





Answer:
Explanation:
From the correct question above:
The reaction can be represented as:

From the above reaction; the ICE table can be represented as:

I (mol/L) 0.086 0.28 0 0
C -4x -3x +2x +6x
E 0.086 - 4x 0.28 - 3x +2x +6x
At equilibrium;
The water vapor = 


![\text{equilibrium constant} ({k_c}) = \dfrac{ [N_2]^2 [H_2O]^6 }{ [[NH_3]^4] [O_2]^3 }](https://tex.z-dn.net/?f=%5Ctext%7Bequilibrium%20constant%7D%20%20%28%7Bk_c%7D%29%20%3D%20%20%5Cdfrac%7B%20%5BN_2%5D%5E2%20%5BH_2O%5D%5E6%20%7D%7B%20%5B%5BNH_3%5D%5E4%5D%20%5BO_2%5D%5E3%20%7D)

Replacing the value of x, we have:


The first reaction is a synthesis reaction (A+B -> AB)
Answer:conducts ,steel, polystyrene, thermal insulation
Explanation: