Answer:
Explanation:
First, we find the mass of the air originally in the tank.
Density is given as mass divided by volume. It is given as:

Therefore, mass is:

Density of air =
; Volume of the tank = 

The mass of the air initially in the tank is 7 kg.
After air is allowed to enter, the mass changes.
New density = 
The new mass will be:

We can now find the mass of air that has entered the tank:
Mass of air that entered tank = New mass of air - Original mass of air
M = 22.75 - 7.0 = 15.75 kg
The mass of air that entered the tank is 15.75 kg.
Answer:
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Answer:
Object-Oriented Software Engineering Using UML, Patterns, and Java, 3e, shows readers how to use both the principles of software engineering and the practices of various object-oriented tools, processes, and products.
(a) The number of vacancies per cubic centimeter is 1.157 X 10²⁰
(b) ρ = n X (AM) / v X Nₐ
<u>Explanation:</u>
<u />
Given-
Lattice parameter of Li = 3.5089 X 10⁻⁸ cm
1 vacancy per 200 unit cells
Vacancy per cell = 1/200
(a)
Number of vacancies per cubic cm = ?
Vacancies/cm³ = vacancy per cell / (lattice parameter)³
Vacancies/cm³ = 1 / 200 X (3.5089 X 10⁻⁸cm)³
Vacancies/cm³ = 1.157 X 10²⁰
Therefore, the number of vacancies per cubic centimeter is 1.157 X 10²⁰
(b)
Density is represented by ρ
ρ = n X (AM) / v X Nₐ
where,
Nₐ = Avogadro number
AM = atomic mass
n = number of atoms
v = volume of unit cell
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