Mr(BaSO4) = Ar(Ba) + Ar(S) + 4*Ar(O) = 137 + 32 + 4*16 = 233
ω(Ba) = (Ar(Ba)/Mr(BaSO4))*100% = (137/233)*100% ≈ 58,8%
ω(S) = (Ar(S)/Mr(BaSO4))*100% = (32/233)*100% ≈ 13,73%
ω(O) = (4*Ar(O)/Mr(BaSO4))*100% = (4*16/233)*100% ≈ 27,47%
Ответ: ω(Ba) = 58,8%, ω(S) = 13,73%, ω(O) = 27,47%
Mr(BaSO4) = Ar(Ba) + Ar(S) + 4*Ar(O) = 137 + 32 + 4*16 = 233
ω(Ba) = (Ar(Ba)/Mr(BaSO4))*100% = (137/233)*100% ≈ 58,8%
ω(S) = (Ar(S)/Mr(BaSO4))*100% = (32/233)*100% ≈ 13,73%
ω(O) = (4*Ar(O)/Mr(BaSO4))*100% = (4*16/233)*100% ≈ 27,47%
Ответ: ω(Ba) = 58,8%, ω(S) = 13,73%, ω(O) = 27,47%