The given question is incomplete. The complete question is:What is the relative atomic mass of a hypothetical element that consists isotopes in the indicated natural abundances.

Isotope mass amu Relative abundance

1 77.9 14.4

2 81.9 14.3

3 85.9 71.3

Express your answer to three significant figures and include the appropriate units.

Answer: 84.2 amu

Explanation:

Mass of isotope 1 = 77.9

% abundance of isotope 1 = 14.4% = $\frac{14.4}{100}=0.144$

Mass of isotope 2 = 81.9

% abundance of isotope 2 = 14.3% = $\frac{14.3}{100}=0.143$

Mass of isotope 3 = 85.9

% abundance of isotope 2 = 71.3% = $\frac{71.3}{100}=0.713$

Formula used for average atomic mass of an element :

$\text{ Average atomic mass of an element}=\sum(\text{atomic mass of an isotopes}\times {{\text { fractional abundance}})$

$A=\sum[(77.9\times 0.144)+(81.9\times 0.143)+(85.9\times 0.713)]$

$A=84.2amu$

Therefore, the average atomic mass of a hypothetical element that consists isotopes in the indicated natural abundances is 84.2 amu

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What values can you draw from this lesson?​

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