A small portion of a crystal lattice is sketched below. What is the name of the unit cell of this lattice? Your answer must be a
word, a very short phrase, or a standard abbreviation. Spelling counts!
1 answer:
This is an incomplete question, the given sketch is shown below.
Answer : The name of given unit cell is, FCC (face-centered cubic unit cell)
Explanation :
Unit cell : It is defined as the smallest 3-dimensional portion of a complete space lattice which when repeated over the and again in different directions produces the complete space lattice.
There are three types of unit cell.
- SCC (simple-centered cubic unit cell)
- BCC (body-centered cubic unit cell)
- FCC (face-centered cubic unit cell)
In SCC, the atoms are arranged at the corners.
The number of atoms of unit cell = Z = 1
In BCC, the atoms are arranged at the corners and the body center.
The number of atoms of unit cell = Z = 2
The given unit cell is, FCC because the atoms are arranged at the corners and the center of the 6 faces.
The number of atoms of unit cell = Z = 4
Thus, the name of given unit cell is, FCC (face-centered cubic unit cell)
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Answer:
The amount of heat to absorb is 6,261 J
Explanation:
Calorimetry is in charge of measuring the amount of heat generated or lost in certain physical or chemical processes.
The total energy required is the sum of the energy to heat the ice from -20 ° C to ice of 0 ° C, melting the ice of 0 ° C in 0 ° C water and finally heating the water to 60 ° C.
So:
- Heat required to raise the temperature of ice from -20 °C to 0 °C
Being the sensible heat of a body the amount of heat received or transferred by a body when it undergoes a temperature variation (Δt) without there being a change of physical state (solid, liquid or gaseous), the expression is used:
Q = c * m * ΔT
Where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation (ΔT=Tfinal - Tinitial).
In this case, m= 10 g, specific heat of the ice= 2.1 and ΔT=0 C - (-20 C)= 20 C
Replacing: Q= 10 g*2.1 *20 C and solving: Q=420 J
- Heat required to convert 0 °C ice to 0 °C water
The heat Q necessary to melt a substance depends on its mass m and on the called latent heat of fusion of each substance:
Q= m* ΔHfusion
In this case, being 1 mol of water= 18 grams: Q= 10 g*= 3.333 kJ= 3,333 J (being kJ=1,000 J)
- Heat required to raise the temperature of water from 0 °C to 60 °C
In this case the expression used in the first step is used, but being: m= 10 g, specific heat of the water= 4.18 and ΔT=60 C - (0 C)= 60 C
Replacing: Q= 10 g*4.18 *60 C and solving: Q=2,508 J
Finally, Qtotal= 420 J + 3,333 J + 2,508 J
Qtotal= 6,261 J
<u><em> The amount of heat to absorb is 6,261 J</em></u>
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