This table compares magnitude and depth of four earthquakes occurring between 1955 - 2017. The surface changes that occur as a r
esult of earthquakes begin in Earth's crust. What conclusion can be drawn about earthquake activity within Earth's crust based on the data shown?
A) Earthquake depth has continuously increased throughout the past century.
B) Higher-magnitude earthquakes do not always happen deeper in Earth's crust.
C) The higher an earthquake's magnitude, the deeper in Earth's crust it occurs.
D) Earthquake magnitude has continuously decreased throughout the past century.
A) Earthquake depth has continuously increased throughout the past century.
B) Higher-magnitude earthquakes do not always happen deeper in Earth's crust.
C) The higher an earthquake's magnitude, the deeper in Earth's crust it occurs.
D) Earthquake magnitude has continuously decreased throughout the past century.
1 answer:
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Answer:
a
The radius of an impurity atom occupying FCC octahedral site is
b
The radius of an impurity atom occupying FCC tetrahedral site is .
Explanation:
In order to get a better understanding of the solution we need to understand that the concept used to solve this question is based on the voids present in a unit cell. Looking at the fundamentals
An impurity atom in a unit cell occupies the void spaces. In FCC type of structure, there are two types of voids present. First, an octahedral void is a hole created when six spheres touch each other usually placed at the body center. On the other hand, a tetrahedral void is generated when four spheres touch each other and is placed along the body diagonal.
Step 1 of 2
(1)
The position of an atom that fits in the octahedral site with radius is as shown in the first uploaded image.
In the above diagram, R is the radius of atom and a is the edge length of the unit cell.
The radius of the impurity is as follows:
The relation between radius of atom and edge length is calculated using Pythagoras Theorem is shown as follows:
Consider as follows:
Substitute as
and
as a and
as a in above equation as follows:
Substitute value of aa in equation (A) as follows:
The radius of an impurity atom occupying FCC octahedral site is
Note
An impure atom occupies the octahedral site, the relation between the radius of atom, edge length of unit cell and impure atom is calculated. The relation between the edge length and radius of atom is calculated using Pythagoras Theorem. This further enables in finding the radius of an impure atom.
Step 2 of 2
(2)
The impure atom in FCC tetrahedral site is present at the body diagonal.
The position of an atom that fits in the octahedral site with radius rr is shown on the second uploaded image :
In the above diagram, R is the radius of atom and a is the edge length of the unit cell.
The body diagonal is represented by AD.
The relation between the radius of impurity, radius of atom and body diagonal is shown as follows:
In
For calculation of AD, AB is determined using Pythagoras theorem.
Substitute as a and
as a in above equation as follows:
Also,
Substitute value of for
in equation (1) as follows:
Therefore, the length of body diagonal is calculated using Pythagoras Theorem in as follows:
Substitute as
a and
as a in above equation as follows:
For calculation of radius of an impure atom in FCC tetrahedral site,
Substitute value of AD in equation (B) as follows:
Substitute a as in above equation as follows:
The radius of an impurity atom occupying FCC tetrahedral site is .
Note
An impure atom occupies the tetrahedral site, the relation between the radius of atom, edge length of unit cell and impure atom is calculated. The length of body diagonal is calculated using Pythagoras Theorem. The body diagonal is equal to the sum of the radii of two atoms. This helps in determining the relation between the radius of impure atom and radius of atom present in the unit cell.
☃️ Chemical formulae ➝
For solving this question, We need to know how to find moles of solution or any substance if a certain weight is given.
Atomic weight of elements:
Ca = 40
C = 12
O = 16
❍ Molecular weight of
= 40 + 12 + 3 × 16
= 52 + 48
= 100 g/mol
❍ Given weight: 10 g
Then, no. of moles,
⇛ No. of moles = 10 g / 100 g mol‐¹
⇛ No. of moles = 0.1 moles
☄ No. of moles of Calcium carbonate in that substance = <u>0.1 moles</u>
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