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[tex]\displaystyle TERMOKIMIA \ [XI] \\ \\ n \ LiCl \ = \frac{4,25 \ g}{42,5 \ g/mol} \ = 0,1 \ mol \\ \\ Assume \ \rho = 1 \ g/cm^{3} \\ m = 4.25 \ g \ + \ 395,75 \ g \ = \ 400 \ g \\ \\ q \ = m \ . \ c \ . \ \delta T \\ - \frac{\delta H}{n} = 400 \ . \ c \ . (27,5\°C - 25\°C) \\ - \frac{- \ 40}{0,1} = 1000 \ . \ c \\ c \ = \ 0,4 \ J.g^{-1}.\°C^{-1} \ (B) \ -\ \textgreater \ \ Jawab[/tex]