# who can help with math please?

a chemistry teacher needs to make 150L of 42% sulfuric acid solution. the acid solutions available are 30% and 50% sulfuric acid, by volume. How many litres pf each solution must be mixed to make the 42% solution? round your answer to the nearest tenth of a L.

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Let the quantity of 50% acid solution mixed be x L

THEN , the quantity of 30% acid solution mixed WILL BE ( 150- x) L

0.50x + 0.30 ( 150 -x ) = 0.42 *150

0.50x + 45 - 0.30x = 63

0.20x = 18

x = 90

ANSWER 90L of 50% AND 60L of 30% acid solution

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• The total amount of sulfuric acid will be 42% of 150L and this comes from the two contributions.

Let required volume of 30% be V.

Since the total volume will be 150L, the volume of 50% must be 150 - V

Try to learn this method.

0.3V + 0.5(150 - V) = 0.42 * 150L

75 - 63 = 0.2V

V = 12*5

Volume of 30% required is V = 60L

This mixes with 90L of 50% sulfuric acid

Regards - Ian

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• After teaching for the past 9 years, I have discovered these percent-mixture problems are the hardest algebra problems for students to solve.

The easiest way (I believe) to come up with the proper equation to solve is by creating a table...

on the first row (horizontal) would be 3 entries:

" %, Amount, Total "

where % is the percent of the mixture, Amount is the amount of that mixture, and Total is the product of (% and Amount).

.....................................

%, Amount, Total

30, x, 0.3(x)

50, 150 - x, 0.5(150 - x)

42, 150, 0.42(150)

the equation formed is:

0.3x + 0.5(150 - x) = 0.42(150)

- 0.2x + 75 = 63

- 0.2x = - 13

x = 65

so 65 L of the 30% sulfuric acid, and

(150 - 65)

85 L of the 50% sulfuric acid.

Source(s): by the way, notice that Amount (1) + Amount (2) is the total Amount of the combined mixture in each of these. here's another set of problems just like this with different variations: http://answers.yahoo.com/question/index?qid=201309...
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