Explanation:
Sodium has atomic number of 11 and its electronic configuration is given by:
![[Na]=1s^22s^22p^63s^1](https://tex.z-dn.net/?f=%5BNa%5D%3D1s%5E22s%5E22p%5E63s%5E1)
The nearest stable electronic configuration to sodium is of the neon. So, in order to attain stability of noble gas it will loose its single electron.

![[Na^+]=1s^22s^22p^63s^0](https://tex.z-dn.net/?f=%5BNa%5E%2B%5D%3D1s%5E22s%5E22p%5E63s%5E0)
Sodium has single valency that is 1.
Let nbe the valency of the ion 'X'
By criss-cross method, the oxidation state of the ions gets exchanged and they form the subscripts of the other ions. This results in the formation of a neutral compound.

So, the formulas for all the possible compounds that sodium can form with the other ions will be:

Answer:
A) 31.22
Explanation:
The reaction of sulfuric acid with NaOH is:
H₂SO₄ + 2 NaOH → Na₂SO₄ + 2H₂O
To solve this problem we need to determine the moles of acid that will react, and, using the chemical equation we can determine the moles of NaOH and the volume that a 0.2389M NaOH solution would require to neutralize it.
<em>Moles H₂SO₄ (Molar mass: 98.08g/mol):</em>
0.9368g * 39.04% = 0.3657g H₂SO₄ * (1mol / 98.08g) =
3.7289x10⁻³moles H₂SO₄
And moles of NaOH that you require to neutralize the acid are:
3.7289x10⁻³moles H₂SO₄ * (2 moles NaOH / 1 mole H₂SO₄) =
7.4578x10⁻³ moles NaOH
Using a 0.2389M NaOH solution:
7.4578x10⁻³ moles NaOH * (1L / 0.2389mol) = 0.03122L = 31.22mL
Right answer is:
<h3>A) 31.22
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