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Zolol [24]
3 years ago
11

Predict the species that will be oxidized first if the following mixture of molten salts undergoes electrolysis:

Chemistry
1 answer:
Ede4ka [16]3 years ago
4 0

Answer:

Explanation:

the species that will be oxidized first will depend upon their oxidation potential .  

oxidation potential are given as follows

F⁻ =  -  2.87 V   , Br⁻ = - 1.09 V , Cl⁻ = -1.36 V, Mg⁺² = -2.37 V, Cu⁺² = +.34 V

Higher the oxidation potential , higher the tendency to be oxidised .

so Cu⁺²  is easiest to be oxidised  . F⁻ is to be oxidised most difficult.

The order from easiest to most difficult as follows

Cu⁺² >  Br⁻ > Cl⁻ > Mg⁺² > F⁻

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Explanation:

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Do you think placing an Alka-Seltzer in water causes a chemical reaction? <br>Explain Why.
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3 years ago
I NEED HELP PLEASE, THANKS! :)
krok68 [10]

Answer:

\large \boxed{\text{2.20 g Pb}}

Explanation:

They gave us the masses of two reactants and asked us to determine the mass of the product.

This looks like a limiting reactant problem.

1. Assemble the information

We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.

Mᵣ:       239.27   32.00        207.2

            2PbS   +   3O₂   ⟶  2Pb   +   2SO₃

m/g:      2.54        1.88

2. Calculate the moles of each reactant

\text{Moles of PbS} = \text{2.54 g PbS } \times \dfrac{\text{1 mol PbS}}{\text{239.27 g PbS}} = \text{0.010 62 mol PbS}\\\\\text{Moles of O}_{2} = \text{1.88 g O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{32.00 g O}_{2}} = \text{0.058 75 mol O}_{2}

3. Calculate the moles of Pb from each reactant

\textbf{From PbS:}\\\text{Moles of Pb} =  \text{0.010 62 mol PbS} \times \dfrac{\text{2 mol Pb}}{\text{2 mol PbS}} = \text{0.010 62 mol Pb}\\\\\textbf{From O}_{2}:\\\text{Moles of Pb} =\text{0.058 75 mol O}_{2} \times \dfrac{\text{2 mol Pb}}{\text{3 mol O}_{2}}= \text{0.039 17 mol  Pb}\\\\\text{PbS is the $\textbf{limiting reactant}$ because it gives fewer moles of Pb}

4. Calculate the mass of Pb

\text{ Mass of Pb} = \text{0.010 62 mol Pb} \times \dfrac{\text{207.2 g Pb}}{\text{1 mol Pb}} = \textbf{2.20 g Pb}\\\\\text{The reaction produces $\large \boxed{\textbf{2.20 g Pb}}$}

5 0
3 years ago
Why does ΔG for a reaction scale with reaction quantity but E does not? For example, ΔG0 rxn for the combustion of 1 mol of hydr
wariber [46]

Answer:

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ΔG is extensive and E is Intensive. The explanation is as given below

Explanation:

Basically both ΔG and the cell potential or the electromotive force (E.M.F) has some disparity especially in their spontaneity, for spontaneous reaction ΔG = -ve while E = +ve and vice versa. But the most important disparity is their state function i.e one is intensive and the other is extensive property.

ΔG is an example of an extensive property, they are properties whose value is dependent on the volume or the size of the system. other examples are mass, volume etc.

E on the other hand is an intensive property, they are properties whose value is not dependent on the size of the system. As such, this differences explains why ΔG for a reaction scale with a reaction quantity and E does not.

4 0
3 years ago
Te second period at the bottom of the table is called
katovenus [111]

Answer:

The second row elements are called Actinidies . abbreviated as Ac

Explanation:

5 0
4 years ago
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