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Go to solution Solved by Real_PhillBert,
2 hours ago, TVwazhere said:

I think OP is asking for 1CM of atoms in length, not volume. 

Well fuck a duck.

 

EDIT: I guess in that case you can simply take the 3rd root of 1.133x10^23, which gives ~48,388,000 carbon atoms in one linear cm.

How many atoms is in 1 CM ? Like to many atoms stacked on top of each other to make 1 CM ?

 

It is late, I need to know. I can't sleep. Help.

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Status updates my dude, status updates. 

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2 minutes ago, Zando Bob said:

Status updates my dude, status updates. 

What ? 

 

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Depends on the element.

 

I remember questions like this from Chemistry back in the day, it's really not complex to figure out. 

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3 minutes ago, Real_PhillBert said:

Depends on the element.

 

I remember questions like this from Chemistry back in the day, it's really not complex to figure out. 

I dont know. Carbon ? Helium ? 

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2 hours ago, George_Bud said:

I dont know. Carbon ? Helium ? 

Depends on many factors such as element, temperature, pressure. 

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Atom is mostly empty spaces, there is vast amount of spaces between nuclei and it's orbiting electrons. Pretty sure you can stick a lot of them together if you can overcome repusive charges. 

Sudo make me a sandwich 

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14 hours ago, George_Bud said:

I dont know. Carbon ? Helium ? 

Since no one has helped with this yet, I'll give it a go but be warned, I took chemistry 10 years ago and got a C...

 

As others have said, the answer is not super clear cut, temperature and pressure play into it, as well as the specific element. I'll just be lazy and assume standard temp and pressure. 

 

A quick google search reveals that the density of carbon in it's graphite form is 2.26g/cm^3.

 

https://periodictable.com/Properties/A/Density.al.html

 

We also know the molar mass of Carbon to be 12.011g/mol. 

 

https://periodictable.com/Elements/006/data.html

 

With this info, we can find how many moles are within 1cm^3, (2.26g) / (12.011g/mole) ~= .188mol

 

A mol is simply a mass of a known number of atoms or molecules, 6.022x10^23.

 

Then we can simply use the known number of mols (.1882) and multiply it by the number of atoms per mol (6.022x10^23) and that will give us the number of atoms in 2.26g, which is the mass of 1cm^3. (.188) * (6.022x10^23) ~= 1.133x10^23 atoms.

 

Again, it's been a really long time since I've done any chemistry stuff so I may very well be wrong, but I figured I'd give it a go. 

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21 minutes ago, Real_PhillBert said:

Since no one has helped with this yet, I'll give it a go but be warned, I took chemistry 10 years ago and got a C...

 

-Science Ensues-

I think OP is asking for 1CM of atoms in length, not volume. 

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2 hours ago, TVwazhere said:

I think OP is asking for 1CM of atoms in length, not volume. 

Well fuck a duck.

 

EDIT: I guess in that case you can simply take the 3rd root of 1.133x10^23, which gives ~48,388,000 carbon atoms in one linear cm.

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