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∙ 14y agoIn order to reduce the percentage of acid from 80% to 50%, you would need to add another 36 kg of diluent (e.g. water).
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∙ 14y agoAnswer: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
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.
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.
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
16 2/3 liters
10 liters.
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%