Answer:
A. 444.5 pm
Explanation:
We know that:

i.e.


in a face-centered cubic crystal, the number of atoms per unit cell is (n) = 4
The molar mass of manganese (II) oxide ![[Mn(11)O] = 70.93 \ g/mol](https://tex.z-dn.net/?f=%5BMn%2811%29O%5D%20%3D%2070.93%20%5C%20g%2Fmol)
Density
is given as 5.365 g/cm³
Avogadro constant
= 6.023 × 10²³ atoms/mol
∴

Making th edge length "a" the subject, we get:



![a= \sqrt[3]{8.78 \times 10^{-23} \ cm^3}](https://tex.z-dn.net/?f=a%3D%20%5Csqrt%5B3%5D%7B8.78%20%5Ctimes%2010%5E%7B-23%7D%20%5C%20cm%5E3%7D)
a = 4.445 × 10⁻⁸ cm
a = 444.5 pm
I think b would make the most sense.
Answer:
![[NO]=\frac{k_{-1}}{k_1} [N_2O_2]](https://tex.z-dn.net/?f=%5BNO%5D%3D%5Cfrac%7Bk_%7B-1%7D%7D%7Bk_1%7D%20%5BN_2O_2%5D)
Explanation:
Hello!
In this case, since the reaction may be assumed in chemical equilibrium, we can write up the rate law as shown below:
![r=-k_1[NO]+k_{-1}[N_2O_2]](https://tex.z-dn.net/?f=r%3D-k_1%5BNO%5D%2Bk_%7B-1%7D%5BN_2O_2%5D)
However, since the rate of reaction at equilibrium is zero, due to the fact that the concentrations remains the same, we can write:
![0=-k_1[NO]+k_{-1}[N_2O_2]](https://tex.z-dn.net/?f=0%3D-k_1%5BNO%5D%2Bk_%7B-1%7D%5BN_2O_2%5D)
Which can be also written as:
![k_1[NO]=k_{-1}[N_2O_2]](https://tex.z-dn.net/?f=k_1%5BNO%5D%3Dk_%7B-1%7D%5BN_2O_2%5D)
Then, we solve for the concentration of NO to obtain:
![[NO]=\frac{k_{-1}}{k_1} [N_2O_2]](https://tex.z-dn.net/?f=%5BNO%5D%3D%5Cfrac%7Bk_%7B-1%7D%7D%7Bk_1%7D%20%5BN_2O_2%5D)
Best regards!