QuestionJune 14, 2025

Silver has an atomic mass of 107.868 amu. The Ag-109 isotope (108.905 amu) is 48.16% What is the amu of the other isotope? 107.1 amu 106.9 amu 106.7 amu 107.3 amu

Silver has an atomic mass of 107.868 amu. The Ag-109 isotope (108.905 amu) is 48.16% What is the amu of the other isotope? 107.1 amu 106.9 amu 106.7 amu 107.3 amu
Silver has an atomic mass of 107.868 amu. The Ag-109 isotope (108.905 amu) is 48.16%  What is the amu of the other isotope?
107.1 amu
106.9 amu
106.7 amu
107.3 amu

Solution
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Answer

106.7 amu Explanation 1. Define the equation for average atomic mass Use the formula for average atomic mass: \text{Average Atomic Mass} = (x_1 \cdot m_1) + (x_2 \cdot m_2) , where x_1 and x_2 are the fractional abundances, and m_1 and m_2 are the masses of the isotopes. 2. Set up the equation with given values Let m_2 be the mass of the unknown isotope. Given m_1 = 108.905 amu and x_1 = 0.4816, then x_2 = 1 - x_1 = 0.5184. The average atomic mass is 107.868 amu. 3. Solve for the unknown isotope mass Substitute into the equation: 107.868 = (0.4816 \cdot 108.905) + (0.5184 \cdot m_2). Calculate: 107.868 = 52.448048 + 0.5184 \cdot m_2. Rearrange: 0.5184 \cdot m_2 = 107.868 - 52.448048. Solve: m_2 = \frac{55.419952}{0.5184}.

Explanation

1. Define the equation for average atomic mass<br /> Use the formula for average atomic mass: $ \text{Average Atomic Mass} = (x_1 \cdot m_1) + (x_2 \cdot m_2) $, where $x_1$ and $x_2$ are the fractional abundances, and $m_1$ and $m_2$ are the masses of the isotopes.<br />2. Set up the equation with given values<br /> Let $m_2$ be the mass of the unknown isotope. Given $m_1 = 108.905$ amu and $x_1 = 0.4816$, then $x_2 = 1 - x_1 = 0.5184$. The average atomic mass is 107.868 amu.<br />3. Solve for the unknown isotope mass<br /> Substitute into the equation: $107.868 = (0.4816 \cdot 108.905) + (0.5184 \cdot m_2)$.<br /> Calculate: $107.868 = 52.448048 + 0.5184 \cdot m_2$.<br /> Rearrange: $0.5184 \cdot m_2 = 107.868 - 52.448048$.<br /> Solve: $m_2 = \frac{55.419952}{0.5184}$.
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