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已知在0.5mol/LH2SO4介质中,φθ(Ce4+/Ce3+)=1.44V,φθ(Fe3+/Fe2+)=0.68V。计算此条件下以0.100mol/LCe4+滴定0.100mol/LFe2+至化学计量点时,反应物及滴定产物的浓度()

  • A、[Ce4+]=[Fe2+]=1.8×10-8mol/L,[Ce3+]=[Fe3+]≈0.050mol/L
  • B、[Ce4+]=[Fe3+]≈0.050mol/L,[Ce3+]=[Fe2+]=2.0×10-8mol/L
  • C、[Ce4+]=[Fe2+]=0.047mol/L,[Ce3+]=[Fe3+]=0.003mol/L
  • D、[Ce4+]=[Fe3+]=0.047mol/L,[Ce3+]=[Fe2+]=0.003mol/L

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更多 “已知在0.5mol/LH2SO4介质中,φθ(Ce4+/Ce3+)=1.44V,φθ(Fe3+/Fe2+)=0.68V。计算此条件下以0.100mol/LCe4+滴定0.100mol/LFe2+至化学计量点时,反应物及滴定产物的浓度()A、[Ce4+]=[Fe2+]=1.8×10-8mol/L,[Ce3+]=[Fe3+]≈0.050mol/LB、[Ce4+]=[Fe3+]≈0.050mol/L,[Ce3+]=[Fe2+]=2.0×10-8mol/LC、[Ce4+]=[Fe2+]=0.047mol/L,[Ce3+]=[Fe3+]=0.003mol/LD、[Ce4+]=[Fe3+]=0.047mol/L,[Ce3+]=[Fe2+]=0.003mol/L” 相关考题
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