there are two naturally occurring isotopes of boron.\n$^{10}$b has a mass of 10.0129 u.\n$^{11}$b has a mass…

Explanation:

Step1: Let the abundance of $^{10}\text{B}$ be $x$, then the abundance of $^{11}\text{B}$ is $1 - x$.

Let the abundance of $^{10}\text{B}$ be $x$ (as a decimal), and since the sum of abundances of all isotopes is 1, the abundance of $^{11}\text{B}$ is $1 - x$.

Step2: Use the formula for average atomic mass.

The average atomic mass of boron is approximately 10.81 u. The formula for average atomic mass $A$ is $A=m_1x_1 + m_2x_2$, where $m_1$ and $m_2$ are the masses of the isotopes and $x_1$ and $x_2$ are their abundances. So, $10.81=10.0129x+11.0093(1 - x)$.

Step3: Expand and simplify the equation.

$10.81 = 10.0129x+11.0093-11.0093x$. Combine like - terms: $10.81-11.0093=10.0129x - 11.0093x$. So, $- 0.1993=-0.9964x$.

Step4: Solve for $x$.

$x=\frac{0.1993}{0.9964}\approx0.20$.

Step5: Calculate the abundances as percentages.

The abundance of $^{10}\text{B}$ as a percentage is $0.20\times100 = 20%$. The abundance of $^{11}\text{B}$ is $(1 - 0.20)\times100=80%$.

Answer:

$^{10}\text{B}$ abundance: 20% $^{11}\text{B}$ abundance: 80%