The formation of polyethylene from calcium carbide takes place as foll

1.	The formation of polyethylene from calcium carbide takes place as follows: CaC2+2H2O → Ca(OH)2 + C2H2 C2H2 + H2 → C2H4 nC2H4 → (−CH2−CH2−)n the amount of polyethylene obtained from 64 kg of CaC2 is:
Answer» The formation of polyethylene from calcium carbide takes place as follows: $C a C_{2} + 2 H_{2} O$ $\to$ $C a {(O H)}_{2}$ + $C_{2} H_{2}$ $C_{2} H_{2}$ + $H_{2}$ $\to$ $C_{2} H_{4}$ $n C_{2} H_{4}$ $\to$ ${(- C H_{2} - C H_{2} -)}_{n}$ the amount of polyethylene obtained from 64 kg of $C a C_{2}$ is:

The formation of polyethylene from calcium carbide takes place as follows: CaC2+2H2O → Ca(OH)2 + C2H2 C2H2 + H2 → C2H4 nC2H4 → (−CH2−CH2−)n the amount of polyethylene obtained from 64 kg of CaC2 is:

Answer»

The formation of polyethylene from calcium carbide takes place as follows:

$C a C_{2} + 2 H_{2} O$ $\to$ $C a {(O H)}_{2}$ + $C_{2} H_{2}$
$C_{2} H_{2}$ + $H_{2}$ $\to$ $C_{2} H_{4}$
$n C_{2} H_{4}$ $\to$ ${(- C H_{2} - C H_{2} -)}_{n}$
the amount of polyethylene obtained from 64 kg of $C a C_{2}$ is:

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The formation of polyethylene from calcium carbide takes place as follows: CaC2+2H2O → Ca(OH)2 + C2H2 C2H2 + H2 → C2H4 nC2H4 → (−CH2−CH2−)n the amount of polyethylene obtained from 64 kg of CaC2 is:

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