Add an 0.4 L of solvent to 0.6 L solvent:
50% * 0.4 L = 20% * (0.4 +0.6) L
Take 100 grams of 5% solution and do one of the following:Mix 95 g of it with 5 grams of sugar to end up with 100 g of 10% solution, or, when you are short of sugar:Evaporate 50 grams of water from 100 g of the 5% solution to end up with 50 g of 10% solution.
Let x represent the gallons of 10% ammonia solution. The total volume of the mixture is x + 50 gallons. The equation for the mixture is: 0.10x + 0.30(50) = 0.15(x + 50). Solving this equation gives x = 50 gallons of the 10% ammonia solution needed.
First you must either know how much 50% NaOH you want to start with or know how much 2% NaOH you need at the end. Let's say you want 1L of the 2% NaOH. Use this equation, and solve for x, where in this case, x is the volume of 50% NaOH needed. (50% NaOH)x = (2%) * 1000 ml 0.5x = 0.02(1000) x = 20/0.5 = 40 Therefore, 40 ml of 50% NaOH in 960 ml water will produce 2% NaOH. Use a 1000 ml volumetric flask to be most precise.
A 50% NaOH aqueous solution means that the solution contains 50% sodium hydroxide (NaOH) by weight and the rest is water. This concentration indicates that for every 100 grams of the solution, 50 grams is NaOH.
To make 10 gallons of a 50% sodium hydroxide solution, you would need 10 pounds of sodium hydroxide. This is because the percentage indicates the weight of sodium hydroxide in the solution. Hence, in a 50% solution, half of the weight of the solution is sodium hydroxide.
120liters
120
To make a 10 percent solution, you would need to dilute the 50 percent solution by adding 4 ml of solvent to 1 ml of the 50 percent solution. This will result in a total volume of 5 ml with a 10 percent concentration.
50
50 Liters of the 60% solution.
To create a 50% chlorinated solution from the 60% and 40% solutions, the chemist will need to mix the two in equal amounts. Therefore, 50 L of the 60% solution and 50 L of the 40% solution are needed to make a 100 L solution that is 50% chlorinated.
Take 100 grams of 5% solution and do one of the following:Mix 95 g of it with 5 grams of sugar to end up with 100 g of 10% solution, or, when you are short of sugar:Evaporate 50 grams of water from 100 g of the 5% solution to end up with 50 g of 10% solution.
How much 50 percent antifreeze solution and 40 percent antifreeze solution should be combined to give 50 gallons of 46 percent antifreeze solution?
Let x = ounces of 50% solution, and y = ounces of 1% solution. So that we have: 0.5x + 0.01y = 8(0.2) which is a linear equation in two variables, meaning there are infinitely many choices of mixing those solutions.
mary mixed 2l of an 80% acid solution with 6l of a 20% acid solution. what was the percent of acid in the resulting mixture
4.84
Let x = the amount of 20% solution Let x + 10 = the amount of the final solution. So we have: (.20)x + (.50)(10) = (.40)(x + 10) .20x + 5 = .40x + 4 .20x = 1 x = 5 liters of 20% solution of saline.