The rate law for the decomposition of phosphine () is It takes 135. s for 1.00 M to decrease to 0.250 M. How much time is required for 3.00 M to decrease to a concentration of 0.450 M

189.71 secs

Explanation:

We know that decomposition is a first order reaction;

So;

ln[A] = ln[A]o - kt

But;

[A]o = 1.00 M

[A] = 0.250 M

t =135 s

Hence;

ln[A] -  ln[A]o = kt

k = ln[A] -  ln[A]o/t

k = ln(1) - ln(0.250)/135

k =0 - (-1.386)/135

k = 1.386/135

k= 0.01

So time taken now will be;

ln[A] -  ln[A]o = kt

t = ln[A] -  ln[A]o/k

t = ln (3) - ln(0.450)/0.01

t = 1.0986 - (-0.7985)/0.01

t = 1.0986 + 0.7985/0.01

t = 189.71 secs

what's the

c-14 is carbon-14, or radiocarbon.   it is a radioactive isotope of carbon with an atomic nucleus containing 6 protons and 8 neutrons.