Answer:
C
Explanation:
This group of starts creates a pattern
Electrochemical cell representation for above reaction is,
Br-/Br2//I2/I-
Reaction at Anode: Br2 + 2e- → 2Br- (1)
Reaction at Cathode: 2I- → I2 + 2e- (2)
Standard reduction potential for Reaction 1 = Ered(anode) = 1.066 v
Standard reduction potential for Reaction 2 = Ered(cathode) = 0.535 v
Eo cell = Ered(cathode) - Ered(anode)
= 0.535 - 1.066
= -0.531v
Now, we know that ΔGo = -nF (Eo cell) ..............(3)
Also, ΔGo = RTln(K) ..........(4)
Equation 3 and 4 we get,
ln (K) = nF (Eo cell) / RT
= 2 X 96500 X (-0.531)/ (8.314 X 298)
∴ K = 1.085 X 10^-18.
Long wave I think is the correct answer
Assuming that the ammonium sulfide formula is (NH4)2S then you can see that there are 2 nitrogen, 8 hydrogen and 2 sulfur atoms for every ammonium sulfide. If the amount of ammonium sulfide is 8.9 moles, then the number of hydrogen atoms should be: 8/1 * 8.9 mol= 71.2 moles
Kinda hard to draw on here...but there are 11 electrons.....2 on first shell, 8 on second and one on outer shell