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.
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.
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.
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.
How much 50 percent antifreeze solution and 40 percent antifreeze solution should be combined to give 50 gallons of 46 percent antifreeze solution?
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
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.
4.84