The Density is 7.87g/cm^3 , and the average mass for a single iron atom is 9.28*10^-23g .
How many iron atoms are present in an iron bar whose dimensions are 2.00 cm * 3.00 cm * 8.00 cm ?
SOLUTION This solution is FREE courtesy of BrainMass!
Mass/Volume = Density
The volume of the iron bar is V = 2 * 3 * 8 cm^3
V = 48 cm^3
Since the density of iron is given to be 7.87g/cm^3 we can calculate the mass of the iron bar as,
mass = density * volume = 48 * 7.87 = 377.76 g
This is the total mass of all the iron atoms in the bar. But each iron atom has a mass, m = 9.28 * 10^-23g
We have 377.76g as the mass of N atoms in the bar
ie, N*m = 377.76
Now, N = 377.76/9.28 * 10^-23 = 40.7*10^23
The number of iron atoms present in the bar is 40.7*10^23 (ans)© BrainMass Inc. brainmass.com December 24, 2021, 4:47 pm ad1c9bdddf>