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5.6 quarts
A pharmacist mixed a 20 percent solution with a 30 percent solution to obtain 100 liters of a 24 percent solution. How much of the 20 percent solution did the pharmacist use in the mixture (in liters).
10
The purpose is to obtain pure compounds from a mixture.
This question can be answered through the application a little bit of algebra. Allow me to demonstrate: First, assign variable (essentially letter) values to the amounts of each type of milk. In this case, X represents the quarts of 4% butterfat milk, and Y represents the quarts of 1% butterfat milk. We know that we need 75 quarts of 1% butterfat milk, so no matter how much of each type we mix, they must add up to 75 quarts. Thus... X+Y=75 That's our first equation. To solve this, we're going to need one more. To obtain the objective percentage of butterfat, we must convert all percentages into decimal format. Thus 4%=0.04, 1%=0.01, and 3%=0.03. Now, we know that X has 4% butterfat, thus the butterfat content contributed by X milk is represented by 0.04X And we know that Y milk has 1% butterfat, thus the butterfat content contributed by Y milk is represented by 0.01Y To calculate the percentage of butterfat in the entire mixture, one must divide the sum of the concentrations by the total volume of 75 quarts, meaning that the beginning of our our second equation would look like this: (0.04X+0.01Y)/75 And since we want our objective mixture to have a 3% butterfat concentration, the equation would finish out like this: (0.04X+0.01Y)/75=0.03 Now we have a system of equations. X+Y=75 (0.04X+0.01Y)/75=0.03 There are many ways to solve this, but one of the most visually demonstrable methods is the method of substitution. This means getting one equation in terms of one variable. The best way to do this would be to set the first equation equal to Y. Thus Y=75-X Now, every time that we see Y appear in the second equation, we replace it with (75-X). Like this (0.04X+0.01(75-X))/75=0.03 Now we can solve this equation for X. The following equation demonstrates multiplying the answer by the denominator of the fraction and the distribution of the 1% 0.04X+0.75-0.01X=2.25 Now we combine like terms and subtract 0.75 from the answer to get 0.03X=1.5 All that's left is to divide the answer by 0.03 to know what X equals X=50 This means that we're going to need 50 quarts of 4% butterfat milk. Now, to solve for the 1% butterfat milk. We can simply take the value we found for X and plug it into the modified version of our first equation to get that Y=75-50 or Y=25 This means that we will need 25 quarts of 1% butterfat milk mixed with 50 quarts of 4% butterfat milk to obtain 75 quarts of 3% butterfat milk. This form of algebraic computation can be used to solve any similar problem.
x=gallons of 25% y=gallons of 3.5% .25x + .035y = .125(50) y=50-x using substitution for a system of equations: .25x + .035(50-x) = 6.25 .25x + 1.75 - .035x = 6.25 .215x = 4.5 x=20.93 gallons of cream y=29.07 gallons of milk
Here is a procedure for isolating a neutral compound from a mixture containing basic impurities: Begin by adding an acid to the mixture containing the neutral compound and basic impurities. This will convert the basic impurities into their protonated forms which can be extracted with the aqueous layer. Perform liquid-liquid extraction using a solvent that is immiscible with water. The neutral compound will partition into the organic solvent while the protonated basic impurities will remain in the aqueous layer. Separate the organic layer containing the neutral compound and evaporate the solvent to obtain the pure compound. Further purify the compound using techniques such as recrystallization or chromatography if needed.
Fractional distillation of the liquefied mixture is one possible method.
- To obtain the useful components from a mixture.- To remove the unwanted components from a mixture. - To separate 2 or more useful components from a mixture.
10% of 150 is 15
To obtain copper sulfate crystals from a mixture with sand, you can dissolve the mixture in water. The copper sulfate will dissolve, while the sand will not. You can then filter the solution to separate the sand from the copper sulfate solution. By evaporating the water from the copper sulfate solution, you can obtain copper sulfate crystals.
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