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.
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!
To obtain a 12% alcohol solution, you would need to mix 12ml of alcohol with 48ml of water. This would give you a total volume of 60ml, with 12% of it being alcohol.
6 litres of 50% + 4 litres of 25%
Most de-icer sprays for windshields and car locks contain methanol (methyl alcohol) and propylene glycol (antifreeze). These can be harmful to humans and pets, so should be used sparingly and wiped up after use.
You don't need alcohol at all. It's not necessary.
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.