Final answer:
The mass of silver nitrate that contains 3.05×10²⁵ oxygen atoms is approximately 8.54 × 10³ grams.
Step-by-step explanation:
In order to determine the mass of silver nitrate that contains a certain number of oxygen atoms, we need to use the molar mass and stoichiometry of the compound. The molar mass of silver nitrate (AgNO3) is 169.88 g/mol. From the balanced chemical equation for the formation of silver nitrate, we can see that each molecule of silver nitrate contains one oxygen atom. Therefore, to calculate the mass of silver nitrate that contains 3.05×1025 oxygen atoms, we can use the following calculation:
(3.05×1025 oxygen atoms) × (1 molecule AgNO3)/(6.022×1023 oxygen atoms/mol) × (169.88 g AgNO3)/(1 molecule AgNO3) = X g AgNO3
Simplifying the calculation we find:
X = (3.05×1025 × 169.88)/(6.022×1023) ≈ 8.54 × 103 g AgNO3
Therefore, the mass of silver nitrate that contains 3.05×1025 oxygen atoms is approximately 8.54 × 103 grams.