Answer :
Final answer:
In 38.1 g of barium chloride (BaCl2), there are approximately 3.30 × 10²³ atoms. This is calculated by first converting the mass to moles using the molar mass of BaCl2, accounting for the number of atoms of each element in a molecule of BaCl2, and then multiplying by Avogadro's number.
Explanation:
To answer the question of how many atoms are in 38.1 g of barium chloride (BaCl2), we first need to calculate the number of moles of BaCl2 in 38.1 g. The molar mass of barium chloride, BaCl2, is approximately 208.3 g/mol (137.3 g/mol for Ba and 2*35.5 g/mol for Cl). Therefore, there are 38.1 g / 208.3 g/mol = 0.183 moles of BaCl2 in 38.1 g.
Each formula unit of BaCl2 consists of three atoms (one Ba and two Cl), so in one mole of BaCl2, there are 3 moles of atoms. Therefore, there are 0.183 moles * 3 = 0.549 moles of atoms in 38.1 g of BaCl2.
Finally, to find the number of atoms, we multiply the number of moles by Avogadro's number, 6.022 × 10²³ atoms/mol. Hence, there are 0.549 moles * 6.022 × 10²³ atoms/mol = 3.30 × 10²³ atoms in 38.1 g of BaCl2.
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