In order to reduce the percentage of acid from 80% to 50%, you would need to add another 36 kg of diluent (e.g. water).
Answer:12mL of 50% solution and 24mL of 20% solution must be mixed to produce 36mL of a 30% acid solutionLet x = 50% acid solutiony = 20% acid solution Equations:x + y = 36mL ----equation (1)0.5x + 0.2y = 0.3 * 360.5x + 0.2y = 10.8multiplying by 105x + 2y = 108 ----equation (2)eliminating equations (1) and (2)-2(x + y = 36)-2-2x -2y = -725x +2y = 108=========3x = 36x=12substitute x=12 to equation (1)12 + y = 36y = 36 - 12y = 24thus 12mL of 50% solution and 24mL of 20% solution must be mixed to produce 36mL of a 30% acid solution
x=45
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%
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
Mixing 80 liters of 15% solution and 520 liters of 90% solution will give 600 liters of 80% solution.
133.33
144liters
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
10 liters
The chemist will use 100 liters of the 80% acid solution and 100 liters of the 30% acid solution to make a 200-liter solution that is 62% acid. The amount of acid in the 80% solution will be 0.8 * 100 = 80 liters, and in the 30% solution, it will be 0.3 * 100 = 30 liters.
To create a 400 L solution that is 62% acid, you would need 200 L of the 80% acid solution and 200 L of the 30% acid solution. This would result in a final solution with the desired concentration.
16 2/3 liters
10 liters.
Answer:12mL of 50% solution and 24mL of 20% solution must be mixed to produce 36mL of a 30% acid solutionLet x = 50% acid solutiony = 20% acid solution Equations:x + y = 36mL ----equation (1)0.5x + 0.2y = 0.3 * 360.5x + 0.2y = 10.8multiplying by 105x + 2y = 108 ----equation (2)eliminating equations (1) and (2)-2(x + y = 36)-2-2x -2y = -725x +2y = 108=========3x = 36x=12substitute x=12 to equation (1)12 + y = 36y = 36 - 12y = 24thus 12mL of 50% solution and 24mL of 20% solution must be mixed to produce 36mL of a 30% acid solution
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.
x=45
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%