A sample of crude sodium iodide was analyzed by the following balanced reaction. The oxidation number of S in SO₄²⁻ is +6.
8 I⁻ + 6 H₂O + SO₄²⁻ → 4 I₂ + H₂S + 10 OH⁻
Let's consider the following unbalanced redox reaction.
I⁻ + SO₄²⁻ → I₂ + H₂S
- The oxidation number of I goes from -1 (I⁻) to 0 (I₂) so it is oxidized.
- The oxidation number of S goes from +6 (SO₄²⁻) to -2 (H₂S) so it is reduced.
The corresponding half-reactions are:
I⁻ → I₂
SO₄²⁻ → H₂S
We will perform the mass balance adding OH⁻ and H₂O where appropriate.
2 I⁻ → I₂
6 H₂O + SO₄²⁻ → H₂S + 10 OH⁻
Then, we will perform the charge balance adding electrons where appropriate.
2 I⁻ → I₂ + 2 e⁻
8 e⁻ + 6 H₂O + SO₄²⁻ → H₂S + 10 OH⁻
Finally, we will multiply the first half-reaction by 4 and the second by 1, and add them.
4 × (2 I⁻ → I₂ + 2 e⁻)
1 × (8 e⁻ + 6 H₂O + SO₄²⁻ → H₂S + 10 OH⁻)
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8 I⁻ + 6 H₂O + SO₄²⁻ → 4 I₂ + H₂S + 10 OH⁻
A sample of crude sodium iodide was analyzed by the following balanced reaction. The oxidation number of S in SO₄²⁻ is +6.
8 I⁻ + 6 H₂O + SO₄²⁻ → 4 I₂ + H₂S + 10 OH⁻
Learn more: brainly.com/question/2671074
For the data set 1, 1, 2, 5, 6, 6, 9 the median is 5. For the data set 1, 1, 2, 6, 6, 9 the median is 4.
hope it helps you
Answer:
3
Explanation:
The first election shell can only hold 2 electrons, but the next one can hold up to 8
Answer:
47.5 g of water can be formed
Explanation:
This is the reaction:
CH₄ + 2O₂ → CO₂ + 2H₂O
Methane combustion.
In this process 1 mol of methane react with 2 moles of oxygen to produce 2 moles of water and 1 mol of carbon dioxide.
As ratio is 1:2, I will produce the double of moles of water, with the moles of methane I have.
1.320 mol .2 = 2.64 moles
Now, we can convert the moles to mass (mol . molar mass)
2.64 mol . 18g/mol = 47.5 g
