chemistry

An Ampere is one coulomb of electrons per second.

So o.o250 amperes for 300 seconds (5 min) is 7.5 coulombs of electrons

Now for the electrolysis of water--the anode collects oxygen, the cathode hydrogen by the reaction H2O(l)-> H2(g)+1/2O2(g) by the two half cell reactions

2 H+(aq) + 2e− → H2(g) (cathode)
and
2 H2O(l) → O2(g) + 4 H+(aq) + 4e− (anode)

1 molecule of O2 needs 4 electrons---1 coulombs is about 6.2E+18 electrons, so 7.5 coulombsis about 4.65E+19 electrons--will a 100% anode efficiency this would produce 1.16E+19 molecules of O2(g) at the anode--

Convert this to moles (6.023E+23) molecules (1.16E+19/6.023E +23=1.9E-5 moles)

Finally the gas calculation. 1 mole of an ideal gas at STP (1 atm, 25C) has a volume of 22.4 liter so this amount of gas (1.9E-5 mole) is 4.4E-4 liter (0.42 ml) at STP. At 19C this is 0.41ml

Rap

your question must have a error in it

thanks

sorry, here is the anode equation:
2H2O(l) ----> O2(g)+4H+(aq)+4e-

Thats right i will post you the answer ver soon