Asked by Amanda
How many atoms of aluminum [Al] are present in a sample of aluminum carbonate [AlCar] weighing 3.45E6 amu?
Answers
Answered by
DrBob222
mass Al2(CO3)3 = 3.45E6 amu x (1.66E-24 g/amu) = ? grams.
mols Al2(SO4)3 = grams/molar mass = ?
mols Al = mols Al2(CO3)3 x 2 since there are two mols Al in 1 mol Al2(SO4)3.
#atoms = mols Al x (6.022E23 atoms/mol) =?
mols Al2(SO4)3 = grams/molar mass = ?
mols Al = mols Al2(CO3)3 x 2 since there are two mols Al in 1 mol Al2(SO4)3.
#atoms = mols Al x (6.022E23 atoms/mol) =?
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