**Isotope Abundance Problem***Problem Statement:*An unknown element is a mixture of isotopes \(^{129}\text{X}\) and \(^{132}\text{X}\). The average atomic mass of X is 130.35 amu. What is the percent abundance of \(^{132}\text{X}\)?*Solution Approach:*To solve this problem, you need to use the formula for the average atomic mass of isotopes:\[\text{Average Atomic Mass} = (f_1 \times m_1) + (f_2 \times m_2)\]where:- \(f_1\) and \(f_2\) are the fractional abundances of isotopes \(^{129}\text{X}\) and \(^{132}\text{X}\), respectively.- \(m_1\) and \(m_2\) are the masses of \(^{129}\text{X}\) and \(^{132}\text{X}\), which are 129 amu and 132 amu.Also, \(f_1 + f_2 = 1\).Let's set \(f_2 = x\). Then \(f_1 = 1 - x\).Substitute these values into the equation:\[130.35 = (1-x) \times 129 + x \times 132\]Simplify and solve for \(x\) to find the percent abundance of \(^{132}\text{X}\).

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Answered: An unknown element is a mixture of…

An unknown element is a mixture of sotopes 129X and 132X. The average atomic mass of X is 130.35 amu. What is the percent abundance of 32X?

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Question

**Isotope Abundance Problem**

*Problem Statement:*

An unknown element is a mixture of isotopes \(^{129}\text{X}\) and \(^{132}\text{X}\). The average atomic mass of X is 130.35 amu. What is the percent abundance of \(^{132}\text{X}\)?

*Solution Approach:*

To solve this problem, you need to use the formula for the average atomic mass of isotopes:

\[
\text{Average Atomic Mass} = (f_1 \times m_1) + (f_2 \times m_2)
\]

where:
- \(f_1\) and \(f_2\) are the fractional abundances of isotopes \(^{129}\text{X}\) and \(^{132}\text{X}\), respectively.
- \(m_1\) and \(m_2\) are the masses of \(^{129}\text{X}\) and \(^{132}\text{X}\), which are 129 amu and 132 amu.

Also, \(f_1 + f_2 = 1\).

Let's set \(f_2 = x\). Then \(f_1 = 1 - x\).

Substitute these values into the equation:

\[
130.35 = (1-x) \times 129 + x \times 132
\]

Simplify and solve for \(x\) to find the percent abundance of \(^{132}\text{X}\).

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