Answer to Question #148215 in General Chemistry for Key

Question #148215

3.088 x 10^-19g Ag= ? atoms of Ag

Expert's answer

number of atoms is n*Na = m(Ag)/Ar(Ag)*Na = (3.088*10^-19 g)/(107.8682 g/mol)*(6.02*10^23 1/mol) = 1723 atoms (Na is the Avogadro constant)

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