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denis-greek [22]
3 years ago
14

While there are many ways to solve this problem, one strategy is to calculate the volume of any metal's unit cell given its theo

retical density (Equation 3.8) and atomic weight. What is the volume of the zirconium unit cell in cubic meters?
Engineering
1 answer:
IgorC [24]3 years ago
8 0

Answer:

V_{c}=1.396x10^{-28}  m^{3} /unit.cell

Explanation:

z = number of atoms

M = Molar mass of zirconium

N = Avogadro’s number

Vc = volume of zirconium unit cell

d = density

z=12x\frac{1}{6}+2x\frac{1}{2}+3=6

z = 6 atoms per unit cell

M = 91.224 g/mol

N = 6.023x10^{23}  atoms/mol

d = 6.51g/cm^{3}

V_{c}=\frac{zxM}{dxN}

V_{c}=\frac{6x(91.224g/mol)}{(6.51g/cm^{3}) x(6.023x10^{23}atoms/mol) }

V_{c}=1.396x10^{-22}  cm^{3} /unit.cell

V_{c}=1.396x10^{-28}  m^{3} /unit.cell

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The domain of discourse is the members of a chess club. The predicate B(x, y) means that person x has beaten person y at some po
satela [25.4K]

Answer:A. No one has ever beat Nancy.

Explanation:

The dormain of discourse in a simple language is the set of entities upon which our discussions are based when discussing about something.

The dormain of discourse is also known simply as universe, can also be said to be a set of entities o

upon which certain variables of interest in some formal treatment may range.

The dormain of discourse is generally attributed to Augustus De Morgan, it was also extensively used by George Boole in his Laws of Thought.

THE LOGICAL UNDERSTANDING OF THE THE QUESTION IS THAT NO ONE HAS EVER BEAT NANCY.

8 0
3 years ago
A piece of aluminum wire is 500 ft long and has a diameter of 0.03 inches. What is the resistance of the piece of wire?​
dexar [7]

Answer:

8.85 Ω

Explanation:

Resistance of a wire is:

R = ρL/A

where ρ is resistivity of the material,

L is the length of the wire,

and A is the cross sectional area.

For a round wire, A = πr² = ¼πd².

For aluminum, ρ is 2.65×10⁻⁸ Ωm, or 8.69×10⁻⁸ Ωft.

Given L = 500 ft and d = 0.03 in = 0.0025 ft:

R = (8.69×10⁻⁸ Ωft) (500 ft) / (¼π (0.0025 ft)²)

R = 8.85 Ω

5 0
3 years ago
Let f(t) be an arbitrary signal with bandwidth Ω. Determine the minimum sampling frequencies ωs needed to sample the following a
disa [49]

Answer:

See explaination

Explanation:

We can describr Aliasing as a false frequency which one get when ones sampling rate is less than twice the frequency of your measured signal.

please check attachment for the step by step solution of the given problem.

7 0
3 years ago
A 35-ft³ rigid tank has propane at 25 psia, 540 R and is connected by a valve to another tank of 20 ft³ with propane at 40 psia,
gulaghasi [49]

Answer:

final pressure = 200KPa or 29.138psia

Explanation:

The detailed step by step calculations with appropriate conversion factors applied are as shown in the attachment.

8 0
3 years ago
3.94 x 105) + (2.04 x 105)
Flura [38]
627.9 is the answer
6 0
2 years ago
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