Asked by Ben
Naturally occuring boron is 80.20% boron-11 and 19.80% of some other isotopic form of boron. What must be the atomic mass of this second isotope be in order to account for the 10.81 amu average atomic mass for boron?
Answers
Answered by
DrBob222
Do you have the mass of B-11?
Let X = mass B-X , then,
0.8020(mass B-11) + 0.1980(X) = 10.81
Substitute mass B-11 and solve for X
Let X = mass B-X , then,
0.8020(mass B-11) + 0.1980(X) = 10.81
Substitute mass B-11 and solve for X
Answered by
miguel
10.0amu
Answered by
Dick Ducker
b is for b and d is for ak
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