<u>Given:</u>
Concentration of HNO3 = 7.50 M
% dissociation of HNO3 = 33%
<u>To determine:</u>
The Ka of HNO3
<u>Explanation:</u>
Based on the given data
[H+] = [NO3-] = 33%[HNO3] = 0.33*7.50 = 2.48 M
The dissociation equilibrium is-
HNO3 ↔ H+ + NO3-
I 7.50 0 0
C -2.48 +2.48 +2.48
E 5.02 2.48 2.48
Ka = [H+][NO3-]/HNO3 = (2.48)²/5.02 = 1.23
Ans: Ka for HNO3 = 1.23
The product of prime polynomials is equivalent to 36x3 – 15x2 – 6x is letter B which is 3x(3x – 2)(4x 1). Below is the solution.
3x(3x - 2) (4x + 1)
= 9x2 - 6x (4x + 1)
= 36x3 + 9x2 + - 24x2 - 6x
= 36x3 - 15x2 - 6x
Emission spectrum results from the movement of an electron from a higher to a lower energy level. The frequency of the photon is 5.5 * 10^14 Hz.
From the formula;
E = hc/λ
h = Plank's constant =
Js
c = speed of light= 
λ = wavelength =
m
E = 
E =
J
Also;
E =hf
Where;
h = Planks's constant
f = frequency of photon
f = E/h
f = 
f =
Hz
Learn more: brainly.com/question/18415575
Answer: The body produce NADPH a reducing equivalents as antioxidants.
Explanation:
During the pentose phosphate pathway reducing equivalents like NADPH is produced which help to prevent oxidative stress. It reduces glutathione via glutathione reductase which converts reactive H2O2 to H2O by glutathione peroxidase.