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

Question

anonymous · Answer

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

anonymous · Answer

ask away...

anonymous · Answer

just like this question idk where to even start

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

no:
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