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Murljashka [212]
2 years ago
5

The purpose of cracking in the production of ethene is to

Chemistry
1 answer:
faltersainse [42]2 years ago
7 0

The purpose of cracking in the production of ethene is to convert large molecules into small molecules.

Ethene can be produced from the thermal cracking of ethane and propene.

In thermal cracking, high temperature of about 750-850°C and pressure of up to about 70 atm is used to break carbon-carbon bonds. Free radicals are formed in the process.

More about petroleum refining processes:

Petroleum refinery can be classified into two broad categories, namely the main refining processes and conversion processes. Main refining processes includes distillation, absorption and adsorption, while conversion processes involves thermal cracking, catalytic cracking and hydrocracking.

As the name implies, cracking is generally the <u>breaking up</u> of larger molecules into smaller, more useful molecules, such as those with 5 to 12 carbons.

To better comprehend cracking processes, we can use the analogy of dropping a ceramic mug.

When this ceramic mug is dropped, it breaks into smaller ceramic pieces. This represents smaller molecules being formed from a larger molecule. The force used to break them into smaller molecules would differ in the various types of cracking processes. For example, thermal cracking mainly uses heat while catalytic cracking mainly uses a zeolite catalyst.

Note that for these cracking processes to occur, there are other conditions involved.

Additional:

To learn more about cracking, do check out the following!

  • brainly.com/question/13565152
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Answer:

The approximate molar enthalpy of combustion of this substance is -66 kJ/mole.

Explanation:

First we have to calculate the heat gained by the calorimeter.

q=c\times \Delta T

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c = Specific heat = 2.68 kJ/^oC

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Now put all the given values in the above formula, we get:

q=2.68 kJ/^oC\times 3.08^oC

q=8.2544 kJ

Now we have to calculate molar enthalpy of combustion of this substance :

\Delta H_{comb}=-\frac{q}{n}

where,

\Delta H_{comb} = enthalpy change = ?

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n = number of moles methane = \frac{\text{Mass of methane}}{\text{Molar mass of methane }}=\frac{2.00 g}{16.042 g/mol}=0.1247 mole

\Delta H_{comb}=-\frac{8.2544 kJ}{0.1247 mole}=-66.21 kJ/mole\approx -66 kJ/mole

Therefore,  the approximate molar enthalpy of combustion of this substance is -66 kJ/mole.

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Answer:

this is answers to the second picture

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