Answer to Question #325731 in General Chemistry for Kayla Cook

General Chemistry

Question #325731

How many aluminum atoms are in 4.22 g of aluminum?

Expert's answer

N = n * N_A

n(Al) = m(Al)/M(Al)

n(Al) = 4.22 g / 27g/moll = 0.156 moll

N = 0.156 * 6.02*10²³ = 9.39*10²³atoms

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