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3 years ago

HELP !!

Chemistry

1 answer:

rusak2 [61]3 years ago

4 0

It would be volume.

Volume is not an intensive property because it <em>does</em> change as the amount of substance increases or decreases. The rest of the properties are constant no matter the amount of substance.

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Which transformation could take place at the anode of an electrochemical cell? Which transformation could take place at the anod

mr_godi [17]

Answer:

None of the above could take place at the anode.

Explanation:

In a cell, a redox reaction takes place, so one substance must reduce (gain electrons), and others must oxidize (lose electrons). At the anode, the substance oxidizes, and at the cathode the substance reduces.

Thus, we need to find what transformation oxidation is happening. To this, let's calculate the oxidation number (nox) of the atoms. If it's increasing, and oxidation is happening, if it's decreasing, a reduction is happening.

O₂ to H₂O:

O2 is a simple substance, so it has nox = 0. In a compound, O has nox -2 and H +1, so the nox of O is decreasing.

Cr₂O₇⁻² → Cr⁺²:

O has nox -2, so let's call the nox of Cr as x:

2x +7*(-2) = -2

2x -14 = -2

x = +6

And Cr⁺² has nox +2, so it's reducing.

F₂ to F⁻:

F₂ is a simple substance, so F has nox 0, and F⁻ has nox -1, then it's reducing.

HAsO₂ to As:

H has nox +1, and O has nox -2, so calling x the nox of As:

+1 + x + 2*(-2) = 0

1 + x - 4 = 0

x = +3

And As is a simple substance, that has nox 0, the it's reducing.

Thus, none of them is oxidizing, and none of them could take place at the anode.

6 0

3 years ago

H 2(g)+I 2(g)→2 H I(g)

dlinn [17]

Answer:

The images for the 3 sub-questions have been presented in the first, second and third attached image to this solution.

Explanation:

The reaction for the reaction between Hydrogen gas and Iodine to give Hydrogen iodide

H₂ (g) + I₂ (g) → 2 H I(g)

ΔH∘rxn = −9.5 kJ/mol r x n

a) From the information provided, the heat of reaction being negative shows that the reaction is an exothermic reaction with the products for an exothermic reaction having a lower energy content/level as the reactants for this reaction.

This shows that the energy diagram will have the products at a lower level than fe reactants.

The sketch of the energy diagram for this exothermic reaction is presented in the first attached image to this solution.

The heat of reaction is shown on the energy diagram, together with the activation energy, Ea, the heat content/energy level for the reactants (Hr) & products (Hr) and finally the ΔH∘rxn.

ΔH∘rxn = Hp - Hr

b) When a catalyst that speed up a chemical reaction is introduced to the chemical reaction, the catalyst goes about making the reaction faster by lowering the activation energy of the reaction.

The activation energy is the minimum energy that the reactants must possess to the able to form products. On the graph, it is denoted as Ea, the difference in energy between the peak of the curve and the emergy level of the reactants.

So, a catalyst will cause this activation energy to be lowered to a new level of Ea₁.

The energy diagram of this is presented in the second attached image to this answer.

The ΔH∘rxn remains unchanged for this introduction of catalyst.

The broken lines are used to represent the lowered activation energy action of the catalyst.

c) The reverse reaction will have opposite the energy properties of the initial reaction.

The initially exothermic reaction with higher heat content of reactants than products is now an endothermic reaction with heat content/energy level of the products higher than that of the reactants.

The energy diagram is presented in the third attached image.

The activation energy is bigger and the ΔH∘rxn is now positive and equal to 9.5 kJ/mol, the direct additive inverse of the ΔH∘rxn for the initial exothermic reaction.

Hope this Helps!!!