Answer:
The answer to your question is 3.69 x 10²¹ atoms of gold
Explanation:
Data
density = 19.32 g/cm³
dimensions = 2.5 cm x 2.5 cm x 0.1 mm
number of atoms of gold = x
Process
1.- Find the volume of the foil
Volume = length x height x width
Volume = 2.5 x 2.5 x 0.01
Volume = 0.0625 cm³
2.- Calculate the mass of the foil
Density = mass/volume
mass = density x volume
mass = 19.32 x 0.0625
mass = 1.208 g
3.- Calculate the number of atoms
Atomic number of Gold = 197 g
197 g -------------------- 6.023 x 10²³ atoms
1.208 g --------------- x
x = (1.208 x 6.023 x 10²³)/197
x = 3.69 x 10²¹ atoms of gold
Assuming that the ammonium sulfide formula is (NH4)2S then you can see that there are 2 nitrogen, 8 hydrogen and 2 sulfur atoms for every ammonium sulfide. If the amount of ammonium sulfide is 8.9 moles, then the number of hydrogen atoms should be: 8/1 * 8.9 mol= 71.2 moles
