Question #a351e

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Question #a351e

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Answer 1 · 2017-12-04T01:02:21

$3.299 \cdot 10^{22}$ $3.299 * 10^(22)$ ${atoms cm}^{- 3}$ $"atoms cm"^(-3)$

Explanation:

The trick here is to recognize that a unified atomic mass unit, or $u$ $"u"$ , is equivalent to ${1 g mol}^{- 1}$ $"1 g mol"^(-1)$ . This means that you have

$207.2 u = {207.2 g mol}^{- 1}$ $"207.2 u" = "207.2 g mol"^(-1)$

which tells you that $1$ $1$ mole of lead has a mass of $207.2 g$ $"207.2 g"$ . Since you know that lead has a density of ${11.35 g cm}^{- 3}$ $"11.35 g cm"^(-3)$ , you can say that ${1 cm}^{3}$ $"1 cm"^3$ of lead will contain

$11.35 color(red)(cancel(color(black)("g"))) * "1 mole Pb"/(207.2color(red)(cancel(color(black)("g")))) = "0.0548 moles Pb"$

Now all you have to do is to use the fact that you have

$color(blue)(ul(color(black)("1 mole Pb" = 6.022 * 10^(23)color(white)(.)"atoms Pb"))) ->$ Avogadro's constant

to find the number of atoms present in $"1 cm"^3$ of lead.

$0.05478 color(red)(cancel(color(black)("moles Pb"))) * (6.022 * 10^(23)color(white)(.)"atoms Pb")/(1color(red)(cancel(color(black)("mole Pb")))) = 3.299 * 10^(22)color(white)(.)"atoms Pb"$

Therefore, you can say that the density of lead in atoms per cubic centimeter is equal to

$3.299 * 10^(22)color(white)(.)"atoms cm"^(-3)$

The answer is rounded to four sig figs.

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Question #a351e

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