Question #43923

1 Answer
Oct 24, 2017

#2.1 * 10^(23)#

Explanation:

The thing to remember here is that in order to have #1# mole of copper, you need to have #6.022 * 10^(23)# atoms of copper #-># this is known as Avogadro's constant.

What this tells you is that every time you have #6.022 * 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 * 10^(23)# atoms of copper.

More specifically, the sample contains

#0.35 color(red)(cancel(color(black)("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.