none.... 'cause the acid would eat through the tank and seep into the water supply.
10 liters
6 litres of 50% + 4 litres of 25%
The wrong amino acid will be used
some of the acid has dissociated APEX
7 liters of a 20% acid solution consists of 1.4 liters of acid (20% of the total volume) mixed with 5.6 liters of water (80% of the total volume). The amount of acid isn't going to change in the new solution. You are just going to add enough water to make it a 10% solution instead of a 20% solution. So it will be more dilute. That means that 1.4 liters of acid will represent 1/10 of the volume of the new solution. So the total volume of the new solution will be 10 x 1.4 or 14 liters. The amount of water in the new solution will be 14 - 1.4 = 12.6 liters. That is a difference of 12.6 - 5.6 = 7 liters from the amount of water you started with. So you need to add 7 liters of water to the original 20% solution to make it a 10% solution. This makes sense because if you double the amount of the mixture from 7 liters to 14 liters and the amount of acid is unchanged, the solution will be half as strong.
It makes no difference what's in the 16 gallons, or whether they're empty.(16 gallons) x (231 cubic inches per gallon) / (1,728 cubic inches per cubic foot) = 2.1389 cubic feet (rounded)
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.
It will not be accurate, as mixing 1 gallon of acid with one of water will not make 2 gallons. But approximately 7714 gallons of 100% acid would be needed to make a 30% volume/volume acid solution. Your problem now is that very few liquid acids come as 100%, and most that do are very dangerous around water.
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
The answer will depend on the acid, which do not have the same density.
about 5000 gallons to 200 gallons they are all different sizes
Also the concentration of HCl and NaOH are needed to be known (not only the amounts) to answer this question.
x = amount of acid to add. y = final volume. 5 gal + x = y original amount of acid + acid added = final amount of acid (5 X .2) + x = O.5y Subtract the second equation from the first one. 5 - 1 + x - x = y - 0.5y 4 = 0.5y 8 = y Therefore the final volume is 8 gallons. 5 gal + x = 8 x = 3 gal. the amount of pure acid to add. Check the answer 5 X .2 = 1 gal of acid in original solution. 3 gallons added = 4 gallons total acid in solution. 4 gallons total acid in final solution of 8 gallons total solution = 50% acid.
You need 17 gallons for a 15% volume/volume mixture.
100,000 gallons a year
One pound of phosphoric acid is equal to 0.07111 gallons. It is also equal to 9.103 fluid ounces or 1.139 cups or 0.2854 quarts.
Depending on the concentration (and the density) of sulfuric acid. For sulfuric acid 30 % the volume is approx. 3,7 gallons.