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How many gallons of a solution that is 75 antifreeze must be mixed with 4 gallons of a 30 solution to obtain a mixture that is 50 antifreeze?
Wiki User
∙ 6y ago
You will need 3.2 gallons.
Wiki User
∙ 6y ago
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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 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 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 ounces of a 16 alcohol solution and a 24 alcohol solution must be combined to obtain 64 of a 19 solution?
40 fl oz of the 16% solution and 24 of the other.
How many liters of water must be added to 7 liters of a 20 percent acid solution to obtain a 10 percent acid solution?
7 liters of a 20% acid solution consists of 1.4 liters of acid (20% of the total volume) mixed with 5.6 liters of water (80% of the total volume). The amount of acid isn't going to change in the new solution. You are just going to add enough water to make it a 10% solution instead of a 20% solution. So it will be more dilute. That means that 1.4 liters of acid will represent 1/10 of the volume of the new solution. So the total volume of the new solution will be 10 x 1.4 or 14 liters. The amount of water in the new solution will be 14 - 1.4 = 12.6 liters. That is a difference of 12.6 - 5.6 = 7 liters from the amount of water you started with. So you need to add 7 liters of water to the original 20% solution to make it a 10% solution. This makes sense because if you double the amount of the mixture from 7 liters to 14 liters and the amount of acid is unchanged, the solution will be half as strong.
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 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 3 gallons of pure acid in order to obtain the 15 acid solution?
You need 17 gallons for a 15% volume/volume mixture.
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 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 gallons of a 9 percent salt solution must be mixed with 25 gallons of a 31 percent solution to obtain a 20 percent solution?
25 gallons
How many gallons of 30 percent antifreeze should be mixed with 10 percent gallons of 93 percent antifreeze to obtain a 65 percent antifreeze mixture?
Suppose there are G gallons of the 30% mix. Then G gallons of 30% contain 0.3*G gallons of the active ingredient. Also 10% gallons = 0.1 gallons of 93% contain 0.093 gallons of the active ingredient. Therefore, the total volume is G+0.1 gallons which contains 0.3*G + 0.093 gallons of the active ingredient. So its strength is (0.3*G + 0.393)/(G+0.1) which is 65% or 0.65 Thus 0.3*G + 0.093 = 0.65*G + 0.065 So that 0.028 = 0.35G Or G = 0.08 gallons
How much of water should be add to obtain a solution that is twenty percent antifreeze?
80% water
How many gallons of 3 percent salt solution must be mixed with 50 gallons of 7 percent salt solution to obtain a 5 percent salt solution?
50 gallons @ 3% 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
