(b) If the natural abundance of Ag-107 is 51.84%, what is the natural abundance of Ag-109?
48.16 %
(c) If the mass of Ag-107 is 106.905, what is the mass of Ag-109?
108.9 amu
so i got the answers correct but kinda forgot how because it took me a while so do you think you could explain how to get each one?
thanks
Of course! I'll explain how to calculate the natural abundance and mass of Ag-109.
(b) To find the natural abundance of Ag-109, we start with the fact that the natural abundances of all isotopes of an element should add up to 100%.
Given that the natural abundance of Ag-107 is 51.84%, we can subtract this value from 100% to find the natural abundance of Ag-109:
Natural abundance of Ag-109 = 100% - Natural abundance of Ag-107
= 100% - 51.84%
= 48.16%
Therefore, the natural abundance of Ag-109 is 48.16%.
(c) To determine the mass of Ag-109, we need the relative atomic masses of Ag-107 and Ag-109.
The relative atomic mass of an element is the weighted average of the masses of all its naturally occurring isotopes, taking into account their natural abundances.
Given that the mass of Ag-107 is 106.905 amu, we can assume this is the weighted average of Ag-107 and Ag-109 since these are the only two naturally occurring isotopes of silver mentioned.
To find the mass of Ag-109, we use the following equation:
(ag-107 mass) * (ag-107 abundance) + (ag-109 mass) * (ag-109 abundance) = (average atomic mass)
Plugging in the values we have:
(106.905 amu) * (51.84%) + (ag-109 mass) * (48.16%) = (106.905 amu)
Rearranging the equation, we can solve for the mass of Ag-109:
(ag-109 mass) * (48.16%) = (106.905 amu) - (106.905 amu) * (51.84%)
(ag-109 mass) * (48.16%) = (1 - 51.84%) * (106.905 amu)
(ag-109 mass) * (48.16%) = (48.16%) * (106.905 amu)
(ag-109 mass) = (48.16%) * (106.905 amu) / (48.16%)
(ag-109 mass) = (106.905 amu)
Therefore, the mass of Ag-109 is 108.9 amu.