How many ammonium ions and how many sulfate ions are present in an 0.400 mol sample of (nh4)2so4? how many atoms of n, h, s, and o are contained in this sample?

In every 1 mole of (NH₄)₂SO₄, there are 2 moles of ammonium (NH₄) ions and 1 mole of sulfate (SO₄) ions. Also, in every 1 mole of (NH₄)₂SO₄, there are 2 moles of N, 8 moles of H, 1 mole of S and 4 moles of O. Then, we use the Avogadro's number, 6.022×10²³ particles/mol, to find the number of ions and atoms.

1. Number of ammonium ions
0.4 mol(2 mol NH₄/1 mol (NH₄)₂SO₄)(6.022×10²³ ions/mol) = 4.82×10²³ ammonium ions

2. Number of sulfate ions
0.4 mol(1 mol SO₄/1 mol (NH₄)₂SO₄)(6.022×10²³ ions/mol) = 2.41×10²³ sulfate ions

3. Atoms of N
0.4 mol(2 mol N/1 mol (NH₄)₂SO₄)(6.022×10²³ atoms/mol) =4.82×10²³ nitrogen atoms

4. Atoms of H
0.4 mol(8 mol H/1 mol (NH₄)₂SO₄)(6.022×10²³ atoms/mol) =1.927×10²⁴ hydrogen atoms

5. Atoms of S
0.4 mol(1 mol S/1 mol (NH₄)₂SO₄)(6.022×10²³ atoms/mol) =2.41×10²³ sulfur atoms

6. Atoms of O
0.4 mol(4 mol O/1 mol (NH₄)₂SO₄)(6.022×10²³ atoms/mol) =9.635×10²³ oxygen atoms

-122 J/K

Explanation:

Let's consider the following balanced reaction.

N₂(g) + 2 O₂(g) ⇒ 2 NO₂(g)

We can calculate the standard reaction entropy (ΔS°) using the following expression.

ΔS° = Σ ηp × Sf°p - Σ ηr × Sf°r

where,

η: stoichiometric coefficients of products and reactantsSf°r: entropies of formation of products and reactants

ΔS° = 2 mol × 240.06 J/K.mol - 1 mol × 191.61 J/K.mol - 2 mol × 205.14 J/K.mol

ΔS° = -121.77 J/K ≈ -122 J/K