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UkoKoshka [18]
3 years ago
5

CuO+H₂ → Cu +H₂O redox equation

Chemistry
2 answers:
Agata [3.3K]3 years ago
8 0

Answer:

Cu   H₂O

Explanation:

CuO + H₂  -----> Cu + H₂O

Oxidation : Gaining of oxygen OR Losing of hydrogen

Reduction : Gaining of Hydrogen OR Losing of Oxygen

Here ,

CuO gets reduced [ it loses oxygen ] and become Cu

H₂ gets oxidized  [ it gains oxygen ] and becomes H₂O

Substance that gets oxidised = H₂

Substance that gets reduced = CuO

Also ,

Oxidising agent = substance that gets reduced = CuO

Reducing agent = substance that gets oxidized = H₂

Allisa [31]3 years ago
8 0

Answer:

Cu⁺² + 2e⁻ → Cu

Explanation:

A redox reaction is the half-equation where the element is reduced (gains electrons).

Compounds always have to have oxidation numbers that add to 0.

Oxygen's oxidation number is -2.

Stand alone elements always have an oxidation number of 0.

Reactants

Because Copper's oxidation number + Oxygen's oxidation number = 0, and Oxygen's oxidation number = -2, we know Copper's oxidation number is +2.

H₂ has to have a net oxidation of 0, so Hydrogen's oxidation number is 0.

Products

Copper's oxidation number is 0 because it's a standalone element.

2 * Hydrogen's oxidation number + Oxygen's oxidation number = 0, so that means Oxygen's oxidation number is -2, and Hydrogen's is +1.

Because Copper's oxidation number went down from +2 to 0, that means it gained electrons and therefore was reduced. So your answer is:

Cu⁺² + 2e⁻ → Cu

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A catalyst lowers the activation energy for both the forward and the reverse reactions in an equilibrium system, so it has no ef
rosijanka [135]

Answer:

True.

Explanation:

  • Catalyst increases the rate of the reaction without affecting the equilibrium position.
  • Catalyst increases the rate via lowering the activation energy of the reaction.
  • This can occur via passing the reaction in alternative pathway (changing the mechanism).
  • The activation energy is the difference in potential energies between the reactants and transition state (for the forward reaction) and it is the difference in potential energies between the products and transition state (for the reverse reaction).
  • in the presence of a catalyst, the activation energy is lowered by lowering the energy of the transition state, which is the rate-determining step, catalysts reduce the required energy of activation to allow a reaction to proceed and, in the case of a reversible reaction, reach equilibrium more rapidly.
  • with adding a catalyst, both the forward and reverse reaction rates will speed up equally, which allowing the system to reach equilibrium faster.
  • Kindly, see the attached image.

5 0
3 years ago
Read 2 more answers
Consider the following equilibrium systems: A⇌2B ΔH o =20.0 kJ/mol Reaction 1 A+B⇌C ΔH o =−5.4 kJ/mol Reaction 2 A⇌B ΔH o =0.0 k
galina1969 [7]

Answer:

Reaction 1: Kc increases

Reaction 2: Kc decreases

Reaction 3: The is no change

Explanation:

Let us consider the following reactions:

Reaction 1: A ⇌ 2B ΔH° = 20.0 kJ/mol

Reaction 2: A + B ⇌ C ΔH° = −5.4 kJ/mol

Reaction 3: 2A⇌ B ΔH° = 0.0 kJ/mol

To predict what will happen when the temperature is raised we need to take into account Le Chatelier Principle: when a system at equilibrium suffers a perturbation, it will shift its equilibrium to counteract such perturbation. This means that <em>if the temperature is raised (perturbation), the system will react to lower the temperature.</em>

Reaction 1 is endothermic (ΔH° > 0). If the temperature is raised the system will favor the forward reaction to absorb heat and lower the temperature, thus increasing the value of Kc.

Reaction 2 is exothermic (ΔH° < 0). If the temperature is raised the system will favor the reverse reaction to absorb heat and lower the temperature, thus decreasing the value of Kc.

Reaction 3 is not endothermic nor exothermic (ΔH° = 0) so an increase in the temperature will have no effect on the equilibrium.

8 0
3 years ago
What group is the element barium(Ba) In?<br>A. 1<br>B. 6<br>C. 56<br>D. 4​
LenaWriter [7]
Ummmmmmmmmmm 56 yes 56
7 0
2 years ago
Calculate the heat required to convert 5.0 g of ice at 0.0 c to steam at 100
Fiesta28 [93]
Heat= latent heat of fusion+sensible heat+ latent heat of vapourization
=(79.7*5)+(5*100*1)+(540*5)
=3598.5 cal
6 0
3 years ago
1 C8H10(l) +21/2O2(g) → 8CO2(g) + 5H2O(g), Hcomb= ? Hf for C8H10(l) = +49.0kJ/mol C8H10(l) Use the balanced combustion reaction
nikklg [1K]

Answer:

H_{comb}=-4406kJ/mol

Explanation:

Hello,

In this case, the enthalpy of combustion is understood as the energy released when one mole of fuel, in this case octene, is burned in the presence of oxygen and is computed with the enthalpies of formation of the fuel, carbon dioxide and water as shown below (oxygen is circumvented as it is a pure element):

H_{comb}=8*\Delta _fH_{CO_2}+5\Delta _fH_{H_2O}-\Delta _fH_{C_8H_{10}}

Thus, since we already know the enthalpy of combustion of the fuel, for carbon and water we have -393.5 and -241.8 kJ/mol respectively, thereby, the enthalpy of combustion turns out:

H_{comb}=8*(-393.5kJ/mol)+5(-241.8kJ/mol)-49.0kJ/mol\\\\H_{comb}=-4406kJ/mol

Best regards.

4 0
3 years ago
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