Chemistry Question: Calculating Number of Sulfer Atoms

How many atoms of sulfur are in 4.28 x 10²³ units of bismuth (V) Sulfate?

The number of sulfur atoms in 4.28 x 10²³ units of Bismuth (V) Sulfate is 4.28 x 10²³, since each molecule of Bismuth (V) Sulfate contains one atom of sulfur.

Answer:

The subject of this question involves the concept of molar equivalence in chemistry. Bismuth (V) Sulfate is a compound containing Bismuth, Sulfate (SO4), and sulfur. One molecule of Bismuth (V) Sulfate contains one atom of sulfur. Therefore, to calculate the number of sulfur atoms in 4.28 x 10²³ molecules of Bismuth (V) Sulfate, one must multiply the total number of molecules by the number of sulfur atoms in one molecule.

Explanation:

4.28 x 10²³ molecules of Bismuth (V) Sulfate contain 4.28 x 10²³ sulfur atoms. This is because each molecule of Bismuth (V) Sulfate contains one sulfur atom. Therefore, the number of sulfur atoms is directly proportional to the number of molecules of Bismuth (V) Sulfate.

To better understand sulfur atoms and their role in chemistry, it is recommended to explore further resources and research on this topic.