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18% of 330 = 59.4
So there are 59.3ml of pure acid in 330ml of 18% acid solution.
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∙ 13y ago
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How much pure acid should be mixed with 5 gallons of 70 percent acid solution in order to get a 90 percent acid solution?
x=45
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 2 gallons of a 70 percent acid solution in order to get a 90 percent acid solution?
Set up. .70(2 gal) + X = .90(X + 2 gal) distribute 1.4 + X = .90X + 1.8 X = .90X + .4 .1X = .4 X = 4
How much water should be added to six liters of thirty percent acid solution to dilute twelve percent solution?
6 litres of 30% solution would contain 6*30/100 = 1.8 litres of pure acid.1.8 litres of pure acid, in a 12% solution means the total volume is 1.8*(100/12) = 15 litres in total. So you need to add 9 litres of pure water.6 litres of 30% solution would contain 6*30/100 = 1.8 litres of pure acid.1.8 litres of pure acid, in a 12% solution means the total volume is 1.8*(100/12) = 15 litres in total. So you need to add 9 litres of pure water.6 litres of 30% solution would contain 6*30/100 = 1.8 litres of pure acid.1.8 litres of pure acid, in a 12% solution means the total volume is 1.8*(100/12) = 15 litres in total. So you need to add 9 litres of pure water.6 litres of 30% solution would contain 6*30/100 = 1.8 litres of pure acid.1.8 litres of pure acid, in a 12% solution means the total volume is 1.8*(100/12) = 15 litres in total. So you need to add 9 litres of pure water.
How many ml of 40 percent acid should be added to pure acid to make 75 ml of 65 percent acid?
The total volume of the new acid solution is 75 mL, and this solution is 65%, so the amount of acid in this solution is (0.65)*75mL = 48.75 mLYou have two equations and two unknowns:x + y = 75(0.40)*x + (1)*y = 48.75 {0.40 represents 40% and 1 represents 100% pure}Solve and get x = 43.75 and y = 31.25, so you need 43.75 mL of the 40% acid and 31.25 mL of the pure acid.
How much pure acid should be mixed with 5 gallons of 70 percent acid solution in order to get a 90 percent acid solution?
x=45
How much pure acid should be mixed with 2 gallons of a 50 percent acid solution in order to get an 80 percent acid solution?
pH less than 7
How much pure acid must be added to 12 ounces of a 40 percent acid solution in order to produce 60 percent 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.
If 120 liters of an acid solution is 75 percent acid how much pure acid is there in the mixture?
There are 90 liters of pure acid in the mixture because 120 x 0.75 = 90.
How much pure acid is in 860 milliliters of a 19 solution?
To calculate the amount of pure acid in the solution, you can use the formula: amount of pure acid = volume of solution * concentration of acid. In this case, it would be 860 mL * 0.19 = 163.4 mL of pure acid in the 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 many liters of pure acid are there in 8 liters of a 20 percent solution?
1.6
How much pure acid is in 440 milliliters of 12 solutions?
There are 52.8 milliliters of pure acid in 440 milliliters of a 12% acid solution.
How much pure acid should be mixed with 4 gallons of a 50 percent acid sollution in order to get an 80 percent acid solution?
Assuming that you mean that your pure acid is twice as concentrated as your 50% acid. Pretending your pure acid is at 1 mol/gallon and the 50% acid is 0.5 mol/gallon1 molgallon-1 * x gallons = x mol0.5 molgallon-1 * 4 gallons = 2 mol(x + 2) mol for (x+4) gallons = 0.8(x+2) mol / (x+4) gallons= 0.8 molgallon-1x + 2 = 0.8x + 3.20.2x = 1.2x = 6add 6 gallons of pure acid to the 4 gallons to make 10 gallons of 80% acid solution
How much pure acid should be mixed with 2 gallons of a 70 percent acid solution in order to get a 90 percent acid solution?
Set up. .70(2 gal) + X = .90(X + 2 gal) distribute 1.4 + X = .90X + 1.8 X = .90X + .4 .1X = .4 X = 4
How much pure water must be mixed with 10 liters of a 25 percent acid solution to reduce it to a 10 percent acid solution?
To reduce a 25% acid solution to 10%, the water needed can be calculated using the formula: (10 x (25-10))/(10) = 1.5 liters. So, 1.5 liters of pure water must be mixed with the 10 liters of 25% acid solution to reduce it to 10%.
How much water should be added to six liters of thirty percent acid solution to dilute twelve percent solution?
6 litres of 30% solution would contain 6*30/100 = 1.8 litres of pure acid.1.8 litres of pure acid, in a 12% solution means the total volume is 1.8*(100/12) = 15 litres in total. So you need to add 9 litres of pure water.6 litres of 30% solution would contain 6*30/100 = 1.8 litres of pure acid.1.8 litres of pure acid, in a 12% solution means the total volume is 1.8*(100/12) = 15 litres in total. So you need to add 9 litres of pure water.6 litres of 30% solution would contain 6*30/100 = 1.8 litres of pure acid.1.8 litres of pure acid, in a 12% solution means the total volume is 1.8*(100/12) = 15 litres in total. So you need to add 9 litres of pure water.6 litres of 30% solution would contain 6*30/100 = 1.8 litres of pure acid.1.8 litres of pure acid, in a 12% solution means the total volume is 1.8*(100/12) = 15 litres in total. So you need to add 9 litres of pure water.
