Write the molecilar formula of the 2nd and 3rd member of the homologous series whose first member is methane.
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the answer is c2h6 and c3h8
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The homologous series of straight-chained alkanes begins methane (CH4), ethane (C2H6), propane (C3H8), butane (C4H10), and pentane (C5H12). In that series, successive members differ in mass by an extra methylene bridge (-CH2- unit) inserted in the chain. Thus the molecular mass of each member differs by 14 atomic mass units.
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