ); document.write( "0.550248=x \n" ); document.write( "x=0.550248\r \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Mixture \n" ); document.write( "Amount =(81+121)=202 \n" ); document.write( "Concentration =x = 0.550248 = 0.550248/100 = 55.0248%\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Chemist will have 202 of 55.0248% acid mixture.\r \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |