Answer:
1. The oxidation half-reaction is: Mn(s) ⇄ Mn²⁺(aq) + 2e⁻
2. The reduction half-reaction is: Ag⁺(aq) + 1e⁻ ⇄ Ag(s)
Explanation:
Main reaction: 2Ag⁺(aq) + Mn(s) ⇄ 2Ag(s) + Mn²⁺(aq)
In the oxidation half reaction, the oxidation number increases:
Mn changes from 0, in the ground state to Mn²⁺.
The reduction half reaction occurs where the element decrease the oxidation number, because it is gaining electrons.
Silver changes from Ag⁺ to Ag.
1. The oxidation half-reaction is: Mn(s) ⇄ Mn²⁺(aq) + 2e⁻
2. The reduction half-reaction is: Ag⁺(aq) + 1e⁻ ⇄ Ag(s)
To balance the hole reaction, we need to multiply by 2, the second half reaction:
Mn(s) ⇄ Mn²⁺(aq) + 2e⁻
(Ag⁺(aq) + 1e⁻ ⇄ Ag(s)) . 2
2Ag⁺(aq) + 2e⁻ ⇄ 2Ag(s)
Now we sum, and we can cancel the electrons:
2Ag⁺(aq) + Mn(s) + 2e⁻ ⇄ 2Ag(s) + Mn²⁺(aq) + 2e⁻
Answer:-
The reaction of 2-bromopropane reacts with sodium iodide in acetone is an example of Sn2 reaction.
The I - attacks from backside to give the transition state for both.
If we compare the transition state for cyclobromopropane 2-bromopropane then we see in case of cyclobromopropane transition state, one of the H is very close to the incoming I -.
This results in steric strain and less stability of the transition state. Hence 2-bromopropane reacts with sodium iodide in acetone over 104 times faster than bromocyclopropane.
Answer:
Dude im not 100% sure but I think its b and c im sorry if im wrong its just that im not really sure which ones are.
Explanation:
Answer:
According to the following position vs. Time graph the bicyclist was : D
D. Not moving for the first two seconds, then begin moving at the constant speed.
Explanation:
Along Y-axis = Distance in meters
Along X-axis = Time taken in second
- For first 2 seconds the The object has not changed its position.There is Zero distance covered.So , Speed = 0
- After 2 second,The Object changed the distance by equal amount in equal interval of time.(Uniform Speed).Hence at each point after 2 second, there is same value speed
