Answer :
Answer:
B. 1.84 × 10^24
Explanation:
[tex]75.0 g H2SO4(\frac{1 mol H2SO4}{98.1 g})(\frac{4 mol O}{1 mol H2SO4})(\frac{6.023 x 10^23 atoms}{1 mol O} )[/tex] = 1.84 x 10^24 atoms
Final answer:
In 75.0 g of sulfuric acid, there are 1.84 * 10^24 atoms of oxygen. This is calculated by converting the given mass to moles, multiplying by the number of oxygen atoms per molecule, and then converting moles to atoms using Avogadro's number.
Explanation:
To calculate the number of oxygen atoms in 75.0 g of sulfuric acid (H2SO4), we first need to convert the mass of H2SO4 to moles. We use its molar mass of 98.1 g/mol for this conversion, so 75.0 g H2SO4 * (1 mol/98.1 g) = 0.764 moles H2SO4. Next, we need to consider the chemical formula of sulfuric acid, which shows us that there are 4 moles of oxygen for every 1 mole of H2SO4. Therefore, 0.764 moles H2SO4 * (4 mol O/1 mol H2SO4) = 3.06 moles of O. Finally, we convert moles of O to atoms using Avogadro's number, 6.02 * 10^23 atoms/mol. Thus, 3.06 moles O * (6.02 * 10^23 atoms/1 mol) = 1.84 * 10^24 atoms of oxygen. The correct answer is B, 1.84 * 10^24.
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