A 3.50 g of an alloy which contains only lead and tin is dissolved in hot HNO3. Excess sulfuric acid is added to this solution and 1.57 g of PbSO4(s) is obtained. (a) Write the net ionic equation for the formation of PbSO4.

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Identify the ions involved in the formation of PbSO4. Lead(II) ions (Pb^{2+}) and sulfate ions (SO_4^{2-}) are the key players in this reaction.

Write the dissociation equations for the reactants: Pb^{2+} (aq) and SO_4^{2-} (aq) are already in their ionic forms.

Combine the ions to form the solid product: Pb^{2+} (aq) + SO_4^{2-} (aq) \rightarrow PbSO_4 (s).

Ensure the equation is balanced. In this case, the charges and atoms are balanced as written.

The net ionic equation for the formation of lead(II) sulfate is: Pb^{2+} (aq) + SO_4^{2-} (aq) \rightarrow PbSO_4 (s).

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The arsenic in a 1.22-g sample of a pesticide was converted to AsO₄^3- by suitable chemical treatment. It was then titrated using Ag⁺ to form Ag₃AsO₄ as a precipitate. (b) Name Ag₃AsO₄ by analogy to the corresponding compound containing phosphorus in place of arsenic.

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