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.
Initially, there are 36 liters of milk and 4 liters of water in the mixture. To increase the water content to 20%, you would need to add 16 liters of water to make a total of 20 liters of water in the new mixture.
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.
The human body does not need any percentage of alcohol for optimal health. Consuming alcohol is a personal choice and should be done in moderation to avoid negative health consequences.
Distilling hand sanitizer to make edible alcohol is not recommended. Hand sanitizer contains ingredients that are toxic if ingested, such as isopropyl alcohol. It is best to purchase food-grade alcohol for consumption, as it is manufactured specifically for this purpose and is safe for human consumption.