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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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soldier1979 [14.2K]

Answer:

Determine A) The Volume Of The Tank (ft^3) Later A Pump Is Used To Extract ... A rigid, sealed cylinder initially contains 100 lbm of water at 70 degrees F and atmospheric pressure. ... Later a pump is used to extract 10 lbm of water from the cylinder. The water remaining in the cylinder eventually reaches thermal equilibrium ...

3 0
3 years ago
How did the universe begin
stepan [7]

Answer:

big bang

Explanation:

the universe began from big bang

5 0
3 years ago
to determine cam ring speed you must use a ___________________,____________________or a________________
saul85 [17]

Answer:

=> The total numbers of cylinders on the engine.

=> Total number of lobes in the cam ring.

=> The direction at which the cam ring rotates.

Explanation:

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5 0
3 years ago
Suppose that the cost of electrical energy is $0.15 per kilowatt hour and that your electrical bill for 30 days is $80. Assume t
mr_godi [17]

Answer:

a) 740 W b) 6.2 A c) 8.1%

Explanation:

We need first to get the total energy spent during the 30 days, that can be calculated as follows:

1 month = 30 days = 720 hr

If the total cost amounts $80 (for 720 hr), and the cost per kwh is 0.15, we have:

80 $/mo  =  0.15 $/Kwh*x Kwh/mo

Solving for the total energy spent in the month:

x (kwh/mo) = \frac{80}{0.15} = 533.3 kwh

Assuming that the power delivered is constant over the entire 30 days, as power is the rate of change of energy, we can find the power as follows:

P = \frac{E}{t} = \frac{533.3 kWh}{720 h}  = 0.74 kW = 740 W

b) If the power is supplied by a voltage of 120 V, we can find the current I as follows:

I =\frac{P}{V} =\frac{740W}{120V} = 6.2A

c) If part of the electrical load is a 60-W light, we can substract this power from the one we have just found, as follows:

P = 740 W - 60 W = 680 W

The new value of the energy spent during the entire month will be as follows:

E = 0.68 kW*24(hr/day)*30(days/mo) = 490 kWh/mo

The reduction in percentage regarding the total energy spent can be calculated as follows:

ΔE = \frac{533.3-490}{533.3} = 0.081*100= 8.1%

⇒ ΔE(%) = 8.1%

%(E) =\frac{533.3-490}{533.3}  = 0.081 * 100 = 8.1%

4 0
4 years ago
Phân tích phương pháp gia công plasma
Irina-Kira [14]

Answer:

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