Dell
Nov9-09, 03:26 PM
find the length of the cube "a" of the lattice of iron Fe at room temperature if the density of iron is 7.87g/cm3
i know that 7.87=m/V = m/a^3
a=(m/7.87)^1/3
Mw=55.845
a=[(n*55.845)/(NA*7.87)]^1/3
have i done this correctly?
but how do i know how many atoms there are, what kind of lattice is FE? BCC FCC SC ...??
i know that 7.87=m/V = m/a^3
a=(m/7.87)^1/3
Mw=55.845
a=[(n*55.845)/(NA*7.87)]^1/3
have i done this correctly?
but how do i know how many atoms there are, what kind of lattice is FE? BCC FCC SC ...??