A
nswer:170mL of 5% acid solution and 30mL of 25% solution must be mixed to produce 200mL of a 8% acid solution
Let x = 5% acid solution
y = 25% acid solutionEquations:
x + y = 200mL ----equation (1)
0.05x + 0.25y = 0.08 * 200
0.05x + 0.25y = 16
multiplying by 100
5x + 25y = 1600 ----equation (2)
eliminating equations (1) and (2)
-5(x + y = 200)-5
-5x -5y = -1000
5x +25y = 1600
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20y = 600
y=30
substitute x=30 to equation (1)
x + 30 = 200
x = 200 - 30
x = 170
thus 170mL of 5% solution and 30mL of 25% solution must be mixed to produce 200mL of a 8% acid solution
20%
75 litres
Answer: 170ml of 5% acid must be added to30 ml of 25% to get 200ml of 8% acidLet x = wt of 5% acidy = wt of 25% acidEquations:x + y = 200ml ----equation 10.05x + 0.25y = 0.08(200)0.05x + 0.25y = 16 -----equation 2multiply equation 2 by 1005x + 25y = 1600 -------equation 3multiply equation 1 by -5-5x -5y = -1000 ------- equation 4eliminateequations 3 and 45x + 25y = 1600-5x - 5y = -1000=====================0 +20y = 600divide by 20y = 30substitute y = 30 in equation 1x + 30 = 200x = 200 - 30x = 170
No. The reulting concentration (percent) must be between the two components. So, with the two acids you are mixing, you cannot get an acid that is less than 10% or more than 40%
How much must be added to 29602 to make 3412 is 6000
6 ounces
About 80ml of water must be added to 40ml of a 25 percent by weight solution to make a 2 percent by weight solution.
About 80ml of water must be added to 40ml of a 25 percent by weight solution to make a 2 percent by weight solution.
20%
4 litres
75 litres
Answer: 170ml of 5% acid must be added to30 ml of 25% to get 200ml of 8% acidLet x = wt of 5% acidy = wt of 25% acidEquations:x + y = 200ml ----equation 10.05x + 0.25y = 0.08(200)0.05x + 0.25y = 16 -----equation 2multiply equation 2 by 1005x + 25y = 1600 -------equation 3multiply equation 1 by -5-5x -5y = -1000 ------- equation 4eliminateequations 3 and 45x + 25y = 1600-5x - 5y = -1000=====================0 +20y = 600divide by 20y = 30substitute y = 30 in equation 1x + 30 = 200x = 200 - 30x = 170
The original mixture contains 41.4 ounces of glycol. for this to be 30 percent of the mixture, the total mixture must be 138 ounces, so 46 ounces of water must be added.
Let us assume that x pint water must be mixed with 4 pints of 50 percent acetic acid to produce a mixture that is 33 percent acetic acid. Therefor 4 pint*50 %=(4+x)pint*33% 200 =132 pints + x pints 200-132 = x*33 pints 68 /33 = x pints So x = 2.06 pints Answer : 2.06 pint water must be mixed with 4 pints of 50 percent acetic acid to produce a mixture that is 33 percent acetic acid.
12*(90%)=10.8=X*.25 10.8/.25=X=43.2 43.2-12=31.2 31.2 grams of water must be added
No. The reulting concentration (percent) must be between the two components. So, with the two acids you are mixing, you cannot get an acid that is less than 10% or more than 40%
mary mixed 2l of an 80% acid solution with 6l of a 20% acid solution. what was the percent of acid in the resulting mixture