A hypothetical metal crystallizes with the face-centered cubic unit cell. The radius of the metal atom is 198 picometers and its molar mass is 195.08 g/mol. Calculate the density of the metal in g/cm3.

7.38 g/cm³ is the density of the metal

Explanation:

In a Face-centered cubic unit cell you have 4 atoms. Also, the edge length is √8×r (r is radius of the atom).

To solve this problem, we need first to calculate the volume of the unit cell and then, with molar mass calculate the mass of 4 atoms. As density is the ratio between mass and volume we can obtain this value.

Volume = a³

a = √8×r

(r = 198x10⁻¹²m)

a = 5.6x10⁻¹⁰ m

Volume = 1.756x10⁻²⁸ m³

1m = 100cm → 1m³ = (100cm)³:

1.756x10⁻²⁸ m³× ((100cm)³ / 1m³) =

There are 4 atoms of gold:

4 atoms × (1mol / 6.022x10²³ atoms) = 6.64x10⁻²⁴ moles of gold

As 1 mole weighs 195.08g:

6.64x10⁻²⁴ moles of gold × (195.08g / mol) =

1.296x10⁻²¹g / 1.756x10⁻²² cm³ =

first, consult an activity series chart to determine if al is more reactive than fe. it is, so it will replace the fe. now, one can continue with the problem.

next, make sure you have a correctly balanced equation.

3feo + 2al ==> 3fe + al2o3

now, determine which reactant (feo or al) is in limiting supply.

moles feo present = 125 g x 1 mole/71.8 g = 1.74 moles

moles al present = 25.0 g x 1 mole/26.98 g = 0.927 moles

since from the balanced equation, it take 3 mol feo for every 2 mol al, al will be limiting.

you can determine this simply by dividing moles of each by their coefficient in the balanced equation. for example: 1.74/3 = 0.58 and 0.927/2 = 0.464. al is less, so it is limiting.

finally, using the stoichiometry in the balanced eq. determine mass of fe that can be produced from 0.927 moles of al.

0.927 moles al x 3 moles fe/2 moles al x 55.8 g/mole = 77.6 g

answer: they produce more slow than unicellular organisms.