MCQ
The De-broglie $\lambda$ of electron in the $2^{st}$ Bohr orbit is
- A$4\pi r_1$
- ✓$\pi r_1$
- C$2\pi r_1$
- D$6\pi r_1$
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$\begin{array}{*{20}{c}}
{C{H_3} - CH - CH - C{H_2} - C{H_3}} \\
{\,\,\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{\,\,CN\,\,\,\,\,\,\,\,\,CH = C{H_2}\,\,}
\end{array}$
$CH _{3} COOH + NaOH \rightarrow CH _{3} COONa + H _{2} O$ $(\Delta H =-55.3\,kJ\,mol ^{-1})$ The enthalpy of ionization of $CH _{3} COOH$ as calculated by the student is $kJ mol ^{-1}$. (nearest integer)
(image) $\xrightarrow[{KOH}]{{KMn{O_4}}}\,B\,\xrightarrow[{FeC{L_3}}]{{B{r_2}}}\,C\,\xrightarrow[{{H^ + }}]{{{C_2}{H_5}OH}}\,D$
$D$ would be