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how much carbon is in 4.87*10^22

 Dec 10, 2014

Best Answer 

 #3
avatar+109 
+5

Assuming that you're speaking in units of atoms, you can convert them like so:

 

$$Grams \Leftrightarrow Mols \Leftrightarrow Atoms$$

 

Using Avogadro's Constant, we know that 1 mol = 6.022 x 10^23 atoms, allowing us to convert from atoms to mols:

 

$$\left({\mathtt{4.87}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{{\mathtt{22}}}{\mathtt{\,\times\,}}{\mathtt{atomsC}}\right){\mathtt{\,\times\,}}\left({\frac{\left({\mathtt{1}}{\mathtt{\,\times\,}}{\mathtt{molC}}\right)}{\left({\mathtt{6.022}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{{\mathtt{23}}}{\mathtt{\,\times\,}}{\mathtt{atomsC}}\right)}}\right) = {\mathtt{0.808\: \!7}}{\mathtt{\,\times\,}}{\mathtt{molsC}}$$

 

To get grams, we convert again only from mols, to grams. Knowing that 1 molC = 12.00gC we get:

 

$${\mathtt{0.808\: \!7}}{\mathtt{\,\times\,}}{\mathtt{molsC}}{\mathtt{\,\times\,}}\left({\frac{\left({\mathtt{12}}{\mathtt{\,\times\,}}{\mathtt{gC}}\right)}{\left({\mathtt{1}}{\mathtt{\,\times\,}}{\mathtt{molC}}\right)}}\right) = {\mathtt{9.077}}{\mathtt{\,\times\,}}{\mathtt{gC}}$$

 Dec 11, 2014
 #1
avatar+4164 
0

wow Scroll down to what Sorasyn said I belive that is correct lol

 Dec 10, 2014
 #2
avatar+270 
0

That's chemistry...

make me ill *-*

 Dec 10, 2014
 #3
avatar+109 
+5
Best Answer

Assuming that you're speaking in units of atoms, you can convert them like so:

 

$$Grams \Leftrightarrow Mols \Leftrightarrow Atoms$$

 

Using Avogadro's Constant, we know that 1 mol = 6.022 x 10^23 atoms, allowing us to convert from atoms to mols:

 

$$\left({\mathtt{4.87}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{{\mathtt{22}}}{\mathtt{\,\times\,}}{\mathtt{atomsC}}\right){\mathtt{\,\times\,}}\left({\frac{\left({\mathtt{1}}{\mathtt{\,\times\,}}{\mathtt{molC}}\right)}{\left({\mathtt{6.022}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{{\mathtt{23}}}{\mathtt{\,\times\,}}{\mathtt{atomsC}}\right)}}\right) = {\mathtt{0.808\: \!7}}{\mathtt{\,\times\,}}{\mathtt{molsC}}$$

 

To get grams, we convert again only from mols, to grams. Knowing that 1 molC = 12.00gC we get:

 

$${\mathtt{0.808\: \!7}}{\mathtt{\,\times\,}}{\mathtt{molsC}}{\mathtt{\,\times\,}}\left({\frac{\left({\mathtt{12}}{\mathtt{\,\times\,}}{\mathtt{gC}}\right)}{\left({\mathtt{1}}{\mathtt{\,\times\,}}{\mathtt{molC}}\right)}}\right) = {\mathtt{9.077}}{\mathtt{\,\times\,}}{\mathtt{gC}}$$

Sorasyn Dec 11, 2014

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