document.write( "Question 158793: 1)How many liters of pure acid must be added to 40 L of a 20% acid solution to make a solution which is 36% acid?\r
\n" ); document.write( "\n" ); document.write( "2)A laboratory technician has a solution that is 25% acid. He wishes to increase the concentration to 40% by adding pure acid. If he has 800 mL of solution to begin with, how much acid must be added to make the desired concentration?\r
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Algebra.Com's Answer #116953 by checkley77(12844)\"\" \"About 
You can put this solution on YOUR website!
x+.2*40=.36(40+x)
\n" ); document.write( "x+8=14.4+.36x
\n" ); document.write( "x-.36x=14.4-8
\n" ); document.write( ".64x=6.4
\n" ); document.write( "x=6.4/.64
\n" ); document.write( "x=10 liters of pur acid is needed.
\n" ); document.write( "Proof:
\n" ); document.write( "10+.2*40=.36(40+10)
\n" ); document.write( "10+8=.36*50
\n" ); document.write( "18=18
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\n" ); document.write( ".25*800+x=.4(800+x)
\n" ); document.write( "200+x=320+.4x
\n" ); document.write( "x-.4x=320-200
\n" ); document.write( ".6x=120
\n" ); document.write( "x=120/.6
\n" ); document.write( "x=200 ml of pure acid is needed.
\n" ); document.write( "Proof:
\n" ); document.write( ".25*800+200=.4(800+200)
\n" ); document.write( "200+200=.4(1,000)
\n" ); document.write( "400=400
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