Base on your question where a concentration cell consist of two SN/SN2+half cells. The solution in one half cell A is 0.13M SN(NO3)2 and is 0.87 M Sn(NO3)2 in the other half cell to get the cell potential at 25 degree the answer is 0.059/2 log0.13/0.87
So I’m not 100% sure what you’re asking but I’m going to give it a go. The elimination reaction is a term used in organic chemistry that describes a type of reactions. The name kinda tells you what’s going to happen. Something is going to be removed/eliminated from initial reactant/substrate and as a result, an alkene (double bond containing compound) will form.
In elimination reactions a hydrogen atom is first removed (as a H+) from the beta carbon. As a result, the left behind electrons create a pi bond between the beta carbon and the neighboring alpha carbon. This promotes the electronegative atom, on the alpha carbon, to leaves the substrate taking both electrons from the shared sigma bond with the alpha carbon.
Answer:
this is what i got
Explanation:
α-decay: When a radioactive nucleus disintegrates by emitting an αα-particle, the atomic number decreases by two and mass number decreases by four. Example: 88Ra226→86Rn222+2He4.
Your answer would be they have mobile ions in solution.
Yes, the atomic radius increases as you move down a group of elements.
this is true
going down leads to valence electrons that are further away from nucleus -> less electrostatic attraction -> less pull towards nuc. -> greater radius/volume taken
