# Question #a0043

Apr 6, 2016

$8.4 \cdot {10}^{23} \text{atoms}$

#### Explanation:

Your goal here is to use a conversion factor that can take you from moles of zinc to atoms of zinc.

The idea is that if you know how many atoms you get in one mole of zinc, you can cause this to calculate how many atoms you'd get in $1.4$ moles of zinc.

Now, one mole of any element is said to contain $6.022 \cdot {10}^{23}$ atoms of that element $\to$ this is known as Avogadro's number.

You can use Avogadro's number as a way to go from moles to atoms or vice versa.

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \text{1 mole" = 6.022 * 10^(23)"atoms} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

So, if one mole of zinc contains $6.022 \cdot {10}^{23}$ atoms, it follows that $1.4$ moles will contain

$1.4 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles Zn"))) * overbrace((6.022 * 10^(23)"atoms")/(1color(red)(cancel(color(black)("mole Zn")))))^(color(purple)("Avogadro's number")) = color(green)(|bar(ul(color(white)(a/a)8.4 * 10^(23)"atoms} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

The answer is rounded to two sig figs, the number of sig figs you have for the number of moles of zinc.