Question #ccd7d

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Answer 1 · 2014-04-23T03:07:52

The equation to solve this problem is as follows:

$N a$ $Na$ ( $\frac{number of atoms}{particles}$ $"number of atoms"/"particles"$ ) =

$n$ $n$ ( $no. of moles)$ $"no." "of moles")$ $\cdot$ $*$ $N(Avogadro's number)$ $"N(Avogadro's number)"$

In your example, you would sub in 1 mole into n, then sub in Avogadro's number as such:

$N a$ $Na$ = $(1 mol) \cdot (6.02 x 10^{23})$ $(1" mol")*(6.02x10^23)$
$N a$ $Na$ = $6.02 \cdot 10^{23}$ $6.02 * 10^23$

Now, because you know that there is 3 times as many chlorine atoms "Cl3”, you multiply this number by 3:

$(6.02 x 10^{23}) \cdot (3)$ $(6.02 x 10^23)*(3)$ = $1.81 x 10^{24} atoms$ $1.81 x 10^24 " atoms"$

Hope this helps! :)