Answer : The oxidizing element is N and reducing element is O.
is act as an oxidizing agent as well as reducing agent.
Explanation :
An Oxidizing agent is the agent which has ability to oxidize other or a higher in oxidation number.
Reducing agent is the agent which has ability to reduce other or lower in oxidation number.
The given reaction is :
act as an oxidizing agent.
The oxidation number of N in is calculated as:
(+1)+(x)+3(-2) = 0
x = +5
And the oxidation number of N in 
is calculated as:
(+1)+(x)+2(-2) = 0
x = +3
From the oxidation number method, we conclude that the oxidation number reduced this means itself get reduced to and it can act as an oxidizing agent.
act as a reducing agent.
The oxidation number of O in is calculated as:
(+1)+(+5)+3(x) = 0
x = -2
The oxidation number of O in 
is Zero (o).
Now, we conclude that the oxidation number increases this means itself get oxidized to and it can act as reducing agent.
The chalk particles embed themselves into the small pores on the surface.
Although a chalkboard seems smooth to the touch, it is quite rough at the microscopic level, with <em>pores</em> that reach below the surface.
When you drag chalk across the board, friction causes small particles of chalk to rub off onto the surface.
If you leave the markings for a long time, some of the chalk particles will work their way into the pores.
A brush will remove the surface particles, but <em>it will not be able to get at the particles in the pores</em>.
Answer:
both spheres have a positive charge
Metals on the left side, metalloids on the staircase, nonmetals on right side
H2SO.Mgslfurmobile phase in this experiment
