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2 years ago

The polymer formed from the monomer ch2=ch–cn is select one:

Chemistry

1 answer:

podryga [215]2 years ago

5 0

Answer:
None of the given options show polymer made up of H₂C=CH-CN (Acrylonitrile).

Explanation:
Acrylonitrile (H₂C=CH-CN) which is a monomer on self linkage results in a large chain polymer called as Polyacrylonitrile.
The structure of Polyacrylonitrile is as follow,

--(H₂C-CHCN-)n--

Where n shows the number of Acrylonitrile units joined together in the formation of Polyacrylonitrile. This polymerization reaction can take place by different mechanisms including free radical mechanism, acid catalyzed addition or base catalyzed addition reaction.

The polymerization is shown below,

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What is formula for chromium (III) phosphate trihydrate and cobalt (II) phosphate octahydrate

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The formula for chromium (III) phosphate trihydrate is CrPO4- 3H20. This compound if in the anhydrous state, exists as a green crystal whereas a hydrated form violet crystal. The formula for cobalt(II) phosphate octahydrate is Co3(PO4)2•8H2O.

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2 years ago

What is the hybridization of the central atom in each of the following? 1. Beryllium chloride 2. Nitrogen dioxide 3. Carbon tetr

Lina20 [59]

Answer :

(1) The hybridization of central atom beryllium in $BeCl_2$ is, $sp$

(2) The hybridization of central atom nitrogen in $NO_2$ is, $sp^2$

(3) The hybridization of central atom carbon in $CCl_4$ is, $sp^3$

(4) The hybridization of central atom xenon in $XeF_4$ is, $sp^3d^2$

Explanation :

Formula used :

$\text{Number of electron pair}=\frac{1}{2}[V+N-C+A]$

where,

V = number of valence electrons present in central atom

N = number of monovalent atoms bonded to central atom

C = charge of cation

A = charge of anion

Now we have to determine the hybridization of the following molecules.

(1) The given molecule is, $BeCl_2$

$\text{Number of electrons}=\frac{1}{2}\times [2+2]=2$

The number of electron pair are 2 that means the hybridization will be $sp$ and the electronic geometry of the molecule will be linear.

(2) The given molecule is, $NO_2$

$\text{Number of electrons}=\frac{1}{2}\times [4]=2$

If the sum of the number of sigma bonds, lone pair of electrons and odd electrons present is equal to three then the hybridization will be, $sp^2$ .

In nitrogen dioxide, there are two sigma bonds and one lone electron pair. So, the hybridization will be, $sp^2$ .

(3) The given molecule is, $CCl_4$

$\text{Number of electrons}=\frac{1}{2}\times [4+4]=4$

The number of electron pair are 4 that means the hybridization will be $sp^3$ and the electronic geometry of the molecule will be tetrahedral.

(4) The given molecule is, $XeF_4$

$\text{Number of electrons}=\frac{1}{2}\times [8+4]=6$

Bond pair electrons = 4

Lone pair electrons = 6 - 4 = 2

The number of electrons are 6 that means the hybridization will be $sp^3d^2$ and the electronic geometry of the molecule will be octahedral.

But as there are four atoms around the central xenon atom, the fifth and sixth position will be occupied by lone pair of electrons. The repulsion between lone and bond pair of electrons is more and hence the molecular geometry will be square planar.

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What is the mole fraction of potassium dichromate, K2Cr2O7, in a solution prepared from 24.42 g of potassium dichromate and 240.

Harrizon [31]

Answer:

$6.19\times 10^{-3}$ is the mole fraction of potassium dichromate.