Question

A chemist measures 8 grams of a substance that

Answer 1

Answer:

$y(t) =8 (b)^t$ Where b is the decay rate for this case. Using the condition given we have:

$4 = 8 b^6$

$\frac{1}{2}= b^6$

$b = (\frac{1}{2})^{1/6}$

And our model would be given by:

$y(t) = 8 (\frac{1}{2})^{t/6}$

And replacing the value of t=15 we got:

$y(15) = 8 (\frac{1}{2})^{15/6} = 1.414 grams$

Step-by-step explanation:

For this case since the half life is 6 hours we have the following condition:

$y(6) = \frac{1}{2}A_o$

Where $A_o=4$ is the initial amount

Our model for this case is given by this expression:

$y(t) =8 (b)^t$ Where b is the decay rate for this case. Using the condition given we have:

$4 = 8 b^6$

And solving for b we got:

$\frac{1}{2}= b^6$

And solving for b we got:

$b = (\frac{1}{2})^{1/6}$

And our model would be given by:

$y(t) = 8 (\frac{1}{2})^{t/6}$

And replacing the value of t=15 we got:

$y(15) = 8 (\frac{1}{2})^{15/6} = 1.414 grams$