There are 3 litres of alcohol in your starting mixture of 4 litres. If you add 6 litres of water you will have 3 litres of alcohol in a total of 10 litres. This is the required strength.
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.
684 ml
x = volume of 95% y = volume of 30% so (x+y) is volume of 70% 0.95 x + 0.30 y = 0.70(x+y) 0.95 x + 0.30 y = 0.70x + 0.70y 0.25x = 0.40y x/y = 0.40/0.25 = 8/5 for every 8 volumes of 95% mix in 5 volumes of 30%the volumes can be ml, liters, cups, pints etc.
The question is best solved using basic algebra. You need 20 gallons of 32% alcohol. This will contain 0.32*20 = 6.4 gallons of pure alcohol. Now suppose you have X gallons of 25% alcohol in the mixture. That contains 0.25X gallons of pure alcohol. Also, since you have 20 gallons in total, you must have 20-X gallons of the 35% alcohol. This will contain 0.35*(20-X) = 7 - 0.35X gallons of pure alcohol. Then, the total amount of pure alcohol is 0.25X + 7 - 0.35X = 7 - 0.1X gallons. So you have 7 - 0.1X = 6.4 or 0.6 = 0.1X or X = 6. So the answer is 6 gallons of 25% alcohol and 14 gallons of the stronger stuff!
Let x = ounces of 50% solution, and y = ounces of 1% solution. So that we have: 0.5x + 0.01y = 8(0.2) which is a linear equation in two variables, meaning there are infinitely many choices of mixing those solutions.
If you're talking about 10 liters of water and not percent, then 10 liters. But then you'll have 60 liters of mixture. It would be 2/5 or 0.4 x 50 = 20 this would make a mix of 20/50. Waterman
10 liters.
2/3 of 70% and 1/3 of 10%
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.
88 ml
684 ml
70%
x = volume of 95% y = volume of 30% so (x+y) is volume of 70% 0.95 x + 0.30 y = 0.70(x+y) 0.95 x + 0.30 y = 0.70x + 0.70y 0.25x = 0.40y x/y = 0.40/0.25 = 8/5 for every 8 volumes of 95% mix in 5 volumes of 30%the volumes can be ml, liters, cups, pints etc.
The question is best solved using basic algebra. You need 20 gallons of 32% alcohol. This will contain 0.32*20 = 6.4 gallons of pure alcohol. Now suppose you have X gallons of 25% alcohol in the mixture. That contains 0.25X gallons of pure alcohol. Also, since you have 20 gallons in total, you must have 20-X gallons of the 35% alcohol. This will contain 0.35*(20-X) = 7 - 0.35X gallons of pure alcohol. Then, the total amount of pure alcohol is 0.25X + 7 - 0.35X = 7 - 0.1X gallons. So you have 7 - 0.1X = 6.4 or 0.6 = 0.1X or X = 6. So the answer is 6 gallons of 25% alcohol and 14 gallons of the stronger stuff!
The answer varies in each different liquor, and should be listed on the bottle as "proof". The "Proof" is twice the alcohol percentage, so 40 proof is 20% alcohol, 180 proof is 90% alcohol.
Add 0,8 L of water.
Wikipedia has an article on Alcohol Rubs here: http://en.wikipedia.org/wiki/Hand_sanitizer They typically use ethanol (standard drinking alcohol mixed with something to render it undrinkable) or isopropanol ("rubbing alcohol" which is poisonous — not in a good way — if swallowed). A concentration of 60-70 percent is needed. Standard rubbing alcohol should do the trick, as would a 120+ proof liquor, although both will dry out your skin. You could mix in a small amount of glycerine for moisturizing, and possibly a thickener, but you risk diluting the alcohol to where the mixture is no longer effective. A mixture of nine parts 70% rubbing alcohol to one part glycerine should give you a good (but runny) substitute at a lower price.