How many moles of Ag contain #4.49 * 10^23# atoms Ag?

Answer 1

Avery Adams

Recall the constant that Avogadro

#1"mol" = 6.02*10^23"units"#

where a unit can be anything. For example, it can be compared to a dozen.

Therefore,

#4.49*10^23"atoms" * "mol"/(6.02*10^23"atoms") approx 0.746"mol"#

will have that many atoms in it.

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Answer 2

Mateo Aguilar

Avogadro's number, which is $6.022 \times 10^{23}$ atoms per mole, can be used to calculate how many moles of Ag contain $4.49 \times 10^{23}$ atoms of Ag. Accordingly, $4.49 \times 10^{23}$ atoms of Ag is equal to: $\frac{4.49 \times 10^{23} \text{ atoms Ag}}{6.022 \times 10^{23} \text{ atoms/mol}}$

$= 0.746 \text{ moles of Ag}$

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.