How many grams of Cl are in 38.0 g of each sample of chlorofluorocarbons (CFCs)?
1 answer:
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Answer:
ΔG° = -118x10³ J/mol
Explanation:
The two half-reactions in the cell are:
Oxidation half-reaction:
Co(s) → Co²⁺(aq) + 2e⁻; E° = -0,28V
Reduction half-reaction:
Cu²⁺(aq)+2e⁻ → Cu(s); E° = 0,34V
The E° of the cell is defined as:
Replacing:
0,34V - (-0,28V) = 0,62V
It is possible to obtain the keq from E°cell with Nernst equation thus:
nE°cell/0,0592 = log (keq)
Where:
E°cell is standard electrode potential (0.62 V)
n is number of electrons transferred (2 electrons, from the half-reactions)
Replacing:
0,62V×2/0,0592 = log (keq)
20,946 = log keq
keq = 8,83x10²⁰≈ 5,88x10²⁰
ΔG° is defined as:
ΔG° = -RT ln Keq
Where R is gas constant (8,314472 J/molK) and T is temperature (298K):
ΔG° = -8,314472 J/molK×298K ln5,88x10²⁰
<em>ΔG° = -118x10³ J/mol</em>
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I hope it helps!
Answer : 1721.72 g/qt are in 18.2 g/cL
Explanation :
As we are given: 18.2 g/cL
Now we have to convert 18.2 g/cL to g/qt.
Conversions used are:
(1) 1 L = 100 cL
(2) 1 L = 1000 mL
(3) 1 qt = 946 qt
The conversion expression will be:
Therefore, 1721.72 g/qt are in 18.2 g/cL