Answer:
The correct answer is:
An electron will be emitted in the second experiment, but it cannot be determined whether it will reach the second plate.
Explanation:
In fact, violet has higher frequency than green light. This means that photons on violet carry more energy than photons of green light (remember that the energy of a photon is proportional to it's frequency:

, so when they hit the surface of the metal, more energy is transferred to the electrons. The electron was already emitted with green light, so it must be emitted with also violet light, given the more energy transferred.
Pretty sure it's b but not an definitely
Explanation:
The question pretty much requires us to find the amount of moles of each compounds based on the number of moles of O given.
H2SO4
1 mol of H2SO4 contains 4 mol of O
x mol of H2SO4 would contain 3.10 mol of O
x = 3.10 * 1 / 4 = 0.775 mol of H2SO4
C2H4O2
1 mol of C2H4O2 contains 2 mol of O
x mol of C2H4O2 would contain 3.10 mol of O
x = 3.10 * 1 / 2 = 1.55 mol of C2H4O2
NaOH
1 mol of NaOH contains 1 mol of O
x mol of NaOH would contain 3.10 mol of O
x = 3.10 * 1 / 1 = 3.10 mol of NaOH
Your answer is False I think