How many atoms are present in a 1.25m length of 20gauge copper wire? A 20gauge copper wire has a diameter of 0.03196in, and the density of copper is 8.92g/cm^3.
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Let's treat the wire as a cylinder and calculate its volume. V = pi x r^2 x h . V = 3.14 x (.03196 in)^2 x 1.25 m. Before we do the math lets get the radius squared factor converted into cm instead of in. 2.54 cm = 1 in so .03196 in x 2.54 cm/in = 0.08118 cm. Now substitute it back and also lets convert the length of 1.25-m to cm. 1.25 m x 100 cm / 1 m = 125 cm. Now:
V = 3.14 x (.08118 cm)^2 x 125 cm = 2.587 cm^3. Let's now find its mass. Density = m/V and mass = density x Volume. m = 8.92-g/cm^3 x 2.587 cm^3 = 23.08-g
OK now that we have the mass of the copper we can get number of atoms by converting mass to moles and then moles x Avogadro's number to get atoms:
23.08-g Cu x ( 1 mol Cu / 63.5-g Cu) x 6.02 x 10^23 atoms / 1 mol Cu) = 2.19 X 10^23 atoms of Cu
V = 3.14 x (.08118 cm)^2 x 125 cm = 2.587 cm^3. Let's now find its mass. Density = m/V and mass = density x Volume. m = 8.92-g/cm^3 x 2.587 cm^3 = 23.08-g
OK now that we have the mass of the copper we can get number of atoms by converting mass to moles and then moles x Avogadro's number to get atoms:
23.08-g Cu x ( 1 mol Cu / 63.5-g Cu) x 6.02 x 10^23 atoms / 1 mol Cu) = 2.19 X 10^23 atoms of Cu
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