Calculate the unit cell edge length for an 87 wt% Ag- 13 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. nm

Explanation:

   Since the crystal structure is FCC , every unit cell will have 4 atoms . If a be the edge length

average density = 4 x average atomic no / volume of unit cell x N

N is avogadro no or 6.02 x 10²³

average atomic no

= 100 / (87/107.9 +13/106.4) = 107.72

average density = 100 / (87/10.49 + 13/12.02)

= 10.667

10.667 = (4 x 107.72 )/ (a³x 6.02 x 10²³)

a³ = 4 x 107.72 / (10.667 x  6.02 x 10²³)

= 6.71 x 10⁻²³

a = 4.063 x 10⁻⁸ m

question: ray weighs 100 kilograms, his wrestling partner, bobby, weighs 50 kilograms. if ray moves at 4 m s , how fast will bobby have to run at him to apply an equal amount of kinetic energy?

answer: 8 m/s he needs to run twice as fast to keep up so he has some work to do

explanation:   if the mass has units of kilograms and the velocity of meters per second, the kinetic energy has units of kilograms-meters squared per second squared.

question answered by

(jacemorris04)

under a microscope, cancer cells may look very different from normal cells. they often have different sizes and some may be larger than normal while others are smaller. cancer cells are often abnormally shaped and the control centre of the cell (the nucleus) may have an abnormal appearance.

i think? ?