Question

Question #3fa99

Answer 1

`1.48xx10^24` `"molecules NH"_3`

Explanation:

We're asked to find the number of molecules of `"NH"_3` present in `2.45` `"mol NH"_3`.

To do this, we can use the fact that one mole of any substance contains `underbrace(6.022xx10^23)_"Avogadro's number"` particles of that substance.

Therefore, we have

`2.45cancel("mol NH"_3)((6.022xx10^23color(white)(l)"molecules NH"_3)/(1cancel("mol NH"_3)))`

`= color(red)(ulbar(|stackrel(" ")(" "1.48xx10^24color(white)(l)"molecules NH"_3" ")|)`

This goes for any substance as well. That is to say, `2.45` moles of carbon dioxide also contains `color(red)(1.48xx10^24color(white)(l)"molecules"` of `"CO"_2`.