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 average kinetic energy and thermal energy decreases, and the temperature decreases as well.
Answer:
Colors of transition metal compounds are due to two types of electronic transitions. Due to the presence of unpaired d electrons, transition metals can form paramagnetic compounds. Transition metals are conductors of electricity, possess high density and high melting and boiling points.
Explanation:
Δmc
2
For one reaction:
Mass Defect =Δm
=2(m
H
)−m
He
−m
n
=2(2.015)−3.017−1.009
=0.004 amu
1 amu=931.5 MeV/c
2
Hence,
E=0.004×931.5 MeV=3.724 MeV
E=3.726×1.6×10
−13
J=5.96×10
−13
J
For 1 kg of Deuterium available,
moles=
2g
1000g
=500
N=500N
A
=3.01×10
26
Energy released =
2
N
×5.95×10
−13
J
=8.95×10
13