When cool, dense air from over the water flows inland, it's called a sea breeze.
God bless!
Answer:
The balanced equation for the dissociation of KI is
KI → K⁺ + I⁻
Explanation:
KI is the potassium iodide.
K⁺ comes from the KOH, a strong base, so the cation is the conjugate weak acid and in water it does not react.
I⁻ comes from HI, a strong acid, so the anion is the conjugate weak base and in water it does not react.
K⁻ + H₂O ← KOH + H⁺
I⁻ + H₂O ← HI + OH⁻
That's why the arrow in the reaction is in the opposite direction.
By Boyle's Law pV = constant, halfing the volume would require doubling the pressure to maintain the constant. Hence, pressure doubles.
Answer:
A) ΔG° = -3,80x10⁵ kJ
B) E° = 2,85V
Explanation:
A) It is possible to answer this problem using the standard ΔG's of formation. For the reaction:
Mg(s) + Fe²⁺(aq) → Mg²⁺(aq) + Fe(s)
The ΔG° of reaction is:
ΔG° = ΔGFe(s) + ΔGMg²⁺(aq) - (ΔGFe²⁺(aq) + ΔGMg(s) <em>(1)</em>
Where:
ΔGFe(s): 0kJ
ΔGMg²⁺(aq): -458,8 kJ
ΔGFe²⁺(aq): -78,9 kJ
ΔGMg(s): 0kJ
Replacing in (1):
ΔG° = 0kJ -458,8kJ - (-78,9kJ + okJ)
<em>ΔG° = -3,80x10² kJ ≡ -3,80x10⁵ kJ</em>
B) For the reaction:
X(s) + 2Y⁺(aq) → X²⁺(aq) + 2Y(s)
ΔG° = ΔH° - (T×ΔS°)
ΔG° = -629000J - (298,15K×-263J/K)
ΔG° = -550587J
As ΔG° = - n×F×E⁰
Where n are electrons involved in the reaction (<em>2mol</em>), F is faraday constant (<em>96485 J/Vmol</em>) And E° is the standard cell potential
Replacing:
-550587J = - 2mol×96485J/Vmol×E⁰
<em>E° = 2,85V</em>
I hope it helps!
