There are 6 Nitrogen atoms
Explanation:
The chemical equation is as follows.

And, the given enthalpy is as follows.
;
= 102.5 kJ
Cl-Cl = 243 kJ/mol, O=O = 498 kJ/mol
Since, the bond enthalpy of Cl-Cl is not given so at first, we will calculate the value of Cl-Cl as follows.
102.5 = ![[(\frac{1}{2})x + 498] - [(2)(243)]](https://tex.z-dn.net/?f=%5B%28%5Cfrac%7B1%7D%7B2%7D%29x%20%2B%20498%5D%20-%20%5B%282%29%28243%29%5D)
102.5 = 
102.5 - 12 = 
x = 181 kJ
Now, total bond enthalpy of per mole of ClO is calculated as follows.

x = ![[(\frac{1}{2})181 + (\frac{1}{2})498] - 243](https://tex.z-dn.net/?f=%5B%28%5Cfrac%7B1%7D%7B2%7D%29181%20%2B%20%28%5Cfrac%7B1%7D%7B2%7D%29498%5D%20-%20243)
= 339.5 - 243
= 96.5 kJ
Thus, we can conclude that the value for the enthalpy of formation per mole of ClO(g) is 96.5 kJ.
Answer:
The standard enthalpy of combustion of solid urea ((CO(NH2)2) is -632 kJ mol-1 at 298 K and its standard molar entropy is 104.60 J K-1 mol-1
Explanation:
1) divide each percentage by the relative atomic mass of the element
2) divide all results by the smallest number
3)multiply by a whole number to get the simplest whole number ratio (if necessary)
that is to say:
Na S O
32.37÷23 22.58÷32 45.05÷16
= 1.407 = 0.7056 = 2.816 (to 4 significant figures)
the smallest number here is 0.7056 so:
1.407÷0.7056 0.7056÷0.7056 2.816÷0.7056
=1.99 approx.2 = 1 3.99 approx. 4
here there is no need to carry out step 3 as ratio obtained is already a simplest whole number ratio
so empirical formula is: Na₂SO₄