Answer:
b. erosion.
I believe that this is the best answer
Answer:
58.8%
Explanation:
n= Experimental/Theoretical ·100
n= 5.50/(your answer to part A) 9.36·100 = 58.8%
When battery discharge / delivering current the lead at the anode is oxidized
that is ;
pb---->pb+ 2e-
since the lead ions are in presence of aquous sulfate in insoluble lead sulfate precipitate onto the electrode
the overall reaction at the anode is therefore
Pb + SO4^2- ---> PbSO4 + 2e-
8moles of SO_2 need 11mol O_2
Moles of sO_2

So
Moles of O_2
We know
- 1mol at STP=22.4L
- 1.4mol=0.14(22.4)=3.13L O_2
Unsafe to dono because it’s way too high