Aradioactive substance decays exponentially. a scientist begins with 190 milligrams of a radioactive substance. after 17 hours, 95 mg of the substance remains. how many milligrams will remain after 21 hours

To solve this item, we need write first all the necessary data to solve the problem.  Governing principle of this problem is basically exponential decay of a certain substance.

Given:

Let t represents time
     N represents mass of substance

Initially,
at t(0), N(0) = 190mg radioactive substance

at t(17), N(17) = 95mg radioactive substance

Required: N(21) 

Solution:

We write first the formula for exponential decay, that is

N(t) = N(0)*e^(-k*t) ;   k = exponential decay constant

We need first to determine the k, substituting the given we have
95 = 190*e^(-k*17)

Dividing by 190, we get
0.5 = e^(-k*17)

To eliminate the exponential e, we take the natural logs of both sides
ln(0.5) = ln(e^(-k*17))
-0.693 = -k*17

Solving for k,
k = 0.04077

To solve for N(21), we need to substitute k to the original equation
N(21) = N(0)e^(-k*21)
N(21) = 190*e^(-0.04077*21)
N(21) = 80.7 mg

ANSWER: 80.7 mg radioactive substance left after 21 hours 

Answersublimation;

The answer to the question: choose the correct statement.  a) when ethane reacts with fluorine gas the product is less stable.b) addition reactions of unsaturated hydrocarbons and halogens requires a high activation energy.  c)the double covalent bond within unsaturated hydrocarbons requires a lot of energy to break.  d)the double covalent bond within unsaturated hydrocarbons can be broken easily.is: a) when ethane reacts with fluorine gas the product is less stable.