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To solve the problem, we first need to convert the mass of an iron atom from kilograms to grams. This step is important because the given density is in grams per cubic centimeter, and consistency in units is crucial for accurate calculations.
Given that 1 kilogram equals 1000 grams, we convert the mass using the formula: \[9.27 \times 10^{-26} \text{ kg} \times \frac{1000 \text{ g}}{1 \text{ kg}} = 9.27 \times 10^{-23} \text{ g}\]
Understanding how to convert units ensures precise results in subsequent steps.

Density links mass and volume, and it's given by the equation \[\text{Density} = \frac{\text{Mass}}{\text{Volume}}\]. In our case, the density of iron is given as 7.87 g/cm³. Knowing the mass of one iron atom (9.27 × 10^-23 g) and using the density, we can calculate the number of atoms in one cubic centimeter of iron.
First, we rearrange the density formula for the number of atoms: \[\text{Number of atoms} = \frac{\text{Density} \times \text{Volume}}{\text{Mass of one atom}}\]. Plugging in the known values, we get: \[\text{Number of atoms} = \frac{7.87 \text{ g/cm}^3 \times 1 \text{ cm}^3}{9.27 \times 10^{-23} \text{ g}} \]. This yields the number of atoms per cubic centimeter of iron. Explaining how density contributes to this calculation reinforces the necessity of unit consistency and formula manipulation.

The volume of one iron atom is determined by dividing the volume of one cubic centimeter by the number of atoms calculated in the previous step. This step illustrates how closely iron atoms are packed within a given volume.
The formula we use is: \[\text{Volume of one atom} = \frac{1 \text{ cm}^3}{\text{Number of atoms}} \]. Substituting the values provides the volume of a single iron atom in cubic centimeters. Understanding the spatial arrangement of atoms helps visualize how their physical properties translate into macroscopic characteristics.

To find the radius of an iron atom, we utilize the volume of the atom, which was previously computed. Assuming atoms are spheres, the volume formula for a sphere \[\text{Volume} = \frac{4}{3} \times \text{Ï€} \times \text{radius}^3\] is rearranged to solve for radius: \[\text{radius} = \bigg(\frac{3 \times \text{Volume}}{4 \times \text{Ï€}}\bigg)^{\frac{1}{3}}\].
Finally, to determine the distance between the centers of adjacent atoms, we consider that in tightly packed spherical atoms, each atom essentially occupies a cube in a face-centered cubic lattice. Hence, the distance between centers equals twice the atomic radius. Understanding the atomic radius and lattice arrangement helps in predicting material properties and atomic interactions.