0
How much water should be added to 3 gallons of pure acid in order to obtain the 15 acid solution?
Add your answer:
Earn +20 pts
How much water should be added to 1 gallon of pure antifreeze to obtain a solution that is 60 percent antifreeze?
2 gallons.
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 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 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 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 35 and 8203 antifreeze?
Approx 1.86 gallons.
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 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 many gallons of a 80 percent acid solution must be mixed with 60 gallons of a 16 percent solution to obtain a solution that is 60 percent?
Let x be the gallons of the 80% acid solution needed. The amount of acid in the 80% solution is 0.8x, and the amount in the 16% solution is 0.16*60=9.6. We want a total of (x+60) gallons of solution with 60% acid, so we have the equation 0.8x + 9.6 = 0.6(x+60). Solving for x gives x = 24 gallons.
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 should be added to 15 g of salt solution to obtain 15 percent salt solution?
Dissolve 15 g salt in 100 mL water.
How much of an 18 percent solution of sulfuric acid should be added to 360 ml of a 10 percent solution to obtain a 15 percent solution?
To determine the amount of 18% solution to add: Let x = volume of 18% solution to be added. 0.18x + 0.10(360) = 0.15(x + 360) Solving for x, you would need to add 75 ml of the 18% solution to the 360 ml of 10% solution to obtain a 15% solution.
How many gallons of pure water must be added to 10 gallons of 30 saltwater to make a solution that is 25 salt water?
I take it that there should be percentage signs in this question. How many gallons of pure water must be added to 10 gallons of 30% saltwater to make a solution that is 25% saltwater?" If 30% of the 10 gallons is saltwater that means that 3 gallons are saltwater. As only pure water will be added, we need to find out how many gallons = 100% if 3 gallons = 25%. The relationship of 100 to 25 is 4 : 1, so there will need to be a total of (3 x 4) gallons at the end of the operation. As 3 x 4 = 12, and as we started with 10 gallons, that means that we need to add 2 gallons of pure water.
What happens when a solution of nitric acid is added to a solution of magnesium?
Will obtain magnisium nitrate{Mg(NO3)2} and H2 gas
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 must be added to 12 L of a 40 percent solution of alcohol to obtain a 30 percent solution?
4 litres
