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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Answer:
PV=nRt
Therefore n(number of moles)=PV/RT
=>(0.49×3.80)/(0.08206×320)
Therefore Number of moles is = 0.071mols
Explanation: By using the Real gas equation..
PV=NRT .
We can solve for the number of moles of Ar by making N the subject..
Always make sure you pressure is In atm, your Volume is in Litres and temperature in degree Kelvin.
Also Recall the universal gas constant R used in this type of questions which is 0.08206.
Hence l, by making N the subject we get our answer as
Given what we know, the ability of water to absorb more heat than the other substances mentioned is a reflection of its high boiling point.
<h3>What do we mean by boiling point?</h3>
This is the temperature at which the substance boils, and subsequently evaporates. Having a higher boiling point means that the substance will be able to absorb much more heat than that of a substance with a lower boiling point.
Therefore, Water molecules have a higher boiling point than molecules of similar size, such as ammonia and methane, reflecting its capacity to absorb large amounts of heat.
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