Respuesta :
Answer:
Unit cells which are present per cubic centimeter of Ag = [tex]=1.45\times 10^{22}\ unit\ cells /cm^3[/tex]
Volume = [tex]68.9\times 10^{-24}\ cm^3[/tex]
Edge length = [tex]4.1\times 10^{-8}\ cm[/tex]
Explanation:
(a)
Given that:-
The density of the solid Ag = 10.5 g/cm³
Molar mass of silver = 107.8682 g/mol
So, Moles present per cm³ of Ag = [tex]\frac{10.5\ g/cm^3}{107.8682\ g/mol}[/tex]=0.0973 mol/cm³
Also, 1 mole = [tex]6.023\times 10^{23}[/tex] atoms.
So,
Atoms present per cm³ of Ag = [tex]0.0973\ mol/cm^3\times 6.023\times 10^{23}\ atoms/mol=5.8\times 10^{22}\ atoms/cm^3[/tex]
Thus, answer = [tex]5.8\times 10^{22}\ atoms/cm^3[/tex]
In FCC, the number of atoms in the unit cell = 4 unit cells
So,
Unit cells which are present per cubic centimeter of Ag = [tex]\frac{5.8\times 10^{22}\ atoms/cm^3}{4}=1.45\times 10^{22}\ unit\ cells /cm^3[/tex]
Unit cells which are present per cubic centimeter of Ag = [tex]=1.45\times 10^{22}\ unit\ cells /cm^3[/tex]
(b)
The reciprocal of the unit cell/cm³ is the volume of the unit cell.
So, [tex]Volume=\frac{1}{1.45\times 10^{22}\ unit\ cells /cm^3}=68.9\times 10^{-24}\ cm^3[/tex]
Volume = [tex]68.9\times 10^{-24}\ cm^3[/tex]
(c)
Also, Volume = [tex]{(Edge\ length)}^3[/tex]
Thus, edge length = [tex]{Volume}^{\frac{1}{3}}[/tex] = [tex]\left(68.9\times \:\:10^{-24}\right)^{\frac{1}{3}}\ cm=4.1\times 10^{-8}\ cm[/tex]
Edge length = [tex]4.1\times 10^{-8}\ cm[/tex]
The answer is: [tex]\bold{5.8633 \times 10^{22} \frac{atoms}{cm^3}, 1.4658 \times 10^{22} \ \frac{unit cells}{cm^3}, 6.822 \times 10^{-23} \ cm^3, \ and\ 4.086 \times 10^{-8}\ cm}[/tex]
density:
[tex]Ag[/tex] density [tex]= 10.5\ \frac{g}{cm^3}[/tex]
So, 10.5 g of Ag is present in 1 cm3 of Ag.
[tex]Ag[/tex] (Atomic weight) [tex]= 107.86 \ \frac{g}{mol}[/tex]
Calculating the moles in [tex]Ag[/tex]:
[tex]\to \frac{10.5\ g}{107.86 \ \frac{g}{mol}} = 0.09735\ mol\\\\[/tex]
Calculating the atoms in 1 mole:
Therefore
atoms in 1 mole = Avogadro number[tex]= 6.023 \times 10^{23}[/tex]
in 0.09735 mol atoms:
[tex]\to 0.09735 \times 6.023 \times 10^{23} \\\\\to 5.8633 \times 10^{22}[/tex]
In a face-centered unit cell,
When there is total [tex]\bold{4\ \ atoms}[/tex] in the unit cell:
[tex]\to \text{corners} = \frac{1}{8} \times 8 = 1 \ \text{atom}\\\\ \to \text{face centers} = \frac{1}{2} \times 6 = 3\ \text{atom}\\\\ \to \text{total}\ 3+1 = 4\ \text{atom}\\\\[/tex]
Calculating the mass:
When
1 [tex]Ag[/tex] atom:
[tex]= \frac{107.86 \ \frac{g}{mol}}{6.023 \times 10^{23}}\\\\= 17.908 \times 10^{-23}\\\\= 1.7908 \times 10^{-22}\ g[/tex]
4 [tex]Ag[/tex] atom:
[tex]= \frac{4 \times 107.86 \ \frac{g}{mol}}{6.023 \times 10^{23}}\\\\= 4\times 17.908 \times 10^{-23}\\\\= 71.63 \times 10^{-23}\\\\ = 7.163 \times 10^{-22}\ g[/tex]
Calculating the unit cell volume:
[tex]= \frac{mass}{density} \\\\= \frac{(7.1632 \times 10^{-22})}{(10.5\ \frac{g}{cm^3})}\\\\ = 6.822 \times 10^{-23}\ \ cm^3[/tex]
[tex]\to \frac{\text{total number of unit cell}}{cm^3} \\\\ \to \frac{1\ cm^3}{6.822 \times 10^{-23}}\ cm^3 \\\\\to 1.4658 \times 10^{22} \ \frac{unit \ cell}{cm^3}[/tex]
The volume of the unit cell [tex]Ag = 6.822 \times 10^{-23}\ cm^3[/tex] (calculation above)
The edge length of the unit cell of Ag,
Formula:
the volume of a cube=[tex]a^3[/tex]
The volume of the unit cell[tex]= 6.822 \times 10^{-23}[/tex]
Calculating the side length:
[tex]\to \sqrt[3]{(6.822 \times 10^{-23})} \\\\\to (6.822 \times 10^{-23})^{(\frac{1}{3})}\\\\ \to 4.086 \times 10^{-8}\ cm[/tex]
Therefore, the final answer is "[tex]\bold{5.8633 \times 10^{22} \frac{atoms}{cm^3}, 1.4658 \times 10^{22} \ \frac{unit cells}{cm^3}, 6.822 \times 10^{-23} \ cm^3, \ and\ 4.086 \times 10^{-8}\ cm}[/tex]".
Find out more about the density here:
brainly.com/question/952755
