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⁻
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Answer:
The valid quantum numbers are l=0, l=-2 and l= 2.
Explanation:
Given that,
n = 3 electron shell
Suppose, the valid quantum numbers are
l = 3
m = 3
l = 0
m = –2
l = –1
m = 2
We know that,
The value of n = 3
Principle quantum number :
Then the principal quantum number is 3. Which is shows the M shell.
So, n = 3
Azimuthal quantum number :
The azimuthal quantum number is l.

Magnetic quantum number :
The magnetic quantum number is

Hence, The valid quantum numbers are l=0, l=-2 and l= 2.
Answer:
C
Explanation:
Metals are good conductors of heat
Pacticles of gas are more compact, but still have the ability to move.