Answer:
The melting and boiling point of water in Fahrenheit scale is divided into 180 equal intervals, that means each division denoting 1°F.
division means their melting and boiling point is divided into 180 equal intervals that means each division indicates 1 degree F.
Answer:
The edge length is 0.4036 nm
Solution:
As per the question:
Density of Ag, ![\rho = 10.49 g/cm^{3}](https://tex.z-dn.net/?f=%5Crho%20%3D%2010.49%20g%2Fcm%5E%7B3%7D)
Density of Pd, ![\rho = 12.02 g/cm^{3}](https://tex.z-dn.net/?f=%5Crho%20%3D%2012.02%20g%2Fcm%5E%7B3%7D)
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}}](https://tex.z-dn.net/?f=%5Crho_%7Ba%7D%20%3D%20%5Cfrac%7Bn%20A_%7Bavg%7D%7D%20%7BV_%7Bc%7D%5Ctimes%20N_%7BA%7D%7D)
where
= 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}}](https://tex.z-dn.net/?f=%5Crho_%7Ba%7D%20%3D%20%3D%20%5Cfrac%7Bn%20A_%7Bavg%7D%7D%20%7BV_%7Bc%7D%5Ctimes%20N_%7BA%7D%7D)
Therefore, the length of the unit cell is given as:
(1)
Average atomic weight is given as:
![A_{avg} = \frac{100}{\frac{C_{Ag}}{A_{Ag}} + \frac{C_{Pd}}{A_{Pd}}}](https://tex.z-dn.net/?f=A_%7Bavg%7D%20%3D%20%5Cfrac%7B100%7D%7B%5Cfrac%7BC_%7BAg%7D%7D%7BA_%7BAg%7D%7D%20%2B%20%5Cfrac%7BC_%7BPd%7D%7D%7BA_%7BPd%7D%7D%7D)
where
= 79 %
= 107
= 21%
= 106
Therefore,
![A_{avg} = \frac{100}{\frac{79}{107} + \frac{21}{106}} = 106.78](https://tex.z-dn.net/?f=A_%7Bavg%7D%20%3D%20%5Cfrac%7B100%7D%7B%5Cfrac%7B79%7D%7B107%7D%20%2B%20%5Cfrac%7B21%7D%7B106%7D%7D%20%3D%20106.78)
In the similar way, average density is given as:
![\rho_{a} = \frac{100}{\frac{C_{Ag}}{\rho_{Ag}} + \frac{C_{Pd}}{\rho_{Pd}}}](https://tex.z-dn.net/?f=%5Crho_%7Ba%7D%20%3D%20%5Cfrac%7B100%7D%7B%5Cfrac%7BC_%7BAg%7D%7D%7B%5Crho_%7BAg%7D%7D%20%2B%20%5Cfrac%7BC_%7BPd%7D%7D%7B%5Crho_%7BPd%7D%7D%7D)
![\rho_{a} = \frac{100}{\frac{79}{10.49} + \frac{21}{12.02}} = 10.78 g/cm^{3}](https://tex.z-dn.net/?f=%5Crho_%7Ba%7D%20%3D%20%5Cfrac%7B100%7D%7B%5Cfrac%7B79%7D%7B10.49%7D%20%2B%20%5Cfrac%7B21%7D%7B12.02%7D%7D%20%3D%2010.78%20g%2Fcm%5E%7B3%7D)
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](https://tex.z-dn.net/?f=a%20%3D%20%28%5Cfrac%7B4%5Ctimes%20106.78%7D%7B10.78%5Ctimes%206.023%20X%2010%5E23%7D%29%5E%7B1%2F3%7D%20%3D%204.036%5Ctimes%2010%5E%7B-%208%7D%20cm%20%3D%200.4036%5Ctimes%2010%5E%7B-%209%7D%20m%20%3D%200.4036%20nm)
There are approximately 3 different types of atoms that are present in one molecule of aluminum hydroxide, AI(OH)3.