Question

How many atoms of aluminum (Al) are in 2.34 mol of aluminum?

Answer 1

Answer: "
`1.41 * 10^(24)` " atoms of aluminum (Al) .

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Step-by-step explanation:

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Using a technique called "dimensional analysis" ; as follows:

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→ "
`(2.34 mol Al)/(1) * (6.022*10^(23)atoms Al)/(1 mol Al) = ?`

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The units "mol Al" ["moles of aluminum"] cancel out; and we have:

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→ "
`[(2.34) * (6.022 * 10^(23))] = ?` atoms of Al [aluminum].

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Note that by definition: 1 (one) "mole" (abbreviated "mol") of anything consists of "
`6.022*10^(23)` " units " of that particular thing.

This number: 6.022*10^{23} —is known as: "Avogadro's number".

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→ [ 2.34 * 6.022 * 10^23] ;

= 1.409148 * 10^24 ;

→ Round to 3 (Three) significant figures;

→ since: "2.34" has 3 (Three) significant figures;

→ to get: "
`1.41 *10^(24)` atoms of aluminum (Al).

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Hope this is helpful to you! Wishing you the best!

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