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1. C20H42
bo C20H(2*20+2)
2. CnH28
C13H28 bo C13H(2*13+2)
3. wzór butanu C4H10
masa C = 12u masa H = 1u
Masa C4H10 = 48 + 10 = 58u
58-100%
48-X%
58x = 4800 / :58 x= 82,75%