Answer:
Make the question more clear for me
Explanation:
The person above me is correct I took a test on this so it’s the right answer
Answer:
Ionic equation:
Na⁺(aq) + OH⁻(aq) + H⁺(aq) + Cl⁻(aq) → H₂O(l) + Na⁺(aq) + Cl⁻ (aq)
Explanation:
Chemical equation:
NaOH(aq) + HCl(aq) → H₂O(l) + NaCl (aq)
Balanced chemical equation:
NaOH(aq) + HCl(aq) → H₂O(l) + NaCl (aq)
Ionic equation:
Na⁺(aq) + OH⁻(aq) + H⁺(aq) + Cl⁻(aq) → H₂O(l) + Na⁺(aq) + Cl⁻ (aq)
Net ionic equation:
OH⁻(aq) + H⁺(aq) → H₂O(l)
The Cl⁻(aq) and Na⁺ (aq) are spectator ions that's why these are not written in net ionic equation. The water can not be splitted into ions because it is present in liquid form.
Spectator ions:
These ions are same in both side of chemical reaction. These ions are cancel out. Their presence can not effect the equilibrium of reaction that's why these ions are omitted in net ionic equation.
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
The more kinetic energy a substance has, the warmer it will be and the faster particles will be moving, which reduces the density of the substance
