<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:
B = basic
Explanation:
Given data:
[OH⁻] = 5.35×10⁻⁴M
pH = ?
Solution:
pOH = -log[OH⁻]
pOH = - [5.35×10⁻⁴]
pOH = 3.272
it is known that,
pH + pOH = 14
pH = 14- pOH
pH = 14 - 3.272
pH = 10.728
The acidic pH is range from zero to less than 7 while 7 pH is neutral and above 7 the pH is basic. So, the given solution is basic.
A reducing agent is one which is oxidised in the reaction itself. When you take into account the oxidation numbers you will see that the Cl- ions are oxidised from an oxidation number of -1 to 0 in Cl2. Therefore Cl- ions are the reducing agent.
Answer:
2x+3y=-6
4x+3y=12
3x-y=5
5x+3y=1
Explanation:
standard form: ax + by = C