I'd say that the answer is erosion
<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 
Love the profile picture tho
Answer:
0.013%
Yes, it does. The answer agrees with the statement.
Explanation:
Both conformers are in equilibrium, and it can be represented by the equilibrium equation K:
K = [twist-boat]/[chair]
The free energy between them can be calculated by:
ΔG° = -RTlnK
Where R is the gas constant (8.314 J/mol.K), and T is the temperature (25°C + 273 = 298 K).
ΔG° = 5.3 kcal/mol * 4.182 kJ/kcal = 22.165 kJ/mol = 22165 J/mol
22165 = -8.314*298*lnK
-2477.572lnK = 22165
lnK = -8.946
K = 
K = 1.30x10⁻⁴
[twist-boat]/[chair] = 1.30x10⁻⁴
[twist-boat] = 1.30x10⁻⁴[chair]
The percentage of the twist-boat conformer is:
[twist-boat]/([twist-boat] + [chair]) * 100%
1.30x10⁻⁴[chair]/(1.30x10⁻⁴[chair] + [chair]) *100%
0.013%
The statement about the conformers is that the chair conformer is more stable, and because of that is more present. So, the answer agrees with it.