Moles of electrons:
The moles of electrons that are transferred are 12F
A balanced equation:
2 moles of Aluminium metal react with excess copper(II) nitrate.
Given:
Moles of Aluminium = 2
As Aluminium goes from 0 to +3 oxidation state
And copper goes from +2 to 0
On balancing the number of electrons we get:
For 1 mole of Al 
is required.
Therefore for 2 moles of Al,
Total 
F mole of electrons
Where F= Faraday's constant= 96500 C
So, 12F moles of electrons are transferred.
Learn more about Faraday's Law here,
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Answer:
The molarity of urea in this solution is 6.39 M.
Explanation:
Molarity (M) is <em>the number of moles of solute in 1 L of solution</em>; that is
To calculate the molality, we need to know the number of moles of urea and the volume of solution in liters. We assume 100 grams of solution.
Our first step is to calculate the moles of urea in 100 grams of the solution,
using the molar mass a conversion factor. The total moles of 100g of a 37.2 percent by mass solution is
60.06 g/mol ÷ 37.2 g = 0.619 mol
Now we need to calculate the volume of 100 grams of solution, and we use density as a conversion factor.
1.032 g/mL ÷ 100 g = 96.9 mL
This solution contains 0.619 moles of urea in 96.9 mL of solution. To express it in molarity, we need to calculate the moles present in 1000 mL (1 L) of the solution.
0.619 mol/96.9 mL × 1000 mL= 6.39 M
Therefore, the molarity of the solution is 6.39 M.
Ten name if this compound is Potassium Oxide
Answer:
0.01M = [H⁺]; 1x10⁻¹²M = [OH⁻]; Ratio is: 1x10¹⁰
Explanation:
pH is defined as -log [H⁺]
For a pH of 2 we can solve [H⁺] as follows:
pH = -log [H⁺]
2 = -log [H⁺]
10^-2 = [H⁺]
<h3>0.01M = [H⁺]</h3>
Using Keq of water:
Keq = 1x10⁻¹⁴ = [H⁺] [OH⁻]
1x10⁻¹⁴ / 0.01M = [OH⁻]
<h3>1x10⁻¹²M = [OH⁻]</h3><h3 />
The ratio is:
[H⁺] / [OH⁻] = 0.01 / 1x10⁻¹² =
<h3>1x10¹⁰</h3>
Answer:
Change in entropy for the reaction is
ΔS° = -268.13 J/K.mol
Explanation:
To calculate the change in entropy for the balanced reaction, we require the natural entropy of all the reactants and products in the reaction.
3 NO₂(g) + H₂O(l) → 2 HNO₃(l) + NO(g)
From Literature.
S°(NO₂) = 240.06 J/K.mol
S°(H₂O) = 69.91 J/K.mol
S°(HNO₃) = 155.60 J/K.mol
S°(NO) = 210.76 J/K.mol
These are the entropies of the reactants and products under standard conditions of 298.15 K and 1 atm.
Note that
ΔS° = Σ nᵢS°(for products) - Σ nᵢS°(for reactants)
Σ nᵢS°(for products) = [2 × S°(HNO₃)] + [1 × S°(NO)]
= (2 × 155.60) + (1 × 210.76) = 521.96 J/K.mol
Σ nᵢS°(for reactants) = [3 × S°(NO₂)] + [1 × S°(H₂O)]
= (3 × 240.06) + (1 × 69.91) =790.09 J/K.mol
ΔS° = Σ nᵢS°(for products) - Σ nᵢS°(for reactants)
ΔS° = 521.96 - 790.09 = -268.13 J/K.mol
Hope this Helps!!
