Answer:
Plutonium is the second transuranium element of the actinide series.
Answer:
where is the answer options because it sounds like I need some
Answer : The value of
is, 0.34 V
Explanation :
Here, copper will undergo reduction reaction will get reduced. Zinc will undergo oxidation reaction and will get oxidized.
The oxidation-reduction half cell reaction will be,
Oxidation half reaction: 
Reduction half reaction: 
Oxidation reaction occurs at anode and reduction reaction occurs at cathode. That means, gold shows reduction and occurs at cathode and chromium shows oxidation and occurs at anode.
The overall balanced equation of the cell is,

To calculate the
of the reaction, we use the equation:


Putting values in above equation, we get:


Hence, the value of
is, 0.34 V
Answer:
V = 12.93 L
Explanation:
Given data:
Number of moles = 0.785 mol
Pressure of balloon = 1.5 atm
Temperature = 301 K
Volume of balloon = ?
Solution:
The given problem will be solve by using general gas equation,
PV = nRT
P= Pressure
V = volume
n = number of moles
R = general gas constant = 0.0821 atm.L/ mol.K
T = temperature in kelvin
Now we will put the values.
V = nRT/P
V = 0.785 mol × 0.0821 atm.L/ mol.K × 301 K / 1.5 atm
V = 19.4 L /1.5
V = 12.93 L
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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