Answer:
Correct choice are C and D (they are both, the same).
Explanation:
Cathode is the positive(+) electrode where a reduction occurs.
Reduction is the chemical reaction where the oxidation state is reduced.
2Ag(s) + 1/2 O2(g) + 2H+(aq) → 2Ag+(aq) + H2O (l)
A. 2H2O (l) → O2 (g) + 4H+ (aq) + 4e-
B. 2Ag (s) → 2Ag+ (aq) + 2e-
C. 1/2 O2 (g) + 2H+ (aq) + 2e- → H2O (l)
D. 1/2 O2 (g) + 2H+ (aq) + 2e- → H2O (l)
C or D, are ok. They are the same equation.
Oxygen from ground state reduce the oxidation state from 0 to -2
I'm in the car now so can't lookup periodic table. But you find molecular weight of: 1 Ca + 2 Cl
Then that
The reaction for the combustion of methane can be expressed as follows.
CH4 + 2O2 --> CO2 + 2H2O
We solve first for the amount of carbon dioxide in moles by dividing the given volume by 22.4L which is the volume of 1 mole of gas at STP.
moles of CO2 = (5.6 L) / (22.4 L/1 mole)
moles of CO2 = 0.25 moles
Then, we can see that every mole of carbon dioxide will need 1 mole of methane
moles methane = (0.25 moles CO2) x (1 moles O2/1 mole CO2)
= 0.25 moles CH4
Then, multiply this by the molar mass of methane which is 16 g/mole. Thus, the answer is 4 grams methane.
