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noname [10]
3 years ago
10

What does it mean when the electronegativity between elements is 0?

Physics
1 answer:
Alecsey [184]3 years ago
3 0
Electronegativity<span> is a relative measure of how strongly an atom </span>will<span> attract the electrons in a bond</span>
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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
ankoles [38]

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

5 0
3 years ago
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