Answer:
The system will change its concentration to shift to a new equilibrium position.
Explanation:
For example in the Haber Process
N2 + 3H2 ⇄ 2NH3
If the pressure is increased the process will move to the right - to have more NH3 and less of the nitrogen and hydrogen.
The first one is substance 3
The second one is Oxygen, Helium, and carbon dioxide
The third one is the oil floats on top of the water
C becuase that is the one that you would have to do becuase that is the only option
Answer:
The value of dissociation constant of the monoprotic acid is
.
Explanation:
The pH of the solution = 2.46
![pH=-\log[H^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%5BH%5E%2B%5D)
![2.46=-\log[H^+]](https://tex.z-dn.net/?f=2.46%3D-%5Clog%5BH%5E%2B%5D)
![[H^+]=0.003467 M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D0.003467%20M)

Initially
0.0144 0 0
At equilibrium
(0.0144-x) x x
The expression if an dissociation constant is given by :
![K_a=\frac{[A^-][H^+]}{[HA]}](https://tex.z-dn.net/?f=K_a%3D%5Cfrac%7B%5BA%5E-%5D%5BH%5E%2B%5D%7D%7B%5BHA%5D%7D)

![x=[H^+]=0.003467 M](https://tex.z-dn.net/?f=x%3D%5BH%5E%2B%5D%3D0.003467%20M)


The value of dissociation constant of the monoprotic acid is
.
The middle nitrogen has two sigma bonds and one pi bond. You know that one p orbital is used in the double bond and two sp2 orbitals are involved in the sigma bond. This leaves one sp2 orbital for the lone pair to occupy.