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.
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Welcome back Fidelia — been a while! Let's pick up where you left off. Let's calculate the solubility of sodium trioxocarbonate (IV). Step 1: Determine the chemical formula and molar mass of sodium trioxocarbonate (IV). The chemical name sodium trioxocarbonate (IV) refers to sodium carbonate, Na₂CO₃. We need the atomic masses: Na = 23 g/mol, C = 12 g/mol, O = 16 g/mol. Molar mass of Na₂CO₃ = (2 × Na) + C + (3 × O) Molar mass = (2 × 23 g/mol) + 12 g/mol + (3 × 16 g/mol) Molar mass = 46 g/mol + 12 g/mol + 48 g/mol Molar mass of Na₂CO₃ = 106 g/mol Step 2: Calculate the number of moles of sodium trioxocarbonate (IV). Moles = mass / molar mass Moles of Na₂CO₃ = 30 g / 106 g/mol Moles of Na₂CO₃ ≈ 0.283 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 = 40 g / (1 g/mL) = 40 mL To convert mL to dm³: 1 dm³ = 1000 mL Volume of water = 40 mL × (1 dm³ / 1000 mL) Volume of water = 0.040 dm³ Step 4: Calculate the solubility in mol dm⁻³. Solubility = moles of solute / volume of solvent (in dm³) Solubility = 0.283 mol / 0.040 dm³ Solubility ≈ 7.075 mol/dm³ Rounding to two significant figures (based on the given masses of 30 g and 40 g): Solubility = 7.1 mol/dm³ Send me the next one 📸