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∙ 13y agoI honestly have no idea at all...but I looked it up and saw that people explained it out to be 15 liters... I have this question on my test...so I'm putting 15.
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∙ 13y agoTo reduce a 25% acid solution to 10%, the water needed can be calculated using the formula: (10 x (25-10))/(10) = 1.5 liters. So, 1.5 liters of pure water must be mixed with the 10 liters of 25% acid solution to reduce it to 10%.
Let x be the liters of the 30% acid solution and y be the liters of the 60% acid solution. We can set up a system of equations: x + y = 50 (total liters) and 0.3x + 0.6y = 0.57*50 (acid content). Solving this system of equations, we find that x = 20 liters of the 30% acid solution and y = 30 liters of the 60% acid solution.
Let x be the number of liters of the 10% solution needed. The amount of silver iodide in the 5-liter solution is 0.04 * 5 = 0.2 liters. The final amount of silver iodide in the mixture would be 5 * 0.06 = 0.3 liters. Set up the equation: 0.1x + 0.2 = 0.3, solve for x, x = 1 liter. Hence, 1 liter of the 10% solution is needed.
The percentage strength of the solution is 2%. This can be calculated by dividing the mass of chemical (4kg) by the total volume of the solution (200 liters) and then multiplying by 100 to get the percentage.
The molarity can be calculated using the formula: moles of solute divided by liters of solution. In this case, the moles of sucrose is 25, and the liters of solution is 50. This gives a molarity of 0.5 M.
Let x represent the amount of 12% solution and (10-x) represent the amount of 20% solution. The equation to solve is: 0.12x + 0.20(10-x) = 0.14(10). Solving for x gives x = 4, so you need 4 gallons of the 12% solution and 6 gallons of the 20% solution to make 10 gallons of the 14% solution.
A pharmacist mixed a 20 percent solution with a 30 percent solution to obtain 100 liters of a 24 percent solution. How much of the 20 percent solution did the pharmacist use in the mixture (in liters).
210Type your answer here...
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
Mixing 80 liters of 15% solution and 520 liters of 90% solution will give 600 liters of 80% solution.
10
80 liters
0.1125% of polymer solution.
6 litres of 50% + 4 litres of 25%
Let L be the liters to be added. Then (L) x (0.2) + (40 x 0.9) = (L + 40) x 0.6 0.2L + 36 = 0.6L + 24 36 - 24 = (0.6 - 0.2)L 12 = 0.4L L = 12/0.4 = 120/4 = 30 Liters
Let x be the liters of the 30% acid solution and y be the liters of the 60% acid solution. We can set up a system of equations: x + y = 50 (total liters) and 0.3x + 0.6y = 0.57*50 (acid content). Solving this system of equations, we find that x = 20 liters of the 30% acid solution and y = 30 liters of the 60% acid solution.
Let x be the number of liters of the 10% solution needed. The amount of silver iodide in the 5-liter solution is 0.04 * 5 = 0.2 liters. The final amount of silver iodide in the mixture would be 5 * 0.06 = 0.3 liters. Set up the equation: 0.1x + 0.2 = 0.3, solve for x, x = 1 liter. Hence, 1 liter of the 10% solution is needed.
133.33