Discuss Bill's answer to: Question about Math-- Word Problems

A mixture of acid and water is 35% acid. If the mixture contains a total of 40L, how many liters of pure acid are in the mixture? How many liters of pure water are in the mixture?

As with all solutions, we measure by weight, not by volume.  However we consider an acid solution (which acid(s)) and I presume deionized water (again a dangerous presumption). 

Already we have problems with the validity of the question.  But let us devise a calculation anyhow.

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A mixture of acid and water is 35% acid. If the mixture contains a total of 40L, how many liters of pure acid are in the mixture? How many liters of pure water are in the mixture?

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40L (+H-O-H+) + (N-acid [35%])

Calculation: 1/35 = .0287...X 40 = 11.48 L (N-acid) and 28.52 L (+H-O-H+).

I hate to make this sort of calculation because we deviate from format.

I changed my buddy icon because Physicalist pirated my screen name. Atheists have a way of being obnoxious, but that is because they lack a case. My new buddy icon is the planet Uranus. It will stay so until Physicalist learns his place -- under the front porch.
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3 Comments About This Answer

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PGroot Thinks this answer is Not Helpful:

Wrong.  .0287 x 40 would be 1.1428.  But 35% is .35, not 1/35 = .0287.

 
John McCann Thinks this answer is Not Helpful:

Bill is a phony, Pgroot and does not have the intimation of an idea here. He just slopped something together in consultation with all the rest of these wackaloon avatars and tries to pass it off as erudite.

 
John McCann Thinks this answer is Not Helpful:

Also, Bill, +H-O-H+ is not entirely correct as this is a polar covalent molecule and the charge is slightly positive on the hydrogen ends and slightly negative on the oxygen end designated by appropriately signed sigma letters.

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