Consider the two reactions. 2nh3(g)+3n2o(g)4nh3(g)+3o2(g)⟶4n2(g)+3h2o(l)⟶2n2(g)+6h2o(l) δ∘=−1010 kjδ∘=1531 kj using these two reactions, calculate and enter the enthalpy change for the reaction below. n2(g)+12o2(g)⟶n2o(g)

The for the reaction is 591.9 kJ.

Explanation:

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.

The chemical reaction for the formation reaction of is:

The intermediate balanced chemical reaction are:

(1)        ( ÷  3)

(2) ( ÷  6)

Reversing Equation 1 and then adding both the equations, we get the enthalpy change for the chemical reaction.

Putting values in above equation, we get:

Hence, the for the reaction is 591.9 kJ.

A) -5

Answer to the 2nd question: the linear factors of the cubic are: A) (x-2)(x-3)(x-5)

Answer to the 3rd question: to solve for the leading coefficient, use: D) -5=a(0-2)(0-3)(0-5)

Answer to the 4th question: A=1/6

Answer to the 5th question: write the function: f(x)=1/6  x^3-=5/3 x^2+=31/6 x-=5