5 examples of gases found in the normal home environment include; oxygen (air), nitrogen (most abundant element found in the air), carbon (air), a slight trace of argon and finally, hydrogen. these are- Nitrogen, oxygen, carbondioxide, carbonmonoxide and SO2.
Nitrogen (N2) and hydrogen (H2) gases react to form ammonia, which requires -99.4 J/K of standard entropy (ΔS°).
What is standard entropy?
The difference between the total standard entropies of the reaction mixture and the summation of the standard entropies of the outputs is the standard entropy change. Each entropy in the balanced equation needs to be compounded by its coefficient, as shown by the letter "n."
Calculation:
Balancing the given reaction following-
1/2 N₂(g) + 3/2 H₂ (g)→ NH₃ (g)
ΔS° = [1 mol x S° (NH₃)g] - [1/2 mol x S° (N₂)g] - [3/2 mol x S°(H₂)g]
Here S° = standard entropy of the system
Insert into the aforementioned equation all the typical entropy values found in the literature:
ΔS° = [1 mol x 192.45 J/mol.K] - [1/2 mol x 191.61 J/mol.K] - [3/2 mol x 130.684 J/mol.K]
⇒ΔS° = - 99.4 J/K
Therefore, the standard entropy, ΔS° is -99.4 J/K.
Learn more about standard entropy here:
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As you have not provided the options, still we can figure out the answer by understanding the key difference between saturated and unsaturated hydrocarbons.
SATURATED HYDROCARBONS are those hydrocarbons which only consist of a carbon carbon single bonds. All the bonds are sigma there are no pi bonds at all. Examples are shown below.
While, UNSATURATED HYDROCARBONS are those hydrocarbons which may contain either a double bond or triple bonds or both of them between the carbon atoms as shown below.
Answer:
As the Bohr's fixed orbit gives precise information about the radial position and momentum of the orbit, it is against the Heisenberg uncertainty principle. Thus it is inferred that the Heisenberg uncertainty principle goes and the concept Bohr's fixed Orbit are opposite to each other.
Explanation:
