88 ml
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.
0.5 to 1 percent starch solution should be used
The final concentration of the hydrochloric acid solution would be 5% after dilution. Therefore, the label should indicate that the solution is a 5% hydrochloric acid solution.
70%
Make an equation: x(.10) = 50gal(.15) Solve algebraically: x = 75 gal
2/3 of 70% and 1/3 of 10%
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.
10 liters.
80% water
684 ml
Let x = ounces of 50% solution, and y = ounces of 1% solution. So that we have: 0.5x + 0.01y = 8(0.2) which is a linear equation in two variables, meaning there are infinitely many choices of mixing those solutions.
2 gallons.
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?
2%
0.25 gallons of water (or 1 quart)
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.