Answer:
what you've asked so I guess the answer will be 10 days
Answer:
<h3>2Al+ Fe2O3 gives 2Fe + Al2O3. The given reaction is a redox reaction. As oxidation and reduction are taking place simultaneously.</h3>
Explanation:
like this...Identify oxidation and reduction with their agents:
<h3>•2Al+ Fe2O3 →2Fe + Al2O3</h3>
<h3>•Fe2O3 is reduced to Fe whereas Al is oxidized to Al2O3</h3>
<h3>In the above reaction:</h3>
<h3>Oxidizing agent:Fe2O3</h3>
<h3>Reducing agent:Al</h3>
I hope it's help you (◠‿・)—☆
Yes, it is a special case of enthalpy of neutralization.
The enthalpy of neutralization (ΔHn) is the change in enthalpy that occurs when one equivalent of an acid and one equivalent of a base undergo a neutralization reaction to form water and a salt.
The standard enthalpy change of neutralization is the enthalpy change when solutions of an acid and an alkali react together under standard conditions to produce 1 mole of water.
If you start with 0.30 m Mn₂ , at 12.5 pH, free Mn₂ concentration be equal to 4.6 x 10⁻¹¹ m
Initial molarity of Mn₂ = 0.30 M
Final molarity of Mn₂ = 4.6 x 10⁻¹¹
pH = ?
Ksp [Mn(OH)₂] = 4.6 x 10⁻¹⁴ (standard value)
Write the ionic equation
Mn(OH)₂ → Mn⁺² + 2OH⁻
[Mn⁺²] = 4.6 x 10⁻¹¹
We will calculate the concentration of OH⁻ by using Ksp expression
Ksp = [Mn⁺²][OH-]²
[Mn⁺²][OH⁻]² = 4.6 x 10⁻¹⁴
[OH⁻]² = 4.6 x 10⁻¹⁴ / 4.6 x 10⁻¹¹
[OH⁻]² = 10⁻³
[OH⁻] = (10⁻³)¹⁽²
[OH⁻] = 0.0316 M
Calculate the pOH
pOH = -log [OH⁻]
pOH = -log [0.0316]
pOH = 1.5
Now calculate pH
pH = 14 - pOH
pH = 14 - 1.5
pH = 12.5
You can also learn about molarity from the following question:
brainly.com/question/14782315
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