To begin with, we'll have to make use of the radioactive decay equation:

$\displaystyle A_{t}=A_{o}e^{-kt}$

Where:

$\displaystyle A_{t}$ = the amount of Compound X after an amount of time, $\displaystyle t$, has passed

$\displaystyle A_{o}$ = the amount of Compound X at the start

$\displaystyle k$ = rate constant for the decay process

$\displaystyle t$ = the amount of time that has passed

Rearranging, we can see that:

$\displaystyle \frac{A_{t}}{A_{o}}=e^{-kt}$

Furthermore, we can convert minutes into seconds.

$\displaystyle 20min\left ( \frac{60s}{1min} \right )=1200s$

Next, we can plug the values we have into the above equation to obtain:

$\displaystyle \frac{A_{t}}{A_{o}}=e^{-kt}=e^{-\left ( 3.7\cdot 10^{-4}s^{-1}\right )(1200s)}=0.64$

From the above expression, we see that:

$\displaystyle \frac{A_{t}}{A_{o}}=0.64$

This means that after a time of 20min has elapsed, there will be $\displaystyle 64\%$ as much of Compound X as there was when at the beginning. Since there is $\displaystyle 64\%$ remaining after this time period, then we can conclude that $\displaystyle 100\%-64\%=36\%$ of Compound X has degraded within this amount of time.

Positron decay (positive beta decay) is where a proton is converted into a neutron, a positron, and an electron neutrino.

$\displaystyle p\rightarrow n+e^++v_e$

Positrons are called $\displaystyle \beta^+$ particles (beta plus particles). They are the antiparticle to $\displaystyle \beta^-$ particles (the antimatter version of electrons). They are used in the medical field for PET (Positron Emission Tomography) scans by highlighting a radioactive substance (called a tracer) ingested earlier to show how organs and tissues are working. In nuclear reactors, they cause the water coolant to give a blue glow called Cherenkov radiation. This is because the positrons move faster than light does through water.

225 years have passed since Rex died. Find the number of half-lives that have elapsed.

$\displaystyle \frac{225}{5730}=0.039267$

To find the proportion of a substance that remains after a certain number of half-lives, use the following equation:

$\displaystyle \left(\frac{1}{2} \right )^n$

Here, $\displaystyle n$ is the number of half lives that have elapsed.

$\displaystyle \left(\frac{1}{2} \right )^{0.039267}=0.973\cdot 100\%=97.3\%$

Use the following equation:

$\displaystyle A=A_{0}e^{-\lambda t}$

The decay constant is defined as: 

$\displaystyle \lambda =\frac{ln(2)}{t_{\frac{1}{2}}}$

From the first equation, we find:

$\displaystyle \lambda=\frac{0.0488}{hr}$

Plug this value into the second equation above and solve.

$\displaystyle t_{\frac{1}{2}}=14.2 hr$

Use the relationship:

$\displaystyle A=A_0e^{-\lambda t}$

Here, $\displaystyle A$ is the activity at a given time, $\displaystyle A_0$ is the initial activity, $\displaystyle \lambda$ is the radioactive decay constant, and $\displaystyle t$ is the time passed since the initial reading.

Rearrange the equation to solve for $\displaystyle \lambda$.

$\displaystyle \frac{(-ln(\frac{A}{A_0}))}{t}=\lambda$

Convert minutes to seconds and plug in values to solve for the radioactive decay constant.

$\displaystyle \frac{(-ln(\frac{220 Bq}{1990 Bq}))}{600s}=\lambda$

$\displaystyle \frac{7.51*10^{-2}}{s}=\lambda$

Use the relationship:

$\displaystyle A=A_0e^{-\lambda t}$

Where $\displaystyle A$ is the activity at a given time, $\displaystyle A_0$ is the initial activity, $\displaystyle \lambda$ is the radioactive decay constant and $\displaystyle t$ is the time passed since the initial reading.

Rearrange to solve for $\displaystyle \lambda$.

$\displaystyle \frac{(-ln(\frac{A}{A_0}))}{t}=\lambda$

Convert minutes to seconds and plug in values to find the decay constant.

