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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 solution that is 62 percent acid?
Wiki User
∙ 15y ago
You have enough information to obtain two equations. 1. You know that when the two liquids are added together, you need 400L total. 2. You also know that the 80% acid times its volume plus 30% acid times its volume must equal 62% times the 400L. 3. Write the two equations. Let x = volume 80% acid & y = volume 30% acid. Therefore from (1) above: x + y = 400; equation A From (2) above: .8x + .3y = .62*400; equation B 4. You have two equations, two unknowns. Solve for x & y. 5. Multiply equation A by -.8 and add equations A & B. Note x adds out to 0. Equation A now becomes -.8x - .8y = -.8*400 or -.8x - .8y = -320. -.8x - .8y = -320 +.8x + .3y = 248 (note .62*400 = 248) So -.5y = -72 and divide both sides by -.5 which yields y = 144. 6. Use equation A to solve for x since we now know "y"; x + 144 = 400 or x = 256. 7. Answer: 256L of 80% acid & 144L of 30% acid.
Wiki User
∙ 15y ago
Let x be the amount of the 30% acid solution needed. The total amount of acid in the final solution is 0.62*400L. Since the amount of acid from the 30% solution is 0.3x and from the 80% solution is 0.8(400-x), you can set up and solve an equation to find x. Solving x = 240L, we need 240L of the 30% solution.
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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.
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A chemist has one solution that is 60 percent chlorinated and another solution that is 40 percent chlorinated How much of the first 60 percent solution is needed to make a 100 L solution that is 5?
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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 400 L solution that is 62 percent acid?
144liters
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50
A chemist has one solution that is 60 percent chlorinated and another that is 40 percent chlorinated How much of each solution is needed to make a 100 liter solution that is 50 percent chlorinated?
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A chemist has one solution that is 60 percent chlorinated and another solution that is 40 percent chlorinated How much of the first 60 percent solution is needed to make a 100 L solution that is 50?
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A chemist has one solution that is 80 percent acid and another solution that is 30 percent acid How much of the first 80 percent solution is needed to make a 400 L solution that is 62 percent acid?
To create a 400 L solution that is 62% acid, you would need 200 L of the 80% acid solution and 200 L of the 30% acid solution. This would result in a final solution with the desired concentration.
When A chemist has one solution that is 80 acid and another solution that is 30 acid. How much of the first (80) solution is needed to make a 400 L solution that is 62 acid?
You need 256 litres.
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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.
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