<u>Answer:</u> The half life of the sample of silver-112 is 3.303 hours.
<u>Explanation:</u>
All radioactive decay processes undergoes first order reaction.
To calculate the rate constant for first order reaction, we use the integrated rate law equation for first order, which is:
![k=\frac{2.303}{t}\log \frac{[A_o]}{[A]}](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B2.303%7D%7Bt%7D%5Clog%20%5Cfrac%7B%5BA_o%5D%7D%7B%5BA%5D%7D)
where,
k = rate constant = ?
t = time taken = 1.52 hrs
= Initial concentration of reactant = 100 g
[A] = Concentration of reactant left after time 't' = [100 - 27.3] = 72.7 g
Putting values in above equation, we get:

To calculate the half life period of first order reaction, we use the equation:

where,
= half life period of first order reaction = ?
k = rate constant = 
Putting values in above equation, we get:

Hence, the half life of the sample of silver-112 is 3.303 hours.
The answer is 1/8.
Half-life is the time required for the amount of a sample to half its value.
To calculate this, we will use the following formulas:
1.

,
where:
<span>n - a number of half-lives
</span>x - a remained fraction of a sample
2.

where:
<span>

- half-life
</span>t - <span>total time elapsed
</span><span>n - a number of half-lives
</span>
The half-life of Sr-90 is 28.8 years.
So, we know:
t = 87.3 years
<span>

= 28.8 years
We need:
n = ?
x = ?
</span>
We could first use the second equation, to calculate n:
<span>If:

,
</span>Then:

⇒

⇒

<span>⇒ n ≈ 3
</span>
Now we can use the first equation to calculate the remained amount of the sample.
<span>

</span>⇒

⇒

<span>
</span>
Converting a liquid to a gas is evaporation.
To answer the question above, we need to c<span>onvert 12.8 g to moles by dividing by 46.07 first.</span>
<span>For every mole you burn, you get 1235 kJ of heat. So multiply your number of moles by 1235. It'll be something in the neighborhood of 500. </span>
<span>Take your kJ that you calculated and divide by 5.65 to get the number of degrees that your calorimeter goes up. Add that to 25.
I hope my answer helped you</span>