Answer:
Explanation:
Cu²⁺ + 2e⁻ → Cu ( copper gets reduced )
Cu → Cu²⁺ + 2e⁻ ( copper gets oxidized )
Oxidation:
Oxidation involve the removal of electrons and oxidation state of atom of an element is increased.
Reduction:
Reduction involve the gain of electron and oxidation number is decreased.
Consider the following reactions.
4KI + 2CuCl₂ → 2CuI + I₂ + 4KCl
the oxidation state of copper is changed from +2 to +1 so copper get reduced.
CO + H₂O → CO₂ + H₂
the oxidation state of carbon is +2 on reactant side and on product side it becomes +4 so carbon get oxidized.
Na₂CO₃ + H₃PO₄ → Na₂HPO₄ + CO₂ + H₂O
The oxidation state of carbon on reactant side is +4. while on product side is also +4 so it neither oxidized nor reduced.
H₂S + 2NaOH → Na₂S + 2H₂O
The oxidation sate of sulfur is -2 on reactant side and in product side it is also -2 so it neither oxidized nor reduced.
A only gets out some of it but not all
C that isn't for alcaholl
D I don't think that even exists
B is the only correct answer
Answer:
KOH is a strong base, and therefore dissociates completely into negative and positive ions in water solution. pH 8.35 is a high pH (> 7), therefore this is a basic solution. 4.
Explanation:
Answer:
The correct answer is Option C (E1) and Option B (carbocation).
Explanation:
- Intramolecular immunity idols are considered as that of the formation mechanism with E1 responses or reactivity.
- Reactants with E1 were indeed obligations of both parties, meaning that an E1 reaction was conducted thru all the two stages known as ionization but rather deprotonation. Involves the absence of either an aromatic ring, a carbocation has been generated throughout the ionization solution.
Some other possibilities offered aren't relevant to the procedure outlined. So the above alternative is accurate.