Answer:
There will not be any ejection of photoelectrons
Explanation:
Energy of the photon= hc/λ
Where;
h= Plank's constant
c= speed of light
λ= wavelength of the incident photon
E= 6.6×10^-34 × 3 ×10^8/488 × 10^-9
E= 4.1 ×10^-19 J
Work function of the metal (Wo)= 2.9 eV × 1.6 × 10^-19 = 4.64 × 10^-19 J
There can only be ejected photoelectrons when E>Wo but in this case, E<Wo hence there will not be any ejection of photoelectrons.
Answer: Dissociation constant of the acid is
.
Explanation: Assuming the acid to be monoprotic, the reaction follows:

pH of the solution = 6
and we know that
![pH=-log([H^+])](https://tex.z-dn.net/?f=pH%3D-log%28%5BH%5E%2B%5D%29)
![[H^+]=antilog(-pH)](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3Dantilog%28-pH%29)
![[H^+]=antilog(-6)=10^{-6}M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3Dantilog%28-6%29%3D10%5E%7B-6%7DM)
As HA ionizes into its ions in 1 : 1 ratio, hence
![[H^+]=[A^-]=10^{-6}M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D%5BA%5E-%5D%3D10%5E%7B-6%7DM)
As the reaction proceeds, the concentration of acid decreases as it ionizes into its ions, hence the decreases concentration of acid at equilibrium will be:
![[HA]=[HA]-[H^+]](https://tex.z-dn.net/?f=%5BHA%5D%3D%5BHA%5D-%5BH%5E%2B%5D)
![[HA]=0.1M-10^{-6}M](https://tex.z-dn.net/?f=%5BHA%5D%3D0.1M-10%5E%7B-6%7DM)
![[HA]=0.09999M](https://tex.z-dn.net/?f=%5BHA%5D%3D0.09999M)
Dissociation Constant of acid,
is given as:
![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%7BHA%7D)
Putting values of
in the above equation, we get


Rounding it of to one significant figure, we get

Answer:
87.5198
Explanation:
(43.65 * 87.05) + (48.25 * 87.93) + (8.11 * 87.50) = 8751.98
8751.98 / 100 = 87.5198
Answer : The total pressure at equilibrium is 0.350 atm
Solution : Given,
Initial pressure of
= 0.350 bar
= 0.016
The given equilibrium reaction is,

Initially 0.350 0 0
At equilibrium (0.350-2x) x x
The total pressure at equilibrium = 
Thus, the total pressure at equilibrium is 0.350 atm