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nika2105 [10]
3 years ago
11

How many moles are in 1.35 x10^23 representative particles of Fluorine?

Chemistry
2 answers:
Scorpion4ik [409]3 years ago
8 0
Atomic weight of fluorine is 19 g/mol.

We know that, 1 mol (19 g) of Fluorine will have Avagadro number (6.022 X 10^23) of Fluorine atoms 

Thus, we have 6.022 X 10^23 atoms of F ≡ 19 g of F
Therefore, 1.35 x10^23 atoms of F ≡ 4.259 g of F

Now, 19 g of Fluorine ≡ 1 mole
∴, 4.259 g of Fluorine ≡ 0.224 mole 

Answer: 0.224 <span>moles are in 1.35 x10^23 representative particles of Fluorine</span>
DochEvi [55]3 years ago
4 0
<span>Answer: 0.224 moles


Explanation:


1) Avogadro's number is 1 mol = 6.022 x 10²³, which means that there are 6.022 x 10²³ particles in 1 mol.


2) Divide the number of of particles by Avogadro's number to find how many moles that is:
</span><span />

<span>(1.35x10²³ / (6.022x10²³) = 0.224 moles</span>


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