Posted by Ben on .
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?
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
b is for b and d is for ak
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