# Question 3a4ee

Aug 18, 2017

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

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

#### Explanation:

We somehow have got a mass of $2 \cdot g$ nitrate ion, $N {O}_{3}^{-}$. 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 $N {O}_{3}^{-}$, there are $7 + 3 \times 8 + 1 = 32 \cdot \text{electrons}$. Why did I add an extra electron.

Now we have a molar quantity of $\frac{2 \cdot g}{62.01 \cdot g \cdot m o {l}^{-} 1} = 3.23 \times {10}^{-} 2 \cdot m o l$

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?