Assume that you have 2.59 mol of aluminum. How many atoms of aluminum do you have?

Answer 1

#2.59# #"mols"# of aluminum ATOMS corresponds to #2.59xxN_A#, where #N_A# is Avogadro's number.

#N_A# #=# #6.022xx10^23#. What are the masses of #1# #mol# of #Al#, and #2.59# #"mol"# aluminum?
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Answer 2

To find the number of atoms of aluminum, you can use Avogadro's constant, which is (6.022 \times 10^{23}) atoms/mol.

(2.59 , \text{mol} \times 6.022 \times 10^{23} , \text{atoms/mol} = 1.56 \times 10^{24}) atoms of aluminum.

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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.

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.

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.

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.

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