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Silver has an atomic mass of 107.9 amu. If 51.84% of Ag exists as Ag-107 (106.9051 amu), what is the identity and the atomic mass of the other isotope?

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Answer:

To determine the identity and atomic mass of the other isotope of silver (Ag), we can use the given information about the percentage abundance and atomic mass of Ag-107.

Let's assume that the other isotope is Ag-x, where x represents the mass number of the isotope. Since the total percentage abundance of both isotopes must add up to 100%, we can calculate the percentage abundance of Ag-x by subtracting the given percentage abundance of Ag-107 from 100%.

Percentage abundance of Ag-x = 100% - 51.84% = 48.16%

Now, we can set up an equation to find the atomic mass of Ag-x. The equation is as follows:

(percentage abundance of Ag-x / 100) * (atomic mass of Ag-x) + (percentage abundance of Ag-107 / 100) * (atomic mass of Ag-107) = atomic mass of silver

Plugging in the known values:

(48.16 / 100) * (atomic mass of Ag-x) + (51.84 / 100) * (106.9051 amu) = 107.9 amu

Simplifying the equation:

0.4816 * (atomic mass of Ag-x) + 0.5184 * (106.9051 amu) = 107.9 amu

0.4816 * (atomic mass of Ag-x) = 107.9 amu - (0.5184 * 106.9051 amu)

0.4816 * (atomic mass of Ag-x) = 107.9 amu - 55.4072 amu

0.4816 * (atomic mass of Ag-x) = 52.4928 amu

Dividing both sides by 0.4816:

(atomic mass of Ag-x) = 52.4928 amu / 0.4816

(atomic mass of Ag-x) ≈ 108.8 amu

Therefore, the other isotope of silver (Ag-x) has an atomic mass of approximately 108.8 amu.

Step-by-step explanation:

User Anjil Panchal
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