At 25°c, the following heats of reaction are known: 2clf(g) + o2(g) → cl2o(g) + f2o(g) δh°rxn = 167.4 kj/mol 2clf3(g) + 2o2(g) → cl2o(g) + 3f2o(g) δh°rxn = 341.4 kj/mol 2f2(g) + o2(g) → 2f2o(g) δh°rxn = –43.4 kj/mol at the same temperature, use hess's law to calculate δh°rxn for the reaction: clf(g) + f2(g) → clf3(g)

-108.7 KJ/mol

Explanation:

Hess's law state that in a multiple stage reaction, the enthalpy change is the sum of all the enthalpy changes.

2CIF (g) + O₂ (g) → Cl₂O(g) + F₂O (g)        ΔH°rxn =167.4  KJ/mol

2ClF₃(g) + 2O₂(g) → Cl₂O(g) + 3F₂O(g) ΔH°rxn = 341.4 kJ/mol

flip the above reaction

Cl₂O(g) + 3F₂O(g) → 2ClF₃(g) + 2O₂(g) ΔH°rxn = -341.4 kJ/mol

2CIF (g) + O₂ (g) → Cl₂O(g) + F₂O (g)        ΔH°rxn = 167.4  KJ/mol

2F₂(g) + O₂(g) → 2F₂O(g) ΔH°rxn = –43.4 kJ/mol

add all the enthalpy change = -341.4 kJ/mol + 167.4  KJ/mol – 43.4 kJ/mol = -217.4

2CIF (g) + 2F₂(g)  → 2ClF₃(g)          ΔH°rxn =  -217.4 KJ/mol

ClF(g) + F2(g) → ClF3(g) =         (ΔH°rxn =  -217.4 KJ/mol) / 2 = -108.7 KJ/mol

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%

the answer is:  

line graphs show patterns in data with variables that change over time; whereas, circle graphs show data as parts out of a whole (such as out of 100%).