Keeping your arm straight in front of you, you rotate 90° to your left, and see the left side of the circle lit while the right side is dark. Half the ball is still lit up, but you can see only part of the lit area. As you continue to rotate, you see a different amount of the ball.
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.
If it’s a negative number and a positive it might be 1.4
Answer:
20L is the new volume
Explanation:
In this case, moles and T° from the gas remain constant. This is the formula we must apply, to solve this:
P₁ . V₁ = P₂ . V₂
5 atm . 10 L = P₂ . 2.5L
P₂ = (5 atm . 10 L) / 2.5L →20L
Answer:
Argon (Ar) is smallest element in 3rd period in periodic table.