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To prepare an 80% glycerol solution, you would mix 80 parts glycerol with 20 parts water. For example, you could mix 80 mL of glycerol with 20 mL of water to make 100 mL of an 80% glycerol solution. Make sure to thoroughly mix the glycerol and water until the solution is homogenous.
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A chemist is making 200 L of a solution that is 62 percent acid He is mixing an 80 percent acid solution with a 30 percent acid solution How much of the 80 percent acid solution will he use?
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.
How can you make 2 percent tween80?
To make a 2% Tween 80 solution, you would typically mix 2g of Tween 80 with 98g of water or another appropriate solvent to make a total of 100g of solution. This will give you a 2% concentration of Tween 80 in the final solution.
A chemist has one solution that is 80 percent acid and another solution that is 30 percent acid how much of the second 30 percent solution is needed to make a 400L that is 62 percent acid?
To find out how much of the 30% acid solution is needed, we can set up a weighted average equation. Let x be the volume of the 30% solution needed. Since the final solution is 62% acid, we can write the equation as: 0.80*(400-x) + 0.30x = 0.62*400. Solving this equation, we find x to be 280 liters. Therefore, 280 liters of the 30% acid solution is needed to make a 400L solution that is 62% acid.
How much of a 6 percent solution and a 12 percent solution would use to make 240 milliliters of a 7 percent electrolyte?
Let x be the volume of the 6% solution and y be the volume of the 12% solution. We have the following system of equations: x + y = 240 (total volume) 0.06x + 0.12y = 0.07 * 240 (electrolyte amount) Solving this system, we find x = 160 mL of the 6% solution and y = 80 mL of the 12% solution to make 240 mL of a 7% electrolyte.
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
