Question

Question 1 ( point) A 64g sample of Germanium-66 is left undisturbed for 12.5 hours. At the end of that period, only 2.0g remain. How many half-lives transpired during this time period? half-lives Answer = Blank 1:​

Answer 1

Answer:

2.5 half-lives

Step-by-step explanation:

Find $t_{h}$ in $N(t)=N_{0}e^{-kt}$ given $N(t) = 2, N_{0} =64,t=\frac{25}{2}$

$N(t)$ is the amount after the time $t$, $N_0$ `is the initial amount, $t_{h}$ is the half-life

We know that after half-life there will be twice less the initial quantity:

$N(t_{h})=\frac{N_{0} }{2}=N_{0}e^{kt_{h} }$.

Simplifying gives $\frac{1}{2} =e^{-kt_h} or k = -\frac{In(\frac{1}{2}) }{t_{h} } .$

Plugging this into the initial equation, we obtain that $N(t)=N_{0}e^\frac{In(\frac{1}{2}) }{t_{h} }$or $N(t)= N_{0} (\frac{1}{2})^\frac{t}{t_h}.$

Finally, just plug in the given values and find the unknown one.

From $2=64(\frac{1}{2})^\frac{\frac{25}{2} }{t_h}$, we have that $t_h$ = $\frac{25 In (2)}{2 In (32)\\}$

So you should get: $t_{h} =\frac{25 In (2)}{2 In (32)\\}=2.5.$

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Answer 2

Answer:

You should def give them brainiest :)