Answer:
Ley.
Explanation:
En la teoría de la ciencia, la regularidad de los procesos en la naturaleza se denomina ley de la naturaleza. Las leyes naturales se diferencian de otras leyes en que los seres humanos no pueden ponerlas en vigor ni anularlas a su discreción. En tal sentido, la composición química del agua es indudablemente una ley natural, en tanto el hombre no puede modificarla sin modificar las características inherentes del agua como tal.
Answer:
d. The gold(III) ion is most easily reduced.
Explanation:
The standard reduction potentials are
Au³⁺ + 3e⁻ ⟶ Au; 1.50 V
Hg²⁺ + 2e⁻ ⟶ Hg; 0.85 V
Zn²⁺ + 2e⁻ ⟶ Zn; -0.76 V
Na⁺ + e⁻ ⟶ Na; -2.71 V
A <em>more positive voltage</em> means that there is a <em>stronger driving force</em> for the reaction.
Thus, Au³⁺ is the best acceptor of electrons.
Reduction Is Gain of electrons and, Au³⁺ is gaining electrons, so
Au³⁺ is most easily reduced.
Answer:
- To increase the temperature as it is a reactant in terms of its endothermicity.
- To remove it will enable more space for the reactant to favor its production.
- To add more reactant in order to increase its equilibrium concentration.
Explanation:
Hello,
The undergoing chemical reaction is:
Thus, in order to intensify the amount of nitrogen as the chemical reaction is endothermic, considering the Le Chatelier's principle we state:
- To increase the temperature as it is a reactant in terms of its endothermicity.
- To remove it will enable more space for the reactant to favor its production.
- To add more reactant in order to increase its equilibrium concentration.
Best regards.
Answer:
SO₃²⁻ is the reducing agent and Cr₂O₇²⁻ is the oxidizing agent.
Explanation:
Oxidation reaction:
3SO₃²⁻ (aq) + 3H₂O (l) → 3SO₄²⁻ (aq) + 6H⁺ (aq) + 6e⁻
Reduction reaction:
Cr₂O₇²⁻ (aq) + 14H⁺ (aq) + 6e⁻ → 2Cr ³⁺ (aq) + 7H₂O (l)
Now, adding the oxidation and the reduction reactions we get the full net reaction:
Cr₂O₇²⁻ (aq) + 3SO₃²⁻ (aq) + 8H⁺ (aq) → 2Cr ³⁺ (aq) + 3SO₄²⁻ (aq) + 4H₂O(l)
Since, the S in SO₃²⁻, present in the +4 oxidation state is oxidized to +6 oxidation state in SO₄²⁻, by the loss of 2e⁻.
<u>Therefore, SO₃²⁻ is the reducing agent. </u>
And, the Cr in Cr₂O₇²⁻, present in the +6 oxidation state is getting reduced to +3 oxidation state, Cr ³⁺, by the gain of 6e⁻.
<u>Therefore, Cr₂O₇²⁻ is the oxidizing agent.</u>
