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2 years ago

What is oxidized and what is reduced C2H4 + 2O2 → 2CO + 2H2O C2H4 + 3O2 → 2CO2 + 2H2O

Chemistry

1 answer:

Anika [276]2 years ago

7 0

C2H4 is oxidized and O2 is reduced in both reactions.

<h3>What is oxidation/reduction?</h3>

Oxidation is defined in several ways. Some of the definitions are:

The addition of oxygen or removal of hydrogen
Increase in the oxidation number of atoms
Addition of electronegative or the removal of electropositive elements

Reduction, on the other hand, is defined as:

Removal of oxygen or addition of hydrogen
Decrease in the oxidation number of atoms
Addition of electropositive elements or the removal of electronegative elements.

In the two reactions, oxygen is being added to C2H4. Thus, C2H4 is being oxidized.

The oxidizing agent is O2. In oxidation reactions, the oxidizing agents usually get reduced. Thus, O2 is reduced in both reactions.

More on oxidation and reduction can be found here: brainly.com/question/3867774

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What observations can you make about this group of people?

DedPeter [7]

Answer:The people are part of a softball team. They appear happy, they are showing number one with their fingers, and they are holding a trophy. They probably won the softball league championship

Explanation:

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3 years ago

Suppose you are titrating a sulfuric acid solution of unknown concentration with a sodium hydroxide solution according to the eq

pogonyaev

Answer:

Explanation:

H ₂ S O ₄ + 2 N a O H ⟶ 2 H ₂ O + N a ₂ S O ₄

29.09 mL of 0.639 M N a O H is mixed with 213.8 mL of H ₂ S O ₄

Let the concentration of H ₂ S O ₄ be S₂ .

In terms of normal or equivalent solution is will be 2 N solution

From the formula S₁ V₁ = S₂ V₂

= 29.09 x .639 = 213.8 x S₂

S₂ = .087 N solution

In terms of molar solution it will be .087 / 2 M

= .0435 M

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3 years ago

An HCl solution has a concentration of 9.8 x 10-5 M. What is the pH of this solution?

Inessa05 [86]

HCl = H⁺ + Cl⁻

c(HCl)=9.8*10⁻⁵ mol/l

pH=-lg[H⁺]

[H⁺]=c(HCl)

pH=-lg{c(HCl)}

pH=-lg{9.8*10⁻⁵}=4.009

pH=4.009

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3 years ago

A 2.89 g sample of a metal chloride, MCl2, is dissolved in water and treated with excess aqueous silver nitrate. The silver chlo

dsp73

Answer:

58.0 g/mol

Explanation:

The reaction that takes place is:

MCl₂ + 2AgNO₃ → 2AgCl + M(NO₃)₂

First we calculate how many moles of silver chloride were produced, using its molar mass:

6.41 g AgCl ÷ 143.32 g/mol = 0.0447 mol AgCl

Then we convert AgCl moles into MCl₂ moles, using the stoichiometric ratio:

0.0447 mol AgCl * $\frac{1molMCl_2}{2molAgCl}$ = 0.0224 mol MCl₂

Now we calculate the molar mass of MCl₂, using the original mass of the sample:

2.86 g / 0.0224 mol = 127.68 g/mol

We can write the molar mass of MCl₂ as:

Molar Mass MCl₂ = Molar Mass of M + (Molar Mass of Cl)*2

127.68 g/mol = Molar Mass of M + (35.45 g/mol)*2

Finally we calculate the molar mass of M:

Molar Mass of M = 57 g/mol

The closest option is 58.0 g/mol.

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3 years ago

What is the mass, in grams, of 0.000142 mol of vitamin C?

tatiyna

Answer:

0.0250 g

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3 years ago