Formation of polyethylene from calcium carbide takes place as follows \[Ca{{C}_{2}}+2{{H}_{2}}O\to Ca{{(OH)}_{2}}+{{C}_{2}}{{H}_{2}}\] \[{{C}_{2}}{{H}_{2}}+{{H}_{2}}\to {{C}_{2}}{{H}_{4}}\] \[n({{C}_{2}}{{H}_{4}})\to {{(-C{{H}_{2}}-C{{H}_{2}}-)}_{n}}\] The amount of polyethylene obtained from 64.1 kg \[Ca{{C}_{2}}\] is [AIIMS 1997]

Formation of polyethylene from calcium carbide takes place as follows

1.	Formation of polyethylene from calcium carbide takes place as follows \[Ca{{C}_{2}}+2{{H}_{2}}O\to Ca{{(OH)}_{2}}+{{C}_{2}}{{H}_{2}}\] \[{{C}_{2}}{{H}_{2}}+{{H}_{2}}\to {{C}_{2}}{{H}_{4}}\] \[n({{C}_{2}}{{H}_{4}})\to {{(-C{{H}_{2}}-C{{H}_{2}}-)}_{n}}\] The amount of polyethylene obtained from 64.1 kg \[Ca{{C}_{2}}\] is [AIIMS 1997]
A.	7 kg
B.	14 kg
C.	21 kg
D.	28 kg
Answer» E.

Discussion

No Comment Found

Related MCQs

In an organic compound of molar mass \[108gmo{{l}^{-1}}C\], H and N atoms are present in 9 : 1 : 3.5 by mass. Molecular formula can be:
Which of the following acid is formed when \[Si{{F}_{4}}\] reacts with water [BHU 2004]
Triple point of water is [AFMC 2004]
Hardness of water is due to presence of salts of [BHU 2005]
\[C{{H}_{3}}C{{H}_{2}}CH=C{{H}_{2}}\] and \[C{{H}_{3}}-CH=CH-C{{H}_{3}}\] show
X and Y are two elements which form \[{{X}_{2}}{{Y}_{3}}\] and \[{{X}_{3}}{{Y}_{4}}\]. If 0.20 mol of \[{{X}_{2}}{{Y}_{2}},\] weighs 32.0 g and 0.4 mol of \[{{X}_{3}}{{Y}_{4}}\] weighs 92.8 g, the atomic weights of X and Y are respectively
How many ml of 1 (M) \[{{H}_{2}}S{{O}_{4}}\] is required to neutralise 10 ml of 1 (M) NaOH solution [MP PET 1998; MNR 1982; MP PMT 1987]
Which element do you think would have been named by Lawrence Berkeley laboratory?
DIRECTION: Read the passage given below and answer the questions that follows: In hexagonal systems of crystals, a frequently encountered arrangement of atoms is described as a hexagonal prism. Here, the top and bottom of the cell are regular hexagons and three atoms sandwiched in between them. A space-of this model of this structure, called hexagonal close-packed (HCP), is constituted of a sphere on a flat surface surrounded in the same plane by six identical spheres as closely as possible. There spheres are then placed over the first layer so that they touch each other and represent the second layer. Each one of these three spheres touches three spheres of the bottom layer. Finally, the second layer is covered with third layer that is identical to the bottom layer in relative position. Assume radius of every sphere to be 'r'. The number of atoms in the HCP unit cell is
Which one of the following oxides is neutral [IIT-JEE 1996]

Discussion

No Comment Found

Related MCQs

Reply to Comment