MCQ
Calculate the $\lambda$ of $CO_2$ molecule moving with a velocity $440\, m/s.$
- A$\lambda = 1.03 \times 10^{-11}$
- B$\lambda = 2.06 \times 10^{-10}$
- C$\lambda = 4.12 \times 10^{-11}$
- ✓$\lambda = 2.06 \times 10^{-11}$
Thus, the wavelength of $C O_{2}$ is $2.06 \times 10^{-11} \mathrm{m}$
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$(I) 0.5\ mole$ of $O_3$ $(II) 0.5\ gm$ atom of oxygen $(III) 3.011 \times 10^{23}$ molecules of $O_2$ $(IV) 5.6\ litre$ of $CO_2$ at $STP$
$Zn_{(s)} + Ag_2O_{(s)} + H_2O_{(l)} \rightleftharpoons $$2Ag_{(s)} + Zn^{2+}_{(aq)}+ 2OH^-_{(aq)}$
If half cell potentials are
$Zn^{2+}_{(aq)} + 2e^- \rightarrow Zn_{(s)}\,;\,\, E^o = - 0.76\, V$
$Ag_2O_{(s)} + H_2O_{(l)} + 2e^- \rightarrow 2Ag_{(s)} + 2OH^-_{(aq)}\,,$$E^o = 0.34\, V$
The cell potential will be ........... $V.$
$(A)\,\, KO_2$
$(B)\,\,$ cis $-[Co(en)_2Cl_2]^+$
$(C)\,\, K_3[Co(Ox)_3]$
$(D)\,\,[CoCl_3]^{2-}$