Hi the answer is actually B
<u>Answer:</u> The concentration of radon after the given time is 
<u>Explanation:</u>
All the radioactive reactions follows first order kinetics.
The equation used to calculate half life for first order kinetics:

We are given:

Putting values in above equation, we get:

Rate law expression for first order kinetics is given by the equation:
![k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B2.303%7D%7Bt%7D%5Clog%5Cfrac%7B%5BA_o%5D%7D%7B%5BA%5D%7D)
where,
k = rate constant = 
t = time taken for decay process = 3.00 days
= initial amount of the reactant = 
[A] = amount left after decay process = ?
Putting values in above equation, we get:
![0.181days^{-1}=\frac{2.303}{3.00days}\log\frac{1.45\times 10^{-6}}{[A]}](https://tex.z-dn.net/?f=0.181days%5E%7B-1%7D%3D%5Cfrac%7B2.303%7D%7B3.00days%7D%5Clog%5Cfrac%7B1.45%5Ctimes%2010%5E%7B-6%7D%7D%7B%5BA%5D%7D)
![[A]=3.83\times 10^{-30}mol/L](https://tex.z-dn.net/?f=%5BA%5D%3D3.83%5Ctimes%2010%5E%7B-30%7Dmol%2FL)
Hence, the concentration of radon after the given time is 
Answer:
D
Explanation:
The high jump of ionization energy indicates that we are trying to remove electron from noble gas configuration state.
The ionization energy data specifies that the Elements are from group 1 at period 3 or greater.
Removing the first electron require 496 kJ and the second ionization energy jump significantly due to the removal of electron from the noble gas configuration which is logical because electron try to maintain the especially stable state.
The mass numbers of the different isotopes of that element are averaged according to their respective abundances in nature.