The reducing agent in the reaction 2Li(s) + Fe(CH₃COO)₂(aq) → 2LiCH₃COO(aq) + Fe(s) is lithium (Li).
The general reaction is:
2Li(s) + Fe(CH₃COO)₂(aq) → 2LiCH₃COO(aq) + Fe(s) (1)
We can write the above reaction in <u>two reactions</u>, one for oxidation and the other for reduction:
Li⁰(s) → Li⁺(aq) + e⁻ (2)
Fe²⁺(aq) + 2e⁻ → Fe⁰(s) (3)
We can see that Li⁰ is oxidizing to Li⁺ (by <u>losing</u> one electron) in the lithium acetate (<em>reaction 2</em>) and that Fe²⁺ in iron(II) acetate is reducing to Fe⁰ (by <u>gaining</u> two <em>electrons</em>) (<em>reaction 3</em>).
We must remember that the reducing agent is the one that will be oxidized by <u>reducing another element</u> and that the oxidizing agent is the one that will be reduced by <u>oxidizing another species</u>.
In reaction (1), the<em> reducing agent</em> is <em>Li</em> (it is oxidizing to Li⁺), and the <em>oxidizing agent </em>is<em> Fe(CH₃COO)₂</em> (it is reducing to Fe⁰).
Therefore, the reducing agent in reaction (1) is lithium (Li).
Learn more here:
I hope it helps you!
Answer:
This is site for English speakers. Этот сайт на английском, поэтому вопрос могут удалить
Explanation:
1. 2)
2. 3)
3. 4) Sr
4. 3)
5. 4)
6. 2)
7. 1)
8. 4)
9. 3)
10. 3)
11. SO3, H2SO4, Na2SO4
12.
A) оксид меди (II) 2) CuO
Б) хлорид меди(II) 4) CuCl2
В) сульфит меди (II) 3) CuSO 3
Г) гидроксид меди (II) 1) Cu(OH)2
13.
1. Fe+HCl= б) FeCl 2 +H 2
2.Fe+O2= в) Fe 3 O 4
3. Fe(OH) 3 = г)Fe 2 O 3 +H 2O
4. FeCl 2 +NaOH= а) Fe(OH) 2 +NaCl
14. 2Ca + O2 = 2CaO
CaO + H2O = Ca(OH)2
Ca(OH)2 + 2HCl = CaCl2 + 2H2O
Answer:
has Two oxygen atoms
Explanation:
Oxygen is a diatomic element hence exists as O2 for majority of its existence in our atmosphere. Although small portion does exist in form of O3 which protects earth from sun's harmful ray, the majority portion of oxygen has O2 which is the oxygen we breathe.
Actually, I strongly believe it is a switch.
I think it would be solubility but I’m not sure