Answer:
ΔS° = -268.13 J/K
Explanation:
Let's consider the following balanced equation.
3 NO₂(g) + H₂O(l) → 2 HNO₃(l) + NO(g)
We can calculate the standard entropy change of a reaction (ΔS°) using the following expression:
ΔS° = ∑np.Sp° - ∑nr.Sr°
where,
ni are the moles of reactants and products
Si are the standard molar entropies of reactants and products
ΔS° = [2 mol × S°(HNO₃(l)) + 1 mol × S°(NO(g))] - [3 mol × S°(NO₂(g)) + 1 mol × S°(H₂O(l))]
ΔS° = [2 mol × 155.6 J/K.mol + 1 mol × 210.76 J/K.mol] - [3 mol × 240.06 J/K.mol + 1 mol × 69.91 J/k.mol]
ΔS° = -268.13 J/K
Answer:
The mass of ammonia (NH3) that contains 
hydrogen atoms is 
g.
Explanation:
As 
atoms of hydrogen = 1 mole of the hydrogen atom
Therefore, 
atoms of hydrogen 
moles of the hydrogen atom.
Now, there are 3 moles of hydrogen atoms in 1 mole of ammonia 
.
As the mass of 1 mole of ammonia is 17g, so
when there are 3 moles of hydrogen atoms, then the mass of ammonia = 17 g
Therefore, when there are 
moles of hydrogen atoms, then the mass of ammonia 
g.
Hence, the mass of ammonia that contains hydrogen atoms is g.
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