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

2. please answer these 2 for me

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The combustion of octane, C8H18,C8H18, proceeds according to the reaction shown.   2C8H18(l)+25O2(g)⟶16CO2(g)+18H2O(l)2C8H18(l)+25O2(g)⟶16CO2(g)+18H2O(l)   If 4.10×102 mol4.10×102 mol of octane combusts, what volume of carbon dioxide is produced at 19.0 ∘C19.0 ∘C and 0.995 atm?0.995 atm?

Consider the following reaction: C3H13(1) + 25/2 O»(g) → 8 CO2(g) + 9 H20(1) If d[CO2]/dt = 0.5OOM/s, determine the value of d[O2]/dt,

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.

mass of the calorimeter   162.2 g capacity of the calorimeter  1545.6 Mass of the calorimeter with 140 ml of water  271.3 g Water temperature   22 °C Temperature of the mixture   3 °C Mass of the calorimeter + mixture   310.3 g Calculate 11- ∆T H2O = (Tm - T cold H2O) = (Tm - T cold H2O) = 12- ∆T of ice melt = (T mixture - 0 °C) = (T mixture - 0 °C) = 13- ∆T ° C = ∆T cold H2O:14- Mass of water = ( mass of water + 140 ml of water - mass of empty calorimeter) =.15- Mass of ice = ( mass of calorimeter + mixture ) - (mass of water + 140 ml of water) =

What is the AH for the reaction below? C2H2(g) +3 H20() → CH4(g) + 02(9) + CH3OH() Use the following information to find AH for the reaction above. CH4(g) +2 02(g) → CO2(g) + 2 H20(I) AH= -890 kJ 2 C2H2(g) + 5 02(g) - 4 CO2(g) + 2 H20(I) AH= -2599 kJ 2 CH3OH() + 3 O2(g) - 2 CO2(g) + 4 H20() AH = -1453 kJ

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