Calculate the radius of a silver atom in cm, given that ag has an fcc crystal structure, a density of 10.5 g/cm3, and an atomic weight of 107.87 g/mol.

For solid cubic crystals, the equation for density is:

ρ = nA/VcNa
where
n is 4 atoms per unit cell of FCC structure
A is atomic weight (107.87 g/mol)
Vc is the volume of the cubic cell
Na is Avogadro's number (6.022×10²³ atoms/mol)

10.5 g/cm³ = [(4)(107.87 g/mol)]/[Vc(6.022×10²³ atoms/mol)]
Solving for Vc,
Vc = 6.823707×10⁻²³ cm³

The volume is equal to (2r√2)³. Thus,
6.823707×10⁻²³ cm³ = (2r√2)³
Solving for r,
r = 1.445×10⁻⁸ cm

A. Increasing the temperature will favor forward reaction and more CaCo3 formed.

B. More CaCo3 will be formed.

C. CaCo3 will decrease and more react ants formed.

D. Less CaCo3 will be formed.

E. Iridium is a catalyst so there is no effect

Explanation:

A. Temperature will increase because it's an endothermic reaction.

B. Adding Cao will favor forward reaction and more CaCo3 formed.

C. Removing methane, more react ants are formed and CaCo3 decreases.

D. Irridi is a catalyst so it has no effect on the CaCo3 but only speeds its rate of reaction.