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disa [49]
3 years ago
9

The lattice parameter of copper is 0.362 nanometer. The atomic weight of copper is 63.54 g/mole. Copper forms a fcc structure. A

nswer the following questions.
a. Volume the unit cell in cubic centimeters in cubic centimeters is:______
b. Density of copper in g/cm^3 is:________
Engineering
1 answer:
Hatshy [7]3 years ago
3 0

Answer:

a) 4.74 * 10^-23 cm^3

b)  8.9 g/cm^3

Explanation:

Given data :

Lattice parameter of copper = 0.362 nM

Atomic weight of copper = 63.54 g/mole

Given that copper forms a fcc structure

<u>a) Calculate the volume of the unit cell</u>

V = a^3

  = ( 0.362 * 10^-7  cm )^3 = 4.74 * 10^-23 cm^3

<u>b) Calculate density of copper in g/cm^3 </u>

Density = ( n*A ) / ( Vc * NA) ----------- ( 1 )

where: NA = Avogadro's number = 6.022 * 10^23  atoms/mol

n = number of atoms per unit cell = 4

A = atomic weight = 63.54 g/mol

Vc = volume of unit cell =  4.74 * 10^-23 cm^3

Input values into equation 1

Density = 8.9 g/cm^3

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A tank has a gauge pressure of 552 psi. The cover of an inspection port on the tank has a surface area of 18 square inches. What
Reptile [31]

Answer:

44197.55 N

Explanation:

From the question,

Pressure of the pressure guage (P) = Total force experienced by the cover (F)/Area of the cover (A)

P = F/A................ Equation 1

make F the subeject of the equation

F = P×A............... Equation 2

Given: P = 552 psi = (552×6894.76) = 3805907.52 N/m², A = 18 square inches = (18×0.00064516) = 0.01161288 m²

Substitute these values into equation 2

F = ( 3805907.52×0.01161288)

F = 44197.55 N

3 0
4 years ago
You are the project manager assigned to construct a new 10-story office building. You are trying to estimate the costs for this
Semmy [17]

Answer:

Bottom-up Estimation

Explanation:

Bottom-up estimation is a type of project cost estimation that considers the cost of individual project activities and finally sums them up or finds the aggregates. The summation gives an idea of what the entire project will cost.

This is an effective way of estimating the cost of a project as it evaluates the costs on a wholistic basis. It also considers the tiniest details during the estimation process. The process moves from the simpler details to the more complicated details.

8 0
3 years ago
A spherical balloon with a diameter of 9 m is filled with helium at 20°C and 200 kPa. Determine the mole number and the mass of
SIZIF [17.4K]

Answer:

<em>number of mole is 31342.36 moles</em>

<em>mass is 125.369 kg</em>

<em></em>

Explanation:

Diameter of the spherical balloon d = 9 m

radius r = d/2 = 9/2 = 4.5 m

The volume pf the sphere balloon ca be calculated from

V = \frac{4}{3} \pi r^3

V = \frac{4}{3}* 3.142* 4.5^3 = 381.75 m^3

Temperature of the gas T = 20 °C = 20 + 273 = 293 K

Pressure of the helium gas = 200 kPa = 200 x 10^3 Pa

number of moles n = ?

Using

PV = nRT

where

P is the pressure of the gas

V is the volume of the gas

n is the mole number of the gas

R is the gas constant = 8.314 m^3⋅Pa⋅K^−1⋅mol^−1

T is the temperature of the gas (must be converted to kelvin K)

substituting values, we have

200 x 10^3 x 381.75 = n x 8.314 x 293

number of moles n  = 76350000/2436 = <em>31342.36 moles</em>

We recall that n = m/MM

or m = n x MM

where

n is the number of moles

m is the mass of the gas

MM is the molar mass of the gas

For helium, the molar mass = 4 g/mol

substituting values, we have

m = 31342.36 x 4

m = 125369.44 g

m =<em> 125.369 kg</em>

3 0
3 years ago
A flux used for welding, brazing, or soldering prevents the formation of, dissolves, or helps remove?
OverLord2011 [107]

Answer:

Helps to remove Oxides formed during brazing.

Explanation:

Flux is needed to dissolve and remove oxides that may form during brazing. Prevent or inhibit the formation of oxide during the brazing process

8 0
2 years ago
Using the AASHTO procedure, determine the thickness required for a base and a surface layer over existing subgrade. The structur
garri49 [273]

Answer:

<u>the thickness required would be 12 inch HMA and granular base layer of 6 inches</u>

Explanation:

structural number = 4.5

stone base course material coefficient = 0.13

hma material layer coefficient = 0.40

drainage coefficient = 0.90

we will use layered analysis procedure to get thickness

D1 >= sN1/a1

when we cross multiply,

sN1 = a1D1 >=sN1

D2 >= -sN2-sN1/a2m2

sN2* + sN1* >= sN2

D3 >= sN3-(sN1*+sN2*)/a2m2

where a1,a2,a3 = layer coefficient

d1 d2 d3 = actual thickness

m2,m3 = coefficient of base

a1 = 0.4

a1 = 0.13

sN = 4.5

m2 = 0.9

D1 >= sN1/a1 = 4.5/0.4

= 11.25

thickness of surface = 12 inches

a1D1 = 0.4x12 = 4.8

we have value of sN2 = 5.5

(5.5 -4.8)/(0.13*0.9)

= 0.7/0.117

= 5.9829 inches

approximately 6 inches

so the pavement will have 12inch HMA surface and 6 inches granular base layer.

7 0
3 years ago
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