In a chemistry lab, you have two vinegars. One is 5% acetic acid and one is 6.5% acetic acid.
You want to make 200 mL of a vinegar with 6% acetic acid. How many milliliters (mL) of each vinegar do you need to mix together?
Let's call the number of milliliters of the 5% acetic acid vinegar x, and the number of milliliters of the 6.5% acetic acid vinegar y.
We know that the total volume should be 200 mL, so x + y = 200 mL.
We also know that the concentration of the resulting vinegar should be 6% acetic acid, so the concentration of acetic acid from the 5% vinegar should be 0.05x, and the concentration of acetic acid from the 6.5% vinegar should be 0.065y.
Now we can create an equation for the acetic acid concentration in the resulting vinegar:
(0.05x + 0.065y) / 200 = 0.06
To isolate y, we can multiply both sides of the equation by 200:
0.05x + 0.065y = 0.06 * 200
0.05x + 0.065y = 12
To solve for y, we can rearrange the equation:
0.065y = 12 - 0.05x
y = (12 - 0.05x) / 0.065
We want to mix equal amounts of each vinegar, so x = y.
Substituting y into the equation, we get:
x = (12 - 0.05x) / 0.065
Multiplying both sides of the equation by 0.065, we get:
0.065x = 12 - 0.05x
0.065x + 0.05x = 12
0.115x = 12
x = 12 / 0.115
x = 104.3478 mL (approx.)
Substituting x back into the equation for y, we get:
y = (12 - 0.05*104.3478) / 0.065
y = (12 - 5.21739) / 0.065
y = 95.7826 mL (approx.)
Therefore, you need approximately 104.35 mL of the 5% acetic acid vinegar and 95.78 mL of the 6.5% acetic acid vinegar to make 200 mL of vinegar with 6% acetic acid.