The radius of a molybdenum atom is 136 pm. how many molybdenum atoms would have to be laid side by side to span a distance of 2.52 mm

A diameter of 1 atom =2 times the radius of the atom. "pico meters" pm is x10^-12 meters.
Then diameter is 136x10^-12(meters) x 2 = 272x10^ -12 (meters)
And the length to span is = 3.08 (millimeters) = 3.08x10^ -3 (meters)
Divide the length to span over the diameter:
(3.08x10^-3) / (272x10^-12)

Answer Link

luiferhoyos luiferhoyos · Answer 2 · 2019-08-30T00:55:45+02:00

Answer:

9264706 molybdenum atoms have to be laid.

Explanation:

The parameter of the atom which will determine the number of atoms needed to span a certain distance is the diameter. Diameter (d) is simply two times radius(r):

[tex]d=2r=2*136pm=272pm[/tex] (eq. 1)

If we express the number of molybdenum atoms needed with the letter x, the spanned distance (2.52mm) has to be equal to the number of atoms (x) times the diameter of these atoms.

[tex]2.52mm=x*d[/tex] (eq 2.)

Before stating an equation, all units have to be consistent (272 picometres expressed in mm, or 2.52mm expressed in pm, either one). Let's convert 272pm, using a conversion factor:

Conversion factor: [tex]1mm=1*10^9pm[/tex]

[tex]272pm=272pm*(\frac{1mm}{1*10^9pm})=2.72*10^-7mm[/tex]

Now, solving for "x" in the second equation:

[tex]x=\frac{2.52mm}{d}[/tex]

Plugging in the "d" value:

[tex]x=\frac{2.52mm}{2.72*10^-7mm}=9264706[/tex]