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Chemistry 17 Online
OpenStudy (anonymous):

what is the atomic weight of an element if 4.0 g of it contain 2.98x10^22 atoms? how do you set that up

OpenStudy (anonymous):

n= N/L n - amount - mol N - amount - number L - avogadro's constant - mol-1 (6,022*10^23) n(x) = 2,98*10^22 / 6,022*10^23 mol-1 = 0,0495 mol ~ 0,05 mol n= m/M m - mass - g M - molar mass - gmol-1 M= m/n = 4 g / 0,05 mol = 80 gmol-1 and now you know atom is Br

OpenStudy (anonymous):

ask away...

OpenStudy (anonymous):

just like this question idk where to even start

OpenStudy (anonymous):

well you have this simple equation n=N/L and it states that one mole of any atom/molecule has 6,022*10^23 atoms/molecules

OpenStudy (anonymous):

does carbon has an oxidation state of zero in this compound C2H2

OpenStudy (anonymous):

no: |dw:1340940399437:dw|

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