Answer: they will meet and make a biger wave then seperat
Explanation: When two or more waves meet, they interact with each other. The interaction of waves with other waves is called wave interference. Wave interference may occur when two waves that are traveling in opposite directions meet. The two waves pass through each other, and this affects their amplitude.
Answer:
The equilibrium shifts to produce more reactants.
Explanation:
According to the Le- Chatelier principle,
At equilibrium state when stress is applied to the system, the system will behave in such a way to nullify the stress.
The equilibrium can be disturb,
By changing the concentration
By changing the volume
By changing the pressure
By changing the temperature
Consider the following chemical reaction.
Chemical reaction:
2SO₂ + O₂ ⇄ 2SO₃
In this reaction the equilibrium is disturb by increasing the concentration of Product.
When the concentration of product is increased the system will proceed in backward direction in order to regain the equilibrium. Because when product concentration is high it means reaction is not on equilibrium state. As the concentration of SO₃ increased the reaction proceed in backward direction to regain the equilibrium state and more reactant is formed.
Answer : The correct option is, (b) +0.799 V
Solution :
The values of standard reduction electrode potential of the cell are:
![E^0_{[H^{+}/H_2]}=+0.00V](https://tex.z-dn.net/?f=E%5E0_%7B%5BH%5E%7B%2B%7D%2FH_2%5D%7D%3D%2B0.00V)
![E^0_{[Ag^{+}/Ag]}=+0.799V](https://tex.z-dn.net/?f=E%5E0_%7B%5BAg%5E%7B%2B%7D%2FAg%5D%7D%3D%2B0.799V)
From the cell representation we conclude that, the hydrogen (H) undergoes oxidation by loss of electrons and thus act as anode. Silver (Ag) undergoes reduction by gain of electrons and thus act as cathode.
The half reaction will be:
Reaction at anode (oxidation) :
Reaction at cathode (reduction) :
The balanced cell reaction will be,

Now we have to calculate the standard electrode potential of the cell.

![E^o_{cell}=E^o_{[Ag^{+}/Ag]}-E^o_{[H^{+}/H_2]}](https://tex.z-dn.net/?f=E%5Eo_%7Bcell%7D%3DE%5Eo_%7B%5BAg%5E%7B%2B%7D%2FAg%5D%7D-E%5Eo_%7B%5BH%5E%7B%2B%7D%2FH_2%5D%7D)

Therefore, the standard cell potential will be +0.799 V