ОЧ СРОЧНО, ДАЮ ВСЕ БАЛЫ
Ответы
Ответ:
Объяснение:
СО2 PCl5 NaBr Fe2O3 SF4
Mr(СО2) = Ar(С) + 2Ar(О) = 12 + 2 * 16 = 44
w(C) = Ar(C)/Mr(СО2) * 100% = 12/44 * 100% = 27%
w(О) = 2Ar(О)/Mr(СО2) * 100% = 2* 16/44 * 100% = 73%
Mr(PCl5) = Ar(P) + 5Ar(Cl) = 31 + 5 * 35,5 = 208,5
w(P) = Ar(P)/Mr(PCl5) * 100% = 31/208,5 * 100% = 14,87%
w(Cl) = 2Ar(Cl)/Mr(PCl5) * 100% = 5 * 35,5/208,5 * 100% = 85,13%
Mr(NaBr) = Ar(Na) + Ar(Br) = 23 + 80 = 103
w(Na) = Ar(Na)/Mr(NaBr) * 100% = 23/103 * 100% = 22,33%
w(Br) = Ar(Br)/Mr(NaBr) * 100% = 80/103 * 100% = 77,67%
Mr(Fe2O3) = 2Ar(Fe) + 3Ar(О) = 2 * 56 + 3 * 16 = 160
w(Fe) = Ar(Fe)/Mr(Fe2O3) * 100% = 2 * 56/160 * 100% = 70%
w(О) = 2Ar(О)/Mr(Fe2O3) * 100% = 3* 16/160 * 100% = 30%
Mr(SF4 ) = Ar(S) + 4Ar(F) = 32 + 4 * 19 = 108
w(S) = Ar(S)/Mr(SF4 ) * 100% = 32/108 * 100% = 29,63%
w(F) = 2Ar(F)/Mr(SF4 ) * 100% = 4 * 19/108 * 100% = 70,37%