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Leya [2.2K]
3 years ago
6

Calculate the unit cell edge length for an 79 wt% Ag- 21 wt% Pd alloy. All of the palladium is in solid solution, and the crysta

l structure for this alloy is FCC. Room temperature densities for Ag and Pd are 10.49 g/cm3 and 12.02 g/cm3, respectively, and their respective atomic weights are 107.87 g/mol and 106.4 g/mol. Report your answer in nanometers.
Physics
1 answer:
ankoles [38]3 years ago
5 0

Answer:

The edge length is 0.4036 nm

Solution:

As per the question:

Density of Ag, \rho = 10.49 g/cm^{3}

Density of Pd, \rho = 12.02 g/cm^{3}

Atomic weight of Ag, A = 107.87 g/mol

Atomic weight of Pd, A' = 106.4 g/mol

Now,

The average density, \rho_{a} = \frac{n A_{avg}} {V_{c}\times N_{A}}

where

V_{c} = a^{3}  = Volume of crystal lattice

a = edge length

n = 4 = no. of atoms in FCC

Therefore,

\rho_{a} = = \frac{n A_{avg}} {V_{c}\times N_{A}}

Therefore, the length of the unit cell is given as:

a = (\frac{nA_{avg}}{\rho_{a}\times N_{a}})^{1/3}            (1)

Average atomic weight is given as:

A_{avg} = \frac{100}{\frac{C_{Ag}}{A_{Ag}} + \frac{C_{Pd}}{A_{Pd}}}

where

C_{Ag} = 79 %

A_{Ag} = 107

C_{Pd} = 21%

A_{Pd} = 106

Therefore,

A_{avg} = \frac{100}{\frac{79}{107} + \frac{21}{106}} = 106.78

In the similar way, average density is given as:

\rho_{a} = \frac{100}{\frac{C_{Ag}}{\rho_{Ag}} + \frac{C_{Pd}}{\rho_{Pd}}}

\rho_{a} = \frac{100}{\frac{79}{10.49} + \frac{21}{12.02}} = 10.78 g/cm^{3}

Therefore, edge length is given by eqn (1) as:

a = (\frac{4\times 106.78}{10.78\times 6.023 X 10^23})^{1/3} = 4.036\times 10^{- 8} cm = 0.4036\times 10^{- 9} m = 0.4036 nm

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The left side of the periodic table has elements that have less number of electrons in the valence shell.

These elements loose electrons easily.These elements appear as metals or metalloids in nature.These are hard solids.Their inter molecular forces are very strong.

The right side of the periodic table has elements that have more number of electrons in the valence shell.

These elements gain electrons easily.These elements appear as non metals most of which are gases.Their inter molecular forces are weak.

8 0
2 years ago
If the object represented by the FBD below has a mass of 2.5 kg, what is the acceleration of the object?
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Answer:

4 m/s² down

Explanation:

We'll begin by calculating the net force acting on the object.

The net force acting on the object from the left and right side is zero because the same force is applied on both sides.

Next, we shall determine the net force acting on the object from the up and down side. This can be obtained as follow:

Force up (Fᵤ) = 15 N

Force down (Fₔ) = 25 N

Net force (Fₙ) =?

Fₙ = Fₔ – Fᵤ

Fₙ = 25 – 15

Fₙ = 10 N down

Finally, we shall determine the acceleration of the object. This can be obtained as follow:

Mass (ml= 2.5 Kg

Net force (Fₙ) = 10 N down

Acceleration (a) =?

Fₙ = ma

10 = 2.5 × a

Divide both side by 2.5

a = 10 / 2.5

a = 4 m/s² down

Therefore, the acceleration of the object is 4 m/s² down

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3 years ago
g A box having mass 0,5kg is placed in front of a 20 cm compressed spring. When the spring released, box having mass m1, collide
vagabundo [1.1K]

Answer:Expression given below

Explanation:

Given mass of spring\left ( m_1\right )=0.5 kg

Compression in the spring\left ( x\right )=20 cm

Let the spring constant be K

Using Energy conservation

potential energy stored in spring =Kinetic energy of Block\left ( m_1\right )

\frac{1}{2}Kx^2=\frac{1}{2}m_1v^2

v=x\sqrt{\frac{k}{m_1}}

now conserving momentum

m_1v=\left ( m_1+m_2\right )v_0

v_0=\frac{m_1}{m_1+m_2}v

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3 years ago
The current supplied by a battery as a function of time is I(t) = (0.88 A) e^(-t*6 hr). What is the total number of electrons tr
Degger [83]

Answer:

e. 1.2 x 10²³

Explanation:

According to the problem, The current equation is given by:

I(t)=0.88e^{-t/6\times3600s}

Here time is in seconds.

Consider at t=0 s the current starts to flow due to battery and the current stops when the time t tends to infinite.

The relation between current and number of charge carriers is:

q=\int\limits {I} \, dt

Here the limits of integration is from 0 to infinite. So,

q=\int\limits {0.88e^{-t/6\times3600s}}\, dt

q=0.88\times(-6\times3600)(0-1)

q = 1.90 x 10⁴ C

Consider N be the total number of charge carriers. So,

q = N e

Here e is electronic charge and its value is 1.69 x 10⁻¹⁹ C.

N = q/e

Substitute the suitable values in the above equation.

N= \frac{1.9\times10^{4} }{1.69\times10^{-19}}

N = 1.2 x 10²³

8 0
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Two 10cm diameter metal disks separated by a 0.63mm thick piece of pyrex glass are charged to a potential difference of 1000V. D
inessss [21]
Parallel-plate capacitor has there fore formula is

<span>C=(<span>ϵ0</span>A)/d
putting values</span>C=(8.85*10^-12*pi*.05^2)/.00063
=1.1*10^-10F then Q=CV=1.1*10^-10*1000=1.1*10^-7C 
as
<span>η=Q/A</span><span>therefore
(1.1*10^-7)/(pi*.05^2)
=1.4*10^-5C/m^ our answer
hope this helps</span>
5 0
3 years ago
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