Brian M. Scott
science forum Guru
Joined: 10 May 2005
Posts: 332
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 Posted: Wed May 24, 2006 12:01 am Post subject:
Re: Acid problem.
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On Tue, 23 May 2006 19:43:48 EDT, Megan
<summernightsOx@yahoo.com> wrote in
<news:31712939.1148427858858.JavaMail.jakarta@nitrogen.mathforum.org>
in alt.math.undergrad:
| Quote: | How do I do this mixture problem? I'm reviewing for finals
and I forgot my book so I can't look up how to do this
kind of problem...
How much water must be added to 3 quarts of a 30% acid
solution to form a 20% acid solution?
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You have 3 qt of solution, 30% of which is acid, so you have
0.3 * 3 = 0.9 qt of acid and 0.7 * 3 = 2.1 qt of non-acid
(presumably water). You want to add x quarts of water to
make a 20% acid solution. This means that after you've
added the water, the 0.9 qt of acid that you already have
will be only 20% of the total. The total will be 3 + x
quarts, and 20% of that is 0.2 * (3 + x) = 0.2x + 0.6, so
you want 0.9 = 0.2x + 0.6. From here you should have no
trouble.
| Quote: | Is the equation .9=.6+.2n? Grrr.
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Yes, if n is the number of quarts of water to be added.
Brian |
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