Question

PLEASE HELP!!!

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

Answer 1

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.

Answer 2

The half life of the substance is 6.0 days.

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 =
`(33)/(e^(kt) )`

cross multiply, we have;

`e^(kt)` =
`(33)/(16.5)`

`e^(kt)` = 2

Find the natural logarithm of both sides,

ln
`e^(kt)` = ln2

kt = ln2

⇒ t =
`(ln2)/(k)`

t =
`(0.6932)/(0.1142)`

t = 6.07

The half life of the substance is 6.0 days.