Congratulations! You have discovered a new element called Bhauzium (Bz). The average atomic mass of Bz was found to be 348.72 amu and its atomic number is 123.
a) if the atomic masses of three isotopes of Bhsuzium are 342.92 amu, 356.43 amu (6.52%), and 351.29 amu, what is the relative abundance (%) of each isotope?

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  1. Let X = fraction of 342.92
    356.43 is 6.52% or 0.0652 as a fraction.
    351.29 = (1-X-0.0652) = 0.9348-X

    Sum those values = 348.72

    342.92X + 356.43(0.0652) + 351.29(0.9348-x) = 348.72
    Solve for x which is the fraction of the first isotope and 0.9348-x is the fraction of the third isotope listed. The percentages should add to 100.0%.

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