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daser333 [38]
3 years ago
7

Which of the following is an oxidation-reduction reaction?

Chemistry
2 answers:
polet [3.4K]3 years ago
5 0

<u>Answer:</u> The correct answer is Option a.

<u>Explanation:</u>

Oxidation-reduction reaction or redox reaction is defined as the reaction in which oxidation and reduction reactions occur simultaneously.

Oxidation reaction is defined as the reaction in which a substance looses its electrons. The oxidation state of the substance increases.

Reduction reaction is defined as the reaction in which a substance gains electrons. The oxidation state of the substance gets reduced.

For the given options:

<u>Option A:</u> Fe_2O_3+3CO\rightarrow 2Fe+3CO_2

On reactant side:

Oxidation state of iron = +3

Oxidation state of carbon = +2

On product side:

Oxidation state of iron = 0

Oxidation state of carbon = +4

The oxidation state of iron reduces from +3 to 0, it is getting reduced. Thus, it is getting reduced and it undergoes reduction reaction. The oxidation state of carbon increases from +2 to +4. Thus, it is getting oxidized and it undergoes oxidation reaction.

Thus, it is considered as redox reaction.

<u>Option B:</u> CuSO_4+2NaOH\rightarrow Cu(OH)_2+Na_2SO_4

The above reaction is double displacement reaction because here exchange of ions takes place.

<u>Option C:</u> 2NaOH+H_2CO_3\rightarrow Na_2CO_3+2H_2O

The above reaction is a neutralization reaction because an acid reacts with base to produce salt and water.

<u>Option D:</u> Pb(NO_3)2+Na_2SO_4\rightarrow 2NaNO_3+PbSO_4

The above reaction is double displacement reaction because here exchange of ions takes place.

Hence, the correct answer is Option A.

likoan [24]3 years ago
3 0
A) Fe⁺³₂O₃ + 3C⁺²O -> 2Fe⁰ + 3C⁺⁴O₂
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8 0
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Read 2 more answers
Barium nitrate and sodium sulfate solutions can be used to precipitate barium sulfate. How many grams
-Dominant- [34]

Answer:

40.8g of sodium sulfate must be added

Explanation:

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That means, for a complete reaction of an amount of barium nitrate you must add the same amount in moles of sodium sulfate. To solve this problem we need to convert the mass of barium nitrate to moles = Moles of sodium sulfate that must be added:

<em>Moles Ba(NO₃)₂ -Molar mass: 261.3g/mol-:</em>

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0.287 moles * (142.04g / mol) =

<h3>40.8g of sodium sulfate must be added</h3>
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