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A pharmacist needs to obtain a 70 percent alcohol solution How many ounces of a 30 percent alcohol solution must be mixed with 40 ounces of an 80 percent alcohol solution to obtain 70 percent alcohol?
Wiki User
∙ 14y ago
(Note: This answer assumes that the "ounces" specified are avoirdupois or other weight ounces and that percentages are by weight; otherwise possible volume changes on dilution must by considered.)
The weight of pure alcohol in each solution is the product of the percentage and the total weight of the solution. Therefore, designating the unknown weight of 30 % alcohol as w, from the problem statement 0.30w + 0.80(40) = 0.70(w + 40), or
0.30w + 32 = 0.70w + 28, or 32 - 28 = w(0.70 - 0.30) or w = 4/0.40 = 10 ounces of 30 % alcohol.
Wiki User
∙ 14y ago
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How many ounces of a 16 alcohol solution and a 24 alcohol solution must be combined to obtain 64 of a 19 solution?
40 fl oz of the 16% solution and 24 of the other.
How many liters of 50 percent alcohol solution and 20 percent alcohol solution must be mixed to get obtain 18 liters of 30 percent alcohol solution?
The total volume of solution equals 18 L. So if we let X be the 50% solution & Y be the 20% solution we can solve a set of two equations with 2 unknowns: X + Y = 18 (total volume of solution is 18 L) 0.5X + 0.2Y = 18*(0.30) (the total amount of alcohol in the new 18L solution) Solving simultaneously, multiply the 1st equation by 0.5 then substract to solve for Y. X + Y = 18 {*0.5} => 0.5X + 0.5Y = 9 0.5X + 0.2Y = 5.4 => 0.5X + 0.2Y = 5.4 ------------------------------------------------------- 0.3Y = 3.6 Y = 12 Then substitute for Y back into the 1st equation & solve to get X=6. So you need 12 L of the 20% solution & 6 L of the 50% solutions to mix together to get 18 L of a 30% alcohol 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 many liters of pure antifreeze must be added to 30L of 60 percent antifreeze solution to obtain a 75 percent solution?
60% solution contains 6/10 x 30 ie 18 litres 18 + A = 3/4 (30 + A) 72 + 4A = 90 + 3A 4A - 3A = 90 - 72 A = 18 ie add another 18 litres, giving 36 litres out of 48 which is the required 75%.
How many gallons of a solution that is 75 antifreeze must be mixed with 4 gallons of a 30 solution to obtain a mixture that is 50 antifreeze?
You will need 3.2 gallons.
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?
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).
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?
10
How much water must be added to 12 L of a 40 percent solution of alcohol to obtain a 30 percent solution?
4 litres
How much water should be mixed to 12ml of alcohol to obtain a 12 percent of alcohol solution?
88 ml
What fraction of 10 percent alcohol should be mixed with 70 percent alcohol to obtain 50 percent alcohol?
2/3 of 70% and 1/3 of 10%
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?
600ml.
How many gallons of a 9 percent salt solution must be mixed with 25 gallons of a 31 percent solution to obtain a 20 percent solution?
25 gallons
How many ounces of a 16 alcohol solution and a 24 alcohol solution must be combined to obtain 64 of a 19 solution?
40 fl oz of the 16% solution and 24 of the other.
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?
10 liters.
How many gallons of 3 percent salt solution must be mixed with 50 gallons of 7 percent salt solution to obtain a 5 percent salt solution?
50 gallons @ 3% must be added.
How many grams of sugar must be added to 60 grams of a solution that is 32 percent sugar to obtain a solution that is 50 percent sugar?
40.8 grams
How many liters of 50 percent alcohol solution and 20 percent alcohol solution must be mixed to get obtain 18 liters of 30 percent alcohol solution?
The total volume of solution equals 18 L. So if we let X be the 50% solution & Y be the 20% solution we can solve a set of two equations with 2 unknowns: X + Y = 18 (total volume of solution is 18 L) 0.5X + 0.2Y = 18*(0.30) (the total amount of alcohol in the new 18L solution) Solving simultaneously, multiply the 1st equation by 0.5 then substract to solve for Y. X + Y = 18 {*0.5} => 0.5X + 0.5Y = 9 0.5X + 0.2Y = 5.4 => 0.5X + 0.2Y = 5.4 ------------------------------------------------------- 0.3Y = 3.6 Y = 12 Then substitute for Y back into the 1st equation & solve to get X=6. So you need 12 L of the 20% solution & 6 L of the 50% solutions to mix together to get 18 L of a 30% alcohol solution.
