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Romashka-Z-Leto [24]
3 years ago
15

In a first order decomposition, the constant is 0.00813 sec-1. What percentage of the compound has decomposed after 44.1 seconds

Chemistry
1 answer:
Zarrin [17]3 years ago
4 0

Answer:

69.8%

Explanation:

k = 0.00813 sec-1

t = 44.1 s

What percentage of the compound has decomposed?

We can obtain this using the formular;

[A] = [A]o e^(-kt)

where;

Final Concentration = [A]

Initial concentration = [A]o

[A] / [A]o = e^(-kt)

[A] / [A]o  =  e^(-0.00813*44.1)

[A] / [A]o = e^(-0.359)

[A] / [A]o = 0.698

Percentage decomposed = Final Concentration / Initial concentration * 100

Percentage decomposed = [A] / [A]o   * 100%

Percentage decomposed = 0.698 * 100 = 69.8%

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Answer:

There is one single covalent bond between two carbon atoms.

Explanation:

We know that sharing of electrons form covalent bonds.

If we look upon K,L,M ,N shells of the carbon and hydrogen atoms.

We found that Hydrogen is having only  1 electron in K shell.

And Carbon on the other hand is having 2 electrons in K shell and 4 electrons in L shell.

So carbon have  4 valence electrons,and it can share 4 bonds with any relevant atom to complete its octet.

And Hydrogen requires  1 electron to complete its doublet.

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For ethane n=2

C_{2}H_{4+2} = C_{2}H_{6}

1 Carbon atom is shared by  3 Hydrogen.

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An image of the sharing of electrons attached below,

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6 0
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