Ncert Solutions Chemistry Class 11th

Moles of N₂O = $\frac{2.2}{44}=\frac{1}{20}$  

$\Delta H=n{C}_p\Delta T=\frac{1}{20}\times 100\left(-40\right)=-200J$  

$\Delta U=q_p+w$

$w=-P_{ext}.\Delta V$  

$w=-1\frac{\left(167.75-217.1\right)}{1000}\times 101.3J=+5J$  

$\Delta U=-200+5=195J$

Pentavalent oxides of group – 15 elements, E₂O₅ is more acidic than trivalent oxides, E₂O₃ of the same element.

 Acidic strength of trivalent oxides decreases down the group as metallic strength increases.

N₂ gas is obtained by thermal decomposition of Ba (N₃)₂ as,

$Ba{\left(N_3\right)}_2\underset{}{\overset{\Delta }{\to }}Ba+3N_2\uparrow$

$BaC{l}_2+S{O}_3^{2-}\stackrel{}{\to }\underset{White}{BaS{O}_3}\downarrow \stackrel{dil.HCl}{\to }\underset{\begin{array}{l}burningsulphur\\ likesmell\end{array}}{S{O}_2\uparrow }$

$CuO+H_2\to Cu+H_{2}O$

Borax forms basic buffer solution in aqueous medium.

Baking soda = NaHCO₃

Washing soda = Na₂CO₃. 10H₂O

Caustic soda = NaOH

% of C in organic compound

              $=\frac{12}{14}\times \frac{WC{O}_2}{W.O.C.}\times 100$  

              $=\frac{952.56}{21.648}=44\mathrm{\%}$  

              $\mathrm{\%}ofH=\frac{2}{18}\times \frac{W{H}_{2}O}{W.O.C.}\times 100$  

              = $\frac{20}{18}\times \frac{0.4428}{0.492}\times 100$  

  &nb

$C\mathcal{l}{F}_3$ ® T-shaped (sp³d)

 IF₇ ® Pentagonal bipyramidal (sp³d³)

BrF₅ ® Square pyramidal (sp³d²)

BrF₃ ® T-Shaped (sp³d)

I₂Cl₆ ® Triangular bipyramidal (sp³d)

$I{F}_5\to$  Square Pyramidal (sp³d²)

ClF ® (sp³)

ClF₅ ® Square Pyramidal (sp³d²)

Br₅, IF₅ & ClF₅ ® Square Pyramidal

$2B{F}_3+6NaH\to B_2{H}_6+6NaF$

${\mathrm{B}}_2{\mathrm{H}}_6+2\stackrel{?}{\mathrm{N}}{\left(\mathrm{M}\mathrm{e}\right)}_3\to 2{\mathrm{H}}_3\mathrm{B}\leftarrow :\mathrm{N}{\left(\mathrm{M}\mathrm{e}\right)}_3$