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How much of a 50 percent solution should be added to 1 liter to make a 25 percent solution?
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.
A chemist needs 90 milliliters of a 77 percent solution but has only 75 percent and 84 percent solutions available Find how many milliliters of each that should be mixed to get the desired solution?
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.
How much water should be added to 16ml of 9 percent alcohol solution to reduce the concentration to 8 percent?
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).
What conversion factor should be used to convert from gigaliters and liters?
Multiply by one billionth.
What are some steps after coming up with a solution?
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.
How many liters of a 20 percent solution of saline should a nurse mix with 10 liters of a 50 percent solution to obtain a mixture that is 40 percent saline?
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.
A chemist has one solution which is 50 percent acid and a second solution which is 25 percent acid how much of each should be mixed to make a 10 liters of a 40 percent acid solution?
6 litres of 50% + 4 litres of 25%
A solution of 30 percent acid and a solution of 60 percent acid are to be mixed to produce 50 liters of 57 percent acid How many liters of each solution should be used?
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.
How much of an 18 percent solution of sulfuric acid should be added to 360 ml of a 10 percent solution to obtain a 15 percent 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.
How many liters of a 7 percent solution of salt should be added to a 19 percent solution in order to obtain 456 liters of a 15 percent 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).
How much water should be added to 1 gallon of pure antifreeze to obtain a solution that is 60 percent antifreeze?
2 gallons.
How many liters of a 30 solution should be added to a 90 solution to make 18 liters of a 75 solution?
4.5 litres of a 30% solution to the appropriate quantity of the 90% solution.
How much water should be added to 15 g of salt solution to obtain 15 percent salt solution?
Dissolve 15 g salt in 100 mL water.
Jorge has four liters of 30 percent acid solution how much water should he add to make 25 percent acid 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.
How much 50 percent antifreeze solution and 40 percent antifreeze solution should be combined to give 50 gallons of 46 percent antifreeze solution?
How much 50 percent antifreeze solution and 40 percent antifreeze solution should be combined to give 50 gallons of 46 percent antifreeze solution?
A chemist wants to make a 10 solution of fertilizer. how much water and how much of a 30 solution should the chemist mix to get 30 L of a 10 solution?
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 of a ten percent solution should be added to two cups of twenty percent solution to make a twelve percent solution?
2%
