The answer is Al.
If it is a main group element with 3 electrons in its Lewis dot structure, it must be in group 3A. If it is in the 3p orbital section, then it must be in period 3, since the p orbital is a valence orbital and the number that preceeds it is the principal quantum number. Therefore, your answer is the element in period 3 and group 3A, which is aluminum.
I believe it would be 7.731. x 10^5? That is if they want you to evaluate.
Answer:


Explanation:
first write the equilibrium equaion ,
⇄ 
assuming degree of dissociation
=1/10;
and initial concentraion of
=c;
At equlibrium ;
concentration of
![[C_3H_5O_3^{-} ]= c\alpha](https://tex.z-dn.net/?f=%5BC_3H_5O_3%5E%7B-%7D%20%20%5D%3D%20c%5Calpha)
![[H^{+}] = c\alpha](https://tex.z-dn.net/?f=%5BH%5E%7B%2B%7D%5D%20%3D%20c%5Calpha)

is very small so
can be neglected
and equation is;

= 
![P_H =- log[H^{+} ]](https://tex.z-dn.net/?f=P_H%20%3D-%20log%5BH%5E%7B%2B%7D%20%5D)





composiion ;
![c=\frac{1}{\alpha} \times [H^{+}]](https://tex.z-dn.net/?f=c%3D%5Cfrac%7B1%7D%7B%5Calpha%7D%20%5Ctimes%20%5BH%5E%7B%2B%7D%5D)
![[H^{+}] =antilog(-P_H)](https://tex.z-dn.net/?f=%5BH%5E%7B%2B%7D%5D%20%3Dantilog%28-P_H%29)
![[H^{+} ] =0.0014](https://tex.z-dn.net/?f=%5BH%5E%7B%2B%7D%20%5D%20%3D0.0014)


Test tube of ammonium chloride (NH4Cl) being heated over a bunsen burner flame. Ammonium chloride decomposes readily when heated, but condenses in the cooler area at the top of the test tube. This is a reversible reaction, where the ammonium chloride decomposes into the gases ammonia (NH3) and hydrogen chloride (HCl).