A student wants to calculate the average atomic mass of neon in nature. Information about neon is in the table below.
Isotopes in Nature Atomic Mass (amu) Natural Abundance (%)
neon-20 19.992 90.48
neon-21 20.993 .27
neon-22 21.991 9.25
Which of the following is a correct way for the student to set up the calculation?
(1 point)
(19.992 amu×.0027)+(20.993 amu ×.9048)+(21.991 amu ×.0925)(19.992 amu×.0027)+(20.993 amu ×.9048)+(21.991 amu ×.0925)
(19.992 amu×.9048)+(20.993 amu ×.0027)(19.992 amu×.9048)+(20.993 amu ×.0027)
(19.992 amu×.9048)×(20.993 amu ×.0027)×(21.991 amu ×.0925)(19.992 amu×.9048)×(20.993 amu ×.0027)×(21.991 amu ×.0925)
(19.992 amu×.9048)+(20.993 amu ×.0027)+(21.991 amu ×.0925)(19.992 amu×.9048)+(20.993 amu ×.0027)+(21.991 amu ×.0925)
I don't like any of them. Since there are only 3 isotopes, it should be
(19.992 amu ×.9048) + (20.993 amu ×.0027) + (21.991 amu ×.0925)
The correct way for the student to set up the calculation is:
(19.992 amu × 0.0027) + (20.993 amu × 0.9048) + (21.991 amu × 0.0925)
The correct way for the student to set up the calculation is:
(19.992 amu × 0.027) + (20.993 amu × 0.9048) + (21.991 amu × 0.0925)
The student wants to calculate the average atomic mass of neon in nature. The average atomic mass is calculated by multiplying the mass of each isotope by its natural abundance, and then summing up the results.
Looking at the table, we have three isotopes of neon: neon-20, neon-21, and neon-22. We are given their respective atomic masses and natural abundances.
To set up the calculation, we multiply the atomic mass of each isotope by its natural abundance, and then sum up the results.
So, the correct way to set up the calculation is:
(19.992 amu × 0.0027) + (20.993 amu × 0.9048) + (21.991 amu × 0.0925)
This represents multiplying the atomic mass of each isotope by its natural abundance and then adding the results together.
Therefore, the correct answer is:
(19.992 amu × 0.0027) + (20.993 amu × 0.9048) + (21.991 amu × 0.0925)