Answer:
Rate of the reaction is 0.2593 M/s
-0.5186 M/s is the rate of the loss of ozone.
Explanation:
The rate of the reaction is defined as change in any one of the concentration of reactant or product per unit time.

Rate of formation of oxygen : 
Rate of the reaction(R) =![\frac{-1}{2}\frac{d[O_3]}{dt}=\frac{1}{3}\frac{d[O_2]}{dt}](https://tex.z-dn.net/?f=%5Cfrac%7B-1%7D%7B2%7D%5Cfrac%7Bd%5BO_3%5D%7D%7Bdt%7D%3D%5Cfrac%7B1%7D%7B3%7D%5Cfrac%7Bd%5BO_2%5D%7D%7Bdt%7D)
![R=\frac{1}{3}\frac{d[O_2]}{dt}](https://tex.z-dn.net/?f=R%3D%5Cfrac%7B1%7D%7B3%7D%5Cfrac%7Bd%5BO_2%5D%7D%7Bdt%7D)
Rate of formation of oxygen=3 × (R)

Rate of the reaction(R): 
Rate of the reaction is 0.2593 M/s
Rate of disappearance of the ozone:
![R=-\frac{1}{2}\frac{d[O_3]}{dt}](https://tex.z-dn.net/?f=R%3D-%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7Bd%5BO_3%5D%7D%7Bdt%7D)
![\frac{d[O_3]}{dt}=-2\times R=-2\times 0.2593\times M/s=-0.5186M/s](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5BO_3%5D%7D%7Bdt%7D%3D-2%5Ctimes%20R%3D-2%5Ctimes%200.2593%5Ctimes%20M%2Fs%3D-0.5186M%2Fs)
-0.5186 M/s is the rate of the loss of ozone.
The answer is b since on a it says H2 but on the right side there is no H idk if you forgot to put H there so im guessing b. Have a good day
Answer: The entropy change of the surroundings will be -17.7 J/K mol.
Explanation: The enthalpy of vapourization for 1 mole of acetone is 31.3 kJ/mol
Amount of Acetone given = 10.8 g
Number of moles is calculated by using the formula:

Molar mass of acetone = 58 g/mol
Number of moles = 
If 1 mole of acetone has 32.3 kJ/mol of enthalpy, then
0.1862 moles will have = 
To calculate the entropy change for the system, we use the formula:

Temperature = 56.2°C = (273 + 56.2)K = 329.2K
Putting values in above equation, we get
(Conversion Factor: 1 kJ = 1000J)
At Boiling point, the liquid phase and gaseous phase of acetone are in equilibrium. Hence,


There are 9 square meters in one square yard so divide 27.99 by 9
27.99/9=3.11
one square meter is $3.11
I hope I've helped!