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valentinak56 [21]
3 years ago
12

Polar molecules have _____.

Chemistry
2 answers:
Tatiana [17]3 years ago
5 0
An unequal number of positive and negative particles, polar molecules have a positive and a negative side, so they have to be unequal
nika2105 [10]3 years ago
4 0

Answer: Polar molecules have an equal number of positive and negative particles.

Explanation: There are two types of molecules:

1) Polar molecules: This molecule is formed when two different atoms having different electronegativities combine together. The more electronegative atom attracts the shared pair of electron towards itself, thus acquiring a partial negative charge and other atom acquiring partial positive charge. This results in the formation of dipole. In this equal number of positive and negative particles.

For example: HCl. As this molecule is covalent in nature and is also a polar molecule.

2) Non-polar molecules: This is formed when two atoms of same electronegativity combine together. For Example: H_2,O_2 etc.

Hence, Polar molecules have an equal number of positive and negative particles.

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<u>D. It will decrease by a factor of 4</u>

Explanation:

According to the question , the equation follows :

A+B\rightarrow C+D

Rate law : This states the rate of reaction is directly proportional to concentration of reactants with each reactant raised to some power which may or may not be equal to the stoichiometeric coefficient.

Rate\ \alpha [A]^{a}[B]^{b}

r=[A]^{a}[B]^{b}.................(1)

STEP": First, find out the power "a" and "b"

a+b = 3 (because it is given that the reaction follow 3rd order-kinetics)

According to question, <u><em>doubling the concentration of the first reactant causes the rate to increase by a factor of 2 means,</em></u>

r' = 2r if [A'] = 2[A]

Here [B] is uneffected means [B']=[B]

hence new rate =

r'=[A']^{a}[B']^{b}

Put the value of [A'] , [B'] and r' in the above equation:

2r=[2A]^{a}[B]^{b}...........(2)

Divide equation (1) by (2) we , get

\frac{2r}{r}=\frac{[2A]^{2}[B]^{b}}{[A]^{a}[B]^{b}}

2= 2(\frac{A}{A})^{a}\times (\frac{B}{B})^{b}

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B and B cancel each other

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2= 2^{a}\times 1^{b}

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2= 2^{a}

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b = 2

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r=[A]^{a}[B]^{b}

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Look in the question now, it is asked to calculate the concentration of [B],if  cut in half

Hence

[B']=1/2[B]

Insert the value of [B'] in equation (3)

r'=[A]^{1}[B']^{2}

r'=[A]^{1}(\frac{1}{2}[B])^{2}

r'=\frac{1}{4}[A]^{1}[B]^{2}............(a)

But

r=[A]^{a}[B]^{b}..............(b)

Compare equation (a) and (b) , we get

new rate r' =

<u>r' = 1/4 r</u>

7 0
3 years ago
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