Answer: a) %(C/Co) = (e^(-0.027t)) × 100
b) t1/2 = 25.67years
c) 5.872%
Explanation:
a) Radioactive reactions always follow a first order reaction dynamic
Let the initial concentration of Strontium-90 be Co and the concentration at any time be C
The rate of decay will be given as:
(dC/dt) = -KC (Minus sign because it's a rate of reduction)
The question provides K = 2.7% per year = 0.027/year
(dC/dt) = -0.027C
(dC/C) = -0.027dt
∫ (dC/C) = -0.027 ∫ dt
Solving the two sides as definite integrals by integrating the left hand side from Co to C and the Right hand side from 0 to t.
We obtain
In (C/Co) = -0.027t
(C/Co) = (e^(-0.027t))
In percentage, %(C/Co) = (e^(-0.027t)) × 100
(Solved)
b) Half life of a first order reaction (t1/2) = (In 2)/K
K = 0.027/year
t1/2 = (In 2)/0.027 = 25.67 years
c) percentage that remains after 105years,
%(C/Co) = (e^(-0.027t)) × 100
t = 105
%(C/Co) = (e^(-0.027 × 105)) × 100 = 5.87%
I think the answer most be d
To eliminate the carbon dioxide pollutants, the factory manager should employ unit operations for CO2 capture. One example is through absorption process by using commercial CO2 absorbents like soda lime.
What are the options tell me from comment of this answer
