Answer: 0.113 moles of NaCl are created as a result of decomposing 12 grams of
.
Explanation:
To calculate the moles :

The balanced chemical equation for decomposition of
is:
According to stoichiometry :
2 moles of
give = 2 moles of 
Thus 0.113 moles of
give =
of 
Thus 0.113 moles of NaCl are created as a result of decomposing 12 grams of
.
11×2=22
(11×2)+(6)+(8×3)=52
22/52=0.4230
0.4230×100=42.3%
Answer: -105 kJ
Explanation:-
The balanced chemical reaction is,

The expression for enthalpy change is,
![\Delta H=\sum [n\times B.E(reactant)]-\sum [n\times B.E(product)]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5Csum%20%5Bn%5Ctimes%20B.E%28reactant%29%5D-%5Csum%20%5Bn%5Ctimes%20B.E%28product%29%5D)
![\Delta H=[(n_{N_2}\times B.E_{N_2})+(n_{H_2}\times B.E_{H_2}) ]-[(n_{NH_3}\times B.E_{NH_3})]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B%28n_%7BN_2%7D%5Ctimes%20B.E_%7BN_2%7D%29%2B%28n_%7BH_2%7D%5Ctimes%20B.E_%7BH_2%7D%29%20%5D-%5B%28n_%7BNH_3%7D%5Ctimes%20B.E_%7BNH_3%7D%29%5D)
![\Delta H=[(n_{N_2}\times B.E_{N\equiv N})+(n_{H_2}\times B.E_{H-H}) ]-[(n_{NH_3}\times 3\times B.E_{N-H})]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B%28n_%7BN_2%7D%5Ctimes%20B.E_%7BN%5Cequiv%20N%7D%29%2B%28n_%7BH_2%7D%5Ctimes%20B.E_%7BH-H%7D%29%20%5D-%5B%28n_%7BNH_3%7D%5Ctimes%203%5Ctimes%20B.E_%7BN-H%7D%29%5D)
where,
n = number of moles
Now put all the given values in this expression, we get
![\Delta H=[(1\times 945)+(3\times 432)]-[(2\times 3\times 391)]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B%281%5Ctimes%20945%29%2B%283%5Ctimes%20432%29%5D-%5B%282%5Ctimes%203%5Ctimes%20391%29%5D)

Therefore, the enthalpy change for this reaction is, -105 kJ
The kinetic molecular theory<span> of gases is stated in the following four </span>principles<span>: The space between gas </span>molecules<span> is much larger than the </span>molecules<span> themselves. Gas </span>molecules<span> are in constant random motion. The average </span>kinetic<span> energy is determined solely by the temperature.
I got this from my notes from my chemistry class last semester
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