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:
6.78 × 10⁻³ L
Explanation:
Step 1: Write the balanced equation
Mg₃N₂(s) + 3 H₂O(g) ⇒ 3 MgO(s) + 2 NH₃(g)
Step 2: Calculate the moles corresponding to 10.2 mL (0.0102 L) of H₂O(g)
At STP, 1 mole of H₂O(g) has a volume of 22.4 L.
0.0102 L × 1 mol/22.4 L = 4.55 × 10⁻⁴ mol
Step 3: Calculate the moles of NH₃(g) formed from 4.55 × 10⁻⁴ moles of H₂O(g)
The molar ratio of H₂O to NH₃ is 3:2. The moles of NH₃ produced are 2/3 × 4.55 × 10⁻⁴ mol = 3.03 × 10⁻⁴ mol.
Step 4: Calculate the volume corresponding to 3.03 × 10⁻⁴ moles of NH₃
At STP, 1 mole of NH₃(g) has a volume of 22.4 L.
3.03 × 10⁻⁴ mol × 22.4 L/mol = 6.78 × 10⁻³ L
26.25g
Explanation:
Given parameters:
Percentage of NaCl in ocean water = 3.5%
Mass of given seawater = 750g
Unknown:
Mass of NaCl in the seawater = ?
Solution:
This is a simple percentage problem:
We have been given that ocean water contains 3.5% of NaCl;
Therefore;
Mass of NaCl in seawater = Percentage of NaCl in ocean water x Mass of seawater
Mass of NaCl in seawater = 
x 750g = 26.25g
Learn more:
Percentage brainly.com/question/2917248
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