Use the reactions given below to determine the ∆H^o rxn in kJ/mol rxn. ½ H2(g) + ½ I2(s) → HI(g) ∆H = 26 kJ/mol rxn
½ H2(g) + ½ I2(g) → HI(g)∆H = −5.0 kJ/mol rxn

i believe it is c.) 1575 j.

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%

pie chart because it is in percents ( % )