Answer:
3.27·10²³ atoms of O
Explanation:
To figure out the amount of oxygen atoms in this sample, we must first evaluate the sample.
The chemical formula for sodium sulfate is Na₂SO₄, and its molar mass is approximately 142.05[tex]\frac{g}{mol}[/tex].
We will use stoichiometry to convert from our mass of Na₂SO₄ to moles of Na₂SO₄, and then from moles of Na₂SO₄ to moles of O using the mole ratio; then finally, we will convert from moles of O to atoms of O using Avogadro's constant.
19.3g Na₂SO₄ · [tex]\frac{1 mol Na^2SO^4}{142.05g Na^2SO^4}[/tex] · [tex]\frac{4 mol O}{1 mol Na^2SO^4}[/tex] ·[tex]\frac{6.022x10^2^3}{1 mol O}[/tex]
After doing the math for this dimensional analysis, you should get a quantity of approximately 3.27·10²³ atoms of O.
The number of oxygen atoms in 19.3 g of sodium sulfate (Na₂SO₄) is 3.27×10²³ atoms
We'll begin by calculating the number of mole in 19.3 g of sodium sulfate (Na₂SO₄).
Mass of Na₂SO₄ = 19.3 g
Molar mass of Na₂SO₄ = (23×2) + 32 +(16×4)
= 46 + 32 + 64
= 142 g/mol
Mole = mass / molar mass
Mole of Na₂SO₄ = 19.3 / 142
1 mole of Na₂SO₄ contains 4 moles of O.
Therefore,
0.136 mole of Na₂SO₄ will contain = 0.136 × 4 = 0.544 mole of O
Finally, we shall determine the number of atoms in 0.544 mole of O.
From Avogadro's hypothesis,
1 mole of O = 6.02×10²³ atoms
Therefore,
0.544 mole of O = 0.544 × 6.02×10²³
0.544 mole of O = 3.27×10²³ atoms
Thus, 19.3 g of sodium sulfate (Na₂SO₄) contains 3.27×10²³ atoms of oxygen.
Learn more: https://brainly.com/question/25115547
