Answer: Δn for the following equation in relating Kc to Kp is 2.
Explanation:
The balanced chemical equation is:
Relation of 
with 
is given by the formula:
Where,
= equilibrium constant in terms of partial pressure
= equilibrium constant in terms of concentration
R = Gas constant
T = temperature
= change in number of moles of gas particles = 
Putting values in above equation, we get:
Thus the value of Δn for the following equation in relating Kc to Kp is 2.
Is This it? I’m sorry if it wasn’t
Answer:
45
Explanation:
kasi madali lng sya buhatun dpat paning kamotan nimo na sili kay magsalig ka sa brainly dahh
The acid dissociation constant is defined as Ka = [H+][A-]/[HA] where [H+], [A-] and [HA] are the concentrations of protons, conjugate base, and acid in solution respectively. Assuming this is a weak acid as the pH is quite high for a 1.35 M solution, we can assume that the change in [HA] is negligible and therefore [HA] = 1.35 M.
To calculate [H+] we can use the relationship pH = -log[H+], rearranging to give: [H+] = 10^(-pH) = 10^(-2.93) = 1.17 x 10^(-3).
Since the acid is relatively concentrated we can assume therefore that [H+] = [A-] as for each proton dissociated, a conjugate base is formed.
Therefore, we can calculate Ka as:
Ka = [H+]^2/[HA] = (1.17 x 10^-3 M)^2/1.35 = 1.01 x 10^-6 M
