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^^(m)^(0) for NaCl, HCl and NaAc are 126...

`^^_(m)^(0)` for NaCl, HCl and NaAc are `126.4, 425.0 and 91.0 S cm^(2) mol ^(-1)` respectively. Calculate `^^(0)` for Hac.

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`^^_(m(HAc))^(0) = lambda_(H^(+))^(o) + lambda_(Ac^(-))^(o) = lambda_(H^(+))^(o) + lambda_(Cl^(-))^(o)- lambda_(Ac^(-))^(o) + lambda_(Na^(+))^(o) - lambda_(Cl^(-))^(o) - lambda_(Na^(+))^(o)`
` = ^^_(m)^(0)(HCl) + ^^_(m(NaAc))^(0) - ^^_(m(NaCl))^(0)`
` = (425.9 + 91.0 - 126.4) S cm^(2) mol^(-1)`
` = 390.5 S cm^(2) mol`
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  2. i) Arrange the following metals in the order in which they displace ea...

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