141.95
The quantity of rubbing alcohol present at any moment ina bottle labeled "50-milliliters", expressed in milliliters, is(50) - (number of milliliters previously used for rubbing)
200 milliliters
28 milliliters
Need to know how much of the solution you have.
Make an equation: x(.10) = 50gal(.15) Solve algebraically: x = 75 gal
4 ounces
210Type your answer here...
That all depends on what you want the final concentration of alcohol to be.
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.
The answer is 500 mL.
mary mixed 2l of an 80% acid solution with 6l of a 20% acid solution. what was the percent of acid in the resulting mixture
Let a be the number of ounces of 25% alcohol required. Then, 25a + (30x9) = 28(9 + a) 25a + 270 = 252 + 28a 3a = 18 a = 6 Then 6 ounces of 25% alcohol + 9 ounces of 30% alcohol produces 15 ounces of 28% alcohol.
684 ml
You need 50 g of this drug.
473.17
You will have to assume that the 2 % is a volume fraction, then the volume of copper sulfate in the solution would be 11.5 milliliter(575 ml*(0.02). If it were a weight fraction, then you would have to have more information on the solution density.
The quantity of rubbing alcohol present at any moment ina bottle labeled "50-milliliters", expressed in milliliters, is(50) - (number of milliliters previously used for rubbing)