/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 49 Calcium has a cubic closest pack... [FREE SOLUTION] | 91影视

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Chapter 10: Problem 49

Calcium has a cubic closest packed structure as a solid. Assuming that calcium has an atomic radius of \(197 \mathrm{pm}\), calculate the density of solid calcium.

Short Answer

Expert verified
The density of solid calcium with a cubic closest packed structure and an atomic radius of 197 pm is approximately 1.56 g/cm鲁.

Step by step solution

01

Determine the side length of the unit cell

In a cubic closest packed structure, atoms are tightly packed together such that they touch each other along the face diagonals of the unit cell. To find the side length, we have to consider that the face diagonal contains 4 atomic radii. Let a be the side length of the unit cell and r be the atomic radius. Using Pythagorean theorem, the relationship between the side length and the atomic radius can be given as: \(a^2 + a^2 = (4r)^2\) Dividing both sides by 2, we get: \(a^2 = 2(2r)^2\) Now we can solve for the side length a.
02

Calculate the side length a of the unit cell

Given the atomic radius r = 197 pm, we can substitute this into our equation to find the side length a. \(a^2 = 2(2(197\,pm))^2\) Now, find the square root of the right side to get the value for a: \(a = \sqrt{2(2(197\,pm))^2}\) \(a = 556\,pm\)
03

Calculate the volume of the unit cell

Now that we have the side length a, we can calculate the volume V of the unit cell by raising a to the power of 3. \(V = a^3\) \(V = (556\,pm)^3\) \(V = 1.71 脳 10^8\,pm^3\) Since 1 pm = 10^-12 m, we can convert the volume to cubic meters: \(V = 1.71 脳 10^8\,pm^3 脳 (10^{-12\,m/pm})^3\) \(V = 1.71 脳 10^{-16}\,m^3\)
04

Calculate the number of atoms in the unit cell

In a cubic closest packed structure, there are 4 calcium atoms per unit cell as it consists of one atom at each corner and one at the center of each face of the cube. Therefore, the number of atoms n in the unit cell is equal to 4.
05

Determine the molar mass of calcium and use it to find the mass of the atoms within the unit cell

The molar mass of calcium is 40.08 g/mol. To calculate the mass of the 4 calcium atoms within a unit cell, we convert this to an atomic weight in grams for one calcium atom by dividing by Avogadro's number (6.022 x 10^23 atoms/mol): 40.08 g/mol 梅 (6.022 x 10^23 atoms/mol) 鈮� 6.65 x 10^-23 g Next, multiply this atomic mass by the number of atoms (n=4) in the unit cell to get the mass of the atoms within the unit cell: mass = n 脳 atomic mass mass = 4 脳 6.65 x 10^-23 g mass = 2.66 x 10^-22 g
06

Calculate the density of solid calcium

Finally, we can calculate the density 蟻 of solid calcium by dividing the mass of the atoms within the unit cell by the volume of the unit cell: 蟻 = mass / volume 蟻 = (2.66 脳 10^-22 g) 梅 (1.71 脳 10^-16 m^3) To convert the density to g/cm^3, use the conversion factor 1 g/cm^3 = 1000 kg/m^3: 蟻 = (2.66 脳 10^-22 g 梅 1.71 脳 10^-16 m^3) 脳 (1000 kg/m^3 梅 1 g/cm^3) 蟻 鈮� 1.56 g/cm^3 Thus, the density of solid calcium is approximately 1.56 g/cm^3.

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Most popular questions from this chapter

Amino acids are the building blocks of the body's worker molecules called proteins. When two amino acids bond together, they do so through the formation of a peptide linkage, and a dipeptide is formed. Consider the following tripeptide formed when three alanine amino acids bond together: What types of interparticle forces could be present in a sample of this tripeptide?

Rationalize the differences in physical properties in terms of intermolecular forces for the following organic compounds. Compare the first three substances with each other, compare the last three with each other, and then compare all six. Can you account for any anomalies? $$ \begin{array}{|lccc|} \hline & \text { bp }\left({ }^{\circ} \mathrm{C}\right) & \operatorname{mp}\left({ }^{\circ} \mathrm{C}\right) & \Delta H_{\text {vap }}(\mathbf{k J} / \text { mol }) \\ \hline \text { Benzene, } \mathrm{C}_{6} \mathrm{H}_{6} & 80 & 6 & 33.9 \\ \text { Naphthalene, } & & & \\ \mathrm{C}_{10} \mathrm{H}_{8} & 218 & 80 & 51.5 \\ \text { Carbon tetra- } & & & \\ \text { chloride } & 76 & -23 & 31.8 \\ \text { Acetone, } & & & \\ \mathrm{CH}_{3} \mathrm{COCH}_{3} & 56 & -95 & 31.8 \\ \text { Acetic acid, } & & & \\ \mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H} & 118 & 17 & 39.7 \\ \text { Benzoic acid, } & & & \\ \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H} & 249 & 122 & 68.2 \\\ \hline \end{array} $$

Mn crystallizes in the same type of cubic unit cell as Cu. Assuming that the radius of \(\mathrm{Mn}\) is \(5.6 \%\) larger than the radius of \(\mathrm{Cu}\) and the density of copper is \(8.96 \mathrm{~g} / \mathrm{cm}^{3}\), calculate the density of \(\mathrm{Mn}\).

The compounds \(\mathrm{Na}_{2} \mathrm{O}, \mathrm{CdS}\), and \(\mathrm{Zr} \mathrm{I}_{4}\) all can be described as cubic closest packed anions with the cations in tetrahedral holes. What fraction of the tetrahedral holes is occupied for each case?

The Group \(3 \mathrm{~A} /\) Group 5 A semiconductors are composed of equal amounts of atoms from Group \(3 \mathrm{~A}\) and Group \(5 \mathrm{~A}-\) for example, InP and GaAs. These types of semiconductors are used in lightemitting diodes and solid-state lasers. What would you add to make a p-type semiconductor from pure GaAs? How would you dope pure GaAs to make an n-type semiconductor?
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