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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
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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
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just like this question idk where to even start
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
does carbon has an oxidation state of zero in this compound C2H2
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