chemistry
posted by Ben on .
Naturally occuring boron is 80.20% boron11 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 B11?
Let X = mass BX , then,
0.8020(mass B11) + 0.1980(X) = 10.81
Substitute mass B11 and solve for X 
10.0amu

b is for b and d is for ak

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