boron has an average atomic mass of 10.81. one isotope of boron has a mass of 10.012938 and a relative abundance of 19.80 percent. the other isotope has a relative abundance of 80.20 percent. what is the mass of that isotope? report to two decimal places. amu done

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Accepted Answer

# Explanation:
## Step1: Set up the formula for average atomic mass
The formula for average atomic mass $M_{avg}=\sum_{i}(M_i\times x_i)$, where $M_i$ is the mass of isotope $i$ and $x_i$ is its relative abundance (in decimal form). Let $M_1 = 10.012938$, $x_1=0.1980$, $x_2 = 0.8020$, and $M_{avg}=10.81$. We need to find $M_2$.
So, $M_{avg}=M_1\times x_1+M_2\times x_2$.
## Step2: Rearrange the formula to solve for $M_2$
$$
\begin{align*}
M_2\times x_2&=M_{avg}-M_1\times x_1\
M_2&=\frac{M_{avg}-M_1\times x_1}{x_2}
\end{align*}
$$
Substitute the values: $M_1 = 10.012938$, $x_1 = 0.1980$, $x_2=0.8020$, $M_{avg}=10.81$
$$
\begin{align*}
M_2&=\frac{10.81-(10.012938\times0.1980)}{0.8020}\
&=\frac{10.81 - 1.982561724}{0.8020}\
&=\frac{8.827438276}{0.8020}\
&\approx11.01
\end{align*}
$$

# Answer:
$11.01$ amu

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QUESTION IMAGE

Question

Explanation:

Step1: Set up the formula for average atomic mass

Step2: Rearrange the formula to solve for \(M_2\)

Answer: