The unit cell of a metallic element of atomic mass (`108 g //"mole"`) and density `10.5 g//cm^(3)` is a cube with edge length of 409 pm. The structure of the crystal lattice would be:

The unit cell of an element of atomic mass 108 and density '10.5 . g cm^-3' is a cube with edge 'length '409 pm'. Find the structure of the.crystal lattice 'N_A=6.022 xx 10^33'

The unit cell of an element having atomic mass 100 and density 12 g cm^(-3) is a cube with edge length 300 pm, the structure of crystal lattice is

The unit of an element of atomic mass 96 and density 10.3 g cm^(-3) is a cube with edge length of 314 pm. Find the structure of the crystal lattice (simple cubic, FCC or BCC) (Avogadro's constant , N_0 = 6.023 xx 10^(23) mol^(-1))

The unit cell of an element of atomic mass 108 u and density 10.5 g cm^(-3) is cube with edge length 409 pm. Find the type of unit cell of the crystal. (Given : Avogadro's constant = 6.023 xx 10^(23) mol)

The unit cell of an element of atomic mass 96 and density 10.3 gm cm^-3 and edge length of 314 pm. Find the structure of crystal lattice (simple cube, fcc or bcc) (NA-6.023xx10^23 mol^-1)

An element has atomic mass 93 g mol^(-1) and density 11.5 g cm^(-3) . If the edge length of its unit cell is 300 pm, identify the type of unit cell.

The density of an cubic unit cell is 10.5 g cm^(-3) . If the edge length of the unit cell is 409 pm. Find the structure of the crystal lattice (Atomic mass = 108).