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The atomic masses of the two stable isotopes of beam

The atomic masses of the two stable isotopes of beam-example-1

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Answer:

10.81amu

Explanations:

In order to get the average atomic mass of an element, we need the following parameters:

• Natural Abundance (NA),: The percentage of atoms for an element that is a specific isotope.

• Mass (m) ,of each isotope

For the given element (Boron-10 and Boron-11), the natural abundances are 19.78% and 80.22% respectively.

The atomic masses of Boron-10 and Boron-11 are 10.0129amu and 11.0093amu respectively

The formula for calculating the average atomic mass of the element is expressed as:


AAM=(NA_a* m_a)+(NA_b* m_b)

Substitute the given parameters into the formula to have:


A\mathrm{}A\mathrm{}M=(0.1978*10.0129)+(0.8022*11.0093)

Simplify the resulting expression to have:


\begin{gathered} A\mathrm{}A\mathrm{}M=1.98055162+8.83166046 \\ A\mathrm{}A\mathrm{}M=10.81221208 \\ A\mathrm{}A\mathrm{}M\approx10.81amu \end{gathered}

Therefore the average atomic mass of Boron is 10.81amu to two decimal places.

User Jari Oksanen
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