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).
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When naming an ionic compound, write the name of the cation, which is the metal first. Then, write the name of the anion, which is the nonmetal. However, you remove the last 2-3 letters and replace suffixes.
1. RbF --> Rubidium Fluoride
Change fluorine to fluoride
2. CuO --> Copper (II) Oxide
Change oxygen to oxide. Oxide has a charge of -2. Since no subscripts are written, it means they have the same opposite charge. So, we use Copper (II).
<span>3. (NH</span>₄<span>)</span>₂<span>C</span>₂<span>O</span>₄ ---> Ammonium Oxalate
NH₄ is ammonia, but we change it to ammonium for polyatomic ions.
Answer:
The answer to your question is: letter A
Explanation:
A combination reaction is when there are two reactants that gives only one product.
a. 2SO2 + O2—> 2SO3 This is a combination reaction,
2 reactants gives one product.
b. Zn + Cu(NO3)2–>Zn(NO3)2 + Cu This is not a combination reaction,
it's a single replacement reaction.
c. 2H2O2–> 2H2O+O2 This is a decomposition reaction
d. AgNO3 + NaCl → AgCl+NaNO3 THis is a double replacement reaction.
Answer:
V= 0.031L
Explanation:
P= 0.97atm, V= ?, n= 0.12/98 =0.00122mol, R= 0.082, T= 22.4+273= 295.4
Applying
PV=nRT
0.970×V = 0.00122×0.082×295.4
Simplify the above equation
V= 0.031L