Calculate the unit cell edge length for an 78 wt% Ag- 22 wt% Pd alloy. All of the palladium is in solid solution, and the crystal 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.

The unit cell edge length for the alloy is 0.405 nm

Explanation:

Given;

concentration of Ag, = 78 wt%

concentration of Pd, = 22 wt%

density of Ag = 10.49 g/cm³

density of Pd = 12.02 g/cm³

atomic weight of Ag, = 107.87 g/mol

atomic weight of iron, = 106.4 g/mol

Step 1: determine the average density of the alloy

Step 2: determine the average atomic weight of the alloy

Step 3: determine unit cell volume

for a FCC crystal structure, there are 4 atoms per unit cell; n = 4

Step 4: determine the unit cell edge length

Vc = a³

= 0.405 nm

Therefore, the unit cell edge length for the alloy is 0.405 nm

a  solution  is always transparent, light passes through with no scattering from solute particles which are molecule in size. the  solution  is homogeneous and does not settle out. a colloid is intermediate between asolution  and a  suspension. while asuspension  will separate out a colloid will not.

gas

explanation:

smoke is the matter of a gas