87 is a really good grade! anyway the answer is homogeneous mixture :)
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:c c c c c c c c c c c c c c c c c c c c c c c c c c cc c c c c c c c c
cc
Explanation:
1.204 x 10 ( 24 ) atoms of magnesium are present.
In order to go from mass of magnesium there have to do 2 things
1. convert mass of Mg to moles of Mg using the molar mass of Mg as converted to a factor
2. convert moles of Mg to atoms of Mg using the Avogadro's number ( 6.02 x 10 (23) as a conversion factor
Hope my it helped a little :) <span />
<span>2Al + 3Br2 -------------> 2AlBr3
</span>3 g Al = 0.11 mol Al.
<span>6 g Br2 = 0.0375 mole bromine (it is diatomic). </span>
<span>moles of aluminium will take part in reaction = 0.0375 X (2/3) = 0.025. </span>
<span>Gram-mole of AlBr3 will be produced = 0.025 mole = 6.6682 g.
</span>moles of Al left = 0.11 - 0.025 = 0.086<span>
</span>