Gold has a density of 19.32 g>cm3. A piece of gold foil is 2.50 cm on each side and 0.100 mm thick. How many atoms of gold are in this piece of gold foil

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

hey there!

mass = 6.75 g

volume =   9 cm³

therefore:

d = mass / volume

d   = 6.75 / 9

d = 0.7 g/cm³

answer b

5.6⋅10−^19  j

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