Answer:
Approximately 
, assuming that this acid is monoprotic.
Explanation:
Assume that this acid is monoprotic. Let 
denote this acid.
.
Initial concentration of without any dissociation:
![[{\rm HA}] = 0.730\; \rm mol \cdot L^{-1}](https://tex.z-dn.net/?f=%5B%7B%5Crm%20HA%7D%5D%20%3D%200.730%5C%3B%20%5Crm%20mol%20%5Ccdot%20L%5E%7B-1%7D)
.
After 
of that was dissociated, the concentration of both 
and 
(conjugate base of this acid) would become:
.
Concentration of in the solution after dissociation:
.
Let ![[{\rm HA}]](https://tex.z-dn.net/?f=%5B%7B%5Crm%20HA%7D%5D)
, ![[{\rm H}^{+}]](https://tex.z-dn.net/?f=%5B%7B%5Crm%20H%7D%5E%7B%2B%7D%5D)
, and ![[{\rm A}^{-}]](https://tex.z-dn.net/?f=%5B%7B%5Crm%20A%7D%5E%7B-%7D%5D)
denote the concentration (in 
or 
) of the corresponding species at equilibrium. Calculate the acid dissociation constant 
for , under the assumption that this acid is monoprotic:
![\begin{aligned}K_{\rm a} &= \frac{[{\rm H}^{+}] \cdot [{\rm A}^{-}]}{[{\rm HA}]} \\ &= \frac{(0.09125\; \rm mol \cdot L^{-1}) \times (0.09125\; \rm mol \cdot L^{-1})}{0.63875\; \rm mol \cdot L^{-1}}\\[0.5em]&\approx 1.30 \times 10^{-2} \end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7DK_%7B%5Crm%20a%7D%20%26%3D%20%5Cfrac%7B%5B%7B%5Crm%20H%7D%5E%7B%2B%7D%5D%20%5Ccdot%20%5B%7B%5Crm%20A%7D%5E%7B-%7D%5D%7D%7B%5B%7B%5Crm%20HA%7D%5D%7D%20%5C%5C%20%26%3D%20%5Cfrac%7B%280.09125%5C%3B%20%5Crm%20mol%20%5Ccdot%20L%5E%7B-1%7D%29%20%5Ctimes%20%280.09125%5C%3B%20%5Crm%20mol%20%5Ccdot%20L%5E%7B-1%7D%29%7D%7B0.63875%5C%3B%20%5Crm%20mol%20%5Ccdot%20L%5E%7B-1%7D%7D%5C%5C%5B0.5em%5D%26%5Capprox%201.30%20%5Ctimes%2010%5E%7B-2%7D%20%5Cend%7Baligned%7D)
.
Answer:
Inside the nucleus, the attractive strong nuclear force between protons outweighs the repulsive electromagnetic force and keeps the nucleus stable. Outside the nucleus, the electromagnetic force is stronger and protons repel each other.
Explanation:
c group A is the correct answer
