You are about to watch a video about a scientist who studies jellies like the one shown in this image. One thing she investigate
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Answer:
a. 123.9°C
b.
c.
Explanation:
Hello, I'm attaching a picture with the numerical development of this exercise.
a. Since the steam is overheated vapour, the specific volume is gotten from the corresponding table. Then, as it became a saturated vapour, we look for the interval in which the same volume of state 1 is, then we interpolate and get the temperature.
b. Now, at 80°C, since it is about a rigid tank (constant volume for every thermodynamic process), the specific volume of the mixture is 0.79645 m^3/kg as well, so the specific volume for the liquid and the vapour are taken into account to get the quality of 0.234.
c. Now,since this is an isocoric process, the heat transfer per kg of steam is computed as the difference in the internal energy, considering the initial condition (showed in a. part) and the final one computed here.
** The thermodynamic data were obtained from Cengel's thermodynamics book 7th edition.
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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!