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A 33 gram sample of a substance that's used for drug research has a k-value of 0.1142. Fine the substance's half-life, in days. Round to the nearest tenth. N0=initial mass (at time t=0) N=mass at time t k=positive constant that depends on the substance itself and on the units used to measure time t=time,in days

Verified Answer :

wolf1728

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.

View image wolf1728
olayemiolakunle65

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.

 

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