164030 The equation is correctly matched for
$\alpha =\frac{D-d}{(x-1)d}$
Where $\mathbf{D}=$ Theoretical vapour density
d = Observed vapour density

164031 The following gaseous equilibrium are given
${\mathrm{N}}_2+3{\mathrm{H}}_2\rightleftharpoons 2{\mathrm{NH}}_3-\cdots \cdot -\cdots -{\mathrm{K}}_1$
${\mathrm{N}}_2+{\mathrm{O}}_2\rightleftharpoons 2\mathrm{NO}\text{---.--.--.--- }{\mathrm{K}}_2$
${\mathrm{H}}_2+1/2{\mathrm{O}}_2\rightleftharpoons {\mathrm{H}}_2\mathrm{O}\text{..........-. }{\mathrm{K}}_3$
$$The equilibrium constant of the reaction
$2{\mathrm{NH}}_{3(\mathrm{g})}+5/2{\mathrm{O}}_{2(\mathrm{g})}\rightleftharpoons 2\mathrm{NO}(\mathrm{g})+3{\mathrm{H}}_2\mathrm{O}(\mathrm{g})$, in terms of $K_1,K_2$ and $K_3$ is:

164035 If the concentration of ${\mathrm{OH}}^{-}$ions in the reaction $\mathrm{Fe}(\mathrm{OH}{)}_3(\mathrm{s})\rightleftharpoons {\mathrm{Fe}}^{+3}$ (aq.) $+3{\mathrm{OH}}^{-}$(aq.) is decreased upto $1/4$ times, then equilibrium concentration of ${\mathrm{Fe}}^{+3}$ will increase upto:

164030 The equation is correctly matched for
$\alpha =\frac{D-d}{(x-1)d}$
Where $\mathbf{D}=$ Theoretical vapour density
d = Observed vapour density

164031 The following gaseous equilibrium are given
${\mathrm{N}}_2+3{\mathrm{H}}_2\rightleftharpoons 2{\mathrm{NH}}_3-\cdots \cdot -\cdots -{\mathrm{K}}_1$
${\mathrm{N}}_2+{\mathrm{O}}_2\rightleftharpoons 2\mathrm{NO}\text{---.--.--.--- }{\mathrm{K}}_2$
${\mathrm{H}}_2+1/2{\mathrm{O}}_2\rightleftharpoons {\mathrm{H}}_2\mathrm{O}\text{..........-. }{\mathrm{K}}_3$
$$The equilibrium constant of the reaction
$2{\mathrm{NH}}_{3(\mathrm{g})}+5/2{\mathrm{O}}_{2(\mathrm{g})}\rightleftharpoons 2\mathrm{NO}(\mathrm{g})+3{\mathrm{H}}_2\mathrm{O}(\mathrm{g})$, in terms of $K_1,K_2$ and $K_3$ is:

164032 The dissociation equilibrium of a gas $A{B}_2$ can be represented as: $2{\mathrm{AB}}_2(\mathrm{g})\rightleftharpoons 2\mathrm{AB}(\mathrm{g})+{\mathrm{B}}_2(\mathrm{g})$
The degree of dissociation is ' $x$ ' and is small compared to 1 . The expression relating the degree of dissociation ( $\mathbf{x}$ ) with equilibrium constant $\mathrm{Kp}$ and total pressure $\mathbf{P}$ is:

164035 If the concentration of ${\mathrm{OH}}^{-}$ions in the reaction $\mathrm{Fe}(\mathrm{OH}{)}_3(\mathrm{s})\rightleftharpoons {\mathrm{Fe}}^{+3}$ (aq.) $+3{\mathrm{OH}}^{-}$(aq.) is decreased upto $1/4$ times, then equilibrium concentration of ${\mathrm{Fe}}^{+3}$ will increase upto:

164030 The equation is correctly matched for
$\alpha =\frac{D-d}{(x-1)d}$
Where $\mathbf{D}=$ Theoretical vapour density
d = Observed vapour density

