Respuesta :
Answer:
a) Four atoms.
b) 9.23 x 10⁻²³ cm³ .
c) 1.34 x 10⁻²² g .
d) 80.7 amu .
Explanation:
a) In the face-centered cubic cell, there are spheres (which represent atoms or molecules) at the center of each of the six faces of the cube, in addition to eight corner spheres. A face-centered cubic cell contains the equivalent of four complete spheres or atoms—three from the six face-centered atoms and one from the eight shared corner spheres.
b) The volume of a cube is [tex]V = a³[/tex], a being the edge length.
Therefore V= (4.52 x 10⁻⁸ cm)³ = 9.23 x 10⁻²³ cm³
c) We know that δ= m ÷ v
In the problem we are given the density and we just calculated the volume. We rearrange the equation to give:
m= δ × v = 9.23 × 10⁻²³ × 1.45 g/cm⁻³ = 1.34 x 10⁻²² g
d) We know the molar mass of carbon-12 is 12.00 g and there are 6.022 × 10²³ carbon-12 atoms in 1 mole of the substance; therefore, the mass of one carbon-12 atom is given by :
12.00 g carbon-12 atoms ÷ 6.022 × 10²³ carbon-12 atoms= 1.993 x 10⁻²³ g
We can use this result to determine the relationship between atomic
mass units and grams. Because the mass of every carbon-12 atom is exactly 12 amu, the number of atomic mass units equivalent to 1 gram is
amu/gram= 12 amu/1 carbon-12 atom × 1 carbon-12 atom/ 1.993 x 10⁻²³ g
amu/ gram= 6.022 x 10²³ amu/g
Thus, 1 g= 6.022 x 10²³ amu, and 1 amu = 1.661 x10⁻²⁴ g
To calculate the approximate atomic mass, we use the last conversion factor:
1.34 x 10⁻²² g × 1 amu/ 1.661 x10⁻²⁴ g = 80.7 amu
In the given fcc element, a. the number of atoms is 4. b. The volume of a unit cell is [tex]\rm 9.23\;\times\;10^-^2^3\;cm^3[/tex]. c. Mass of unit cell is [tex]\rm 1.34\;\times\;10^-^2^2\;g[/tex]. d. The approximate atomic mass of the element is 80.7 amu.
- The face-centered cubic lattice has 3 atoms from the 6 faces, and 1 atom from the eight corners. Thus, the total atoms in the face-centered lattice are four.
- The face-centered lattice has been a cube.
The volume of cube = [tex]\rm (edge)^3[/tex]
The volume of unit cell = [tex]\rm (4.52\;\times\;10^-^8\;cm)[/tex]
The volume of unit cell = [tex]\rm 9.23\;\times\;10^-^2^3\;cm^3[/tex]
- The mass of a unit cell can be calculated from density. Mass can be defined as the ratio of volume to density.
Mass = [tex]\rm \dfrac{volume}{density}[/tex]
Mass of unit cell = [tex]\rm \dfrac{9.23\;\times\;10^-^2^3\;cm^3}{1.45\;g\;cm^-^3}[/tex]
Mass of unit cell = [tex]\rm 1.34\;\times\;10^-^2^2\;g[/tex].
- The approximate atomic mass of the element can be calculated by the mass of the carbon atom.
Mass of 1 carbon atom = [tex]\rm \dfrac{mass\;of\;1\;mole\;carbon}{number\;of\;atoms\;in\;1\;mole\;Carbon}[/tex]
Mass of 1 carbon atom = [tex]\rm \dfrac{12}{6.023\;\times\;10^2^3}[/tex]
Mass of 1 carbon atom = 1.992 [tex]\rm \times\;10^-^2^3[/tex] grams.
atomic mass unit per gram can be given as;
amu/gram = [tex]\rm \dfrac{12}{1.992\;\times\;10^-^2^3}[/tex]
amu/gram = [tex]\rm 6.022\;\times\;10^2^3[/tex] amu/gram
1 gram = [tex]\rm 6.022\;\times\;10^2^3[/tex] amu
1 amu = 1.661 [tex]\rm \times\;10^-^2^4[/tex] gram.
The average atomic mass = mass of unit cell [tex]\times[/tex] amu\gram
= [tex]\rm 1.34\;\times\;10^-^2^2\;g[/tex]. [tex]\times[/tex] 1 amu/ 1.661 [tex]\rm \times\;10^-^2^4[/tex] gram.
= 80.7 amu.
In the given fcc element, a. the number of atoms is 4. b. The volume of a unit cell is [tex]\rm 9.23\;\times\;10^-^2^3\;cm^3[/tex]. c. Mass of unit cell is [tex]\rm 1.34\;\times\;10^-^2^2\;g[/tex]. d. The approximate atomic mass of the element is 80.7 amu.
For more information about the face-centered cubic lattice, refer to the link:
https://brainly.com/question/14578576
