Question #3a4ee

1 Answer
Aug 18, 2017

Answer:

Well let's see.....I make it.....

#3.23xx10^-2*molxx6.022xx10^23*mol^-1xx32*"electrons"=?#

Explanation:

We somehow have got a mass of #2*g# nitrate ion, #NO_3^-#. We look at our Periodic Table, and we find that nitrogen #Z=7#, has 7 electrons, and oxygen, #Z=8#, has 8 electrons.

So per formula unit of #NO_3^-#, there are #7+3xx8+1=32*"electrons"#. Why did I add an extra electron.

Now we have a molar quantity of #(2*g)/(62.01*g*mol^-1)=3.23xx10^-2*mol#

And so there are #3.23xx10^-2*molxx6.022xx10^23*mol^-1xx32*"electrons"=?#

But hang on, #Z# specifies the number of protons in the nucleus. How can I use #Z# to define the number of electrons?