$Ca{C_2} + 2{H_2}O \to Ca{(OH)_2} + {C_2}{H_2}$
${C_2}{H_2} + {H_2} \to {C_2}{H_4}$
$n({C_2}{H_4}) \to {( - C{H_2} - C{H_2} - )_n}$
$64.1\, kg$ $Ca{C_2}$માંથી મેળવેલ પોલિઇથિલિનનો જથ્થો ......$kg$ છે.
\({C_2}{H_2} + {H_2} \to \mathop {{C_2}{H_4}}\limits_{{\rm{28}}\,g} \)
\(64\,g\) of \(Ca{C_2}\) gives \(28\,g\) of ethylene
\(\therefore \) \(64\,kg\) of \(Ca{C_2}\) will give \(28\,kg\) of polyethylene
$C{H_2} = C{H_2}\xrightarrow[{oxid}]{{Hypochloro}}$ $M\xrightarrow{R}\begin{array}{*{20}{c}}
{C{H_2} - OH} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{C{H_2} - OH}
\end{array}$
$\mathop C\limits_6 {H_3} - \mathop C\limits_5 H = \mathop C\limits_4 H - \mathop C\limits_3 {H_2} - \mathop C\limits_2 \equiv \mathop C\limits_1 H$
કાર્બન્સ $1, 3$ અને $5$ ના સંકરણની સ્થિતિ નીચેના ક્રમમાં છે, તો સાચો ક્રમ શોધો.