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Naturally occurring boron is 80.20% boron-11 (atomic mass 11.01 amu) and 19.80% of some other isotope. What must the atomic mass of this second isotope be in order to account for the 10.81 amu average atomic mass of boron?
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remember the formula I showed you earlier? Use the same thing but solve it like algebra. this time the mass number of the second isotope is unkown 10.81 = (0.8020 * 11.01) + ( 0.1980x) solve for x I think you do that algebra if you're in chem
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