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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

2 Answers

3 votes
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.
PLEASE HELP!!! A 33 gram sample of a substance that's used for drug research has a-example-1
User Dave Shinkle
by
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2 votes

Answer:

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.

User Akshay Kumar
by
6.6k points