document.write( "Question 347694: I can't figure this out: Pure acid is to be added to a 10% acid solution to obtain 54 L of a 20% acid solution. What amounts of each should be used? \n" ); document.write( "
Algebra.Com's Answer #248544 by stanbon(75887)![]() ![]() ![]() You can put this solution on YOUR website! Pure acid is to be added to a 10% acid solution to obtain 54 L of a 20% acid solution. What amounts of each should be used? \n" ); document.write( "----------------------- \n" ); document.write( "Equation: \n" ); document.write( "acid + acid = acid \n" ); document.write( "x + 0.10(54-x) = 0.20*54 \n" ); document.write( "--- \n" ); document.write( "Multiply thru by 100 and solve for \"x\": \n" ); document.write( "100x + 10*54 - 10x = 20*54 \n" ); document.write( "90x = 10*54 \n" ); document.write( "x = 6 liters (amt. of pure acid needed in the mixture) \n" ); document.write( "============================= \n" ); document.write( "Cheers, \n" ); document.write( "Stan H. \n" ); document.write( "=========== \n" ); document.write( " |