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Calculate Lambda_(m)^(oo) for AgCl given that : Lambda_(m)^(oo)AgNO_(3)=133.4" ohm"^(-1)cm^(2)"equiv"^(-1) Lambda_(m)^(oo)KCl=149.9" ohm"^(-1)cm^(2)"equiv"^(-1) Lambda_(m)^(oo)KNO_(3)=145.1" ohm"^(-1)cm^(2)"equiv"^(-1) |
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Answer» `Lambda_(m)^(oo)AGCL=[Lambda_(m)^(oo)AgNO_(3)]+[Lambda_(m)^(oo)KCl]-[Lambda_(m)^(oo)KNO_(3)]` `[133.4+149.9-145.1]=138.2" ohm "^(-1)cm^(2)EQUIV^(-1)`. |
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