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

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 

0.224 moles are in 1.35 x10^23 representative particles of Fluorine

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:


(1.35x10²³ / (6.022x10²³) = 0.224 moles

100%sorry if i’m wronghope this good luck: )

They cool it down and aluminum cools easily