Question

Напишите для предложенных соединений уравнения диссоциации, а также в молекулярной и ионной формах уравнения возможных реакций взаимодействия их H2SO4 с и NaOH. при . HCl; Cr(OH)3

Answer 1

H2SO4 + 2NaOH = Na2SO4 +2H2O
2H(+) + SO4(2-) + 2Na(+) +2OH(-) = 2Na(+) + SO4(2-) + 2H2O
2H(+) + 2OH(-) =  2H2O
H(+) + OH(-) =  H2O
3HCL + Cr(OH)3 = CrCL3 + 3H2O
3H(+) +3CL(-) + Cr(OH)3 = Cr)+3) + 3CL+ 3H2O
3H(+)  + Cr(OH)3 = Cr(+3) +  3H2O

Answer 2

Equations of Dissociation

The dissociation equations for the given compounds are as follows:

- H2SO4 (Sulfuric Acid): - Molecular Form: H2SO4 - Ionic Form: 2H^+ + SO4^2-

- NaOH (Sodium Hydroxide): - Molecular Form: NaOH - Ionic Form: Na^+ + OH^-

- HCl (Hydrochloric Acid): - Molecular Form: HCl - Ionic Form: H^+ + Cl^-

- Cr(OH)3 (Chromium(III) Hydroxide): - Molecular Form: Cr(OH)3 - Ionic Form: Cr^3+ + 3OH^-

Possible Reactions with H2SO4 and NaOH

1. H2SO4 with NaOH: - Equation of Reaction: H2SO4 + 2NaOH → Na2SO4 + 2H2O - Explanation: Sulfuric acid reacts with sodium hydroxide to form sodium sulfate and water.

2. HCl with NaOH: - Equation of Reaction: HCl + NaOH → NaCl + H2O - Explanation: Hydrochloric acid reacts with sodium hydroxide to form sodium chloride and water.

3. Cr(OH)3 with HCl: - Equation of Reaction: Cr(OH)3 + 3HCl → CrCl3 + 3H2O - Explanation: Chromium(III) hydroxide reacts with hydrochloric acid to form chromium(III) chloride and water.

These reactions are based on the dissociation of the compounds and their interactions with the respective substances.

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