Given that Ag-107 (51.839% abundance) and the mass ratio: Ag-109/Ag-107=1.0187. Calculate the mass of Ag-107. (Atomic mass of Ag = 107.87 amu)

Atomic mass of Ag-107 = 106.94 amu

Explanation:

Let the mass of Ag-107 be y

Since the mass ratio of Ag-109/Ag-107 is 1.0187, the mass of Ag-109  is 1.0187 times heavier than the mass of Ag-107

Mass of Ag-109  = 1.0187y

Relative atomic mass of Ag = sum of (mass of each isotope * abundance)

Relative atomic mass of Ag = 107.87

Abundance of Ag-107 = 51.839% = 0.51839

Abundance of Ag-109 = 41.161% = 0.48161

107.87 = (y * 0.51839) + (1.018y * 0.48161)

107.87 = 0.51838y + 0.49027898y

107.87 = 1.00865898y

y = 107.87/1.00865898

y = 106.94 amu

Therefore, atomic mass of Ag-107 = 106.94 amu

increased stability in the completion of fluorine's outer shell.