- A$1\, mg$ of $C_4H_{10}$
- B$1\, mg$ of $N_2$
- C$1\, mg$ of $Na$
- ✓$1\, ml$ of $H_2O$
$1 \mathrm{mg}$ of $N_{2}=\frac{2 N \times 10^{-3}}{28}$ atoms
$1 \mathrm{mg}$ of $N a=\frac{N \times 10^{-3}}{23}$ atoms
$1 \mathrm{mL}=1 \mathrm{g} H_{2} O=\frac{3 N}{18}$ atoms
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(Take $pH_2 = 1\,atm$ ), $T = 298\,K$.
(At. No. $Cr = 24, Mn = 25, $$Fe = 26, Ni = 28$)
$\begin{matrix}
O \\
|| \\
H-C-H, \\
\end{matrix}\begin{matrix}
O\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,O\,\,\,\,\,\,\,O\,\,\,\, \\
||\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,||\,\,\,\,\,\,\,\,\,||\,\,\,\,\, \\
H-C-C{{H}_{2}}-C-C-C{{H}_{3}}, \\
\end{matrix}\begin{matrix}
\,\,\,\,\,O\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,O \\
\,\,\,\,\,\,||\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|| \\
C{{H}_{3}}-C-C{{H}_{2}}-C-H \\
\end{matrix}$
aalkene $(A)$ will be ?