Question

question 13
use your answers from the previous three questions (questions 10, 11, and 12)
to answer this question.
if the average atomic mass of hydrogen in nature is 1.0079 amu, what does that tell
you about the percent composition of h - 1 and h - 2 in nature?
o h - 1 has a higher percent abundance than h - 2
o h - 2 has a higher percent abundance than h - 1
o h - 1 and h - 2 are equally abundant in nature

question 14
1 pts
look at boron. it has two possible isotopes. based on the average atomic mass on the
periodic table (10.81 amu), which isotope is most abundant in nature?
o b - 10
o b - 11

Explanation:

Question 13

The average atomic mass is calculated as a weighted average of the masses of its isotopes. 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 fractional abundance (\(x_i=\frac{\text{Percent Abundance}}{100}\)). For hydrogen, \(M_{H - 1}=1\space amu\) and \(M_{H - 2}=2\space amu\). Since \(M_{avg}=1.0079\space amu\) is very close to \(M_{H - 1}\), it means that the isotope with a mass closer to the average atomic mass (H - 1) has a higher percent abundance.

For boron, if we assume two isotopes \(B - 10\) (\(M_1 = 10\space amu\)) and \(B - 11\) (\(M_2=11\space amu\)). Using the average - atomic - mass formula \(M_{avg}=M_1x_1 + M_2x_2\) and \(x_1 + x_2=1\) (where \(x_1\) and \(x_2\) are fractional abundances). Substituting \(M_{avg}=10.81\space amu\), \(M_1 = 10\space amu\), and \(M_2 = 11\space amu\) into \(M_{avg}=M_1x_1+M_2(1 - x_1)\). We get \(10.81=10x_1+11(1 - x_1)=10x_1 + 11-11x_1=11 - x_1\). Solving for \(x_1\): \(x_1=11 - 10.81=0.19\) and \(x_2=1 - 0.19 = 0.81\). Since \(x_2\) (abundance of \(B - 11\)) is higher.

Answer:

H - 1 has a higher percent abundance than H - 2

Question 14