Respuesta :

CarlsenP
M C2F4 = 100 g/mol

In 100 g of C2H4 we have 4 × 6,02 × 10^23 atoms of fluorine.

In 5,85 g of C2F4 we have x atoms of fluorine.

100 g ---- 2,408 × 10^24 atoms
5,85 g --- x atoms

x = 5,85 × 2,408 × 10^24 / 100 = 1,408 × 10^23

5,85 g of C2F4 contains 1,408 × 10^23 atoms of fluorine.

:-) ;-)
sebassandin

Answer:

[tex]1.41x10^{23} atomsF[/tex]

Explanation:

Hello,

At first, molecular mass of dicarbon tetrafluoride is:

[tex]M_{C_2F_4}=12*2+19*4=100g/mol[/tex]

Now, by applying the mole-mass relationship with the Avogradro's number one finds that:

[tex]5.85gC_2F_4*\frac{1molC_2F_4}{100gC_2F_4}*\frac{4molF}{1molC_2F_4} *\frac{6.022x10^{23}atomsF }{1molF} =1.41x10^{23} atomsF[/tex]

Best regards.

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