Asked by Anonymous
The element boron has three naturally occuring isotopes. One has the atomic mass 10.0129(54.4%), another 11.0093(12.75%). What is the atomic mass of the third isotope if the average atomic mass of boron is 10.811?
Answers
Answered by
DrBob222
So the third isotope has an abundance of
100%-54.4%-12.75% = ? = about 32.85% or 0.3285 as a decimal).
If we let Z = atomic mass of the third isotope we have\
10.0129(0.544) + 11.0093(0.1275) + Z(0.3285) = 10.811
Solve for Z.
100%-54.4%-12.75% = ? = about 32.85% or 0.3285 as a decimal).
If we let Z = atomic mass of the third isotope we have\
10.0129(0.544) + 11.0093(0.1275) + Z(0.3285) = 10.811
Solve for Z.
Answered by
Anonymous
Would the correct answer be 12.167?
Answered by
DrBob222
I came up with 12.0556 which I would round to 12.056 and that's too many significant figures. By the way, I don't see but two isotopes for B; i.e., B-10 and B11. This may be a made up problem. See
http://www.webelements.com/boron/isotopes.html
http://www.webelements.com/boron/isotopes.html
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