To solve this question on the half-life, we will use this expression:
[tex]\begin{gathered} G(t)=G_oe^{-kt} \\ \text{where G(t) is the remaining sample at time t.} \\ G_{o\text{ }}\text{ is the original sample} \\ K\text{ is a constant} \\ t\text{ is time} \end{gathered}[/tex]To proceed in solving, we will need to find the value of constant k
[tex]\begin{gathered} G(t)=G_oe^{-kt} \\ G(t)=17.25 \\ G_o=276 \\ t=255 \\ \text{Now substitute the parameters above into the formula:} \\ 17.25=276e^{-k(255)} \\ \frac{17.25}{276}=e^{-k(255)} \end{gathered}[/tex][tex]\begin{gathered} 0.0625=e^{-k255} \\ \ln 0.0625=-255k \\ \frac{\ln 0.0625}{-255}=k \\ 0.0109=k \end{gathered}[/tex]Now to get the half-life in minutes will be to get the time taken for the sample to go from 276g to 138g.
[tex]\begin{gathered} G(t)=G_oe^{-kt} \\ G(t)=138g \\ 138=276e^{-0.0109t} \\ \frac{138}{276}=e^{-0.0109t} \\ 0.5=e^{-0.0109t} \\ \ln 0.5=-0.0109t \\ \frac{\ln 0.5}{-0.0109}=t \\ 63.591\text{minutes = t} \end{gathered}[/tex]The half-life is 63.59 minutes.
The formula for G(t) at time t is:
[tex]G(t)=276e^{-0.0109t}[/tex]The amount of goo that will remain after 68 minutes is calculated using the formula above:
[tex]\begin{gathered} G(t)=276e^{-0.0109t} \\ t=68\text{ minutes} \\ G(t)=276e^{-0.0109(68)} \\ G(t)=276e^{-0.7412} \\ G(t)=131.5255\text{ grams} \\ G(t)\text{ = 131.53 grams (to 2 d.p)} \end{gathered}[/tex]The amount of goo remaining after 68 minutes is 131.53 grams.
