Consider the following reaction in a sealed vessel at equilibrium with concentrations of $$\mathrm{N}_2=3.0 \times 10^{-3} \mathrm{M}, \mathrm{O}_2=4.2 \times 10^{-3} \mathrm{M}$$ and $$\mathrm{NO}=2.8 \times 10^{-3} \mathrm{M}$$.
$$2 \mathrm{NO}_{(\mathrm{g})} \rightleftharpoons \mathrm{N}_{2(\mathrm{~g})}+\mathrm{O}_{2(\mathrm{~g})}$$
If $$0.1 \mathrm{~mol} \mathrm{~L} \mathrm{~L}^{-1}$$ of $$\mathrm{NO}_{(\mathrm{g})}$$ is taken in a closed vessel, what will be degree of dissociation ($$\alpha$$) of $$\mathrm{NO}_{(\mathrm{g})}$$ at equilibrium?
A compound X contains $$32 \%$$ of A, $$20 \%$$ of B and remaining percentage of C. Then, the empirical formula of $$\mathrm{X}$$ is :
(Given atomic masses of A=64 ; B=40 ; C=32 u)
In a vernier callipers, $$(N+1)$$ divisions of vernier scale coincide with $$N$$ divisions of main scale. If $$1 \mathrm{~MSD}$$ represents $$0.1 \mathrm{~mm}$$, the vernier constant (in $$\mathrm{cm}$$) is:
If the monochromatic source in Young's double slit experiment is replaced by white light, then