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LuckyWell [14K]
1 year ago
15

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

Chemistry
1 answer:
Anika [276]1 year 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:

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

Reduction, on the other hand, is defined as:

  1. Removal of oxygen or addition of hydrogen
  2. Decrease in the oxidation number of atoms
  3. 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

#SPJ1

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sukhopar [10]

A kilogram is a measure of mass; 1,000 grams

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How do electrons differ from protons and neutrons? A. They do not move. B. They are larger. C. They are not in a fixed position.
nikdorinn [45]
Hi!

Electrons are particles which basically 'orbit' around the nucleus. Protons and neutrons are condensed, in a fixed position inside the nucleus. 

With this in mind, the answer will be C.

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3 years ago
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How much heat energy is required to convert 48.3 g of solid ethanol at -114.5 degree C to gasesous ethanol at 135.3 degree C? Th
OLEGan [10]

Answer:

7.21 × 10⁴ J

Explanation:

Ethanol is solid below -114.5°c, liquid between -114.5°C and 78.4°C, and gaseous above 78.4°C.

<em>How much heat energy is required to convert 48.3 g of solid ethanol at -114.5°C to gaseous ethanol at 135.3 °C?</em>

<em />

We need to calculate the heat required in different stages and then add them.

The moles of ethanol are:

48.3g.\frac{1mol}{46.07g} =1.05mol

Solid-liquid transition

Q₁ = ΔHfus . n = (4.60 kJ/mol) . 1.05 mol = 4.83 kJ = 4.83 × 10³ J

where,

ΔHfus: molar heat of fusion

n: moles

Liquid: from -114.5°C to 78.4°C

Q₂ = c(l) . m . ΔT = (2.45 J/g.°C) . 48.3g . [78.4°C-(-114.5°C)] = 2.28 × 10⁴ J

where,

c(l): specific heat capacity of the liquid

ΔT: change in the temperature

Liquid-gas transition

Q₃ = ΔHvap . n = (38.56 kJ/mol) . 1.05 mol = 40.5 kJ = 40.5 × 10³ J

where,

ΔHvap: molar heat of vaporization

Gas: from 78.4°C to 135.3°C

Q₄ = c(g) . m . ΔT = (1.43 J/g.°C) . 48.3g . (135.3°C-78.4°C) = 3.93 × 10³ J

where

c(g): specific heat capacity of the gas

Total heat required

Q₁ + Q₂ + Q₃ + Q₄ = 4.83 × 10³ J + 2.28 × 10⁴ J + 40.5 × 10³ J + 3.93 × 10³ J = 7.21 × 10⁴ J

3 0
3 years ago
Phosphoric acid, which is commonly used as rust inhibitor, food additive and etching agent for dental and orthopedic use, can be
Sphinxa [80]

Answer:

P_4_{(I)}+5O_2_{(g)}+6H_2O_{(l)}\rightarrow +4H_3PO_4_{(l)}

Explanation:

The first step is:

P_4_{(I)}+5O_2_{(g)}\rightarrow 2P_2O_5_{(g)}

Second step is:

P_2O_5_{(g)}+3H_2O_{(l)}\rightarrow 2H_3PO_4_{(l)}

Multiplying second step by 2, and adding both the steps, we get that:

P_4_{(I)}+5O_2_{(g)}+2P_2O_5_{(g)}+6H_2O_{(l)}\rightarrow 2P_2O_5_{(g)}+4H_3PO_4_{(l)}

Cancelling common species, we get that:

P_4_{(l)}+5O_2_{(g)}+6H_2O_{(l)}\rightarrow +4H_3PO_4_{(l)}

6 0
3 years ago
How much energy must a 10 gram block of ice gain in order to melt ?
Minchanka [31]

Answer:

the answer is 10 times

Explanation:

because it takes 10 times as much energy -3330 j - to melt 10.0 grams of ice.

8 0
1 year ago
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