$\displaystyle \frac{(-ln(\frac{768 Bq}{1868 Bq}))}{900s}=\lambda$

$\displaystyle \frac{9.86*10^{-4}}{s}=\lambda$

It is then necessary to use the relationship:

$\displaystyle A= \lambda N$

Where $\displaystyle A$ is the activity, $\displaystyle \lambda$ is the decay constant and $\displaystyle N$ is the number of atoms.

Use the initial activity and the calculated decay constant to solve for the number of atoms:

$\displaystyle 1868= \frac{9.84*10^{-4}}{s} N$

$\displaystyle N=1.90*10^6$

Use the relationship:

$\displaystyle A=A_0e^{-\lambda t}$

Where $\displaystyle A$ is the activity at a given time, $\displaystyle A_0$ is the initial activity, $\displaystyle \lambda$ is the radioactive decay constant and $\displaystyle t$ is the time passed since the initial reading.

Rearrange to solve for $\displaystyle \lambda$

$\displaystyle \frac{(-ln(\frac{A}{A_0}))}{t}=\lambda$

Convert minutes to seconds and plug in values.

$\displaystyle \frac{(-ln(\frac{768 Bq}{1868 Bq}))}{900s}=\lambda$

$\displaystyle \frac{9.86*10^{-4}}{s}=\lambda$

Again use the relationship:

$\displaystyle A=A_0e^{-\lambda t}$

Use the new $\displaystyle t$, which is equal to $\displaystyle 18min=1080s$ to plug in and solve for the activity.

$\displaystyle A=1868e^{-9.86*10^{-4} 1080}$

$\displaystyle A=644Bq$

Use the relationship:

$\displaystyle A=A_0e^{-\lambda t}$

Here, $\displaystyle A$ is the activity at a given time, $\displaystyle A_0$ is the initial activity, $\displaystyle \lambda$ is the radioactive decay constant, and $\displaystyle t$ is the time passed since the initial reading.

Rearrange the equation to solve for $\displaystyle \lambda$.

$\displaystyle \frac{(-ln(\frac{A}{A_0}))}{t}=\lambda$

Convert minutes to seconds and plug in values.

$\displaystyle \frac{(-ln(\frac{768 Bq}{1868 Bq}))}{900s}=\lambda$

$\displaystyle \frac{9.86*10^{-4}}{s}=\lambda$

Use the relationship:

$\displaystyle \frac{ln(2))}{\lambda}=t_{\frac{1}{2}}$

Plug in the calculated value for $\displaystyle \lambda$:

$\displaystyle \frac{ln(2))}{9.86*10^{-4}}=t_{\frac{1}{2}}$

$\displaystyle 702s=t_{\frac{1}{2}}$

Use the relationship:

$\displaystyle A=A_0e^{-\lambda t}$

Here, $\displaystyle A$ is the activity at a given time, $\displaystyle A_0$ is the initial activity, $\displaystyle \lambda$ is the radioactive decay constant, and $\displaystyle t$ is the time passed since the initial reading.

Rearrange the equation to solve for $\displaystyle \lambda$.

$\displaystyle \frac{(-ln(\frac{A}{A_0}))}{t}=\lambda$

Convert minutes to seconds and plug in values.

$\displaystyle \frac{(-ln(\frac{768 Bq}{1868 Bq}))}{900s}=\lambda$

$\displaystyle \frac{9.86*10^{-4}}{s}=\lambda$

Use the relationship:

$\displaystyle A=A_0e^{-\lambda t}$

Where $\displaystyle A$ is the activity at a given time, $\displaystyle A_0$ is the initial activity, $\displaystyle \lambda$ is the radioactive decay constant and $\displaystyle t$ is the time passed since the initial reading.

Rearrange to solve for $\displaystyle \lambda$.

$\displaystyle \frac{(-ln(\frac{A}{A_0}))}{t}=\lambda$

Convert minutes to seconds and plug in values.

$\displaystyle \frac{(-ln(\frac{925}{1570}))}{900s}=\lambda$

$\displaystyle \frac{5.88*10^{-4}}{s}=\lambda$

Use the relationship:

$\displaystyle A= \lambda N$

Where $\displaystyle A$ is the activity, $\displaystyle \lambda$ is the decay constant and $\displaystyle N$ is the number of nuclei.

Use the initial activity and the calculated decay constant to solve for the number of nuclei:

$\displaystyle 1570= \frac{5.88*10^{-4}}{s} N$

$\displaystyle N=2.67*10^6$