How many protons are there in 6.0g of nitrogen gas?

Avogadro constant = 6.022 x 10^23 mol-1

Answer is 1.8 x 10^24

How do I get to this?

Avogadro constant = 6.022 x 10^23 mol-1

Answer is 1.8 x 10^24

How do I get to this?

Original post by anactualmess

How many protons are there in 6.0g of nitrogen gas?

Avogadro constant = 6.022 x 10^23 mol-1

Answer is 1.8 x 10^24

How do I get to this?

Avogadro constant = 6.022 x 10^23 mol-1

Answer is 1.8 x 10^24

How do I get to this?

How many protons are there per molecule of N2?

14?

Original post by Pigster

How many protons are there per molecule of N2?

Original post by anactualmess

14?

how many molecules of gas in 6g?

well there's 6.022 x 10^23 molecules of gas in 28g

28/6 = 4.7

so assuming you divide 6.022 x 10^23 by 4.7 to get the number of molecules of gas in 6g? So 1.3 x 10^23?

28/6 = 4.7

so assuming you divide 6.022 x 10^23 by 4.7 to get the number of molecules of gas in 6g? So 1.3 x 10^23?

Original post by sotor

how many molecules of gas in 6g?

Original post by anactualmess

well there's 6.022 x 10^23 molecules of gas in 28g

28/6 = 4.7

so assuming you divide 6.022 x 10^23 by 4.7 to get the number of molecules of gas in 6g? So 1.3 x 10^23?

28/6 = 4.7

so assuming you divide 6.022 x 10^23 by 4.7 to get the number of molecules of gas in 6g? So 1.3 x 10^23?

yep! be careful of rounding errors, you should have 1.29*10^23.

I did (6/28)*(6.02*10^23) but if your way makes sense to you then stick with that

so how many protons in 1.3*10^23?

only thing I can think of is 1.29 x 10^23 x 14? But I don't really understand why that is

Original post by sotor

yep! be careful of rounding errors, you should have 1.29*10^23.

I did (6/28)*(6.02*10^23) but if your way makes sense to you then stick with that

so how many protons in 1.3*10^23?

I did (6/28)*(6.02*10^23) but if your way makes sense to you then stick with that

so how many protons in 1.3*10^23?

Original post by anactualmess

only thing I can think of is 1.29 x 10^23 x 14? But I don't really understand why that is

yep! if you have 14 protons in every molecule, then in 100 molecules you have 100x14 protons

and if you have 1.29*10^23 molecules, you get your answer

does that make sense?

ah okay thank you!

Original post by sotor

yep! if you have 14 protons in every molecule, then in 100 molecules you have 100x14 protons

and if you have 1.29*10^23 molecules, you get your answer

does that make sense?

and if you have 1.29*10^23 molecules, you get your answer

does that make sense?

1.) Work out moles of Nitrogen : 6g/14 = 0.42892.) Workout the number of particles by multiplying the moles by Avogadro's constant : 0.04289 x 6.022x10^23 = 2.58 x10^23.3.) Because there are 7 protons in each particle of N, multiply your anwser by 7 which give syou 1.8 x10^24.

Original post by ZzZzZz121212

1.) Work out moles of Nitrogen : 6g/14 = 0.42892.) Workout the number of particles by multiplying the moles by Avogadro's constant : 0.04289 x 6.022x10^23 = 2.58 x10^23.3.) Because there are 7 protons in each particle of N, multiply your anwser by 7 which give syou 1.8 x10^24.

Wouldn’t you multiply by 14 because it’s N2 so (7+7) = 14 protons

Original post by Jaafar123

Wouldn’t you multiply by 14 because it’s N2 so (7+7) = 14 protons

they worked out the moles of one atom of nitrogen so they did the protons per atom. I suggest doing it the other guy's way where you take into account it's diatomic and that you multiply it by 14 in the end

Original post by sotor

yep! if you have 14 protons in every molecule, then in 100 molecules you have 100x14 protons

and if you have 1.29*10^23 molecules, you get your answer

does that make sense?

and if you have 1.29*10^23 molecules, you get your answer

does that make sense?

Wrong!

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