Question

What do you observe when:
(i) HgCl₂ and KI when rubbed in mortar
(ii) Glauber’s salt exposed to air
(iii) Action of reducing agents on Fe (III) salts
(iv) H₂S is passed through lead nitrate
(v) SO₂ is passed through Ca(OH)₂

Answer 1

Answer:

u can see in labaratory and home

Answer 2

Reactions

Explanation:

1) HgCl₂ + 2KI → HgI₂ + 2KCl

If KI is in excess, HgI₂ will react with it to form K₂HgI₄

2KI + HgI₂ → K₂HgI₄

2) Glauber's salt (sodium sulphate decahydrate - Na 2SO 4·10H 2O) when exposed to dry air will effloresce to form powdery anhydrous sodium sulphate.

3) The addition of reducing agent will accelerate the Fe(III)/Fe(II) redox cycle, leading to a relatively steady Fe(II) concentration and higher production of free radicals.

4) When H2S is passed through lead nitrate solution, nitric acid and lead sulphide are formed.

5) Sulphur dioxide will dissolve in water to produce sulphurous acic (H2SO3). The acid will react with calcium hydroxide particles in lime water to form calcium sulphite and water. This will remove the milkiness of the lime water.