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DeltaH(f)^(@) for HgO is -21.7 kcal//mol...

`DeltaH_(f)^(@)` for `HgO` is `-21.7 kcal//mol` at `25^(@)C`. Calculate the mass of mercury which can be liberated by treatment of excess of `HgO` with `10.0 kcal. At. Wt.` of `Hg=200.6`.

Text Solution

Verified by Experts

Given `Hg+1//2Orarr HgO, DeltaH^(@)=-21.7kcal`
`:. HgOrarr Hg+1//2O,DeltaH^(@)+21.7kcal,`
`21.7kcal` heat gives `200.6g Hg`
`10.0 kcal `heat gives `(200xx10)/(21.7)=92.44g Hg`
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