Calculate the enthalpy change for the reaction ca(s)+n2(g)+3o2(g)⟶ca(no3)2(s) from the following equations: 8ca(s)+ca(no3)2(s)→ca3n2(s)+6cao(s) δh = -3304 kj ca3n2(s)⟶3ca(s)+n2(g) δh = +432 kj 2cao(s)⟶2ca(s)+o2(g) δh = +1270 kj

Flip all three equations:

(4) Ca₃N₂(s)+6CaO(s) → 8Ca(s)+Ca(NO₃)₂(s)       ΔH = +3304 kJ

(5) 3Ca(s)+N₂(g) ⟶ Ca₃N₂(s)                                 ΔH = -432 kJ 

(6) 2Ca(s)+O₂(g) ⟶ 2CaO(s)                                 ΔH = -1270 kJ

Now, multiply the third equation by 3

(7) 6Ca(s)+3O₂(g) ⟶ 6CaO(s)                                 ΔH = -3810 kJ

Since you multiplied all the coefficients by 3, the delta H will also be multiplied by 3.

Now add equations 4, 5, and 7 together. Since we’re adding reactions, the delta H will all be added together

(8) Ca₃N₂(s) + 6CaO(s) + 3Ca(s)+  N₂(g) + 6Ca(s)+ 3O₂(g) ⟶ 8Ca(s) + Ca(NO₃)₂(s) + Ca₃N₂(s) + 6CaO(s)

ΔH = 3304 + (-432) + (-3810) kJ = -938 kJ

Simplify the equation. You will notice Ca₃N₂(s) and 6CaO(s) will cancel out.

3Ca(s)+  N₂(g) + 6Ca(s)+ 3O₂(g) ⟶ 8Ca(s) + Ca(NO₃)₂(s)

Simplify the Ca(s)

9Ca(s)+  N₂(g) + 3O₂(g) ⟶ 8Ca(s) + Ca(NO₃)₂(s)

We can remove 8 Ca(s) from both sides

Ca(s)+  N₂(g) + 3O₂(g) ⟶ Ca(NO₃)₂(s)         ΔH = -938 kJ

This is the equation we wanted the ΔH for. The ΔH for this reaction is -938 kJ.

metals are good conductor of electricity. it has free electrons on their surface. according to electron pool theory, each atom present in a metal crystal losses all of its valence electron and as a result a pool of free is form. these pool of electrons are free to move and   conduct electricity easily   while making metal as good conductor.

the correct answer is a. the ability of electrons to flow throughout the metal.

if it's not useful, feel free to report for mistake in answer.