What is the mass of $4.22 \times 10^{23}$ $4.22xx10^23$ lead atoms?

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What is the mass of $4.22 \times 10^{23}$ $4.22xx10^23$ lead atoms?

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Answer 1 · 2017-04-10T10:42:53

Approx. $138 \cdot g$ $138*g$ ............

Explanation:

The Periodic Table tells me that $Avogadro's number$ $"Avogadro's number"$ of individual lead atoms have a mass of $207.2 \cdot g$ $207.2*g$ .

And $Avogadro's number$ $"Avogadro's number"$ $\equiv$ $-=$ $6.022 \times 10^{23} \cdot m o l^{- 1}$ $6.022xx10^23*mol^-1$ .

And thus the mass of such a noomber of lead atoms is..........

$\frac{4.22 \times 10^{23} \cdot lead atoms}{6.022 \times 10^{23} \cdot lead atoms \cdot m o l^{- 1}} \times 207.2 \cdot g \cdot m o l^{- 1} = ? ? g$ $(4.22xx10^23*"lead atoms")/(6.022xx10^23*"lead atoms"*mol^-1)xx207.2*g*mol^-1=??g$ .

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What is the mass of $4.22 \times 10^{23}$ $4.22xx10^23$ lead atoms?

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