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t = number of liters of 30% acid solution
s = number of liters of 60% acid solution
t+s=57
.30t+.60s=.50*57
t=57-s
.30(57-s)+.60s=.50*57
.30*57 -.30s +.60s = .50*57
.30s = .50*57 - .30*57 = .20*57
s = .20*57/.30 = 38 liters of 30% solution
t = 57 - s = 57-38 = 19 liters of 60% solution
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A chemist is making a solution that is to be 55 chlorine He has one solution that is 40 chlorine and another that is 65 chlorine He wants 100 liters of the final solution How much of each should?
Each should be 70
How would I answer this question How many grams to 0.1 g of ethanol should be dissolved into 135 g of water to produce a 23 percent by mass ethanol solution?
The answer is 31,05 g ethanol.
A mixture of 40 liters of milk and water contains 10 percent water How much water should be added to this so that water may be 20 percent in the new mixture?
If you're talking about 10 liters of water and not percent, then 10 liters. But then you'll have 60 liters of mixture. It would be 2/5 or 0.4 x 50 = 20 this would make a mix of 20/50. Waterman
A chemist needs 4 liters of a 50 percent salt solution All she has available is a 20 percent salt solution and a 70 percent salt solution How much of each of the two solutions should she mix to obtain?
You will need more data about all densities (in kg/Litre)and you must be sure of using mass% = (g solute)/(100 g solution)Solve two equations for both X and Y:4*d50*50 = X*d20*20 + Y*d70*70 (based on salt mass balance in diff. sol'n.)4*d50 = X*d20 + Y*d70 (based on solutions mass balance)In which:dm = density of the 'm'% salt solution in kg/Litre)X and Y = volume of the 20% and 70% salt solutions respectively
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 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 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 chemist has one solution which is 50 percent acid and a second solution which is 25 percent acid how much of each should be mixed to make a 10 liters of a 40 percent acid solution?
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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 many liters of a 30 solution should be added to a 90 solution to make 18 liters of a 75 solution?
4.5 litres of a 30% solution to the appropriate quantity of the 90% solution.
Jorge has four liters of 30 percent acid solution how much water should he add to make 25 percent acid solution'?
Jorge needs to add 2 liters of water to the 30% acid solution to make a 25% acid solution. This can be calculated using a dilution formula: initial acid amount / final total amount = final acid concentration.
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?
A chemist wants to make a 10 solution of fertilizer. how much water and how much of a 30 solution should the chemist mix to get 30 L of a 10 solution?
30 liters of a 10 % solution of fertilizer has .1(30) = 3 liters of fertilizer 1 liter of 30% solution has .3 liter of fertilizer 10 liters of 30% solution has 3 liters of fertilizer so, the chemist needs 10 liters of the 30% solution and 20 liters of water to make 30 liters of a 10% solution.
How much of a ten percent solution should be added to two cups of twenty percent solution to make a twelve percent solution?
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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?
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A chemist is making a solution that is to be 55 chlorine He has one solution that is 40 chlorine and another that is 65 chlorine He wants 100 liters of the final solution How much of each should?
Each should be 70
