Q: How many liters of a 20 percent solution of acid should be added to 10 liters of a 30 percent solution of acid to obtain a 25 percent solution?

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Mix this 50% solution in equal quantities with water(?) to halve it's strength. So use 1 litre of the 50% solution and 1 litre of water of that's what you are diluting it with.

Let v be the capacity in milliliters of the 75% solution required then (90 - v) is the required capacity of the 84% solution needed. 75/100v + 84/100(90 - v) = 77/100 x 90 75v + 84(90 - v) = 77 x 90 75v + 7560 - 84v = 6930 9v = 7560 - 6930 = 630 v = 70 : therefore (90 - v) = 20 The mixed solution requires 70ml of the 75% solution and 20ml of the 84% solution to create 90ml of 77% solution.

The way I would do this is to first determine the amount of alcohol and that would be 0.09 x 16 ml = 1.44 ml then I would ask myself (0.08 * x) = 1.44 Therefore to have an 8% solution we would have a total of 18 ml of solution (by solving the above equation), thus we would have to add 2 ml of water (18 ml - 16 ml).

Multiply by one billionth.

If you mean for a math problem, after coming up with a solution you should usually check the solution in the original equation, to be safe.

Related questions

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.

6 litres of 50% + 4 litres of 25%

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.

To determine the amount of 18% solution to add: Let x = volume of 18% solution to be added. 0.18x + 0.10(360) = 0.15(x + 360) Solving for x, you would need to add 75 ml of the 18% solution to the 360 ml of 10% solution to obtain a 15% solution.

This is an algebra problem. There will be 2 equations and 2 unknowns. Let x = amount of 7% solution, and y = amount of 19% solution. .07x + .19y = .15*456. x + y = 456; or x = 456-y which will substitute into the first equation. .07(456-y) +.19y = .15*456. 31.92-.07y+.19y = 68.4. .12y = 36.48. y = 304L (amount of 19% solution to add). x = 456-y. x = 456 - 304. x = 152L (amount of 7% solution to add).

2 gallons.

Dissolve 15 g salt in 100 mL water.

4.5 litres of a 30% solution to the appropriate quantity of the 90% solution.

Jorge needs to add 2 liters of water to the 30% acid solution to make a 25% acid solution. This can be calculated using a dilution formula: initial acid amount / final total amount = final acid concentration.

30 liters of a 10 % solution of fertilizer has .1(30) = 3 liters of fertilizer 1 liter of 30% solution has .3 liter of fertilizer 10 liters of 30% solution has 3 liters of fertilizer so, the chemist needs 10 liters of the 30% solution and 20 liters of water to make 30 liters of a 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?

To obtain a solution that is twenty percent antifreeze, you would add 4 parts water to 1 part antifreeze. This means that for every 1 part of antifreeze, you would add 4 parts of water. This would result in a total of 5 parts of solution, with 1 part being antifreeze and 4 parts being water, achieving a solution that is twenty percent antifreeze.