For liquids (2% volume/volume): Simply add 2% of the volume to the water. This comes out to 0.06 gallons or 0.96 cups. Note: It is probably more accurate to add 0.06 gallons of whatever to 2.94 gallons of water but honestly errors in measurement will most likely be greater than this small difference. For solids (2% mass/volume): 3 gallons of water are approximately 11.356 liters which is 11356 ml. Since water has a density of roughly 1 g/ml this is equivalent to 11356 g. To make a 2% solution, you would need to add 2% of this weight of whatever you are dissolving which comes out to 0.227 g.
You stare at it
4.84
98 mL
4 gallons Let x be the amount of antifreeze needed to be added. We know that the total amount of antifreeze in the new solution must equal the amount of antifreeze in the old solution + x: .40*(x+12)=x+.20*12 .60x=2.40 x=4 gallons
3/32 gallons = 1.5 cups.
3179.4 g
Let x represent the amount of 12% solution and (10-x) represent the amount of 20% solution. The equation to solve is: 0.12x + 0.20(10-x) = 0.14(10). Solving for x gives x = 4, so you need 4 gallons of the 12% solution and 6 gallons of the 20% solution to make 10 gallons of the 14% solution.
A one percent solution means 1 part beach and 99 parts water. So we can answer the question by solving the equation 1 part bleach/99 parts water=x gallons bleach/20 gallons water. You can cross multiply. This is the same as 99x=20 so x=20/99 or .20 Let's check .20 gallons bleach/ 20 gallons water=.01 which is 1% .2 gallons is the same as 3.2 cups or 25.6 fluid oz.
It will not be accurate, as mixing 1 gallon of acid with one of water will not make 2 gallons. But approximately 7714 gallons of 100% acid would be needed to make a 30% volume/volume acid solution. Your problem now is that very few liquid acids come as 100%, and most that do are very dangerous around water.
To make 10 gallons of a 50% sodium hydroxide solution, you would need 10 pounds of sodium hydroxide. This is because the percentage indicates the weight of sodium hydroxide in the solution. Hence, in a 50% solution, half of the weight of the solution is sodium hydroxide.
About 80ml of water must be added to 40ml of a 25 percent by weight solution to make a 2 percent by weight solution.
Let x represent the gallons of 10% ammonia solution. The total volume of the mixture is x + 50 gallons. The equation for the mixture is: 0.10x + 0.30(50) = 0.15(x + 50). Solving this equation gives x = 50 gallons of the 10% ammonia solution needed.
To make a 1% solution, you would need to mix the pint of herbicide with 7.5 gallons of water.
To make a 9 percent saline solution, start by preparing a 100 percent salt solution. With a bottle of 100 percent salt water, take 9 percent and dilute with distilled water to make a 9 percent saline solution.
70 gallons of 20% solution contains 70*0.2 = 14 gallons of antifreeze. Suppose you need G gallons of the 80% antifreeze solution. This will contain 0.8*G gallons of antifreeze. Total volume of solution = G + 70 gallons Volume of antifreeze required in this solutions to make it a 70% solution is 0.7*(G + 70) = 0.7G + 49 gallons. Volume of antifreeze = 14 + 0.8G gallons So 0.7G + 49 = 14 + 0.8G 0.7G + 35 = 0.8G 35 = 0.1G 350 = G Answer: 350 gallons.
You stare at it
You dillute it with some more of the same solvent used for the 6 percent solution - pressumably water.