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HelloPoly
@HelloPoly
July 2022
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решите 1,2,3,пожалуйста
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cam6
1. Mr(CuSO4) = Ar(Cu) + Ar(S) + 4Ar(O) = 63 + 32 + 2 × 16 = 127
2. -
3. a) Mr(CuSO4) = 127
oмега(такого знака на клавиатуре у меня нет)(Сu) = Ar(Cu)/Mr(CuSO4)×100% = 63/127 ×100% = 49,6%
омега(S)= Ar(S)/Mr(CuSO4)×100% = 32/127×100% = 25,2%
омега(О)= 4Ar(O)/Mr(CuSO4)×100% = 32/127×100% = 25,2%
б) Mr(Fe2O3) = 2Ar(Fe) + 3Ar(O) = 112 + 48 = 160
омега(Fe) = 2Ar(Fe)/Mr(Fe2O3)× 100% = 112/160 × 100% = 70%
омега(О) = 3Ar(O)/Mr(Fe2O3) × 100% = 48/160 × 100% = 30%
в) Mr(HNO3) = Ar(H) + Ar(N) + 3Ar(O) = 1 + 14 + 48 = 63
омега(H) = Ar(H)/Mr(HNO3) × 100% = 1/63 × 100% = 1,6%
омега(N) = Ar(N)/Mr(HNO3) × 100% = 14/63 × 100% = 22,2%
омега(О) = Ar(O)/Mr(HNO3) × 100% = 48/63 × 100% = 76,2%
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Answers & Comments
2. -
3. a) Mr(CuSO4) = 127
oмега(такого знака на клавиатуре у меня нет)(Сu) = Ar(Cu)/Mr(CuSO4)×100% = 63/127 ×100% = 49,6%
омега(S)= Ar(S)/Mr(CuSO4)×100% = 32/127×100% = 25,2%
омега(О)= 4Ar(O)/Mr(CuSO4)×100% = 32/127×100% = 25,2%
б) Mr(Fe2O3) = 2Ar(Fe) + 3Ar(O) = 112 + 48 = 160
омега(Fe) = 2Ar(Fe)/Mr(Fe2O3)× 100% = 112/160 × 100% = 70%
омега(О) = 3Ar(O)/Mr(Fe2O3) × 100% = 48/160 × 100% = 30%
в) Mr(HNO3) = Ar(H) + Ar(N) + 3Ar(O) = 1 + 14 + 48 = 63
омега(H) = Ar(H)/Mr(HNO3) × 100% = 1/63 × 100% = 1,6%
омега(N) = Ar(N)/Mr(HNO3) × 100% = 14/63 × 100% = 22,2%
омега(О) = Ar(O)/Mr(HNO3) × 100% = 48/63 × 100% = 76,2%