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s 50% o 50%
3.82g of SO2 equals 0,06 moles.
Assuming 100 g of compound50.05 g S x 1 mol S/32 g S = 1.56 moles S present49.95 g O x 1 mol O/16 g O = 3.12 moles O presentmole ratio of O to S is 3.12/1/56 = 2 to 1Empirical formula = SO2
1.51x10^24 atoms of O
It is 1 atom of S to 2 of O.
(16.0g + 16.0g)/(32.1g + 16.0g + 16.0g) x 100 %
s 50% o 50%
I assume you mean mass of oxygen % in SO2. This is calculated by finding the molecular weight of S and O (32 and 16 respectively) As there are 2 oxygens, 16 is multiplied by 2, = 32. put 32 over the total, = 32/64. Which equals 50%
Oxygen is represented as the element "O" with an oxidation state of 2.
O=s->o
3.82g of SO2 equals 0,06 moles.
Oxygen is 49.95% of the mass of SO2. The molecular weight is 48.06 g/mol and sulfur's molecular weight is 32.06 g/mol, so oxgyen must make up the other half of the compounds molecular weight.
Assuming 100 g of compound50.05 g S x 1 mol S/32 g S = 1.56 moles S present49.95 g O x 1 mol O/16 g O = 3.12 moles O presentmole ratio of O to S is 3.12/1/56 = 2 to 1Empirical formula = SO2
88.5%
Yes. O::S::O (::=double bond)
1.51x10^24 atoms of O
It is 1 atom of S to 2 of O.