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n NO = 3/30 = 0,1 mol
n O2 = 16/32 = 0,5 mol
0,1/1=0,5/vO2
v O2 = 5 liter
2. n = jmlh atom/6,02 x 10^23 = (3,01 x 10^23)/(6,02 x 10^23) = 0,5 mol
massa = n x Ar = 0,5 x 63,5 = 31,75 gram
3. jumlah molekul = n x 6,02 x 10^23 = (20/100) x 6,02 x 10^23 = 1,204 x10^23 molekul
O2=n2 & V2
ditanya:v2?
jawab
n1/v1=n2/v2
n1 = 3/30 = 0,1 mol
n2 = 16/32 = 0,5 mol
0,1/1=0,5/vO2
v O2 = 5 liter
2. n = X/6,02 . 10^23 = (3,01 . 10^23)/(6,02 . 10^23) = 0,5 mol
gr = n x Ar = 0,5 x 63,5 = 31,75 gram
3. jumlah molekul(X) = n x 6,02 . 10^23 = (20/100) x 6,02 . 10^23 = 1,204 . 10^23 molekul
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