Question deleted. Solution this question has been deleted Given data is : DyCl3(

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this question has been deleted Given data is : DyCl3(s) --> DyCl3(aq) : dH' =-180.06KJ/mol ----(1) Dy (s) + 3HCl -->DyCl3(aq) +3/2H2(g): dH'' = -699.43KJ/mol ----(2) 1/2H2(g) + 1/2Cl2(g) ---> HCl dH'' = -158.31KJ/mol ---(3) heat of formation of a compound is defined as the heat change during the formation of 1 mole of the compound from its constituents. The Equation representing the formation of DyCl3(s) is : Dy (s) + (3/2) Cl2 -----> DyCl3 (s) : dH = ? ---(4) Eq(4) can be obtained from the first three equations as follows : Eq(4) = reverse of Eq(1) + Eq(2) + 3 * Eq(3) So dH = - dH' + d'' + 3 * d''' = -(-180.06 KJ ) + (-699.43) +( 3 * (-158.31KJ)) = -994.30 KJ / mol So the dHf(DyCl3(s)) is = -994.30 KJ / mol

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