Radio active decay reactions follow first order rate kinetics.
a) The half life and decay constant for radio active decay reactions are related by the equation:
Where k is the decay constant
b) Finding out the decay constant for the decay of C-14 isotope:
c) Finding the age of the sample :
35 % of the radiocarbon is present currently.
The first order rate equation is,
![[A] = [A_{0}]e^{-kt}](https://tex.z-dn.net/?f=%20%5BA%5D%20%3D%20%5BA_%7B0%7D%5De%5E%7B-kt%7D%20%20%20)
![\frac{[A]}{[A_{0}]} = e^{-kt}](https://tex.z-dn.net/?f=%20%5Cfrac%7B%5BA%5D%7D%7B%5BA_%7B0%7D%5D%7D%20%3D%20e%5E%7B-kt%7D%20%20)
t = 7923 years
Therefore, age of the sample is 7923 years.
The answer is intermediate, so E.
Answer:
The correct answer is 160.37 KJ/mol.
Explanation:
To find the activation energy in the given case, there is a need to use the Arrhenius equation, which is,
k = Ae^-Ea/RT
k1 = Ae^-Ea/RT1 and k2 = Ae^-Ea/RT2
k2/k1 = e^-Ea/R (1/T2-1/T1)
ln(k2/k1) = Ea/R (1/T1-1/T2)
The values of rate constant k1 and k2 are 3.61 * 10^-15 s^-1 and 8.66 * 10^-7 s^-1.
The temperatures T1 and T2 are 298 K and 425 K respectively.
Now by filling the values we get:
ln (8.66*10^-7/3.61*10^-15) = Ea/R (1/298-1/425)
19.29 = Ea/R * 0.001
Ea = 160.37 KJ/mol
You can answer this question by only searching the element in the periodic table.
The atomic number of iodine, I, is 53. It is placed in the column 17 (this is the Group) and row 5 (this is the Period).
The conclusion is that the iodine is located in Period 5, Group 17, and is classified as a nonmetal.
