Wiki User
∙ 11y agoA
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
=========
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
Wiki User
∙ 11y ago20%
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
How much must be added to 29602 to make 3412 is 6000
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%
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.
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
To find the amount of pure acid to add, set up an equation based on the amount of acid in the original solution and the final solution. Let x be the amount of pure acid to add. The amount of acid in the original solution is 0.4 × 12 = 4.8 ounces. The amount of acid in the final solution is 0.6 × (12 + x) = 4.8 + 0.6x ounces. Therefore, to get a 60% acid solution, you would need to add 10.67 ounces of pure acid.
How much must be added to 29602 to make 3412 is 6000
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
To create a 33% acetic acid solution from 50% acetic acid and pure water, we must use a mixture of 1 part 50% acetic acid to 1 part water. So, to dilute 4 pints of 50% acetic acid to 33%, you would need to add 4 pints of water.