document.write( "Question 248129: how many liters of water must be added to 8 liters of a 40% acid solution to obtain a 10% acid solution?
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document.write( "How many kg of pure salt must be added to 20 kilograms of a 10% salt solution to obtain a 25% salt solution? \n" );
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Algebra.Com's Answer #181094 by ankor@dixie-net.com(22740)![]() ![]() You can put this solution on YOUR website! how many liters of water must be added to 8 liters of a 40% acid solution to obtain a 10% acid solution? \n" ); document.write( ": \n" ); document.write( "Let x = amt of water \n" ); document.write( "a simple \"amt of acid\" equation (the amt of acid does not change only the percent) \n" ); document.write( ".40(8) = .10(x+8) \n" ); document.write( "3.2 = .10x + .8 \n" ); document.write( "3.2 - .8 = .1x \n" ); document.write( "2.4 = .1x \n" ); document.write( "x = \n" ); document.write( "x = 24 liters of pure water required \n" ); document.write( ": \n" ); document.write( "Check for equal acid: \n" ); document.write( ".4(8) = .10(24+8) \n" ); document.write( "3.2 = .1(32) \n" ); document.write( "; \n" ); document.write( "; \n" ); document.write( "How many kg of pure salt must be added to 20 kilograms of a 10% salt solution to obtain a 25% salt solution? \n" ); document.write( ": \n" ); document.write( "Let x = amt of pure salt \n" ); document.write( ".10(20) + x = .25(x+20) \n" ); document.write( "2 + x = .25x + 5 \n" ); document.write( "x - .25x = 5 - 2 \n" ); document.write( ".75x = 3 \n" ); document.write( "x = \n" ); document.write( "x = 4 kg of salt to be added \n" ); document.write( "; \n" ); document.write( ": \n" ); document.write( "Check \n" ); document.write( ".10(20) + 4 = .25(4+20) \n" ); document.write( "2 + 4 = .25(24) \n" ); document.write( "6 = 6 \n" ); document.write( " |