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pogonyaev
3 years ago
14

Rank the following compounds in order of increasing heat of combustion. Be sure to answer all parts. Drag each item to the corre

ct box. A) pentane B) propane C) butane
Chemistry
1 answer:
vfiekz [6]3 years ago
7 0

Answer:

The compounds are arranged from the least heat of combustion to the highest (in order of increasing heat of combustion).

Propane--> butane --> pentane.

Explanation:

Hydrocarbons are compounds which contains carbon and hydrogen only. Alkane is called a saturated hydrocarbon because all four bonds of carbon are single bonds. Examples of alkanes include: pentane, propane and butane. One of the chemical properties of alkanes includes combustion reaction.

In COMBUSTION REACTION, alkanes burns in air (oxygen) to produce carbon dioxide and water.The heat released on complete combustion of one mole of a substance is called HEAT OF COMBUSTION.

The heat of combustion of saturated hydrocarbon ( alkanes) increases with increase in number of carbon atoms in a molecule. This is due to more carbon available for burning and more number of bonds undergoing changes. Therefore in the arrangement of the given compounds, PENTANE has the highest heat of combustion while PROPANE is the least.

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Vera_Pavlovna [14]

Two changes would make this reaction reactant-favored

C. Increasing the temperature

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<h3>Further explanation</h3>

Given

Reaction

2H₂ + O₂ ⇒ 2H₂0 + energy

Required

Two changes would make this reaction reactant-favored

Solution

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If the system temperature is raised, then the equilibrium reaction will reduce the temperature by shifting the reaction in the direction that requires heat (endotherms). Conversely, if the temperature is lowered, then the equilibrium shifts to a reaction that releases heat (exothermic)  

While on the change in pressure, then the addition of pressure, the reaction will shift towards a smaller reaction coefficient  

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As the temperature rises, the equilibrium will shift towards the endothermic reaction, so the reaction shifts to the left towards H₂ + O₂( reactant-favored)

And reducing the pressure, then the reaction shifts to the left H₂ + O₂( reactant-favored)⇒the number of coefficients is greater

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Explanation:

4 0
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yawa3891 [41]

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Naddik [55]

Answer:

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2 years ago
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