To determine the pOH assuming water is the universal solvent take the value of 10 ^ -14 and then divide it by the hydronium concentration and then take the negative logarithm of the final answer that is the solution to the hydroxide ion concentration in the solution.
Answer:
The enthalpy change for the given reaction is 424 kJ.
Explanation:

We have :
Enthalpy changes of formation of following s:



(standard state)
![\Delta H_{rxn}=\sum [\Delta H_f(product)]-\sum [\Delta H_f(reactant)]](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%3D%5Csum%20%5B%5CDelta%20H_f%28product%29%5D-%5Csum%20%5B%5CDelta%20H_f%28reactant%29%5D)
The equation for the enthalpy change of the given reaction is:
=

=


The enthalpy change for the given reaction is 424 kJ.
FeSO₄*7H₂O(s) = FeSO₄(s) + 7H₂O(g)
M(FeSO₄*7H₂O)=278.0 g/mol
M(FeSO₄)=151.9 g/mol
m(FeSO₄*7H₂O)/M(FeSO₄*7H₂O)=m(FeSO₄)/M(FeSO₄)
m(FeSO₄)=M(FeSO₄)m(FeSO₄*7H₂O)/M(FeSO₄*7H₂O)
m(FeSO₄)=151.9*100.0/278.0=54.6 g
m(FeSO₄)=54.6 g
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