How would you calculate the atomic mass of rubidium given the two isotopes of rubidium have atomic masses and relative abundances of 84.91 amu (72.16%) and 86.91 amu (27.84%)?

Answer 1

#"85.47 u"#

The weighted average of the atomic masses of an element's naturally occurring isotopes is used to calculate the element's average atomic mass.

Now, a weighted average just indicates that each isotope adds, in proportion to its percent abundance, to the element's average atomic mass.

#color(blue)("avg. atomic mass" = sum_i ("isotope"_i xx "abundance"_i))#

An isotope's atomic mass will affect the element's average atomic mass more the more abundant it is.

You are aware that rubidium has two stable isotopes in your situation.

When you calculate the average atomic mass, make sure that you use decimal abundance, which is simply percent abundance divided by #100#.

Enter your values now to get

"avg. atomic mass" is equal to "86.91 u" x 0.2784 + "84.91 u" x 0.7216.

#"avg. atomic mass " = " 85.4668 u"#

The response, rounded to four sig figs, is

#"avg. atomic mass " = color(green)(" 85.47 u")#
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Answer 2

For example, to find the atomic mass of rubidium, multiply the atomic mass of each isotope by its relative abundance (84.91 amu * 0.7216) + (86.91 amu * 0.2784). Then, sum the products.

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Answer 3

The atomic mass of rubidium (Rb) can be calculated using the weighted average of the atomic masses of its isotopes, weighted by their respective relative abundances.

Atomic mass of rubidium (Rb) = (Atomic mass of Rb-85) × (Relative abundance of Rb-85) + (Atomic mass of Rb-87) × (Relative abundance of Rb-87)

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Answer 4

To calculate the atomic mass of rubidium ((Rb)), you would use the weighted average formula:

[ \text{Atomic mass of } Rb = (\text{Atomic mass of isotope 1} \times \text{Relative abundance of isotope 1}) + (\text{Atomic mass of isotope 2} \times \text{Relative abundance of isotope 2}) ]

Substituting the given values:

[ \text{Atomic mass of } Rb = (84.91 , \text{amu} \times 0.7216) + (86.91 , \text{amu} \times 0.2784) ]

[ \text{Atomic mass of } Rb = (61.24 , \text{amu}) + (24.20 , \text{amu}) ]

[ \text{Atomic mass of } Rb = 85.44 , \text{amu} ]

Therefore, the atomic mass of rubidium is (85.44 , \text{amu}).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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