0
Find the amount of 14 percent acid solution that should be added to 6 percent acid solution to obtain 50 mL of a 12 percent solution?
Write eqn for weight; x = 14% sol & y = 6% sol which is x + y = 50.
Write eqn for solution: .14x + .06y = .12*50 = 6; 2 equations 2 unknowns.
Solve: Mult 1st eqn by -.14; now -.14x-.14y = -.14*50 or -7
-.14x-.14y=-7
.14x+.06y=6
0-.08y=-1 or y=12.5 or 12.5mL of 6% solution
x+12.5=50 or x=37.5 or 37.5 mL of 14% solution
What else can I help you with?
How many ml of a 20 percent solution should be added to 50 ml of a 40 percent solution to obtain a 25 percent solution?
150 mlLet x be the volume of 20% solution to add. The amount of "stuff" dissolved in the 20% solution plus the amount in the 40% will equal the amount of "stuff" dissolved in the 25% solution.(0.20)(X) + (0.40)(50 ml) = (0.25)(X+50ml)0.2X + 20 = 0.25X + 12.520 - 12.5 = 0.25X - 0.20X7.5 = 0.05XX = 7.5/0.05 = 150 ml
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.
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)
A 280 mL solution is 20 percent salt How much water should be added to make the solution 14 percent salt?
98 mL
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 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 ml of a 20 percent solution should be added to 50 ml of a 40 percent solution to obtain a 25 percent solution?
150 mlLet x be the volume of 20% solution to add. The amount of "stuff" dissolved in the 20% solution plus the amount in the 40% will equal the amount of "stuff" dissolved in the 25% solution.(0.20)(X) + (0.40)(50 ml) = (0.25)(X+50ml)0.2X + 20 = 0.25X + 12.520 - 12.5 = 0.25X - 0.20X7.5 = 0.05XX = 7.5/0.05 = 150 ml
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.
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 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 15 g of salt solution to obtain 15 percent salt solution?
Dissolve 15 g salt in 100 mL water.
How much 50 percent antifreeze solution and 40 percent antifreeze solution should be combined to give 50 gallons of 46 percent antifreeze solution?
How much 50 percent antifreeze solution and 40 percent antifreeze solution should be combined to give 50 gallons of 46 percent antifreeze solution?
How much of a ten percent solution should be added to two cups of twenty percent solution to make a twelve percent solution?
2%
How much of water should be add to obtain a solution that is twenty percent antifreeze?
To obtain a solution that is twenty percent antifreeze, you would add 4 parts water to 1 part antifreeze. This means that for every 1 part of antifreeze, you would add 4 parts of water. This would result in a total of 5 parts of solution, with 1 part being antifreeze and 4 parts being water, achieving a solution that is twenty percent antifreeze.
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 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.
