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:
5.52atm
Explanation:
Using the pressure law formula:
P1/T1 = P2/T2
Where;
P1 = initial pressure (atm)
P2 = final pressure (atm)
T1 = initial temperature (K)
T2 = final temperature (K)
According to the question, the following information were provided;
P1 = 4.72 atm
P2 = ?
T1 = -3.50°C = -3.50 + 273 = 269.5K
T2 = 42°C = 42 + 273 = 315K
Using P1/T1 = P2/T2
4.72/269.5 = P2/315
CROSS MULTIPLY
4.72 × 315 = 269.5 × P2
1,486.8 = 269.5P2
P2 = 1,486.8 ÷ 269.5
P2 = 5.52atm
Answer:
Ok:
Explanation:
So grams = mols*MolarMass. Here, MolarMass (MM) = 105.99g which can be found using the periodic table. mols is given to be 0.802. We can then plug in to get that it corresponds to 85.0g.
Answer:
Glucose also called(Energy)
