Answer:
Initial concentration of HI is 5 mol/L.
The concentration of HI after is 0.00345 mol/L.
Explanation:
Rate Law:
Rate constant of the reaction = k =
Order of the reaction = 2
Initial rate of reaction =
Initial concentration of HI =
Final concentration of HI after t = [A]
t =
Integrated rate law for second order kinetics is given by:
The concentration of HI after is 0.00345 mol/L.
The balance chemical equation is follow,
2 I⁻ + SO₄²⁻ + 4 H⁺ → I₂ + SO₂ + 2 H₂O
According to this reaction, 2 moles of I⁻ reacts with 1 mole of SO₄²⁻ to produce 1 mole of I₂ and 1 mole of SO₂.
Result:
So with the formation of 1 mole of I₂, 1 mole of SO₂ is produced.
POH = 14 - pH
pOH = -log [OH-]
<h3>
Answer:</h3>
3.11 moles
<h3>
Explanation:</h3>
We are given 56 g of water (H₂O)
Required to calculate the number of moles of water.
- The relationship between moles, mass and molar mass is given by;
- Moles = Mass ÷ molar mass
in this case;
- Mass of water = 56 g
- Molar mass of water = 18.02 g/mol
Therefore;
Moles of water = 56 g ÷ 18.02 g/mol
= 3.11 moles
Therefore, moles of water in 56 g will be 3.11 moles