# Question #9afde

##### 1 Answer

#### Explanation:

For starters, the *atomic radius* of rubidium cannot be expressed in *picometers cubed*, **volume**, not for **length**.

My guess would be that you're indeed dealing with the atomic radius of rubidium, which is equal to

If that's the case, the first thing to do here would be to convert the desired length, i.e. *centimeters*, to *picometers*. To do that, go from centimeters to *meters* first, then from meters to picometers

#1.00 color(red)(cancel(color(black)("cm"))) * (1color(red)(cancel(color(black)("m"))))/(10^2color(red)(cancel(color(black)("cm")))) * (10^(12)"pm")/(1color(red)(cancel(color(black)("m")))) = 1.00 * 10^(10)"pm"#

So, you know that the **radius** of a rubidium atom is equal to **diameter** of an atom, which is you know is equal to

#color(purple)(bar(ul(|color(white)(a/a)color(black)("diameter" = 2 xx "radius")color(white)(a/a)|)))#

The diameter of a rubidium atom will thus be

#"diamter" = 2 xx "248 pm" = "496 pm"#

Now all you have to do is figure out how many atoms would fit in that length

#1.00 * 10^(10)color(red)(cancel(color(black)("pm"))) * "1 Rb atom"/(496color(red)(cancel(color(black)("pm")))) = color(green)(bar(ul(|color(white)(a/a)color(black)(2.02 * 10^7"atoms of Rb")color(white)(a/a)|)))#

The answer is rounded to three **sig figs**.