Answer:
The temperature should be higher than 437.9 Kelvin (or 164.75 °C) to be spontaneous
Explanation:
<u>Step 1:</u> Data given
ΔH∘=20.1 kJ/mol
ΔS is 45.9 J/K
<u>Step 2:</u> When is the reaction spontaneous
Consider temperature and pressure = constant.
The conditions for spontaneous reactions are:
ΔH <0
ΔS > 0
ΔG <0 The reaction is spontaneous at all temperatures
ΔH <0
ΔS <0
ΔG <0 The reaction is spontaneous at low temperatures ( ΔH - T*ΔS <0)
ΔH >0
ΔS >0
ΔG <0 The reaction is spontaneous at high temperatures ( ΔH - T*ΔS <0)
<u>Step 3:</u> Calculate the temperature
ΔG <0 = ΔH - T*ΔS
T*ΔS > ΔH
T > ΔH/ΔS
In this situation:
T > (20100 J)/(45.9 J/K)
T > 437.9 K
T > 164.75 °C
The temperature should be higher than 437.9 Kelvin (or 164.75 °C) to be spontaneous
Answer:
Número de moles de cloro en número de moles de NaCl
Explanation:
espero que si sea la correcta
Are there any choices? Because from what the question is it seems like we need choices to help
C: One plate is going underneath the other plate and sinking into the soft rock below.
Explanation:
Where plates are moving towards each other they are said to converging, and are called convergent margins.
The lithosphere is broken into series of slabs called plates. The plates moves on the weak and relatively soft asthenosphere below.
Plates have different motion. At some places, they move apart and they are said to be divergent.
When plates moves towards each other, they are convergent. At a convergent margin, a plate collides with another thereby causing the denser plate usually the oceanic plate to subduct into the asthenosphere. In some other cases, the plates can collide and build upward.
Learn more:
Lithosphere brainly.com/question/9582362
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The balanced equation is 2
AlI
3
(
a
q
)
+
3
Cl
2
(
g
)
→
2
AlCl
3
(
a
q
)
+
3
I
2
(
g
)
.
<u>Explanation:</u>
- Aluminum has a typical oxidation condition of 3+ , and that of iodine is 1- .
Along these lines, three iodides can bond with one aluminum. You get AlI3. For comparable reasons, aluminum chloride is AlCl3.
- Chlorine and iodine both exist normally as diatomic components, so they are Cl2( g ) also, I2( g ), individually. In spite of the fact that I would anticipate that iodine should be a strong.
Balancing the equation, we get:
2AlI
3( aq ) + 3Cl2
( g ) → 2AlCl3
( aq )
+ 3
I
2 ( g )
-
Realizing that there were two chlorines on the left, I simply found the basic numerous of 2 and 3 to be 6, and multiplied the AlCl 3 on the right.
-
Normally, presently we have two Al on the right, so I multiplied the AlI 3 on the left. Hence, I have 6 I on the left, and I needed to significantly increase I 2 on the right.
-
We should note, however, that aluminum iodide is viciously receptive in water except if it's a hexahydrate. In this way, it's most likely the anhydrous adaptation broke down in water, and the measure of warmth created may clarify why iodine is a vaporous item, and not a strong.
