Chemistry

This is a multiple choice answer as classified in NCERT Exemplar

Correct option is (iii)

Hydrogen iodide is a more powerful reducing agent than sulphuric acid, it reduces the amount of iodine in the solution from H₂SO₄ to SO₂ and HI to I₂

When chloride salts are treated with sulfuric acid,  HCl gas is formed, which produces a colourless gas.

NaCl + H₂SO₄→HCl + Na₂SO₄

Violet fumes are produced during the reaction due to the creation of iodine (I₂) gas.

2NaI + H₂SO₄→Na₂SO₄ + 2HI 2H₂O + SO₂ + I₂

This is a short answer type question as classified in NCERT Exemplar

The following reaction with silver nitrate demonstrates phosphinic acid reducing behaviour:

4AgNO₃ + 2H₂O + H₃PO₂→4Ag + 4HNO₃ + H₃PO₄

This is a short answer type question as classified in NCERT Exemplar

PCl₅ + 2Ag→2AgCl + PCl₃

AgCl + 2NH₃ (aq)→ [Ag (NH₃)₂] + Cl^- 

This is a short answer type question as classified in NCERT Exemplar

When nitric acid is mixed with copper metal, it produces distinct oxidation products.

3Cu + 8HNO₃ (dil.)→3Cu (NO₃)₂ + 2NO + 4H₂O

Cu + 4HNO₃ (Conc)→3Cu (NO₃)₂ + 2NO₂ + 2H₂O

This is a short answer type question as classified in NCERT Exemplar

The crystal structure of red phosphorus features a sophisticated network of bonding, whereas white phosphorus is made up of P₄ molecules. To avoid natural combustion, white phosphorus must be stored in water, but red phosphorus is stable in air.

This is a short answer type question as classified in NCERT Exemplar

On interaction with P₄O₁₀ forms an oxide of nitrogen,  N₂O₅, and metaphosphoric acid, 3, nitric acid is formed. HPO₃

HNO₃ + P₄O₁₀→4HPO₃ + 2N₂O₅

This is a short answer type question as classified in NCERT Exemplar

Three oxoacids are

(a) HNO₂, Nitrous acid

(b) HNO₃, Nitric acid

(c) Hyponitrous acid, H₂N₂O₂

3HNO₂ + H₂O + 2NO

This is a short answer type question as classified in NCERT Exemplar

P₄+6Cl₂→4PCl₃…  (i)

PCl₃+3H₂O→H₃PO₃+3HCl *4

On adding eq. (i) and (ii)

P₄+6Cl₂+12H₂O→4H₃PO₃+12HCl

1 mol of white phosphorus produces 12 mol of HCl

62 g of white phosphorus has been taken which is equivalent to $\frac{\mathrm{N}\mathrm{o}.\mathrm{o}\mathrm{f}\mathrm{}\mathrm{M}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{s}\mathrm{ }\mathrm{ }}{\mathrm{V}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{m}\mathrm{e}\mathrm{}\mathrm{i}\mathrm{n}\mathrm{}\mathrm{l}\mathrm{i}\mathrm{t}\mathrm{r}\mathrm{e}\mathrm{s}\mathrm{ }}$ = $\frac{\mathrm{N}\mathrm{o}.\mathrm{o}\mathrm{f}\mathrm{}\mathrm{M}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{s}\mathrm{ }\mathrm{ }}{\mathrm{M}\mathrm{o}\mathrm{l}\mathrm{a}\mathrm{r}\mathrm{i}\mathrm{t}\mathrm{y}}$ mol. Therefore 6 molHCl will be formed.

Mass of 6 molHCl=6*36.5=219.0g HCl

This is a short answer type question as classified in NCERT Exemplar

P₄O₆ + 6H₂O→4H₃PO₃  (i)

Next is neutralization

4 times H₃PO₃ + 2NaOH→Na₂HPO₃ + 2H₂O  (ii)

On addition of 2 equations

P₄O₆+8NaOH→4Na₂HPO₃+2H₂O

P₄O₆ ( mol.mass )= (4*31+16*6)=220

Number of moles of P₄O₆ = $\mathit{?}/\mathit{V}$    = $\mathit{?}$

The product formed is neutralized 8 moles of NaOH

P₄O₆ = 8* $\mathit{?}$  = $\mathit{?}$  molNaOH

Molarity of NaOH in 1 litre  =  0.1M

 Molarity = $\frac{\mathrm{M}\mathrm{o}\mathrm{l}\mathrm{a}\mathrm{r}\mathrm{M}\mathrm{a}\mathrm{s}\mathrm{s}}{\mathrm{G}\mathrm{i}\mathrm{v}\mathrm{e}\mathrm{n}\mathrm{m}\mathrm{a}\mathrm{s}\mathrm{s}}$

 Volume = $\frac{1.1}{220}$  

= $\frac{1.1}{220}$ * $\frac{8.8}{220}$ = 0.4 L

This is a short answer type question as classified in NCERT Exemplar

Due its disintegration into oxygen results in the release of heat ΔH is negative) and an increase in entropy (ΔS is positive), ozone is thermodynamically unstable in comparison to oxygen. These two reactions combine to produce a substantial negative Gibbs energy change (ΔG) for its conversion to oxygen.