# Question #43923

Oct 24, 2017

$2.1 \cdot {10}^{23}$

#### Explanation:

The thing to remember here is that in order to have $1$ mole of copper, you need to have $6.022 \cdot {10}^{23}$ atoms of copper $\to$ this is known as Avogadro's constant.

What this tells you is that every time you have $6.022 \cdot {10}^{23}$ atoms of copper, you can say that the sample contains exactly $1$ mole of copper.

In your case, you know that the sample contains $0.35$ moles of copper, so right from the start, you know that it contains fewer than $6.022 \cdot {10}^{23}$ atoms of copper.

More specifically, the sample contains

$0.35 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles Cu"))) * overbrace((6.022 * 10^(23)color(white)(.)"atoms Cu")/(1 color(red)(cancel(color(black)("mole Cu")))))^(color(blue)("Avogadro's constant")) = color(darkgreen)(ul(color(black)(2.1 * 10^(23)color(white)(.)"atoms Cu}}}}$

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