Question

Помогите сделать химию! Нап. уравнение електролитической дис. елем. H2SO4,Ba(OH)2,NaHSO3~как это сделать?

Answer 1

H2SO4 = 2H(+) + SO4(2-)
Ba(OH)2 = Ba(2+) + 2OH(-)
NaHSO3 = Na(+) + HSO3(-)

Answer 2

Electrolytic dissociation equations for H2SO4, Ba(OH)2, and NaHSO3

To write the electrolytic dissociation equations for the compounds H2SO4, Ba(OH)2, and NaHSO3, we need to understand how these compounds dissociate into ions when they are dissolved in water.

1. H2SO4 (sulfuric acid): Sulfuric acid is a strong acid that dissociates completely in water. It forms two hydrogen ions (H+) and one sulfate ion (SO4^2-) when dissolved in water. The electrolytic dissociation equation for sulfuric acid is:

H2SO4 → 2H+ + SO4^2- [[1]]

2. Ba(OH)2 (barium hydroxide): Barium hydroxide is a strong base that also dissociates completely in water. It forms two barium ions (Ba^2+) and two hydroxide ions (OH-) when dissolved in water. The electrolytic dissociation equation for barium hydroxide is:

Ba(OH)2 → Ba^2+ + 2OH- [[2]]

3. NaHSO3 (sodium bisulfite): Sodium bisulfite is a weak acid salt that partially dissociates in water. It forms sodium ions (Na+) and bisulfite ions (HSO3-) when dissolved in water. The electrolytic dissociation equation for sodium bisulfite is:

NaHSO3 → Na+ + HSO3- [[3]]

Please note that these equations represent the dissociation of the compounds in water and do not take into account any other factors such as concentration or temperature.

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