Oblicz zawartość procentową tlenu w trioleinianie gliceryny.
Masa (C₁₇H₃₃COO)₃C₃H₅ = 3*(17*12u + 33*1u + 16u + 16u) + 3*12u + 5*1u = 848u
Masa 6 O = 6*16u = 96u
848u (C₁₇H₃₃COO)₃C₃H₅ --------stanowi---------100%
to 96u O ------------------------------stanowi----------X%
X= 96u * 100% : 848u ≈ 11,32% tlenu
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Masa (C₁₇H₃₃COO)₃C₃H₅ = 3*(17*12u + 33*1u + 16u + 16u) + 3*12u + 5*1u = 848u
Masa 6 O = 6*16u = 96u
848u (C₁₇H₃₃COO)₃C₃H₅ --------stanowi---------100%
to 96u O ------------------------------stanowi----------X%
X= 96u * 100% : 848u ≈ 11,32% tlenu