MCQ
Product $C$ is
- A
- ✓
- C
- D
✓
Answer
Correct option: B.
b
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
The electronic configuration of element with atomic number $24$ is
→The reaction
→$C{H_3}COH\xrightarrow[{{\text{Zn(Hg)/Conc}}{\text{. HCI}}}]{{\left[ H \right]}}C{H_3}C{H_3}$ is
$\left[ {Co{{\left( {N{H_3}} \right)}_5}N{O_2}} \right]C{l_2}$ and $\left[ {Co{{\left( {N{H_3}} \right)}_5}\left( {ONO} \right)} \right]C{l_2}$ are related to each other as
→$\begin{array}{*{20}{c}}
{\,\,\,\,\,\,\,\,\,O\,\,\,\,\,\,\,\,\,O\,\,\,\,\,\,}\\
{\,\,||\,\,\,\,\,\,\,\,\,\,||}\\
{Ph - C - C - H}
\end{array}$ $\xrightarrow[\Delta ]{{conc.\,NaOH}}{\text{Product}}$
→{\,\,\,\,\,\,\,\,\,O\,\,\,\,\,\,\,\,\,O\,\,\,\,\,\,}\\
{\,\,||\,\,\,\,\,\,\,\,\,\,||}\\
{Ph - C - C - H}
\end{array}$ $\xrightarrow[\Delta ]{{conc.\,NaOH}}{\text{Product}}$
Calculate the half life of the first-order reaction $C_2H_4O(g) \to CH_4(g) + CO(g)$ If the initial pressure of $C_2H_4O(g)$ is $80\, mm$ and total pressure at the end of $20$ $minutes$ is $120\, mm$ .......... $\min$
→The structure of the starting compound $P$ used in the reaction given below is :
→Standard entropy of $X_2, Y_2$ and $XY_3$ are $60, 40$ and $50\,J \,K^{-1}\,mol^{-1},$ respectively. For the reaction,
→$\frac{1}{2}{X_2} + \frac{3}{2}{Y_2} \to X{Y_3},\,\Delta H = - 30\,kJ$ , to be at equilibrium, the temperature will be ............... $\mathrm{K}$
$A + {H_2}O \to B + HCl$
→$B + {H_2}O \to C + HCl$
Compound $(A),\,(B)$ and $(C)$ will be respectively
Using the rules for significant figures, the correct answer for the expression $\frac{0.02858 \times 0.112}{0.5702}$ will be .... .
→When phosphine is passed through aqueous solution of copper sulphate, the product produced is