Ответ:

Номер 3:) CuSO4 Ответ: ω(Cu)=40% ω ( � � ) = 40 % , ω(S)=20% ω ( � ) = 20 % , ω(O)=40% ω ( � ) = 40 % . Дано: Решение CuSO4 � � � � 4 Mr(CuSO4)=Ar(Cu)+Ar(S)+4⋅Ar(O)=64+32+4⋅16=160 а.е.м. � � ( � � � � 4 ) = � � ( � � ) + � � ( � ) + 4 ⋅ � � ( � ) = 64 + 32 + 4 ⋅ 16 = 160 а.е.м. ω(Cu)=100⋅Ar(Cu)Mr(CuSO4)=100⋅64160=40% ω ( � � ) = 100 ⋅ � � ( � � ) � � ( � � � � 4 ) = 100 ⋅ 64 160 = 40 % ω(S)=100⋅Ar(S)Mr(CuSO4)=100⋅32160=20% ω ( � ) = 100 ⋅ � � ( � ) � � ( � � � � 4 ) = 100 ⋅ 32 160 = 20 % ω(O)=100⋅4⋅Ar(O)Mr(CuSO4)=100⋅4⋅16160=40%

Объяснение:

Fe2O3 Ответ: ω(Fe)=70% ω ( � � ) = 70 % , ω(O)=30% ω ( � ) = 30 % . Дано: Решение Fe2O3 � � 2 � 3 Mr(Fe2O3)=2⋅Ar(Fe)+3⋅Ar(O)=2⋅56+3⋅16=160 а.е.м. � � ( � � 2 � 3 ) = 2 ⋅ � � ( � � ) + 3 ⋅ � � ( � ) = 2 ⋅ 56 + 3 ⋅ 16 = 160 а.е.м. ω(Fe)=100⋅2⋅Ar(Fe)Mr(Fe2O3)=100⋅2⋅56160=70% ω ( � � ) = 100 ⋅ 2 ⋅ � � ( � � ) � � ( � � 2 � 3 ) = 100 ⋅ 2 ⋅ 56 160 = 70 % ω(O)=100⋅3⋅Ar(O)Mr(Fe2O3)=100⋅3⋅16160=30% ω ( � ) = 100 ⋅ 3 ⋅ � � ( � ) � � ( � � 2 � 3 ) = 100 ⋅ 3 ⋅ 16 160 = 30 % ω(Fe)=? ω ( � � ) = ? ω(O)=?

HNO3 Ответ: ω(H)=1.6% ω ( � ) = 1.6 % , ω(N)=22.2% ω ( � ) = 22.2 % , ω(O)=76.2% ω ( � ) = 76.2 % . Дано: Решение HNO3 � � � 3 Mr(HNO3)=Ar(H)+Ar(N)+3⋅Ar(O)=1+14+3⋅16=63 а.е.м. � � ( � � � 3 ) = � � ( � ) + � � ( � ) + 3 ⋅ � � ( � ) = 1 + 14 + 3 ⋅ 16 = 63 а.е.м. ω(H)=100⋅Ar(H)Mr(HNO3)=100⋅163=1.6% ω ( � ) = 100 ⋅ � � ( � ) � � ( � � � 3 ) = 100 ⋅ 1 63 = 1.6 % ω(N)=100⋅Ar(N)Mr(HNO3)=100⋅1463=22.2% ω ( � ) = 100 ⋅ � � ( � ) � � ( � � � 3 ) = 100 ⋅ 14 63 = 22.2 % ω(O)=100⋅3⋅Ar(O)Mr(HNO3)=100⋅3⋅1663=76.2% ω ( � ) = 100 ⋅ 3 ⋅ � � ( � ) � � ( � � � 3 ) = 100 ⋅ 3 ⋅ 16 63 = 76.2 % ω(H)=? ω ( � ) = ? ω(N)=? ω ( � ) = ? ω(O)