164031 The following gaseous equilibrium are given
${\mathrm{N}}_2+3{\mathrm{H}}_2\rightleftharpoons 2{\mathrm{NH}}_3-\cdots \cdot -\cdots -{\mathrm{K}}_1$
${\mathrm{N}}_2+{\mathrm{O}}_2\rightleftharpoons 2\mathrm{NO}\text{---.--.--.--- }{\mathrm{K}}_2$
${\mathrm{H}}_2+1/2{\mathrm{O}}_2\rightleftharpoons {\mathrm{H}}_2\mathrm{O}\text{..........-. }{\mathrm{K}}_3$
$$The equilibrium constant of the reaction
$2{\mathrm{NH}}_{3(\mathrm{g})}+5/2{\mathrm{O}}_{2(\mathrm{g})}\rightleftharpoons 2\mathrm{NO}(\mathrm{g})+3{\mathrm{H}}_2\mathrm{O}(\mathrm{g})$, in terms of $K_1,K_2$ and $K_3$ is:

164032 The dissociation equilibrium of a gas $A{B}_2$ can be represented as: $2{\mathrm{AB}}_2(\mathrm{g})\rightleftharpoons 2\mathrm{AB}(\mathrm{g})+{\mathrm{B}}_2(\mathrm{g})$
The degree of dissociation is ' $x$ ' and is small compared to 1 . The expression relating the degree of dissociation ( $\mathbf{x}$ ) with equilibrium constant $\mathrm{Kp}$ and total pressure $\mathbf{P}$ is:

164035 If the concentration of ${\mathrm{OH}}^{-}$ions in the reaction $\mathrm{Fe}(\mathrm{OH}{)}_3(\mathrm{s})\rightleftharpoons {\mathrm{Fe}}^{+3}$ (aq.) $+3{\mathrm{OH}}^{-}$(aq.) is decreased upto $1/4$ times, then equilibrium concentration of ${\mathrm{Fe}}^{+3}$ will increase upto:

164030 The equation is correctly matched for
$\alpha =\frac{D-d}{(x-1)d}$
Where $\mathbf{D}=$ Theoretical vapour density
d = Observed vapour density

164031 The following gaseous equilibrium are given
${\mathrm{N}}_2+3{\mathrm{H}}_2\rightleftharpoons 2{\mathrm{NH}}_3-\cdots \cdot -\cdots -{\mathrm{K}}_1$
${\mathrm{N}}_2+{\mathrm{O}}_2\rightleftharpoons 2\mathrm{NO}\text{---.--.--.--- }{\mathrm{K}}_2$
${\mathrm{H}}_2+1/2{\mathrm{O}}_2\rightleftharpoons {\mathrm{H}}_2\mathrm{O}\text{..........-. }{\mathrm{K}}_3$
$$The equilibrium constant of the reaction
$2{\mathrm{NH}}_{3(\mathrm{g})}+5/2{\mathrm{O}}_{2(\mathrm{g})}\rightleftharpoons 2\mathrm{NO}(\mathrm{g})+3{\mathrm{H}}_2\mathrm{O}(\mathrm{g})$, in terms of $K_1,K_2$ and $K_3$ is:

164032 The dissociation equilibrium of a gas $A{B}_2$ can be represented as: $2{\mathrm{AB}}_2(\mathrm{g})\rightleftharpoons 2\mathrm{AB}(\mathrm{g})+{\mathrm{B}}_2(\mathrm{g})$
The degree of dissociation is ' $x$ ' and is small compared to 1 . The expression relating the degree of dissociation ( $\mathbf{x}$ ) with equilibrium constant $\mathrm{Kp}$ and total pressure $\mathbf{P}$ is:

164035 If the concentration of ${\mathrm{OH}}^{-}$ions in the reaction $\mathrm{Fe}(\mathrm{OH}{)}_3(\mathrm{s})\rightleftharpoons {\mathrm{Fe}}^{+3}$ (aq.) $+3{\mathrm{OH}}^{-}$(aq.) is decreased upto $1/4$ times, then equilibrium concentration of ${\mathrm{Fe}}^{+3}$ will increase upto: