Number of moles:
1.0 mole CH4 -------------- 22.4 L
moles ------------- 30 L
moles CH4 = 30 * 1 / 22.4 = 1.339 moles
Therefore:
1.0 mole CH4 --------------- 6.02x10²³ molecules
1.339 moles ---------------- molecules CH4
molecules CH4 = ( 1.339 ) * ( 6.02x10²³ ) / 1.0
=> 8.6x10²³ molecules CH4
Answer:
Keq = 1.17 × 10²⁰
Explanation:
Let's consider the following redox reaction.
Cu²⁺(aq) + Ni(s) → Cu(s) + Ni²⁺(aq)
We can identify 2 half-reactions.
Cathode (reduction): Cu²⁺(aq) + 2 e⁻ → Cu(s) E°red = 0.337 V
Anode (oxidation): Ni(s) → Ni²⁺(aq) + 2 e⁻ E°red = -0.257 V
The standard cell potential (E°) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.
E° = E°red, cat - E°red, an = 0.337 V - (-0.257V) = 0.594 V
We can calculate the equilibrium constant (Keq) using the following expression.
where,
n are the moles of electrons transferred
2H2 (g) + O2 (g) -->2H2 O(g)
mole ratio of H2:O2=2:1
7.25/2=3.625