Unless you're able to provide a diagram representing the problem, there's not much I can do so solve this problem.
Answer:
30 cm³
Explanation:
Step 1: Given data
- Density of aluminum (ρ): 2.7 g/cm³
- Mass of aluminum (m): 81 g
- Volume occupied by aluminum (V): ?
Step 2: Calculate the volume occupied by aluminum
The density of aluminum is equal to its mass divided by its volume.
ρ = m/V
V = m/ρ
V = 81 g / 2.7 g/cm³
V = 30 cm³
<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
