### Density Calculation of Polonium (Po)**Problem Statement:**The element Polonium (Po) crystallizes in a simple cubic unit cell. If the radius of the Polonium atom is 135 pm, what is the density of Polonium? *(Note: edge length = 2r)*Considering the atomic radius of Polonium to be 135 pm (picometers), the edge length \(a\) of the simple cubic unit cell would be:\[ \text{Edge Length (a)} = 2r \]Given:\[ r = 135 \, \text{pm} \]Therefore:\[ a = 2 \times 135 \, \text{pm} = 270 \, \text{pm} \]Before proceeding to calculate the density, it is necessary to convert the edge length into centimeters for consistency in the unit used in density calculations. 1 pm \(= 1 \times 10^{-12}\) meters = \(1 \times 10^{-10}\) cm.Thus:\[ a = 270 \times 10^{-12} \, \text{m} = 270 \times 10^{-10} \, \text{cm} \]\[ a = 2.70 \times 10^{-8} \, \text{cm} \]### Volume of the Unit Cell:\[ V_{\text{cell}} = a^3 \]\[ V_{\text{cell}} = (2.70 \times 10^{-8} \, \text{cm})^3 \]\[ V_{\text{cell}} = 1.97 \times 10^{-23} \, \text{cm}^3 \]### Mass of One Unit Cell:The molar mass of Polonium (Po) is approximately 208.98 g/mol. Since a simple cubic unit cell contains only one atom:\[ \text{Mass of one unit cell} = \frac{\text{Molar Mass}}{\text{Avogadro's Number}} \]\[ \text{Mass of one Po atom} = \frac{208.98 \, \text{g/mol}}{6.022 \times 10^{23} \, \text{atoms/mol}} \]\[ \text{Mass of one Po atom} \approx 3.47 \times 10^{-22} \, \text{g} \]

Answered: ) The element Polonium (Po)…

) The element Polonium (Po) crystallizes in a simple cubic unit cell. If the radius of the Polonium atom is 135 pm, what is the density of Polonium? (Note: edge length= 2r)

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

See similar textbooks

Similar questions

Chromium has a body-centered cubic unit cell, and a density of 7.19 g/cm3. Calculate the edge length of the unit cell, in pm. (The atomic mass of iron is 51.996 amu. Also, 1 amu = 1.661 × 10–24 g.)
The ionic compound, CsBr, forms a body centered cubic structure (see below) where the anions occupy the corners of the crystal (red) and the cations the body centered position (green). If the ionic radius of Cs +, Br − are 169 pm and 195 pm respectively, what is the density (in g/cm 3)?
Element Xr (99.88 g/mol) has a BCC unit cell and a density of 16.87 g/cm33. Based on this information, what is the radius of an atom of Xr in pm?
Not all unit cells have a cubic structure. Take for example the below image wherein the unit cell takes the shape of a parallelepiped and has lengths a, b, c and angles α, β, γ. This particular unit cell has an atom arrangement where it has one atom at each corner, one central atom at each of the bases, one atom at the center of the unit cell, and one atom at the center of each of the edge lengths across length c. Given a density of 24.19 g/cm3, calculate the length of edge c in angstroms of a theoretical element Ro given the following unit cell dimensions in angstroms and degrees: a = 9.84A b = 6.4A γ = 43o Atomic Mass: Ro: 50.833 g/mol
The ionic compound, CsBr, forms a body centered cubic structure (see below) where the anions occupy the corners of the crystal (red) and the cations the body centered position (green). If the density of the compound is 4.44 g/cm 3 and the ionic radius of Cs + is 169 pm, what is the ionic radius of Br − in pm?
Chromium forms a body-centered cubic unit cell and has a density of 7.18 g/cm3. From this information, determine the radius of a chromium atom. pm
Iron forms a body-centered cubic crystal structure. (a) How many Fe atoms are in a unit cell? (b) What is the mass of a unit cell in grams? (c) If the density of iron is 7.86 g/cm', calculate the volume of unit cell, the edge length of the unit cell in pm, and the atomic radius of iron in pm. (Atomic mass of iron = 55.85 amu; 1 amu = 1.6605 x 10 24 g; 1 pm = 10-10 cm)
A metal has a body-centered cubic lattice with a unit cell edge length of 2.866 Å (1 Å = 10⁻¹⁰ m). The density of the metal is 7.87 g/cm³. What is the mass of an atom of this metal in grams? (1 m = 10¹² pm).
18. Copper crystallizes into a face-centered cubic crystal structure. (a) How many copper atoms are there in one unit cell of copper crystal? (b) If the density of copper is 8.96 g/cm³, what is the edge-length of the unit cell in pm? (c) Calculate the atomic radius of copper in pm. (1 u = 1.6605 x 10-24 g; 1 pm = 10-10 cm) %3D 3
Not all unit cells have a cubic structure. Take for example the below image wherein the unit cell takes the shape of a parallelepiped and has lengths a, b, c and angles α, β, γ. This particular unit cell has an atom arrangement where it has one atom at each corner, one central atom at each of the bases, one atom at the center of the unit cell, and one atom at the center of each of the edge lengths across length c. Given a density of 23.02 g/cm3, calculate the length of edge c in angstroms of a theoretical element Ro given the following unit cell dimensions in angstroms and degrees: a = 9.77A b = 6.68A γ = 44o Atomic Mass: Ro: 54.704 g/mol
A crystalline form of platinum has a density of 21.45 g/cm³. If the radius of platinum atoms is 139.0 pm, what is the platinum unit cell? Choose one: OA simple cubic OB. face-centered cubic OC, body-centered cubic OD. It cannot be determined from the information given.