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Answer: 170ml of 5% acid must be added to

30 ml of 25% to get 200ml of 8% acid


Let x = wt of 5% acidy = wt of 25% acid


Equations:

x + y = 200ml ----equatio

n 1


0.05x + 0.25y = 0.08(200)

0.05x + 0.25y = 16 -----equatio

n 2

multiply equatio

n 2 by 100

5x + 25y = 1600 -------equatio

n 3


multiply equatio

n 1 by -5

-5x -5y = -1000 ------- equatio

n 4


elimi

nate

equatio

ns 3 a

nd 4

5x + 25y = 1600

-5x - 5y = -1000

=====================

0 +20y = 600

divide by 20

y = 30


substitute y = 30 i

n equatio

n 1

x + 30 = 200

x = 200 - 30

x = 170


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Q: How much 5 percent acid must be added to 25 percent acid to get 200ml of acid 8 percent acid?
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How much 5 percent acid must be added to 25 percent acid to get 200 ml of 8 percent acid?

Answer:170mL of 5% acid solution and 30mL of 25% solution must be mixed to produce 200mL of a 8% acid solutionLet x = 5% acid solutiony = 25% acid solution Equations:x + y = 200mL ----equation (1)0.05x + 0.25y = 0.08 * 2000.05x + 0.25y = 16multiplying by 1005x + 25y = 1600 ----equation (2)eliminating equations (1) and (2)-5(x + y = 200)-5-5x -5y = -10005x +25y = 1600=========20y = 600y=30substitute x=30 to equation (1)x + 30 = 200x = 200 - 30x = 170thus 170mL of 5% solution and 30mL of 25% solution must be mixed to produce 200mL of a 8% acid solution


How much pure acid should be mixed with 5 gallons of 70 percent acid solution in order to get a 90 percent acid solution?

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How much pure acid is in 330 mL of a 18 percent acid solution?

18% of 330 = 59.4 So there are 59.3ml of pure acid in 330ml of 18% acid solution.


How much pure acid should be mixed with 6 gallons of a 50 percent acid solution in order to get an 80 percent acid solution?

50% acid in a 6 gallon solution means that 3 gallons are acid. 9 gallons more acid will give you a total of 12 gallons of acid in a 15 gallon solution. 12 is 80% of 15.


How much pure acid should be mixed with 5 gallons of a 20 percent acid solution to get a 50 percent acid solution?

x = amount of acid to add. y = final volume. 5 gal + x = y original amount of acid + acid added = final amount of acid (5 X .2) + x = O.5y Subtract the second equation from the first one. 5 - 1 + x - x = y - 0.5y 4 = 0.5y 8 = y Therefore the final volume is 8 gallons. 5 gal + x = 8 x = 3 gal. the amount of pure acid to add. Check the answer 5 X .2 = 1 gal of acid in original solution. 3 gallons added = 4 gallons total acid in solution. 4 gallons total acid in final solution of 8 gallons total solution = 50% acid.

Related questions

How much 5 percent acid must be added to 25 percent acid to get 200 ml of 8 percent acid?

Answer:170mL of 5% acid solution and 30mL of 25% solution must be mixed to produce 200mL of a 8% acid solutionLet x = 5% acid solutiony = 25% acid solution Equations:x + y = 200mL ----equation (1)0.05x + 0.25y = 0.08 * 2000.05x + 0.25y = 16multiplying by 1005x + 25y = 1600 ----equation (2)eliminating equations (1) and (2)-5(x + y = 200)-5-5x -5y = -10005x +25y = 1600=========20y = 600y=30substitute x=30 to equation (1)x + 30 = 200x = 200 - 30x = 170thus 170mL of 5% solution and 30mL of 25% solution must be mixed to produce 200mL of a 8% acid solution


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