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How much water should be added to 1 gallon of pure antifreeze to obtain a solution that is 35 and 8203 antifreeze?
Approx 1.86 gallons.
How many liters of pure antifreeze must be added to 30L of 60 percent antifreeze solution to obtain a 75 percent solution?
60% solution contains 6/10 x 30 ie 18 litres 18 + A = 3/4 (30 + A) 72 + 4A = 90 + 3A 4A - 3A = 90 - 72 A = 18 ie add another 18 litres, giving 36 litres out of 48 which is the required 75%.
How many ml should be added to 100ml of a 25 percent solution to get produce a 20 percent solution?
add 25ml more of solution x * 20 = 100 * 25 x = 25
How many liters of a 7 percent solution of salt should be added to a 19 percent solution in order to obtain 456 liters of a 15 percent solution?
This is an algebra problem. There will be 2 equations and 2 unknowns. Let x = amount of 7% solution, and y = amount of 19% solution. .07x + .19y = .15*456. x + y = 456; or x = 456-y which will substitute into the first equation. .07(456-y) +.19y = .15*456. 31.92-.07y+.19y = 68.4. .12y = 36.48. y = 304L (amount of 19% solution to add). x = 456-y. x = 456 - 304. x = 152L (amount of 7% solution to add).
What must be added to a to obtain 7a?
UM 6a? Can you be a little more specific maybe? But 6a plus a is 7a
How many quarts of pure antifreeze must be added to 6 quarts of a 10 percent antifreeze solution to obtain a 20 percent antifreeze solution?
To find out how many quarts of pure antifreeze must be added to 6 quarts of a 10% antifreeze solution to obtain a 20% solution, we can set up the equation. The initial amount of antifreeze in the solution is (0.10 \times 6 = 0.6) quarts. Let (x) be the amount of pure antifreeze to add. The final solution will be (6 + x) quarts, and we need the total antifreeze to equal (0.20(6 + x)). Setting up the equation (0.6 + x = 0.20(6 + x)) and solving gives (x = 1.2) quarts. Thus, 1.2 quarts of pure antifreeze must be added.
How many quartz of pure antifreeze must be added to 6 quartz of a 10 percent antifreezesolution to obtain a 20 percent antifreeze solution?
4.2 quarts
A car's radiator has a capacity of 15 quarts and it currently contains 12 quarts of a 30 percent antifreeze solution. How many quarts of pure antifreeze must be added to strengthen the solution 40 per?
The original solution has 3.6 quarts of antifreeze in it. The equation then becomes (3.6 + x)/(12 + x) = 0.40, where x is the amount of antifreeze added. X is then equal to 2.
How many quarts of pure antifreeze must be added to 6 quarts of a 30 percent antifreeze solution to abtain a 40 percent antifreeze solution?
If my math is correct it would take an additional ( .6 of a U.S. quart of antifreeze ) to increase a 30 % antifreeze volume to 40 % if the total volume of the mixture is 6 quarts
How many qt of pure antifreeze must be added to 3 qt of a 10 percent antifreeze solution obtain a 20 percent antifreeze solution?
0.6 of a pint.
How much water should be added to 1 gallon of pure antifreeze to obtain a solution that is 60 percent antifreeze?
2 gallons.
How much water should be added to 1 gallon of pure antifreeze to obtain a solution that is 35 and 8203 antifreeze?
Approx 1.86 gallons.
How much water should be added to 1 gallon of pure antifreeze to obtain a solution that is 80 percent antifreeze?
0.25 gallons of water (or 1 quart)
How much water should be added to 1 gallon of pure antifreeze to obtain a solution that is 45 percent antifreeze?
To create a 45% antifreeze solution from 1 gallon (128 ounces) of pure antifreeze, you need to determine how much of the solution should consist of antifreeze. Since you want 45% antifreeze, you can set up the equation: 0.45 * (128 + x) = 128, where x is the amount of water added. Solving for x, you find that you need to add approximately 104 ounces of water to achieve a 45% antifreeze solution.
How much pure 100 percentage antifreeze should be added to five quarts of 20 percentage solution to raise its concentration to 25 percentage antifreeze solution?
I'd go buy a bottle of pre-mixed antifreeze, and just fill it up with that. Less hassle, works great.
How much antifreeze must be added to 12 gallons of 20 percent antifreeze to make a 40 percent antifreeze solution?
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
How many quarts of pure antifreeze must be added to 5 quarts of 20 percent antifreeze to make a 30 percent antifreeze solution?
In the original solution, you have 5 quarts total, 20% antifreeze. So 20% * 5 = 1. You have 4 qt water and 1 qt antifreeze. You will add an amount A, to this mixture. When you add, the new total volume (in quarts) is 5 + A, and the percentage is:(1 + A)/(5 + A) x 100%. So when you start out at A = zero, you have 1/5 x 100% = 20%.So set up the equation (1 + A)/(5 + A) * 100 = 30, and solve for A.Multiply both sides by (5+A) and isolate the term A. Note that since you are adding, the value of A will never be negative (so you don't have to worry about the denominator (5+A) being zero.Answer = 5/7 (quarts) = 0.714285714285155 quarts.
