найдите массовую долю элементов соединения срочно пж оформлением задачи (если можно)
1. H2SO4
2. NAOH
3. C6H12O6
СРОЧНО ПЖ

Mr(H2SO4) = 2Ar(H) + Ar(S) + 4Ar(O) = 2 * 1 + 32 + 4 * 16 = 98

w(H) = 2Ar(H) / Mr(H2SO4) = 2 / 98 = 0.0204 = 2.04%

w(S) = Ar(S) / Mr(H2SO4) = 32 / 98 = 0.0204 = 0.3265 = 32.65%

w(O) = 4Ar(O) / Mr(H2SO4) = 64 / 98 = 0.6531 = 65.31%

Mr(NaOH) = Ar(Na) + Ar(O) + Ar(H) = 23 + 16 + 1 = 40

w(Na) = Ar(Na) / Mr(NaOH) = 23 / 40 = 0.575 = 57.5%

w(O) = Ar(O) / Mr(NaOH) = 16 / 40 = 0.4 = 40%

w(H) = Ar(H) / Mr(NaOH) = 1 / 40 = 0.025 = 2.5%

Mr(C6H12O6) = 6Ar(C) + 12Ar(H) + 6Ar(O) = 6 * 12 + 12 * 1 + 6 * 16 = 180

w(C) = 6Ar(C) / Mr(C6H12O6) = 72 / 180 = 0.4 = 40%

w(H) = 12Ar(H) / Mr(C6H12O6) = 12 / 180 = 0.0667 = 6.67%

w(O) = 6Ar(O) / Mr(C6H12O6) = 96 / 180 = 0.5333 = 53.33%