Asked by roxanne
                The element Sc has hcp packing with a hexagonal unit cell. The density of scandium is 3.00E3 kg/m3 and the cell volume is 5.00E-26 L. Calculate the value of Avogadro's number to 3 significant figures based on these data. Note: the value may differ from the listed value.
            
            
        Answers
                    Answered by
            DrBob222
            
    I could convert 3.00E3 kg/m^3 to density of ?? g/cc.
Then mass of unit cell is volume x density = xx grams.
The hexagonal close packing produces 2 atoms/unit cell; therefore, that is the mass for 2 atoms. Divide by 2 to obtain the mass of 1 atom.
Then look up the mass of Sc from the periodic table and that mass is the mass of a mole which contains Avogadro's number of atoms.
Then,
N = atomic mass x (1 atom/?? g) = xx atoms. I worked through this very fast and obtained about 6 x 10^23.
Check my thinking. Check my work.
    
Then mass of unit cell is volume x density = xx grams.
The hexagonal close packing produces 2 atoms/unit cell; therefore, that is the mass for 2 atoms. Divide by 2 to obtain the mass of 1 atom.
Then look up the mass of Sc from the periodic table and that mass is the mass of a mole which contains Avogadro's number of atoms.
Then,
N = atomic mass x (1 atom/?? g) = xx atoms. I worked through this very fast and obtained about 6 x 10^23.
Check my thinking. Check my work.
                    Answered by
            roxanne
            
    My answer that i got is 5.99E22
i followed your step all the way till the periodic table..
from there i got Sc atomic mass as 44.96 g
then i multiply it by i atom/7.50E-22
then i got the answer 5.99E22....is my steps and answer correct?
    
i followed your step all the way till the periodic table..
from there i got Sc atomic mass as 44.96 g
then i multiply it by i atom/7.50E-22
then i got the answer 5.99E22....is my steps and answer correct?
                    Answered by
            drbob222
            
    yes. That's the answer I obtaied also. On the others you posted above, I didn't go through any of them in detail. They follow the same general principles. But if you have trouble, post your work and tell us in detail about what you don't understand.
    
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