An unknown substance decays from 14 g to 3 g over a period of 21 days.
3/14 = (1/2)^2.22
so 21 days is 2.22 half-lives
I assume that has something to do with what you want to know...
Good for oobleck. I don't see a question.
To find the rate at which the unknown substance is decaying, we can use the formula for exponential decay:
N(t) = Nā * e^(-kt)
Where:
N(t) = final amount of substance at time t
Nā = initial amount of substance
k = decay constant
t = time
In this case, we want to find the value of k. We are given the initial amount Nā = 14 g, the final amount N(t) = 3 g, and the time t = 21 days.
Substituting these values into the formula, we can solve for k:
3 = 14 * e^(-k * 21)
To isolate e^(-k * 21), divide both sides by 14:
3/14 = e^(-k * 21)
To remove the exponential, take the natural logarithm (ln) of both sides:
ln(3/14) = -k * 21
Now divide both sides by -21 to solve for k:
k = -ln(3/14)/21
Using a calculator, we can evaluate this expression:
k ā 0.0846
Therefore, the decay constant (k) for this unknown substance is approximately 0.0846.