it has been found that ddt in the soil decomposes by a first-order reaction with a half-life of 12.0 years. how long would it take ddt in a soil sample to decompose from a concentration of 195 ppbm (parts per billion by mass) to 15 ppbm?

Answer :

The ddt in the soil with half-life of 12 years needs 44.4 years to decompose.

We need to know about the half-life of the radioactive elements to solve this problem. The radioactive element will decay over time and follow the equation

N = No(1/2)^(t/t'')

where N is the final quantity, No is the initial quantity, λ is the decaying constant, t is time and t'' is the half-life of a radioactive element.

From the question above, we know that

t'' = 12 yr

N = 15 ppbm

No = 195

By substituting the given parameter, we can calculate the required for decomposing

N = No(1/2)^(t/t'')

15 = 195(1/2)^(t/12)

1/13 = (0.5) ^ (t/12)

t/12 = ⁰'⁵log(1/13)

t/12 = 3.7

t = 44.4 years

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