Two cars travel next to each other. One speedometer reads
1 answer:
Answer:
The pair of arrows which represents the relationship of speeds of the two cars is;
The second option as shown in the attached drawing
Explanation:
The given parameters are;
The reading on the speedometer of one car = 20 m/s
The reading on the speedometer of the other car = 72 km/h = 20 m/s
The blue arrow = 20 m/s
Therefore, given that the speeds of both cars are equal (20 m/s = 72 km/h = 20 m/s), the pair of arrows that represent the relationship of speeds of the two cars is two equal length blue arrows which is the second option
The attached diagram showing the pair of arrows that represents the relationship of speeds of the two cars is drawn using Microsoft Visio.
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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:
- Oxidation reaction
Li⁰(s) → Li⁺(aq) + e⁻ (2)
- Reduction reaction
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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