This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.
ScanSolve AI Answer
Let's calculate the solubility of lead(II) trioxonitrate(V). Step 1: Determine the chemical formula and molar mass of lead(II) trioxonitrate(V). The chemical name lead(II) trioxonitrate(V) refers to lead(II) nitrate, Pb(NO₃)₂. We need the atomic masses: Pb = 207 g/mol, N = 14 g/mol, O = 16 g/mol. Molar mass of Pb(NO₃)₂ = Pb + (2 × N) + (6 × O) Molar mass = 207 g/mol + (2 × 14 g/mol) + (6 × 16 g/mol) Molar mass = 207 g/mol + 28 g/mol + 96 g/mol Molar mass of Pb(NO₃)₂ = 331 g/mol Step 2: Calculate the number of moles of lead(II) trioxonitrate(V). Moles = mass / molar mass Moles of Pb(NO₃)₂ = 12.2 g / 331 g/mol Moles of Pb(NO₃)₂ ≈ 0.03686 mol Step 3: Convert the mass of water to volume in dm³. We assume the density of water is 1 g/mL (or 1 g/cm³). Volume of water = mass / density Volume of water = 21 g / (1 g/mL) = 21 mL To convert mL to dm³: 1 dm³ = 1000 mL Volume of water = 21 mL × (1 dm³ / 1000 mL) Volume of water = 0.021 dm³ Step 4: Calculate the solubility in mol dm⁻³. Solubility = moles of solute / volume of solvent (in dm³) Solubility = 0.03686 mol / 0.021 dm³ Solubility ≈ 1.755 mol/dm³ Rounding to two significant figures (based on the given mass of 21 g of water): Solubility = 1.8 mol/dm^3 3 done, 2 left today. You're making progress.