What is the number of moles of beryllium atoms in 36 g of #Be#?

1 Answer
anor277
Mar 14, 2016

Approx. #4# #"moles"#. Thus #4xxN_A# beryllium atoms, where #N_A = "Avogadro's number"#

Explanation:

The atomic mass of berylium #=# #9.012*g*mol^-1#.

Thus #"Avogadro's number"# of beryllium atoms has a mass of just over #9# #g#.

Since #N_A = "Avogadro's number"# #=# #6.022xx10^23# #mol^-1#, the number of beryllium atoms #=#

#(36.0*cancelg)/(9.012*cancelg*cancel(mol^-1)# #xx# #6.022xx10^23*cancel(mol^-1)# #=# #??#

The answer is a dimensionless number AS REQUIRED.

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