Silver has an average atomic mass of 107.9 u and is known to have only two naturally occurring isotopes. If 51.84% of Ag exists as Ag-107 (106.9051 u), what is the identity and the atomic mass of the other isotope?

Ag 107 = 0.5184 = 106.9051

Ag x = 1-0.5184 = 0.4816
-----------------
0.5184*(106.9051) + 0.l4816*X = 107.9
Solve for X.
X will give you the atomic mass, round that to the nearest whole number for the mass number for that Ag isotope.

Thanks

To find the identity and atomic mass of the other isotope of Silver (Ag), we need to consider that the average atomic mass of Silver is 107.9 u. Additionally, we know that there are only two naturally occurring isotopes of Silver.

Let's assign the atomic mass of the other isotope as x.

Given that 51.84% of Ag exists as Ag-107 (atomic mass = 106.9051 u), we can calculate the atomic mass of the other isotope using the following equation:

(51.84/100) * 106.9051 + (48.16/100) * x = 107.9

Simplifying the equation, we have:

0.5184 * 106.9051 + 0.4816 * x = 107.9

55.3787584 + 0.4816x = 107.9

0.4816x = 107.9 - 55.3787584

0.4816x = 52.5212416

x = 52.5212416 / 0.4816

x ≈ 109.109

Therefore, the other isotope of Silver (Ag) is Ag-109, with an atomic mass of approximately 109.109 u.

To determine the identity and atomic mass of the other isotope of silver, we can first find the percentage abundance of the second isotope by subtracting the given percentage abundance of Ag-107 from 100%.

Percentage abundance of Ag-107 = 51.84%
Percentage abundance of the other isotope = 100% - 51.84% = 48.16%

Next, let's assign variables to the atomic masses of the two isotopes. Let x represent the atomic mass of the other isotope.

Atomic mass of Ag-107 = 106.9051 u
Percentage abundance of Ag-107 = 51.84%

Atomic mass of the other isotope = x
Percentage abundance of the other isotope = 48.16%

To find the atomic mass of the other isotope, we need to set up an equation using the average atomic mass of silver:

Average atomic mass of silver (107.9 u) = (Atomic mass of Ag-107 * Percentage abundance of Ag-107) + (Atomic mass of the other isotope * Percentage abundance of the other isotope)

107.9 u = (106.9051 u * 51.84%) + (x * 48.16%)

Simplifying the equation:

107.9 u = 55.4541864 u + 0.4816x

Subtracting 55.4541864 u from both sides of the equation:

52.4458136 u = 0.4816x

Dividing both sides of the equation by 0.4816:

x = 52.4458136 u / 0.4816

x ≈ 108.698 u

Therefore, the identity of the other isotope of silver is Ag-109, and its atomic mass is approximately 108.698 u.