Home > Chemistry > The Mole

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.

By signing up, you agree to our Terms of Service and Privacy Policy

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} ]

By signing up, you agree to our Terms of Service and Privacy Policy

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.