Iron(III) sulfate, Fe2(SO43, decomposes upon heating to make iron(III) oxide and sulfur trioxide by the following reaction: Fe2(SO4l3(s) Fe2O3(s) + 3 SO3s(g) What is the variance of the system? OF= 2 OF=3 OF=1 OF=0

11. Using Hess' Law, calculate the AHOrxn for the following reaction: CH4 (8) + 1/2O2 (8) CO (8) + 2 H2 (8) The following information is given: CH4 + 202 (8) (8) CH4 + CO₂ (g) (8) CH4 + H₂O (g) (g) → → CO2 (g) + 2 H₂O (1) 2 CO (8) 2 H2 (g) + CO(g) + 3 H₂ (g) AH = ΔΗ = - 802 kJ 247 kJ ΔΗ - 206 kJ

Hw.180.

Lauto.d21?ou=2194845&isprv3D&drc=0&qi=2824083&cfql=0&dnb=D0&f Calculate the standard enthalpy change for the reaction 2C8H18(1) + 17O2(g) → 16CO(g) + 18H20(1) Given: 2C3H18(1) + 2502(g) – 16CO2(g) + 18H20(1) AH° = -11020 kJ/mol 2C0(g) + O2(g) – 2CO2(g) AH° = -566 kJ/mol a) 10450 kJ/mol b) -6492 kJ/mol O c) 15550 kJ/mol d) 6492 kJ/mol e) -10450 kJ/mol

Calculate A Hrxn for the following reaction: 2C,H4 (g) + 2H2 (g) →2C2H6 (g) Using the following data and Hess's Law. C2H4 (g) + 302 (g) - 2CO2 (g) + 2H2O () ΔΗ -1411 kJ = 2C2H6 (g) + 702 (g) 4CO2 (g) + 6H2O (1) ΔΗ -1560 kJ = 2H2 (g) + O2 (g)- → 2H20 (g) AH= -285.8 kJ

For the states and processes shown below. Calculate 1 W3 in Joules: 3 Pi= P3=200Kta P= 50KPa 2 V3=1m?

2HNO3 (1) + 6HB1(g) 4H2O(1) + 2NO(g) + 3B12 (1) Standard Entropies at 25°C (J/mol K) HBr(g) 198.6 Br2 (1) HNO3 (1) 152.2 155.6 NO(g) 210.7 H2O(1) 69.9 AS J/K 3C12 (g) + 3H2O(1) → 5C1¯(ag) + Cl03 (aq) + 6H* (ag) Standard Entropies at 25°C (J/mol·K) Cl2 (9) Cl03 (ag) CI (ag) H* (ag) H,O(1) 223.0 162.3 56.5 0.0 69.9 AS° = J/K 2MNO4 (ag) + 6I (ag) + 8H* (ag) 312 (s) + 2MnO2(s) + 4H2O(1) Standard Entropies at 25°C (J/mol K) I (ag) 111.3 I2 (8) 116.1 MnO4 (aq) 191.2 MnO2 (s) 53.0 H* (ag) 0.0 H2O(1) 69.9 AS° J/K

Calculate the average deviation of the following specific heat determinations: 0.18 J/g °C 0.34 J/g °C 0.27 J/g °C

please show me how to answer this using hess law

1) PCI3(1) + Cl2(g) - PCI5(g) Jse equations (2) and (3) to calculate AHrxn of equation (1): 2) P4(s) + 6Cl2(g)→ 4PCI3(1) AH =-1280 kJ 3) P4(s) + 10C12(g) → 4PCI5(g) AH =-1774 kJ