Given the reactions
a.) n2(g)+o2(> 2no(g) delta h= -180.5
b.) n2(g) + 3h2(g) ——> 2nh3(g) delta h= -91.8
c.)2h2(g)+ o2(g) —-> 2h2o(g) delta h= -486.6

what is the delta h for the reaction below
4nh3(g)+5o2(> 4no(g)+6h2o(g)

The answer to your question is: ΔH = 1637.8

Explanation:

Hess' law: This law states that the enthalpy change can be calculated even if it is not calculated directly.

"if a chemical change takes place by several routes, the overall enthalpy change is the same regardless the route".

Process

A) N2(g)+O2(g)—->2NO(g)                              Δ H= -180.5

B) N2(g) + 3H2(g) ——> 2NH3(g)                    Δ H= -91.8

C)2H2(g)+ O2(g) —-> 2H2O(g)                       Δ H= -486.6

The result must be:

                                4NH3(g)+5O2(g)—->4NO(g)+6H2O(g)

Turn letter B and multiply it by 2

                               4NH3   ⇒  2N2  +  6H2       ΔH = 183.6

Multiply letter A by 2

                               2N2 + 2O2 ⇒ 4 NO             ΔH = -361  

Multiply letter C by 3

                              6H2 + 3O2 ⇒ 6H2O            ΔH = -1459.8

Finally we add the equations up and simplify then:

                            4NH3 + 5O2 ⇒ 4NO + 6 H2O

And we add the ΔH = 183.6 - 361 - 1459.8

                                 = -1637.8

answers: to make sure that all of the broken pieces of tablet were added to the water, to allow for fair comparison of all three trials, with different particle sizes to avoid experimental error

the electron configuration that you gave me has 16 electrons ergo it has 16 protons ergo it is the element sulfur.