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3 years ago

Which substance is the oxidizing agent in this reaction? feo+co→fe+co2 express your answer as a chemical formula?

2 answers:

6 0

In the equation given above, the oxidizing agent is FeO.
An oxidizing agent in a chemical reaction is defined as that substance, which has the capacity to oxidize other substances by gaining electrons from them. In a chemical reaction, an oxidizing agent is usually reduced by gaining electrons.
In the chemical equation given above, iron ll oxide oxidizes carbon monoxide to carbon dioxide.

Vlad [161]3 years ago

5 0

In the reaction; FeO + CO → Fe + CO2, FeO is the oxidizing agent.

<h2>Further Explanation</h2><h3>Redox reactions </h3>

Redox reactions are types of reactions that involve both oxidation and reduction occurring simultaneously. In these reactions one of the reactant undergoes oxidation while another undergoes reduction.

<h3>Oxidation </h3>

Oxidation involves the loss of electrons by an element. When an element or an ion undergoes oxidation, it looses electrons.
For example; in the above reaction carbon monoxide undergoes oxidation.
C²⁺ → C⁴⁺ + 2e

<h3>Reduction </h3>

Reduction refers to the gain of electrons by an element. Therefore, when an element or an ion undergoes reduction it gains electrons.

For example; in the above reaction iron (ii) oxide undergoes reduction.
Fe²⁺ + 2e → Fe

<h3>Oxidizing agent</h3>

An oxidizing agent in a chemical reaction is defined as that substance, which has the capacity to oxidize other substances by gaining electrons from them. In a chemical reaction, an oxidizing agent is usually reduced by gaining electrons.
In the chemical equation given above, iron ll oxide oxidizes carbon monoxide to carbon dioxide.

<h3>Reducing agent </h3>

A reducing agent in a chemical reaction is defined as that substance, which has the capacity to reduce other substances by losing electrons from them.
In a chemical reaction, a reducing agent is usually oxidized by loosing electrons.
In the chemical equation given above, carbon monoxide reduces iron (ii) oxide to iron metal.

Keywords: Redox reactions, oxidation, reduction, reducing agent, oxidizing agent

<h3>Learn more about:</h3>

Oxidizing agent: brainly.com/question/2890416
Reducing agent: brainly.com/question/2890416
Redox reactions: brainly.com/question/2890416
Oxidation: brainly.com/question/2890416
Reduction: brainly.com/question/2890416

Level: High school

Subject: Chemistry

Topic: Redox reactions: Oxidation and reduction

Sub-topic: Redox reactions

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EastWind [94]

Answer and Explanation:

The Fremy salt is a chemical compound its chemical name is disodium nitrosodisuphonate

The molecular formula of the Fremy salt is $K_2NO(SO_3)_2$

When phenol is treated with Fremy salt in presence of water then benzoquinone is formed

$C_2H_5OH\rightarrow C_6H_4O_2$ ( in presence of Fremy salt and water)

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3 years ago

How many grams of NaCl are in 2.0 Liters of a 3.0 M NaCl solution?

Naddik [55]

Answer:

350.64g

Explanation:

So first you must know that M is mol/L

Next solve the problem using dimensional analysis

2L NaCl (3 mol/L) = 6 mol NaCl

After you got the number of moles you should look at your periodic table to find the molar mass

I see that it's 58.44g/mol

Use dimensional analysis again!

6 mol (58.44g/mol) = 350.64g

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3 years ago

When 33 g of CaO and 10 g of H2O react, how many grams of calcium hydroxide would you expect to be produced* Explain your answer

MrMuchimi

Answer:

41.44 g

Explanation:

First of all, we must put down the equation of the reaction;

$CaO + H2O ----->Ca(OH)2$

Number of moles of CaO = 33g/56 g/mol = 0.59 moles

Number of moles of H20 = 10g/18 g/mol = 0.56 moles

Since the reaction is in 1:1 mole ration, H2O is the limiting reactant

Hence;

mass of Ca(OH)2 produced = 0.56 moles * 74 g/mol = 41.44 g

7 0

3 years ago

Read 2 more answers

When calcium metal (Ca) reacts with oxygen gas (O2), it produces the ionic compound calcium oxide (CaO). Write the balanced equa

malfutka [58]

The balanced chemical equation for the reaction is

2Ca(s) + O₂(g) → 2CaO(s)

and

The process meets the requirements to be considered a redox reaction.

From the question,

We are to write a balanced chemical equation for the reaction between Calcium metal and oxygen gas

The balanced chemical equation for the reaction is

2Ca(s) + O₂(g) → 2CaO(s)

This is the balanced chemical equation for the reaction

To determine whether the reaction meets the requirements to be considered a redox reaction,

We will define the term <em>redox reaction</em>

A <em>redox reaction</em>, also called oxidation-reduction reaction, is any chemical reaction in which the oxidation number of a participating chemical species changes.

From the balanced equation above we can observe that the oxidation number of the participating species(Calcium and oxygen) changed.

The oxidation number of Calcium changed from 0 to +2

and

The oxidation number of Oxygen changed from 0 to -2

∴ The process meets the requirements to be considered a redox reaction.

Hence, the balanced chemical equation for the reaction is

2Ca(s) + O₂(g) → 2CaO(s)

and

The process meets the requirements to be considered a redox reaction.

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2 years ago

If the sample has a total mass of 5.76 g and contains 1.79 g k, what are the percentages of kbr and ki in the sample by mass

Luden [163]

The percentages of KBr and KI in the sample by mass is 80.68 and 19.32 % respectively.

<h3>What is Molar Mass ?</h3>

Molar mass is defined as the mass contained in 1 mole of sample.

It is given that

the sample has a total mass of 5.76 g contains KBr and KI

and contains 1.79 gm of K is present

what is the percentage of KBr and KI in the sample

Molecular weight of K = 39

Molecular weight of Br = 79.9

Molecular weight of I = 126.9

In KBr the mass percentage of K is 39/(39+79.9) = 32.89%

In KI the mass percentage of K is 39/(39+126.9) = 23.5%

Let the mass of KBr present in the sample is x

K will be 0.3289 x

and let the mass of KI present be y

K will be 0.235y

x +y =5.76

0.3289x+0.235y = 1.79

0.0939y = 0.1045

y = 1.1125 gm

x = 5.76-1.1125

x = 4.6475 gm

% of KBr = (4.6475/5.76 )*100 = 80.68 %

% of KI = (1.1125/5.76) *100 = 19.32%

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2 years ago