<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:

Explanation:
Hello!
In this case, since a dilution process implies that the moles of the solute remain the same before and after the addition of diluting water, we can write:

Thus, since we know the volume and concentration of the initial sample, we compute the resulting concentration as shown below:

Best regards!
Answer:
A compound
Example:
Sugar contains carbon, hydrogen and oxygen. Water (H2O) has two hydrogen atoms (H2) and one oxygen atom (O). So both sugar and water are chemical compounds. Together they are a mix of chemical compounds.