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.
According to Diagram B, look at the 1600 elevation until you see the descending air line touches it. Then look down at the temperature at the bottom of the graph. It is between 0 degrees to 5 degrees.
The only number that is between that range is 2 degrees C.
1a. calcium chloride (CaCl2)
b. 2HCl (aq) + Ca(OH)2 (s) —> CaCl2 (aq) + 2H2O (l)
i’m not sure about the rest but i hope this helped ^^
Empirical formula is the simplest ratio of whole numbers of components in a compound
molecular formula is the actual ratio of components in a compound .
the molecular formula for the compounds given are as follows
ethyne - C₂H₂
ethene - C₂H₄
ethane - C₂H₆
methane - CH₄
the actual ratios of the elements simplified ratio
C : H C : H
ethyne 2:2 1:1
ethene 2:4 1:2
ethane 2:6 1:3
methane 1:4 1:4
the only compound where the actual ratio is equal to the simplified ratio is methane
therefore in methane molecular formula CH₄ is the same as empirical formula CH₄
