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
Each should be 70
The answer is 31,05 g ethanol.
To dilute an 18-liter solution of 14% bleach to a 10% concentration, you need to first determine the amount of pure bleach in the original solution. The 18 liters of 14% bleach contains 2.52 liters of pure bleach (0.14 × 18). To find the total volume needed for a 10% solution, set up the equation: 2.52 liters = 0.10 × (18 + x), where x is the amount of water to add. Solving for x, you find that you need to add 7.2 liters of water.
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
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
10 liters.
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.
6 litres of 50% + 4 litres of 25%
add 25ml more of solution x * 20 = 100 * 25 x = 25
4.5 litres of a 30% solution to the appropriate quantity of the 90% 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?
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.
2%
3 to 4 liters of water
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).
Each should be 70