Ncert Solutions Chemistry Class 11th

Depending on the nature of reducing agent H₂O₂ can act as oxidizing agent in both acidic as well as basic medium.

              Density of D₂O = 1.1 gm/cc

              Den

The reason why bond angle is larger in ${\mathrm{N}\mathrm{H}}_3$ than  ${\mathrm{P}\mathrm{H}}_3$ are given below.

• In ${\mathrm{N}\mathrm{H}}_3$ , there is one lone pair on the nitrogen atom increases repulsion, while the lone pair on phosphorus is in a higher energy orbital and causes less repulsion.
• Nitrogen is a small and electronegative atom whereas phosphorus is larg

(a) Polluted water has low value of dissolved oxygen, but high value of B.O.D., because chemical and organic matter uses dissolved oxygen to decompose.

(b) Eutrophication is result of excessive growth of weed in water bodies, which consumes dissolved oxygen, thus decreases oxygen.

Cu is least reactive metal and it lies below hydrogen in reactivity series.

${\left(N{H}_4\right)}_{2}C{r}_2{O}_7\underset{}{\overset{\Delta }{\to }}\boxed{N_2}+C{r}_2{O}_3+4H_{2}O$

$KMn{O}_4+HC\mathcal{l}\to MnC{\mathcal{l}}_2+KC\mathcal{l}+\boxed{C{\mathcal{l}}_2}+H_{2}O$

$A\mathcal{l}+NaOH+H_{2}O\to Na\left[A\mathcal{l}{\left(OH\right)}_4\right]+\boxed{H_2}$

$NaN{O}_3\to NaN{O}_2+\boxed{O_2}$

(a) low solubility of LiF in H₂O due to high lattice energy.

(b) $O_2^{-}$  has unpaired e^- thus paramagnetic.

(c) Na in liq NH₃ forms electronated ammonia thus conducting in nature.

(d) $d_K<d_{Na}$ .

In clark’s method Ca (OH)₂ is used for softening hard water.

$Ca{\left(HC{O}_3\right)}_2+Ca{\left(OH\right)}_2\to 2CaC{O}_3+2H_{2}O$

$Mg{\left(HC{O}_3\right)}_2+2Ca{\left(OH\right)}_2\to 2CaC{O}_3+Mg{\left(OH\right)}_{2}2H_{2}O$    

Thus CaCO₃ and Mg (OH)₂ are formed in reaction.

Total concentration of H⁺ after the mixing of HCl and H₂SO₄

$\left[H^{+}\right]=\frac{\left(200\times 0.01\right)+\left(400\times 2\times 0.01\right)}{600}$

$\left[H^{+}\right]=\frac{1}{60}$

$\therefore pH=lo{g}_{10}\left(\frac{1}{60}\right)=1.78$

Mass of pure carbon in coal = 0.6 × 1000gm × $\frac{60}{100}=360gm$  

Mass of carbon converted into CO = 6 $\frac{60}{100}\times 360=216gm\Rightarrow Mol=\frac{216}{12}=18$  mole of carbon

Mass of carbon converted into CO₂ = 360 – 216 = 144gm Þ Mole = $\frac{144}{12}=12$  mole of carbon.

C _(s) + O₂ (g) -> CO₂ (g) + 400KJ

Mole – 12 for CO₂ production, Hence total energy produced = 400

$\Delta H=\Delta U+\Delta \left(PV\right)$

= $\Delta U+P\Delta V+V\Delta P$

= $\Delta U+P\Delta V$   (At constant pressure)