<span>Jet streams are the major means of transport for weather systems. A jet stream is an area of strong winds ranging from 120-250 mph that can be thousands of miles long, a couple of hundred miles across and a few miles deep. Jet streams usually sit at the boundary between the troposphere and the stratosphere at a level called the tropopause. This means most jet streams are about 6-9 miles off the ground. Figure A is a cross section of a jet stream.
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The dynamics of jet streams are actually quite complicated, so this is a very simplified version of what creates jets. The basic idea that drives jet formation is this: a strong horizontal temperature contrast, like the one between the North Pole and the equator, causes a dramatic increase in horizontal wind speed with height. Therefore, a jet stream forms directly over the center of the strongest area of horizontal temperature difference, or the front. As a general rule, a strong front has a jet stream directly above it that is parallel to it. Figure B shows that jet streams are positioned just below the tropopause (the red lines) and above the fronts, in this case, the boundaries between two circulation cells carrying air of different temperatures.
Answer:
A. ΔG° = 132.5 kJ
B. ΔG° = 13.69 kJ
C. ΔG° = -58.59 kJ
Explanation:
Let's consider the following reaction.
CaCO₃(s) → CaO(s) + CO₂(g)
We can calculate the standard enthalpy of the reaction (ΔH°) using the following expression.
ΔH° = ∑np . ΔH°f(p) - ∑nr . ΔH°f(r)
where,
n: moles
ΔH°f: standard enthalpy of formation
ΔH° = 1 mol × ΔH°f(CaO(s)) + 1 mol × ΔH°f(CO₂(g)) - 1 mol × ΔH°f(CaCO₃(s))
ΔH° = 1 mol × (-635.1 kJ/mol) + 1 mol × (-393.5 kJ/mol) - 1 mol × (-1206.9 kJ/mol)
ΔH° = 178.3 kJ
We can calculate the standard entropy of the reaction (ΔS°) using the following expression.
ΔS° = ∑np . S°p - ∑nr . S°r
where,
S: standard entropy
ΔS° = 1 mol × S°(CaO(s)) + 1 mol × S°(CO₂(g)) - 1 mol × S°(CaCO₃(s))
ΔS° = 1 mol × (39.75 J/K.mol) + 1 mol × (213.74 J/K.mol) - 1 mol × (92.9 J/K.mol)
ΔS° = 160.6 J/K. = 0.1606 kJ/K.
We can calculate the standard Gibbs free energy of the reaction (ΔG°) using the following expression.
ΔG° = ΔH° - T.ΔS°
where,
T: absolute temperature
<h3>A. 285 K</h3>
ΔG° = ΔH° - T.ΔS°
ΔG° = 178.3 kJ - 285K × 0.1606 kJ/K = 132.5 kJ
<h3>B. 1025 K</h3>
ΔG° = ΔH° - T.ΔS°
ΔG° = 178.3 kJ - 1025K × 0.1606 kJ/K = 13.69 kJ
<h3>C. 1475 K</h3>
ΔG° = ΔH° - T.ΔS°
ΔG° = 178.3 kJ - 1475K × 0.1606 kJ/K = -58.59 kJ
Answer:
I think the answer is D. Temperature affects only the rate of reaction.
Answer:
negative
Explanation:
When something slows down, its acceleration is the opposite of the velocity.
Cryo-EM is used to preserve and characterize cycled positive electrodes. Under regular cycling conditions, there isn't an intimate coating layer like CEI.A small electrical short can cause a stable conformal CEI to form in place. The conformal CEI's chemistry is revealed by EELS and cryo-(S)TEM.
It has been assumed that the intimate coating layer generated on the positive electrode, known as cathode electrolyte interphase (CEI), is crucial. However, there are still numerous questions about CEI. This results from the absence of useful instruments to evaluate the chemical and structural characteristics of these delicate interphases at the nanoscale. Here, using cryogenic electron microscopy, we establish a methodology to maintain the natural condition and directly see the interface on the positive electrode.
Learn more about Cathode electrolyte interphase here:
brainly.com/question/861659
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