Question

Iron has a mass of 7.87 g per cubic centimeter of volume, and the mass of an iron atom is 9.27 × 10-26 kg. If you simplify and treat each atom as a cube, (a) what is the average volume (in cubic meters) required for each iron atom and (b) what is the distance (in meters) between the centers of adjacent atoms?

Answer 1

The average volume required for each iron atom is approximately 1.18 × 10⁻²⁹ m³, and the distance between the centers of adjacent iron atoms, assuming they are cubic in shape, is about 2.28 × 10⁻²ⁱ meters.

Step-by-step explanation:

To solve for the average volume required for each iron atom, we start by converting the density of iron from grams per cubic centimeter (g/cm³) to kilograms per cubic meter (kg/m³). This is done by multiplying the density by 1000 (since there are 1000 grams in a kilogram) and by 1000000 (since there are 1000000 cubic centimeters in a cubic meter), giving us 7.87 g/cm³ × 1000 g/kg × 1000000 cm³/m³ = 7.87×10³ kg/m³. Next, to find the volume that one iron atom occupies, we take the mass of one iron atom (9.27 × 10⁻²⁶ kg) and divide it by the density of iron (7.87×10³ kg/m³), which gives us approximately 1.18 × 10⁻²⁹ m³ per iron atom.

For part (b), to find the distance between adjacent iron atoms, we treat each atom as a cube and use the volume to determine the length of each side of the cube (since volume = side³ for a cube). Taking the cube root of the volume of one atom (1.18 × 10⁻²⁹ m³), we get a side length of approximately 2.28 × 10⁻²ⁱ meters. This distance is also the center-to-center distance between adjacent atoms.

Answer 2

a)
`1.1823* 10^(-29)m^3` is the average volume required for each iron atom.

b) The distance between the centers of adjacent atoms is
`2.2781* 10^(-8) m`.

Step-by-step explanation:

a) Mass of an iron = 7.84

Volume of iron =
`1 cm^3`

Let the x atoms of iron weights 7.84 g and volume of
`1 cm^3`.

Mass of an iron atom =
`9.27* 10^(-26) kg=9.27* 10^(-23) g`

`x* 9.27* 10^(-23) g=7.84 g`

`x =8.4573* 10^(22) atoms`

Volume of x atoms of iron =
`1 cm^3`

Volume of an iron atom = v

`x* v=1 cm^3`

`v=(1 cm^3)/(8.4573* 10^(22) )=1.1823* 10^(-23)cm^3`

`1 cm^3= 10^(-6) m^3`

`v=1.1823* 10^(-29)m^3`

`1.1823* 10^(-29)m^3` is the average volume required for each iron atom.

b) Volume of an iron atom = v

Volume of the cube =
`s^3`

Length of iron cube -= s

`1.1823* 10^(-29)m^3=s^3`

`s=2.2781* 10^(-8) m`

According to question we are treating iron atom as cube. So all the cubic iron atoms will have a same side.

`s=2.2781* 10^(-8) m`

The distance between the centers of adjacent atoms:

Horizontal distance of the center from the one face of the cubic iron atom:

=
`(s)/(2)`

Horizontal distance of the center from the one face of the another cubic iron atom:

=
`(s)/(2)`

`(s)/(2)+(s)/(2)=s`

The distance between the centers of adjacent atoms is
`2.2781* 10^(-8) m`.