the cell structure represented by x is the nucleus
AgF consists of Ag+ and F- ions, which are fully dissociated in aqueous solution. When solving electrolysis problems, it is important to remember that water itself may also be a subject to electrolysis. Therefore, determining which species is oxidized and which species is reduced depends on selecting the processes that are the most energetically favorable. The most preferred reduction reaction will be Ag+ + e- = Ag (Emf=0.7996 V) which will occur at the cathode, on the other hand, the most favorable oxidation reaction will be
2H2O = O2 +4H+ + 4e- (Emf = -1.3 V) that will occur at the anode. Thus, the product at the anode is oxygen gas and at the cathode electrode is silver metal.
Answer:
For me
Explanation:
It's Cu because other compound contains negative radicals
Answer:
272
Explanation:
hope it helps i double checked the math :D
Answer:
Kc for this reaction is 0.43
Explanation:
This is the equilibrium:
N₂(g) + 2H₂O(g) → 2NO(g) +2H₂(g)
And we have all the concentration at equilibrium:
N₂: 0.25M
H₂ : 1.3M
NO: 0.33M
H₂: 1.2M
They are ok, because they are in MOLARITY. (mol/L)
Let's make the expression for Kc
Kc = ( [NO]² . [H₂]² ) / ([N₂] . [H₂O]²)
Kc = (0.33² . 1.2²) / (0.25 . 1.2²)
Kc = 0.4356
In two significant digits. 0.43