Question

the element boron has two stable isotopes: boron - 10 (10.0129 u) and boron - 11 (11.0093 u). the periodic table reports a mass for boron as 10.811. what is the natural abundance of each boron isotope? 1st attempt part 1 (1 point) write your answers with at least one digit following the decimal. b - 10 % part 2 (1 point) b - 11 %

Explanation:

Step1: Let the abundance of B - 10 be $x$.

Then the abundance of B - 11 is $1 - x$.

Step2: Set up the weighted - average equation.

The average atomic mass of boron is given by the sum of the products of the mass of each isotope and its abundance. So, $10.0129x+11.0093(1 - x)=10.811$.

Step3: Expand the equation.

$10.0129x + 11.0093-11.0093x=10.811$.

Step4: Combine like terms.

$(10.0129x-11.0093x)+11.0093 = 10.811$, which simplifies to $- 0.9964x=10.811 - 11.0093$.

Step5: Calculate the right - hand side.

$10.811 - 11.0093=-0.1983$. So, $-0.9964x=-0.1983$.

Step6: Solve for $x$.

$x=\frac{-0.1983}{-0.9964}\approx0.199$.

Step7: Find the abundance of B - 11.

The abundance of B - 11 is $1 - x=1 - 0.199 = 0.801$.

Step8: Convert to percentages.

The abundance of B - 10 is $0.199\times100 = 19.9\%$.
The abundance of B - 11 is $0.801\times100 = 80.1\%$.

Answer:

Part 1: 19.9
Part 2: 80.1