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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 liters of a 20 percent solution of acid should be added to 10 liters of a 30 percent solution of acid to obtain a 25 percent solution?
10 liters.
How much water must be added to 12 L of a 40 percent solution of alcohol to obtain a 30 percent solution?
4 litres
How much of a 50 percent solution should be added to 1 liter to make a 25 percent solution?
Mix this 50% solution in equal quantities with water(?) to halve it's strength. So use 1 litre of the 50% solution and 1 litre of water of that's what you are diluting it with.
How much water should be added to 16ml of 9 percent alcohol solution to reduce the concentration to 8 percent?
The way I would do this is to first determine the amount of alcohol and that would be 0.09 x 16 ml = 1.44 ml then I would ask myself (0.08 * x) = 1.44 Therefore to have an 8% solution we would have a total of 18 ml of solution (by solving the above equation), thus we would have to add 2 ml of water (18 ml - 16 ml).
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
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 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
How many quarts of pure antifreeze must be added to 4 quarts of a 20 antifreeze solution to obtain a 40 antifreeze and solution?
You need 1 1/3 quarts of pure antifreeze.
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 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 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 80 percent antifreeze?
0.25 gallons of water (or 1 quart)
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 water should be added to 30 gallons of a solution that is 70 percent antifreeze in order to get a mixture that is 60 percent antifreeze?
First off do not use tap water. Use only distilled water. You will need to add 3 gallons of distilled water to the solution to get 60% antifreeze and 40% distilled water or a 60/40 mix.
How many gallons of 80 percent antifreeze solution must be mixed with 100 gallons of 10 percent antifreeze to get a mixture that is 70 percent antifreeze?
600 gallons. To solve this think about the amount of antifreeze in the solution. When 10% antifreeze is added to x times as much 80% antifreeze, the resultant percentage antifreeze will be: (10 + 80x)/(1 + x) So to obtain a 70% solution, x will need to solve: (10 + 80x)/(1 + x) = 70 ⇒ 10 + 80x = 70 + 70x ⇒ 10x = 60 ⇒ x = 6 So to 100 gallons of 10% antifreeze solution, 100 x 6 = 600 gallons of 80% antifreeze solution will be needed to make it a 70% solution. Consider adding 100 gallons of 10% antifreeze and 100 gallons of 80% antifreeze together and then taking half the resultant solution, that is so that you have 100 gallons of the mixture; the amount of antifreeze in this 100 gallons is (10 gallons + 80 gallons) ÷ 2 = 45 gallons, that is a 45% mixture. Now consider adding 100 gallons of 10% antifreeze and 200 gallons of 80% antifreeze together and then taking a third of the resultant solution, that is so that you have 100 gallons of the mixture; the amount of antifreeze in this 100 gallons is (10 gallons + 2 x 80 gallons) ÷ 3 = 562/3 gallons, that is a 562/3% mixture. If x times as much of the 80% antifreeze is added to the If 1 gallon of 10% antifreeze was added together with 80% antifreeze, the resultant mixture will still be (10+90)/2 = 45%. So if x times as much of the 80% Consider the amount of antifreeze in the solution. When two solutions of the same volume is added, the amount of antifreeze in the same volume is half the new amount. Consider the ratio of anti-freeze to water in each of the solutions in fraction form of antifreeze/water: 10% is 10/90 70% is 70/30 80% is 80/20 When two ratios are added together, the amount of antifreeze in the solution is added together and the amount of water is added together to give some odd fraction maths: 10% + 80% = (10+80)/(90+20) = 90/110 =
How much of a ten percent solution should be added to two cups of twenty percent solution to make a twelve percent solution?
2%
