Answer:
Ka = [H₃O⁺] [SO₃²⁻] / [HSO₃⁻]
Kb = [OH⁻] [H₂SO₃] / [HSO₃⁻]
Explanation:
An amphoteric substance as HSO₃⁻ is a substance that act as either an acid or a base. When acid:
HSO₃⁻(aq) + H₂O(l) ⇄ H₃O⁺(aq) + SO₃²⁻(aq)
And Ka, the acid dissociation constant is:
<h3>Ka = [H₃O⁺] [SO₃²⁻] / [HSO₃⁻]</h3><h3 />
When base:
HSO₃⁻(aq) + H₂O(l) ⇄ OH⁻(aq) + H₂SO₃(aq)
And kb, base dissociation constant is:
<h3>Kb = [OH⁻] [H₂SO₃] / [HSO₃⁻]</h3>
Answer:
The coefficient of O2 is 11
Explanation:
Step 1:
The equation for the reaction:
FeS2 + O2 → SO2 + Fe2O3
Step 2:
Balancing the equation. The equation can be balance as follow:
FeS2 + O2 → SO2 + Fe2O3
There are 2 atoms of Fe on the right side and 1 atom on the left. It can be balance by putting 2 in front of FeS2 as shown below:
2FeS2 + O2 → SO2 + Fe2O3
There are 4 atoms of S on the left side and 1 atom on the right side. It can be balance by putting 4 in front of SO2 as shown below:
2FeS2 + O2 → 4SO2 + Fe2O3
Now, there are a total of 11 atoms of O on the right side and 2 atoms on the left side. It can be balance by putting 11/2 in front of O2 as shown below:
2FeS2 + 11/2O2 → 4SO2 + Fe2O3
Multiply through by 2 to clear the fraction as shown below:
4FeS2 + 11O2 → 8SO2 + 2Fe2O3
Now the equation is balanced.
The coefficient of O2 is 11
Answer;
= 18.24
Explanation;
The ratio of N and O in the formula NO2 IS 1:2
Mass of nitrogen gas is 0.500 g
Moles of nitrogen will be;
= 0.500/16 = 0.03125 moles
Therefore;
The moles of Oxygen from the ratio will be;
= 0.03125 × 2 = 0.0625 moles
But; 0.0625 moles is equal to 1.140 g of Oxygen
The atomic number (mass in 1 mole) will be;
= 1.140 /0.0625
= 18.24
Thus the atomic number of Oxygen from the data is 18.24
Answer:
The potential wrt. calomel is 1.254 V
Explanation:
Given:
Potential wrt. silver chloride 
V
Potential wrt. saturated silver chloride 
V
Potential wrt. SCE 
V
Now potential wrt. hydrogen is given by,
V
And we find for potential wrt. calomel,
potential wrt. hydrogen + potential wrt. SEC
V
Therefore, the potential wrt. calomel is 1.254 V
= 71.2 g