1)
Mr(C6H12O6) = 6Ar(C) + 12Ar(H) + 6Ar(O) = 6 * 12 + 12 * 1 + 6 * 16 = 180
w(O) = 6Ar(O) / Mr(C6H12O6) = 96 / 180 = 0.53 = 53%
Mr(C12H22O11) = 12Ar(C) + 22Ar(H) + 11Ar(O) = 12 * 12 + 22 * 1 + 11 * 16 = 342
w(O) = 11Ar(O) / Mr(C12H22O11) = 176 / 342 = 0.51 = 51%
Ответ: в C6H12O6 массовая доля кислорода больше
Copyright © 2024 SCHOLAR.TIPS - All rights reserved.
Answers & Comments
1)
Mr(C6H12O6) = 6Ar(C) + 12Ar(H) + 6Ar(O) = 6 * 12 + 12 * 1 + 6 * 16 = 180
w(O) = 6Ar(O) / Mr(C6H12O6) = 96 / 180 = 0.53 = 53%
Mr(C12H22O11) = 12Ar(C) + 22Ar(H) + 11Ar(O) = 12 * 12 + 22 * 1 + 11 * 16 = 342
w(O) = 11Ar(O) / Mr(C12H22O11) = 176 / 342 = 0.51 = 51%
Ответ: в C6H12O6 массовая доля кислорода больше