What is the density of iron if it crystallizes in a body-centered cubic unit cell with an edge length of 287 pm

Answer

Density = 7.87g/cm^3

Explanation:

Density is the ratio of mass of the given object to the volume of the object, in this question iron is the given object, then we make use of atomic number of iron

Given:

Length= 287pm = 287*10^-10cm

Atomic mass of Fe= 56.0u

Z=2(for body centered cubic unit cell)

Avogadro number (N 0)=6.022× 10^23

Density= ZM/a^3 × N

Where

Z= body centered cubic unit cell

Then substitute

N= Avogadro's number

a=Length

Density = (2× 56)/(287*10^-10cm)^3 × (6.022 × 10^23)

Density = 7.87g/cm^3

assuming that your periodic table has eighteen columns, you should highlight   the four elements in the third column from the left of the table.

neutral atoms of elements after calcium contain d electrons. the energy of d orbitals of a particular main shell lies between the energy of the s and p orbitals of the next main shell. for example, electrons in the have energy higher than those in the orbital while lower than those of the orbital.

by the aufbau principle, electrons fill orbitals one at a time, starting from those of the lowest energy. the orbital is therefore filled right after orbitals and before orbitals.  

potassium has electron configuration and calcium . the s orbital of a main shell holds a maximum of two electrons. the next element would thus have the electron configuration and satisfy the requirements.

a neutral atom of the element contains three electrons on top of the most recent noble gas element. it should thus be placed in the third column of the periodic table. both scandium and yttrium would satisfy the conditions. an aufbau diagram indicates that the 4f orbital is filled before the 5d. however, it appears that exceptions exists for both lanthanum and actinium. neutral atoms of the two d-block elements also demonstrate the and should be highlighted as well.