<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
Answer:
Na
Explanation:
because sodium has 1 electrons so it loses it to be stable and so have positive charge of 1
we have,
wavelenght=c/f
where c= 3×10^8 m/s
f=6.3×10^12 s^-1
so wavelength=(3×10^8)/(6.3×10^12)
=0.476×10^-4 m
Answer:
the first statement
Explanation:
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