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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.
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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?
.80x+.30y=400*.62 .80x+.30y=248 This is equation one. x+y=400 x=400-y This is the second equation. .80(400-y)+.30y=248 320-.80y+.30y=248 -.50y=-72 Y=144 X=256 CHECK .80(256)+.30(144)=400(.62) 204.8+43.2=248 248=248
How do you produce 1L of 10 percent ammonia solution from 25 percent ammonia solution?
To produce 1L of 10% ammonia solution from 25% ammonia solution, you need to dilute the 25% solution by adding a calculated amount of water. To do this, you can calculate the volume of the 25% solution needed and the volume of water needed using the formula: C1V1 = C2V2, where C1 is the initial concentration (25%), V1 is the initial volume, C2 is the final concentration (10%), and V2 is the final volume (1L).
How much water must be added to 40ml of a 25 percent by weight solution to make a 2 percent by weight solution?
To make a 2% solution from a 25% solution, you need to dilute it. Let x ml be the amount of water needed. Using the equation for mass balance: 0.25(40) = 0.02(40 + x), solve for x to find that x = 875 ml of water needs to be added.
How do you dilute 25 percent Glucose and Glutamic acid?
To dilute a 25% glucose and glutamic acid solution, you would mix the solution with an appropriate amount of water. The exact amount of water needed will depend on the desired final concentration of the solution. Calculate the amount of water needed based on the volume and concentration of the original solution.
How much of 35 percent HCl would be needed to make 1 percent HCl?
To make a 1% HCl solution from a 35% HCl solution, you would need to dilute the concentrated solution with water. The ratio of concentrated HCl to water would be approximately 1:34. So, to make 1% HCl, you would mix 1 part of the 35% HCl solution with 34 parts of water.
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?
50liters
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
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 is needed to make a 100 L solution that's 50 percent?
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?
50 Liters of the 60% solution.
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?
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.
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.
How many liters of 93 percent acid solution are needed to make a 9 percent solution with 50 liters of water?
4.84
Jerry is experimenting with chemicals in the laboratory He mixes a solution that is 10 percent acid with a solution that is 30 percent acid How much of the 10 percent acid solution will be needed to m?
10 liters
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?
.80x+.30y=400*.62 .80x+.30y=248 This is equation one. x+y=400 x=400-y This is the second equation. .80(400-y)+.30y=248 320-.80y+.30y=248 -.50y=-72 Y=144 X=256 CHECK .80(256)+.30(144)=400(.62) 204.8+43.2=248 248=248
How much of a 4 percent solution is needed for a 100mg dose?
2.5ml
Mary is experimenting with chemicals in the laboratory She mixes a solution that is 10 percent acid with a solution that is 25 percent acid How much of the 10 percent acid solution will be needed to m?
16 2/3 liters
