<span>Oxidation is the loss of electrons and corresponds to an increase in oxidation state. The reduction is the gain of electrons and corresponds to a decrease in oxidation state. Balancing redox reactions can be more complicated than balancing other types of reactions because both the mass and charge must be balanced. Redox reactions occurring in aqueous solutions can be balanced by using a special procedure called the half-reaction method of balancing. In this procedure, the overall equation is broken down into two half-reactions: one for oxidation and the other for reduction. The half-reactions are balanced individually and then added together so that the number of electrons generated in the oxidation half-reaction is the same as the number of electrons consumed in the reduction half-reaction.</span>
Answer:
Model A
Explanation:
Model A represents an atom that is more reactive than the others represented.
Valence electrons actually determine the reactivity of elements. They also determine the properties of elements.
Elements with one valence electron are highly reactive because they need low energy to remove them. They can either gain more electrons to become stable or they share/give out their electrons.
Therefore, Model A is the correct answer because it has one valence electron and its valence electron is farther from the nucleus thereby this makes it more reactive.
Answer:
(-1) is the slope of a graph of In(y+3) on the vertical axis versus In(x-2) on the horizontal axis.
Explanation:

Taking natural logarithm on both the sides:
![\ln [(y+3)]-\ln[2]=\ln [b]-\ln [(x-2)]](https://tex.z-dn.net/?f=%5Cln%20%5B%28y%2B3%29%5D-%5Cln%5B2%5D%3D%5Cln%20%5Bb%5D-%5Cln%20%5B%28x-2%29%5D)
![\ln [(y+3)]=\ln[2]+\ln [b]-\ln [(x-2)]](https://tex.z-dn.net/?f=%5Cln%20%5B%28y%2B3%29%5D%3D%5Cln%5B2%5D%2B%5Cln%20%5Bb%5D-%5Cln%20%5B%28x-2%29%5D)
![\ln [(y+3)]=\ln {[2\times b]-\ln [(x-2)]](https://tex.z-dn.net/?f=%5Cln%20%5B%28y%2B3%29%5D%3D%5Cln%20%7B%5B2%5Ctimes%20b%5D-%5Cln%20%5B%28x-2%29%5D)
Slope intercept form is generally given as:

m = slope, c = intercept on y axis or vertical axis
On rearranging equation:
![\ln [(y+3)]=(-1)\times \ln [(x-2)]+\ln {2b}](https://tex.z-dn.net/?f=%5Cln%20%5B%28y%2B3%29%5D%3D%28-1%29%5Ctimes%20%5Cln%20%5B%28x-2%29%5D%2B%5Cln%20%7B2b%7D)
y = ln [(y+3)], x = ln [(x-2)], m=-1 , c = ln 2b
(-1) is the slope of a graph of In(y+3) on the vertical axis versus In(x-2) on the horizontal axis.
Answer:
Equilibrium constant of the given reaction is 
Explanation:
....
....
The given reaction can be written as summation of the following reaction-


......................................................................................

Equilibrium constant of this reaction is given as-
![\frac{[NOBr]^{2}}{[N_{2}][O_{2}][Br_{2}]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BNOBr%5D%5E%7B2%7D%7D%7B%5BN_%7B2%7D%5D%5BO_%7B2%7D%5D%5BBr_%7B2%7D%5D%7D)
![=(\frac{[NOBr]}{[NO][Br_{2}]^{\frac{1}{2}}})^{2}(\frac{[NO]^{2}}{[N_{2}][O_{2}]})](https://tex.z-dn.net/?f=%3D%28%5Cfrac%7B%5BNOBr%5D%7D%7B%5BNO%5D%5BBr_%7B2%7D%5D%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D%7D%29%5E%7B2%7D%28%5Cfrac%7B%5BNO%5D%5E%7B2%7D%7D%7B%5BN_%7B2%7D%5D%5BO_%7B2%7D%5D%7D%29)

