Answer:
Explanation:
Hello,
Clausius Clapeyron equation is suitable in this case, since it allows us to relate the P,T,V behavior along the described melting process and the associated energy change. Such equation is:
As both the enthalpy and volume do not change with neither the temperature nor the pressure for melting processes, its integration turns out:
Solving for the enthalpy of fusion we obtain:
Finally the entropy of fusion is given by:
Best regards.
_____ are types of active transport.
(A)Diffusion and osmosis
<u>(B)Engulfing and transport proteins
</u>
(C)Osmosis and engulfing
(D)Transport proteins and diffusion
I need to go home and play on the game
Answer:
-122 J/K
Explanation:
Let's consider the following balanced reaction.
N₂(g) + 2 O₂(g) ⇒ 2 NO₂(g)
We can calculate the standard reaction entropy (ΔS°) using the following expression.
ΔS° = Σ ηp × Sf°p - Σ ηr × Sf°r
where,
- η: stoichiometric coefficients of products and reactants
- Sf°r: entropies of formation of products and reactants
ΔS° = 2 mol × 240.06 J/K.mol - 1 mol × 191.61 J/K.mol - 2 mol × 205.14 J/K.mol
ΔS° = -121.77 J/K ≈ -122 J/K
2.47L
Hope this helped have a great day :)
