The spectator ions would be the ions that <em>don't </em>comprise the precipitate in this reaction. You have four ionic species here: Pb²⁺, NO₃⁻, H⁺, and SO₄²⁻. Since Pb²⁺ and SO₄²⁻ combine, as you are told, to form the precipitate, that leaves H⁺ and NO₃⁻ (or, in normal text, H+ and NO3-) to be the spectator ions.
Edit: NO3- might be interpreted ambiguously since it's not immediately clear that the NO3 is a polyatomic ion with a -1 charge and not an NO compound with a 3- charge, so it should be written as [NO3]- or (NO3)-.
The net ionic reaction shows the reaction without the spectator ions. In this case, that would be: Pb²⁺(aq) + SO₄²⁻(aq) → PbSO₄(s).
Answer:
3.44 × 10²⁴ atoms of oxygen
Explanation:
Step 1: Calculate the moles corresponding to 163 g of Al₂(SO₄)₃
The molar mass of Al₂(SO₄)₃ is 342.15 g/mol.
163 g × 1 mol/342.15 g = 0.476 mol
Step 2: Calculate the molecules in 0.476 moles of Al₂(SO₄)₃
We will use Avogadro's number: there are 6.02 × 10²³ molecules of Al₂(SO₄)₃ in 1 mole of Al₂(SO₄)₃.
0.476 mol × 6.02 × 10²³ molecules/1 mol = 2.87 × 10²³ molecules
Step 3: Calculate the atoms of O in 2.87 × 10²³ molecules of Al₂(SO₄)₃
According to the chemical equation, the molar ratio of Al₂(SO₄)₃ to O is 1:12. The atoms of O are 12/1 × 2.87 × 10²³ = 3.44 × 10²⁴.
Answer:
lower price and higher voltage capacity for the weight.
