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How many liters of 50 percent alcohol solution and 20 percent alcohol solution must be mixed to get obtain 18 liters of 30 percent alcohol solution?
The total volume of solution equals 18 L. So if we let X be the 50% solution & Y be the 20% solution we can solve a set of two equations with 2 unknowns:
X + Y = 18 (total volume of solution is 18 L)
0.5X + 0.2Y = 18*(0.30) (the total amount of alcohol in the new 18L solution)
Solving simultaneously, multiply the 1st equation by 0.5 then substract to solve for Y.
X + Y = 18 {*0.5} => 0.5X + 0.5Y = 9
0.5X + 0.2Y = 5.4 => 0.5X + 0.2Y = 5.4
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0.3Y = 3.6
Y = 12
Then substitute for Y back into the 1st equation & solve to get X=6.
So you need 12 L of the 20% solution & 6 L of the 50% solutions to mix together to get 18 L of a 30% alcohol solution.
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A pharmacist needs to obtain a 70 percent alcohol solution How many ounces of a 30 percent alcohol solution must be mixed with 40 ounces of an 80 percent alcohol solution to obtain 70 percent alcohol?
(Note: This answer assumes that the "ounces" specified are avoirdupois or other weight ounces and that percentages are by weight; otherwise possible volume changes on dilution must by considered.) The weight of pure alcohol in each solution is the product of the percentage and the total weight of the solution. Therefore, designating the unknown weight of 30 % alcohol as w, from the problem statement 0.30w + 0.80(40) = 0.70(w + 40), or 0.30w + 32 = 0.70w + 28, or 32 - 28 = w(0.70 - 0.30) or w = 4/0.40 = 10 ounces of 30 % alcohol.
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 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 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 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 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.
A pharmacist mixed a 20 percent solution with a 30 percent solution to obtain 100 liters of a 24 percent solution How much of the 20 percent solution did the pharmacist use in the mixture?
A pharmacist mixed a 20 percent solution with a 30 percent solution to obtain 100 liters of a 24 percent solution. How much of the 20 percent solution did the pharmacist use in the mixture (in liters).
Pharmacist mixed a 20 percent solution with a 30 percent solution to obtain 100 liters of a 24 percent solution How much of the 20 percent solution did the pharmacist use in the mixture?
10
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 water should be mixed to 12ml of alcohol to obtain a 12 percent of alcohol solution?
To obtain a 12% alcohol solution, you would need to mix 12ml of alcohol with 48ml of water. This would give you a total volume of 60ml, with 12% of it being alcohol.
How many liters of a 20 percent solution of saline should a nurse mix with 10 liters of a 50 percent solution to obtain a mixture that is 40 percent saline?
Let x = the amount of 20% solution Let x + 10 = the amount of the final solution. So we have: (.20)x + (.50)(10) = (.40)(x + 10) .20x + 5 = .40x + 4 .20x = 1 x = 5 liters of 20% solution of saline.
A pharmacist needs to obtain a 70 percent alcohol solution How many ounces of a 30 percent alcohol solution must be mixed with 40 ounces of an 80 percent alcohol solution to obtain 70 percent alcohol?
(Note: This answer assumes that the "ounces" specified are avoirdupois or other weight ounces and that percentages are by weight; otherwise possible volume changes on dilution must by considered.) The weight of pure alcohol in each solution is the product of the percentage and the total weight of the solution. Therefore, designating the unknown weight of 30 % alcohol as w, from the problem statement 0.30w + 0.80(40) = 0.70(w + 40), or 0.30w + 32 = 0.70w + 28, or 32 - 28 = w(0.70 - 0.30) or w = 4/0.40 = 10 ounces of 30 % alcohol.
How many liters of 50 percent alcohol solution and 20 percent alcohol solution must be mixed to obtain 3 liters of 40 percent alcohol solution?
Let there be x litres of 50% solution and y% of 20 % solution, then you have two equations by considering the amount of alcohol and the total amount of liquid:50% x + 20% y = 40% of 3 litresx + y = 3 litresThese are two simultaneous equations involving 2 unknowns which can be solved:Double {1} and subtract from {2} to give:x - x + y - 40%y = 3 - 80% of 3→ 60% y = 20% of 3→ y = 20%/60% of 3 = 1/3 of 3 = 1Substitute for y in {2} to get:x + 1 = 3x = 2Therefore you need 2 litres of 50% solution and 1 litre of 20% solution.
What fraction of 10 percent alcohol should be mixed with 70 percent alcohol to obtain 50 percent alcohol?
2/3 of 70% and 1/3 of 10%
A bottle of commercial concentrated aqueous ammonia is labeled 28.89 percent NH3 by mass and density is 0.8960 grams mL. What is the molarity of the ammonia solution?
To find the molarity, first calculate the moles of NH3 in 100g of the solution using the mass percent. Then convert the volume of the solution (1 mL) to liters. Finally, divide moles by liters to obtain the molarity.
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 liters would you need to get 0.5moles if you had 0.1m solution?
To find out how many liters of a 0.1 M solution are needed to obtain 0.5 moles, you can use the formula: [ \text{Molarity (M)} = \frac{\text{moles of solute}}{\text{liters of solution}} ] Rearranging this gives: [ \text{liters of solution} = \frac{\text{moles of solute}}{\text{Molarity (M)}} ] Substituting in the values: [ \text{liters of solution} = \frac{0.5 \text{ moles}}{0.1 \text{ M}} = 5 \text{ liters} ] Therefore, you would need 5 liters of a 0.1 M solution to obtain 0.5 moles.
