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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.
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.
Is carbonic acid a pure substance?
Carbonic acid is not considered a pure substance because it is a weak acid that can dissociate into carbon dioxide and water. As such, it is a solution rather than a pure substance.
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
What is the ph number for sulphric acid?
The pH value of any acidic solution depends on the strength of the acid in the particular solution. In this instance, a dilute sulfuric acid solution shows a higher value of pH whereas the concentrated acid shows a very low value.
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 is in 330 mL of a 18 percent acid solution?
18% of 330 = 59.4 So there are 59.3ml of pure acid in 330ml of 18% acid solution.
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 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.
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
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 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.
Is carbonic acid a pure substance?
Carbonic acid is not considered a pure substance because it is a weak acid that can dissociate into carbon dioxide and water. As such, it is a solution rather than a pure substance.
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.
A flask contains 300.0cm3 of a 0.500 acid solution How many moles of the pure acid must have dissolved?
All of the moles of pure acid will have dissolved in the flask.
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 water should be added to 3 gallons of pure acid in order to obtain the 15 acid solution?
You need 17 gallons for a 15% volume/volume mixture.
