Wiki User
∙ 10y agoLet 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.
Wiki User
∙ 10y ago684 ml
98 mL
x=45
Since the percentages of copper in the two components to be mixed are symmetric about the desired result, the answer is that the same amounts should be used. 600 ounces of the 30% copper.
900 ounces. Since this contains 20% copper, The copper content will be 180 ounces. The original 300 ounces contain 30% copper which is also 180 ounces. Hence in the resulting mixture of 1200 ounces (300+900), the total copper is 360 ounces (180+180). Hence the copper content of resulting mixture is 360/1200 which is 30%
684 ml
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 much 50 percent antifreeze solution and 40 percent antifreeze solution should be combined to give 50 gallons of 46 percent antifreeze solution?
2%
2/3 of 70% and 1/3 of 10%
10 liters.
200 ounces.
rubbing alcohol
The way I would do this is to first determine the amount of alcohol and that would be 0.09 x 16 ml = 1.44 ml then I would ask myself (0.08 * x) = 1.44 Therefore to have an 8% solution we would have a total of 18 ml of solution (by solving the above equation), thus we would have to add 2 ml of water (18 ml - 16 ml).
98 mL
How much of an alloy that is 10% copper should be mixed with 400 ounces of an alloy that is 70% copper in order to get an alloy that is 20%
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.