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
Step 1: Identify the given values and constants. • Type of unit cell: Face-centred cubic (FCC). For an FCC structure, the number of atoms per unit cell (Z) is 4. • Edge length (a) = 408 pm. • Molar mass (M) of gold = 197 g/mol. • Avogadro's number (N_A) = 6.022 × 10^23 mol^-1. Step 2: Convert the edge length from picometers (pm) to centimeters (cm). a = 408 pm × 10^-12 m1 pm × 100 cm1 m a = 408 × 10^-10 cm a = 4.08 × 10^-8 cm Step 3: Use the formula for the density () of a unit cell. = (Z · M)/(a^3 · N_A) Step 4: Substitute the values into the formula and calculate the density. = 4 × 197 g/mol(4.08 × 10^-8 cm)^3 × 6.022 × 10^23 mol^-1 = 788 g(4.08^3 × (10^-8)^3 cm^3) × 6.022 × 10^23 = 788 g(67.917312 × 10^-24 cm^3) × 6.022 × 10^23 = 788 g(67.917312 × 6.022) × 10^-24+23 cm^3 = 788 g408.990 × 10^-1 cm^3 = 788 g40.899 cm^3 ≈ 19.265 g/cm^3 Step 5: Round the density to two decimal places and convert to kg/m^3. ≈ 19.27 g/cm^3 To convert to kg/m^3: 1 g/cm^3 = 1 × 10^-3 kg(10^-2 m)^3 = 1 × 10^-3 kg10^-6 m^3 = 10^3 kg/m^3 = 19.27 × 10^3 kg/m^3 The density of gold is approximately 19.27 g/cm^3 or 19.27 × 10^3 kg/m^3. That's 2 down. 3 left today — send the next one.