Ncert Solutions Chemistry Class 11th

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

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

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

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

(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

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

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

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

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

Mass of carbon converted into CO = 6 $\frac{60}{100}*360=216gm\Rightarrow Mol=\frac{216}{12}=18$  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 * 12 = 4800KJ

$C_{\left(s\right)}+\frac{1}{2}{O}_2\left(g\right)\underset{}{\overset{}{\to }}C{O}_{\left(g\right)}+100KJ$  

Mole = 18 for CO production, Hence energy produced = 100 * 18 = 1800KJ

Total heat = 4800 + 1800 = 6600KJ

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

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

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

An atomic orbital is characterized by $n,\mathcal{l}$ $$ and m.

Po₂ = P_Total -   $P_{N_2}$ $5=\left(\frac{x/32}{x/32+\frac{200}{20}}\right)*25\Rightarrow x=80gm$

= 25 – 20 = 5bar

Po₂ =   $\frac{n_{O_2}}{n_{O_2}+n_{Ne}}*P_{Total}$

$5=\left(\frac{x/32}{x/32+\frac{200}{20}}\right)*25\Rightarrow x=80gm$
 

Let x gm is burnt

Moles = x/280

Heat released by x/280 mol = 2.5 * 0.45 kJ

Heat released by 1 mol =  $\frac{2.5*0.45*280}{x}=kJ$

$\begin{array}{l}\Delta H=\Delta U+\Delta ngRT\\ \Delta H=\Delta U\end{array}$            

$9=\frac{2.5*280*0.45}{x}$            

x = 35 gm