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never [62]
3 years ago
7

Define or give an example for each of the following terms.

Chemistry
1 answer:
irakobra [83]3 years ago
8 0

a) A combound which contains only Carbon and Hydrogen. There are covalent bonds between atoms. Hydrogen form one single bond and Carbon forms four covalent bonds. Carbon bonds can be single, double or triple bonds.

  All hydrocarbons are organic compounds, but organic compound can include atoms of other elements.

b)  Alkyne has a  covalent triple bond between two carbon atoms. Simplest alkyne is ethyne HCCH.  

b) Alkane contains only Carbon and  Hydrogen and there are single bonds

  between atoms. Simplest alkane is methane CH4.

c) An alkene has one double bond between Carbon atoms. Simplest

alkene is ethene  H2C=CH2.

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Naturally occurring element X exists in three isotopic forms: X-28 (27.979 amu, 92.21% abundance), X-29 (28.976 amu 4.70% abunda
bezimeni [28]

<u>Answer:</u> The average atomic mass of X is 28.09 amu

<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)

  • <u>For isotope 1:</u>

Mass of isotope 1 = 27.979 amu

Percentage abundance of isotope 1 = 92.21 %

Fractional abundance of isotope 1 = 0.9212

  • <u>For isotope 2:</u>

Mass of isotope 2 = 28.976 amu

Percentage abundance of isotope 2 = 4.70 %

Fractional abundance of isotope 2 = 0.0470

  • <u>For isotope 3:</u>

Mass of isotope 3 = 29.974 amu

Percentage abundance of isotope 3 = 3.09 %

Fractional abundance of isotope 3 = 0.0309

Putting values in equation 1, we get:

\text{Average atomic mass of X}=[(27.979\times 0.9212)+(28.976\times 0.0470)+(29.974\times 0.0309)]

\text{Average atomic mass of X}=28.09amu

Hence, the average atomic mass of X is 28.09 amu

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