The density of iron is 7.87 g/cm3, and the average mass for one iron atom is 9.28×10-23 g.
How many atoms are there in a rectangular iron ingot measuring 3.00 cm × 3.00 cm × 9.00 cm?
you want to convert volume to #atoms. SO, start plugging in your conversion factors:
3.00cm*3.00cm*9.00cm = 81cm^3 * 7.87g/cm^3 / 9.28*10^-23g/atom = 6.87*10^24 atoms
To find the number of atoms in the rectangular iron ingot, we need to calculate the volume of the ingot and then divide it by the volume occupied by one iron atom.
Step 1: Calculate the volume of the ingot.
Volume = length × width × height = 3.00 cm × 3.00 cm × 9.00 cm = 81.00 cm^3
Step 2: Convert the density of iron to g/cm^3 to find the mass of the ingot.
Density = mass/volume
Mass = Density × Volume
Mass = 7.87 g/cm^3 × 81.00 cm^3 = 637.47 g
Step 3: Calculate the number of iron atoms using the mass of the ingot and the average mass of one iron atom.
Number of atoms = Mass of ingot / Average mass of one iron atom
Number of atoms = 637.47 g / (9.28×10^-23 g) = 6.87 × 10^24 atoms
Therefore, there are approximately 6.87 × 10^24 iron atoms in the rectangular iron ingot measuring 3.00 cm × 3.00 cm × 9.00 cm.