<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 
The answer is A. The pH decreases
Answer:
idfk
Explanation:
cause God made it that way
The kinetic energy of the products is equal to the energy liberated which is 92.2 keV. But let's convert the unit keV to Joules. keV is kiloelectro volt. The conversion that we need is: 1.602×10⁻¹⁹ <span>joule = 1 eV
Kinetic energy = 92.2 keV*(1,000 eV/1 keV)*(</span>1.602×10⁻¹⁹ joule/1 eV) = 5.76×10²³ Joules
From kinetic energy, we can calculate the velocity of each He atom:
KE = 1/2*mv²
5.76×10²³ Joules = 1/2*(4)(v²)
v = 5.367×10¹¹ m/s