/*! 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

Consider the following melting point data: $$ \begin{array}{|llllllll|} \hline \text { Compound: } & \mathrm{NaCl} & \mathrm{MgCl}_{2} & \mathrm{AlCl}_{3} & \mathrm{SiCl}_{4} & \mathrm{PCl}_{3} & \mathrm{SCl}_{2} & \mathrm{Cl}_{2} \\ \operatorname{mp}\left({ }^{\circ} \mathrm{C}\right): & 801 & 708 & 190 & -70 & -91 & -78 & -101 \\ \text { Compound: } & \mathrm{NaF} & \mathrm{MgF}_{2} & \mathrm{AlF}_{3} & \mathrm{SiF}_{4} & \mathrm{PF}_{5} & \mathrm{SF}_{6} & \mathrm{~F}_{2} \\ \operatorname{mp}\left({ }^{\circ} \mathrm{C}\right): & 997 & 1396 & 1040 & -90 & -94 & -56 & -220 \\ \hline \end{array} $$ Account for the trends in melting points in terms of interparticle forces.

The critical point of \(\mathrm{NH}_{3}\) is \(132^{\circ} \mathrm{C}\) and \(111 \mathrm{~atm}\), and the critical point of \(\mathrm{N}_{2}\) is \(-147^{\circ} \mathrm{C}\) and 34 atm. Which of these substances cannot be liquefied at room temperature no matter how much pressure is applied? Explain.

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