Рассчитайте массовые доли для каждого химического элемента в веществах, формулы которых:
а) FeO3, б) SiH4, в) C6H12O6
Плиз, срочно

а) Mr (FeO₃) = Ar (Fe) + 3Ar (O) = 55,80 + 3 * 16,00 = 103,80

ω (Fe) = Ar (Fe) / Mr (FeO₃) * 100% = 55,80 / 103,80 * 100% = 53,76%

ω (O) = 100% - ω (Fe) = 100% - 53,76% = 46,24%

б) Mr (SiH₄) = Ar (Si) + 4Ar (H) = 28,09 + 4 * 1,01 = 32,13

ω (Si) = Ar (Si) / Mr (SiH₄) * 100% = 28,09 / 32,13 * 100% = 87,43%

ω (H) = 100% - ω (Si) = 100% - 87,43% = 12,57%

в) Mr (C₆H₁₂O₆) = 6Ar (C) + 12Ar (H) + 6Ar (O) = 6 * 12,01 + 12 * 1,01 + 6 * 16,00 = 180,18

ω (C) = 6Ar (C) / Mr (C₆H₁₂O₆) * 100% = 6 * 12,01 / 180,18 * 100% = 39,99%

ω (H) = 12Ar (H) / Mr (C₆H₁₂O₆) * 100% = 12 * 1,01 / 180,18 * 100% = 6,73%

ω (O) = 100% - ω (C) - ω (H) = 100% - 39,99% - 6,73% = 53,28%