91Ó°ÊÓ

In a face-centered cubic unit cell, there are atoms at each corner and at the center of each face. The atomic radius can be found by dividing the edge length by 4, multiplied by the square root of 2. Let's denote the edge length as \(a\), and the atomic radius as \(r\): \(r = (a/4) \sqrt{2}\). The given edge length is 3.833 Ã… (angstroms, where 1 Ã… = 10^(-10) m). Now we can calculate the atomic radius: \(r = (3.833/4) \sqrt{2} = 1.9175 \sqrt{2} \approx 2.710 Ã…\) So, the atomic radius of an iridium atom is approximately 2.710 Ã….

Density is defined as mass per unit volume. In a face-centered cubic unit cell, there are four iridium atoms. The mass of one iridium atom can be found using the molar mass of iridium metal and Avogadro's number: Mass of one iridium atom = (Molar mass of iridium)/(Avogadro's number) = (192.22 g/mol)/(6.022 x 10^23 atoms/mol) = 3.194 x 10^(-22) g Now, we will calculate the mass of iridium metal in one unit cell: Mass of iridium metal in one unit cell = 4 x (mass of one iridium atom) = 4 x 3.194 x 10^(-22) g = 1.278 x 10^(-21) g To find the volume of one unit cell, we can use the given edge length, since the volume of a cube is the edge length cubed: Volume of one unit cell = (3.833 Å)^3 x (1 cm/10^8 Å)^3 = 5.623 x 10^(-23) cm^3 Now we can calculate the density of iridium metal: Density = (Mass of iridium metal in one unit cell)/(Volume of one unit cell) = (1.278 x 10^(-21) g)/(5.623 x 10^(-23) cm^3) ≈ 22.72 g/cm^3 Therefore, the density of iridium metal is approximately 22.72 g/cm^3.