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.
10
25 gallons
50 gallons @ 3% must be added.
No. The reulting concentration (percent) must be between the two components. So, with the two acids you are mixing, you cannot get an acid that is less than 10% or more than 40%
4.2 quarts
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).
10
10 liters.
15 litres
Suppose L litres are required.@ 40% it will contain 0.4L of the active ingredient.Then total volume of mixture = L + 100 litresand volume of active ingredient = 0.4L + 25 litresThe strength is (0.4L + 25)/(L + 100) = 30% = 0.3(0.4L + 25) = 0.3*(L + 100)0.4L + 25 = 0.3*L + 300.1L = 5L = 50 litres.
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.
600ml.
25 gallons
4 litres
50 gallons @ 3% must be added.
The purpose is to obtain a homogeneous mixture.
40.8 grams