Ncert Solutions Chemistry Class 11th

(A) Isothermal process $\Rightarrow$ Temperature is constant throughout the process

(B) Isochoric process $\Rightarrow$ Volume is constant throughout the process

(C) Isobaric process $\Rightarrow$ Pressure is constant throughout the process

(D) Adiabatic process $\Rightarrow$ No exchange of heat $\left(q\right)$ between system and surrounding

$2?+C,K_C=4*10^{-3}$

At a given time $t,Q_c$ is to be calculated and been compared with $K_c$ .

$Q_C=\frac{\left[B\right]\left[C\right]}{{[A]}^2}=\frac{\left(2*10^{-3}\right)\left(2*10^{-3}\right)}{{\left(2*10^{-3}\right)}^2}$

$Q_C=1$

BOD is biological oxygen demand which represents the amount of oxygen required to degrade organic matter in water.

Higher the BOD more polluted the water is

So; here water sample with BOD = 3 ppm is the cleanest

Density of metal a very specific and depends upon many factors.

Li -> 0.53 gm/cm³

Na -> 0.97 gm/cm³

K -> 0.86 gm/cm³

Rb ->1.53 gm/cm³

Cs -> 1.90 gm/cm³

Highly polluted water has higher value of BOD ≥17 ppm

Disproportionation reaction is

$\stackrel{+6}{Mg}{O}_4^{--}+4H^{+}\to 2\stackrel{+7}{Mn}{O}_4^{-}+\stackrel{+4}{Mn}{O}_2+2H_{2}O$

Kindly consider the below image

During removal of temporary hardness of water.

$\begin{array}{l}Mg{\left(HC{O}_3\right)}_2\stackrel{Boil}{\to }Mg{\left(OH\right)}_2+2C{O}_2\uparrow \\ Ca{\left(HC{O}_3\right)}_2\stackrel{Boil}{\to }CaC{O}_3+H_{2}O+C{O}_2\end{array}$

B₂H₆ has 4 2 c -2e bonds and 2 3c-2e bonds.

Bridging (B-H) bonds have more value of bond- length then terminal (B – H) bonds

Bridging bonds are in one plane, but terminal bonds are in perpendicular plane.

Due to presence of (3c-2e) bonds, it behaves as electrons deficient and prone to get attached by  lewis      base.

$3NaB{H}_4+4B{F}_3\underset{}{\overset{?}{?}}3NaB{F}_4+2B_2{H}_6$