Can you comment a more specific question? Are you looking for how many liters it is? it would be about 7.57 liters, but I'll need to know more information to give you the volume of the product.
To get a 2% acid solution, you need to dilute the 50% acid solution with water. Since the final volume is 2 gallons, you will need to mix 2 gallons of water with the 2 gallons of 50% acid solution to get a 2% acid solution.
3179.4 g
About 7,570.824 mL is needed to equal two US gallons.
2 gallons.
pH less than 7
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
There are 2 quarts in .5 gallons.
There are 256 fluid ounces in 2 gallons of paint.
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.
2 gallons is 8 quarts
Fill the 6 gallon cylinder half full. Solution Method 3 gallons / 6 gallons = 1/2 full
Let M equal the gallons of the 90% mixing solution. Let F equal the gallons of the final solution. So:90 + M = F.Also, the number of gallons of pure antifreeze in the final will equal the sum of the gallons of antifreeze in the two mixing parts:Original solution: ( 90 gal )*(0.15) = 13.5 gal [0.15 represents 15%]Mixing solution: M*0.90Final solution: F*0.80So 13.5 + M*0.90 = F*0.80Now you have 2 linear equations and 2 unknowns, you can solve for M & F, using your favorite method: M = 585 and F = 675. Add 585 gallons of the 90% to get 675 gallons of 80% solution.