The isotopic masses (amu) and % abundances of the isotopes are given below. Isotope Abundance (%) Mass (amu)159X 30.60 159.37163X 15.79 162.79164X  53.61 163.92 The average atomic mass of the element is __________ amu.Question 1Select one:a.161.75b.162.03c.162.35d.163.15e.33.33

The average atomic mass of Re is 186.207 amu. It is calculated from the two naturally occurring isotopes 185Re and 187Re. If the isotopic abundance of 185Re is 37.4% and the isotopic mass of 185Re is 184.953 amu calculate the mass of 187Re.Group of answer choices

A fictional element has two naturally occurring isotopes with the natural abundances shown here: ATOMIC MASS (AMU) ABUNDANCE1650.0%1850.0% Which statement is TRUE for this element?Group of answer choicesThe atomic mass would be 16.The atomic mass would be less than 16.The atomic mass would be 18.The atomic mass would be more than 18.The atomic mass would be 17.

An element X has four naturally occurring isotopes, with natural abundances and masses outlined in the table below. Mass (a.m.u) Natural Abundance (%)11.00 40.70 %13.00 39.30 %14.00 15.60 %16.00 4.40 %Calculate the atomic mass of element X, to two decimal places.

3. Calculate the average atomic mass of element, X. A sample of X is 62.00% X-121 and 38.00% X-123.121.76 amu122.00 amu121.25 amu123.00 amu

The average atomic mass of a sample of an element X is 16.2 u.What are the percentages of isotopes 168 X and 188 X in thesample?