Calculate the energy for vacancy formation in nickel (ni), given that the equilibrium number of vacancies at 850°c (1123 k) is 4.7 × 1022 m–3. the atomic weight and density (at 850°c) for ni are, respectively, 58.69 g/mol and 8.80 g/cm3.

The energy for vacancy formation can be calculated as:

Here, is equilibrium number of vacancies, N is number of atomic sites per unit vacancies, k is Boltzmann constant, T is temperature.

Here, number of atomic sites per unit vacancies can be calculated as follows:

Here, ρ is density, is Avogadro's number and A is atomic weight.

Putting the values,

Converting to

Since, 1 =

Thus,

Now, the energy for vacancy formation at 850 °C or 1123 K can be calculated using the following equation:

Rearranging,

Putting the values,

Therefore, energy for vacancy formation in nickel is

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

cytokinesis is the splitting of a cell membrane to form 2 identical daughter cells. it's similar to mitosis as both cells have identical copies of dna and cytokinesis is a step in the mitotic cycle.