Question

A 25.0 mm x 40.0 mm piece of gold foil is 0.25 mm thick. The density of gold is 19.32 g/cm?
How many gold atoms are in the sheet? (Atomic weight : Au = 1970)

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Answer 1

Answer:

V= 2.5 * 4 * .025 = .25 cm^3

19.32 g/cm^3 * .25 cm^3 = 4.83 g     mass of the foil

Since the gram-molecular weight of gold is 197

N = 4.83 / 197 = .0245 moles    the number of moles of gold

n = N * A    where A is Avagadro's number in gm-moles

So n = .0245 * 6.02 * 10E23 = 1.48 * 10E22    atoms

Answer 2

Answer

1.47*10^22

HOPE ITS HELP...