Fluorine-20 has a half-life of 11.0 s. If a sample initially contains 49.0 μg of this isotope, how much remains after 66.0 s ?

66s is 6 half-lives, so

49.0 * (1/2)^6 = 0.765 μg

Well, when it comes to half-lives, things can get a bit "half-baked"! Let's break this down.

If Fluorine-20 has a half-life of 11.0 seconds, it means that after every 11.0 seconds, half of it will decay. So, in 66.0 seconds, it will go through 6 half-lives.

Starting with 49.0 μg, after the 1st half-life, we're left with 24.5 μg. After the 2nd, 12.25 μg. After the 3rd, 6.125 μg. After the 4th, 3.0625 μg. After the 5th, 1.53125 μg. And after the 6th half-life, we're left with 0.765625 μg.

So, after 66.0 seconds, there will be approximately 0.765625 μg of Fluorine-20 left. Don't worry, it might be a small amount, but it's still a "fluor-midable" presence!

To find out how much remains after 66.0 s, we can use the concept of half-life. The half-life of fluorine-20 is 11.0 s, which means that after every 11.0 s, half of the sample will decay.

Let's calculate the number of half-lives that have passed in 66.0 s:

Number of half-lives = time elapsed / half-life
Number of half-lives = 66.0 s / 11.0 s
Number of half-lives = 6

Since 6 half-lives have passed, the remaining fraction of the sample can be calculated as (1/2)^6, which is:

Remaining fraction = (1/2)^6
Remaining fraction = 1/64

To find out how much of the sample remains, we can multiply the remaining fraction by the initial amount of the sample:

Remaining sample = Remaining fraction * Initial sample
Remaining sample = (1/64) * 49.0 μg

Let's calculate the remaining sample:

Remaining sample = (1/64) * 49.0 μg
Remaining sample = 0.7656 μg (rounded to four decimal places)

Therefore, approximately 0.7656 μg of the sample remains after 66.0 s.

To determine how much remains after 66.0 s, we need to use the concept of radioactive decay and the formula for exponential decay. The equation for the amount remaining after a certain time can be expressed as:

N(t) = N₀ * (1/2)^(t / T)

Where:
N(t) is the amount remaining after time t,
N₀ is the initial amount,
t is the time elapsed, and
T is the half-life.

Now, let's plug in the given values:
N₀ = 49.0 μg (initial amount)
t = 66.0 s (time elapsed)
T = 11.0 s (half-life)

Using these values, we can now calculate the amount of Fluorine-20 remaining after 66.0 s.

N(66) = 49.0 μg * (1/2)^(66.0 s / 11.0 s)

First, divide 66.0 s by 11.0 s:
N(66) = 49.0 μg * (1/2)^6

Next, simplify the exponential:
N(66) = 49.0 μg * (1/64)

N(66) = 0.7656 μg

Therefore, after 66.0 s, approximately 0.7656 μg of Fluorine-20 remains in the sample.