Question

Question #9afde

Answer 1

`2.02 * 10^(7)"atoms of Rb"`

Explanation:

For starters, the atomic radius of rubidium cannot be expressed in picometers cubed, `"pm"^3`, because that is a unit used for volume, not for length.

My guess would be that you're indeed dealing with the atomic radius of rubidium, which is equal to `"248 pm"`.

If that's the case, the first thing to do here would be to convert the desired length, i.e. `1.00` 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 `"248 pm"`. The thing to look out for here is the fact that you need to use the 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.