floymdiT
2021-10-04
Answered

Pure acid is to be added to a 20% acid solution to obtain 28 L of a 40% acid solution. What amounts of each should be used?

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Nicole Conner

Answered 2021-10-05
Author has **97** answers

Let x be the amount of pure acid to be added so that 28-x is the amount of 20% acid solution, both in L.

In terms of percentage, a pure acid corresponds to 100%. So, we write: 20% of 28-x L+100% of x L = 40% of 28 L

0.2(28-x)+1(x)=0.4(28)

Solve for x: 5.6-0.2x+x=11.2

5.6+0.8x=11.2

0.8x=5.6

x=7

So, we need 7 L of pure acid solution and 21 L of 20% acid solution.

In terms of percentage, a pure acid corresponds to 100%. So, we write: 20% of 28-x L+100% of x L = 40% of 28 L

0.2(28-x)+1(x)=0.4(28)

Solve for x: 5.6-0.2x+x=11.2

5.6+0.8x=11.2

0.8x=5.6

x=7

So, we need 7 L of pure acid solution and 21 L of 20% acid solution.

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