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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.
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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).
How many liters of water must be added to 30 liters of 20 percent ALCOHOL solution to dilute it to a15 percent solution?
You have 6 litres of alcohol in 24 litres of water You need to add x litres to make 6 equal to 15% of 30 + x. 6 is 15% of 40, so x = 10
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 much water must be added to 12 grams of a 90 percent iodine solution to produce a 25 percent iodine solution?
12*(90%)=10.8=X*.25 10.8/.25=X=43.2 43.2-12=31.2 31.2 grams of water must be added
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 many grams of sugar must be added to 60 grams of a solution that is 32 percent sugar to obtain a solution that is 50 percent sugar?
40.8 grams
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 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 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 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 liters of a 40 percent acid solution must be added to 100 liters of a 25 percent solution to obtain a 30 percent solution?
Suppose L litres are required.@ 40% it will contain 0.4L of the active ingredient.Then total volume of mixture = L + 100 litresand volume of active ingredient = 0.4L + 25 litresThe strength is (0.4L + 25)/(L + 100) = 30% = 0.3(0.4L + 25) = 0.3*(L + 100)0.4L + 25 = 0.3*L + 300.1L = 5L = 50 litres.
How many liters of a 20 percent alcohol must be mixed with 40 liters of a 90 percent solution to get a 60 percent solution?
Let L be the liters to be added. Then (L) x (0.2) + (40 x 0.9) = (L + 40) x 0.6 0.2L + 36 = 0.6L + 24 36 - 24 = (0.6 - 0.2)L 12 = 0.4L L = 12/0.4 = 120/4 = 30 Liters
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 kg of pure salt must be added to 8 kg of a 30 percent solution to obtain a 35 percent soltution Round to the nearest tenth?
0.6
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.
