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lbvjy [14]
3 years ago
5

how much water should be mixed with 6 gallons of 55% solution of sulfuric acid to make a 20% solution of sulfuric acid

Mathematics
1 answer:
nevsk [136]3 years ago
6 0
Hello.

For all dilution problems use the dilution equation: 
<span>C1V1 = C2V2 </span>
<span>20*V1 = 6*55 </span>
<span>V1 = 6*55/20 </span>
<span>V1 = 16.5 gallons </span>
<span>The final volume must be 16.5 gallons </span>
<span>You must add 16.5 - 6 = 10.5 gallons water to be added.
</span>
Have a nice day
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<h2><em><u>Answ</u></em><em><u>er</u></em><em><u>:</u></em><em><u>-</u></em></h2>

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⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

<h3>• <u>Given</u><u>:</u><u>-</u></h3>

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<h3>• <u>Solution</u><u>:</u><u>-</u></h3>

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⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

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⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

Therefore, F(x) = 5

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