Question

PLEASE HELP!!!

Answer 1

Answer:

The half life of the substance is 6.0 days.

Step-by-step explanation:

The initial equation for the initial mass and mass at time t is;

                                N = $N_{0}$ $e^{-kt}$

Where N is the mass at time t, $N_{0}$ is the initial mass, k is the constant and t is the half life. After the half life i.e t,

So that at a given time t, $N_{0}$ = 33 grams and N = 16.5 grams

        ⇒       16.5 = 33 $e^{-kt}$

                  16.5 = $\frac{33}{e^{kt} }$

cross multiply, we have;

                $e^{kt}$ = $\frac{33}{16.5}$

                  $e^{kt}$ = 2

Find the natural logarithm of both sides,

          ln $e^{kt}$  = ln2

          kt = ln2

 ⇒     t = $\frac{ln2}{k}$

         t = $\frac{0.6932}{0.1142}$

         t = 6.07

The half life of the substance is 6.0 days.

 

Answer 2

From the attached graphic, Half-life = ln (.5) / k

Half-life =.693147 / 0.1142
= 6.0695884413 days

The value of "k" should be negative and should have units associated with it.