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3 years ago

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Element Y has two naturally occurring isotopes. The most dominant isotope has a mass of 114.3789 amu and a percent abundance of

64.23%. What is
the mass of the second isoptope if the average atomic mass is 128.4359 amu? Remember that the two percentages will have to sum to 100%.
Answer to the correct number of sigtigs including units in your answer.
(to receive full credit, work must be submitted for this question)

1 answer:

olga2289 [7]3 years ago

5 0

Answer:

153.6771 amu

Explanation:

From the question given above, the following data were:

Isotope A:

Mass of A = 114.3789 amu

Abundance (A%) = 64.23%

Isotope B:

Mass of B =.?

Abundance (B%) = 100 – A%

Abundance (B%) = 100 – 64.23

Abundance (B%) = 35.77%

average atomic mass of Element Y = 128.4359 amu

The mass of the 2nd isotope (i.e isotope B) can be obtained as follow:

Average atomic mass = [(Mass of A × A%)/100] + [(Mass of B × B%)/100]

128.4359 = [(114.3789 × 64.23)/100] + [(Mass of B × 35.77) /100]

128.4359 = 73.4656 + (Mass of B × 0.3577)

Collect like terms

128.4359 – 73.4656 = Mass of B × 0.3577

54.9703 = Mass of B × 0.3577

Divide both side by 0.3577

Mass of B = 54.9703 / 0.3577

Mass of B = 153.6771 amu

Therefore, the mass of the 2nd isotope is 153.6771 amu

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<h3>Explanation</h3>

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$\begin{array}{ll}\text{pOH} = \text{pK}_w - \text{pH} \\ \phantom{\text{pOH}} = 14 - 11.20 &\text{True only under room temperature where }\text{pK}_w = 14 \\\phantom{\text{pOH}}= 2.80\end{array}$ .

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