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Allisa [31]
2 years ago
11

2,2-dichlorohexane 4 , 4 Give two reason, explain why 2,2-dichloro-3-methylbutane has structure isomer of 2, 2 dichloropentane.

​
Chemistry
1 answer:
matrenka [14]2 years ago
4 0

Answer:explain why 2,2-dichloro-3-methylbutane has structure isomer of 2, 2 dichloropentane. ​Although there are 12 hydrogen atoms in the molecule, there only 4 possible 1H chemical environments for the hydrogen atoms in 2-methylbutane molecule.

Explanation:

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Chlorine has two naturally occurring isotopes, 35C1 Gsotopic mass 34.9689 amu) and 370 (isotopic mass 36.9659 amu). If chlorine
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<u>Answer:</u> The percentage abundance of _{17}^{35}\textrm{Cl} and _{17}^{37}\textrm{Cl} isotopes are 75.77% and 24.23% respectively.

<u>Explanation:</u>

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i   .....(1)

Let the fractional abundance of _{17}^{35}\textrm{Cl} isotope be 'x'. So, fractional abundance of _{17}^{37}\textrm{Cl} isotope will be '1 - x'

  • <u>For _{17}^{35}\textrm{Cl} isotope:</u>

Mass of _{17}^{35}\textrm{Cl} isotope = 34.9689 amu

Fractional abundance of _{17}^{35}\textrm{Cl} isotope = x

  • <u>For _{17}^{37}\textrm{Cl} isotope:</u>

Mass of _{17}^{37}\textrm{Cl} isotope = 36.9659 amu

Fractional abundance of _{17}^{37}\textrm{Cl} isotope = 1 - x

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Putting values in equation 1, we get:

35.4527=[(34.9689\times x)+(36.9659\times (1-x))]\\\\x=0.7577

Percentage abundance of _{17}^{35}\textrm{Cl} isotope = 0.7577\times 100=75.77\%

Percentage abundance of _{17}^{37}\textrm{Cl} isotope = (1-0.7577)=0.2423\times 100=24.23\%

Hence, the percentage abundance of _{17}^{35}\textrm{Cl} and _{17}^{37}\textrm{Cl} isotopes are 75.77% and 24.23% respectively.

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