Question #3fa99

1 Answer
Aug 20, 2017

Answer:

#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#.