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Question 8 1.2 pts an atom of bromine has a mass about four times greater than that of an atom of neon. which choice makes the correct comparison of the relative numbers of bromine and neon atoms in 1,000 g of each element?

Respuesta :

gvcci
Neon = app. 20.1795 g/mol 
Bromine = app. 79.904 g/mol 
-----> 79.904 g/mol / 20.1795 g/mol = 3.96 (Close to 4) 

Using Moles In 1000 g 

1000 g / 20.1795 g/mol = app. 49.555 mol of Ne 
1000 g / 79.904 g/mol = app. 12.515 mol of Br 
-----> 49.555 mol / 12.515 mol = 3.96 (Close to 4) 

Using Avogadro's Number 

49.555 mol x 6.022x10^23 atoms = app. 2.984x10^25 atoms of Ne 
12.515 mol x 6.022x10^23 atoms = app. 7.537x10^24 atoms of Br 
-----> 2.984x10^25 / 7.537x10^24 = 3.96 (Close to 4) 

So no matter how you look at it or calculate it, the answer is always the same. 

I hope it helps!
okpalawalter8

We have that for the Question "an atom of bromine has a mass about four times greater than that of an atom of neon. which choice makes the correct comparison of the relative numbers of bromine and neon atoms in 1,000 g of each element?"

it can be said that

  • [tex]0.25[/tex] makes the correct comparison of relative numbers of bromine and neon atoms

 

Generally the Atomic mass of Bromine  is given as = [tex]80g/mol[/tex]

The Atomic mass of Neon = [tex]20g/mol[/tex]

Therefore,

[tex]no of moles of bromine atoms in 1000g of bromine element = \frac{1000g}{80g/mol}\\\\ = 12.5mol[/tex]

[tex]no of bromine atoms in the sample\\\\ = 12.5 * 6.022*10^{23}[/tex]

Also,

[tex]no of moles of neon atoms in 1000g of neon element = \frac{1000g}{20g/mol}\\\\ = 50mol[/tex]

[tex]no of neon atoms in the sample\\\\ = 50 * 6.022*10^{23}[/tex]

Therefore,

[tex]\frac{no of bromine atoms}{no of neon atoms} = \frac{12.5 * 6.022*10^{23}}{50 * 6.022*10^{23}}\\\\ = 0.25[/tex]

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