Answer:
0.238 millimoles
Explanation:
Given:
Mass of the table salt = 1.00 lb = 454 g
Copper(I) Iodine (cul) present in the table = 0.0100% by mass
or
Copper(I) Iodine (cul) present in the 1 pound of salt = 0.01% of 454 g
= 0.0454 grams
Molar mass of the CuI = 63.54 grams + 126.90 grams = 190.44 grams
Thus,
the number of moles of CuI present in 1 pound of salt = Mass / Molar mass
= 0.0454 / 190.44
= 2.38 × 10⁻⁴ moles
= 0.238 millimoles
Now help! ASAP!I’ll make you brainly now please
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-
The answer is d
the image shows the moles of gas increasing as well as volume increasing
