I am so lost. Does anyone know how to do this??? Enthalpies of reaction stuff C2H4(g)+3O2(g)→2CO2(g)+2H2O(l) ΔH∘1=?

The combustion of C2H4(g) is represented by the equation above.

(a) Use the enthalpies of formation in the table below to calculate the value of ΔH∘1 for the reaction.

ΔH∘f(kJ/mol)
C2H4(g) 52
CO2(g) −394
H2O(l) −286
O2(g) 0
Please help!!! :(

-1464 kJ/mol

Explanation:

The balanced equation for the reaction is given as;

C2H4(g)+3O2(g)→2CO2(g)+2H2O(l) ΔH∘1=?

The enthalpy of the reaction is given by the equation;

Enthalpy of reaction = Enthalpy of products - Enthalpy of reactants

Products:

2CO2(g)+2H2O(l)

Enthalpy of Products = 2 (−394) + 2(−286)

Enthalpy of Products = -1360 kJ/mol

Reactants:

C2H4(g)+3O2(g)

Enthalpy of Reactants = 2 (52) + 3(0)

Enthalpy of Reactants = 104 kJ/mol

Enthalpy of Reaction =  -1360 - 104 = -1464 kJ/mol

the correct answer is b. a strong acid completely dissociates to its ions when dissolved in water. completely dissociates to and ions. a is incorrect because does not fully ionize to and ions and some remain as the compound . c and d are incorrect because, and do not fall into the category of bronsted lowry acids. a bronsted lowry acid is a proton donor and since both molecules do not have a proton to donate they are not bronsted lowry acids. likewise,   if we use the lewis acid definition, they cannot be acids. a lewis acid is an electron pair acceptor. both and have a complete octet so can not accept any electrons which disqualifies them as acids.

you gotta add the following molecules so we can give u an answer

explanation: