0
How many liters of pure antifreeze must be added to 30L of 60 percent antifreeze solution to obtain a 75 percent solution?
Wiki User
∙ 14y ago
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%.
Wiki User
∙ 14y ago
Add your answer:
Earn +20 pts
How many quarts of pure antifreeze must be added to 4 quarts of a 20 antifreeze solution to obtain a 40 antifreeze and solution?
You need 1 1/3 quarts of pure antifreeze.
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.
How much water should be added to 1 gallon of pure antifreeze to obtain a solution that is 35 and 8203 antifreeze?
Approx 1.86 gallons.
How many liters of water must be added to 7 liters of a 20 percent acid solution to obtain a 10 percent acid solution?
7 liters of a 20% acid solution consists of 1.4 liters of acid (20% of the total volume) mixed with 5.6 liters of water (80% of the total volume). The amount of acid isn't going to change in the new solution. You are just going to add enough water to make it a 10% solution instead of a 20% solution. So it will be more dilute. That means that 1.4 liters of acid will represent 1/10 of the volume of the new solution. So the total volume of the new solution will be 10 x 1.4 or 14 liters. The amount of water in the new solution will be 14 - 1.4 = 12.6 liters. That is a difference of 12.6 - 5.6 = 7 liters from the amount of water you started with. So you need to add 7 liters of water to the original 20% solution to make it a 10% solution. This makes sense because if you double the amount of the mixture from 7 liters to 14 liters and the amount of acid is unchanged, the solution will be half as strong.
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 qt of pure antifreeze must be added to 3 qt of a 10 percent antifreeze solution obtain a 20 percent antifreeze solution?
0.6 of a pint.
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.
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).
How many quartz of pure antifreeze must be added to 6 quartz of a 10 percent antifreezesolution to obtain a 20 percent antifreeze solution?
4.2 quarts
How much water should be added to 1 gallon of pure antifreeze to obtain a solution that is 60 percent antifreeze?
2 gallons.
How much of water should be add to obtain a solution that is twenty percent antifreeze?
80% water
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 should be added to 1 gallon of pure antifreeze to obtain a solution that is 80 percent antifreeze?
0.25 gallons of water (or 1 quart)
How many liters of a 5.0 percent glucose solution would you need to obtain 75 g of glucose?
15 litres
How many quarts of pure antifreeze must be added to 4 quarts of a 20 antifreeze solution to obtain a 40 antifreeze and solution?
You need 1 1/3 quarts of pure antifreeze.
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.
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.
