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How many moles of atoms are in 3.00 g of ##""^13C##?
There are ##3/13xxN_A" carbon atoms"##, where ##N_A="Avogadro's Number, "6.022xx10^23*mol^-1##
In ##13*g## of ##""^(13)C## there are precisely ##"Avogadro's number of atoms"##.You have specified ##3.00*g##, thus there are ##(3.00*g)/(13.00*g*mol)xx6.022xx10^23*mol^-1## ##=## ##??""^13C " atoms"##