Question #2e766

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Question #2e766

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Answer 1 · 2015-10-02T21:05:36

$90 %$ $90%$

Explanation:

So, you know that you're dealing with two isotopes, let's say $^{20} X$ $""^20"X"$ and $^{22} X$ $""^22"X"$ .

The relative atomic mass of the element will be determined by the atomic masses of the two isotopes in proportion to their respective abundances.

$relativ atomic mass = \sum i (isotope_{i} \times abundance_{i})$ $"relativ atomic mass" = sum_i ("isotope"""_i xx "abundance"""_i)$

SInce you're only dealing with two isotopes, you can say that their abundances must add to give 100%.

If you take $x$ $x$ to be the decimal abundance, which is simply the percent abundance divided by $100$ $100$ , of $^{20} X$ $""^20"X"$ , the abundance of $^{22} X$ $""^22"X"$ will be $(1 - x)$ $(1 - x)$ .

This means that you can write

$20 u \cdot x + 22 u \cdot (1 - x) = 20.2 u$ $"20 u" * x + "22 u" * (1-x) = "20.2 u"$

$20 x + 22 - 22 x = 20.2$ $20x + 22 - 22x = 20.2$

$2 x = 1.8 \Rightarrow x = \frac{1.8}{2} = 0.9$ $2x = 1.8 implies x = 1.8/2 = 0.9$

The decimal abundance of $^{22} X$ $""^22"X"$ will thus be $(1 - 0.9) = 0.1$ $(1- 0.9) = 0.1$ .

The percent abundances of the two isotopes are

$^{20} X: 90 %$ $""^20"X: " color(green)(90%)$
$^{22} X: 10 %$ $""^22"X: " 10%$

The result makes sense because the relative atomic mass of the element is much closer to the atomic mass of $^{20} X$ $""^20"X"$ than it is to the atomic mass of $^{22} X$ $""^22"X"$ , which can only imply that $^{20} X$ $""^20"X"$ has a significantly larger percent abundance that $^{22} X$ $""^22"X"$ .

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Question #2e766

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