Final answer:
The amount of fluorine-18 that remains active after 330 minutes is approximately 18.81 mg. Therefore, the amount of fluorine-18 that is still active after 330 min is 0.114 times 165 mg = 18.81 mg.
Step-by-step explanation:
To determine the amount of fluorine-18 that remains active after a given time, we need to use the concept of radioactive decay.
Fluorine-18 has a half-life of 109.7 minutes, which means that after each half-life, the amount of fluorine-18 is halved.
In this case, we need to find out how many half-lives occur within 330 minutes.
To do this, we divide the total time by the half-life: 330 min / 109.7 min per half-life = 3.004 half-lives.
Since each half-life reduces the amount by half, the fluorine-18 will be reduced to 1/2^3.004 or approximately 0.114 times the initial amount.
Therefore, the amount of fluorine-18 that is still active after 330 min is 0.114 times 165 mg = 18.81 mg.