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WITCHER [35]
2 years ago
9

The enthalpy of combustion for octane (C8H18(l)), a key component of gasoline, is -5,074 kJ/mol. This value is the Delta. Hrxn f

or which of the following reactions? 2C8H18(l) 25O2(g) Right arrow. 16CO2(g) 18H2O(g) C8H18(l) 12. 5O2(g) Right arrow. 8CO2(g) 9H2O(g) 16CO2(g) 18H2O(g) Right arrow. 2C8H18(l) 25O2(g) 8CO2(g) 9H2O(g) Right arrow. C8H18(l) 12. 5O2(g).
Chemistry
1 answer:
Inessa [10]2 years ago
6 0

Combustion can be defined as the reaction of a compound with oxygen. The enthalpy of combustion of octane is \Delta H_{\rm rxn} for \rm C_8H_{18}\;+\;25\;O_2\;\rightarrow 8\;CO_2\;+\;9\;H_2O.

<h3>What is the enthalpy of reaction?</h3>

The enthalpy of reaction is the amount of heat energy absorbed or lost by the molecules in the chemical reaction.

The enthalpy of combustion is the amount of heat energy released by the compound in the reaction with oxygen.

The reaction in which heat is liberated with the reaction of a compound with oxygen has an enthalpy of combustion, equivalent to the enthalpy of reaction.

The combustion of octane can be given as:

\rm C_8H_{18}\;+\;25\;O_2\;\rightarrow 8\;CO_2\;+\;9\;H_2O

Thus, the reaction has combustion energy equivalent to the enthalpy of the reaction is \rm C_8H_{18}\;+\;25\;O_2\;\rightarrow 8\;CO_2\;+\;9\;H_2O. Thus, option B is correct.

Learn more about enthalpy of reaction, here:

brainly.com/question/1657608

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How many valence electrons would there be for the element with the<br> following orbital diagram?
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If sodium is burned in chlorine gas, a compound is formed that dissolves in water. What colour will the solution be?
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If sodium is burned in chlorine fuel, a compound is formed that dissolves in water. the solution be: Bright yellow mild

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6 0
10 months ago
​29. A gas has a volume of 1.75 L at -23°C and 150.0 kPa.
arsen [322]

The answer for the following mention bellow.

  • <u><em>Therefore the final temperature of the gas is 260 k</em></u>

Explanation:

Given:

Initial pressure (P_{1}) = 150.0 kPa

Final pressure (P_{2}) = 210.0 kPa

Initial volume (V_{1}) = 1.75 L

Final volume (V_{2}) = 1.30 L

Initial temperature (T_{1}) = -23°C = 250 k

To find:

Final temperature (T_{2})

We know;

According to the ideal gas equation;

P × V = n × R ×T

where;

P represents the pressure of the gas

V represents the volume of the gas

n represents the no of moles of the gas

R represents the universal gas  constant

T represents the temperature of the gas

We know;

\frac{P*V}{T} = constant

\frac{P_{1} }{P_{2} } × \frac{V_{1} }{V_{2} } = \frac{T_{1} }{T_{2} }

Where;

(P_{1}) represents the initial pressure of the gas

(P_{2}) represents the final pressure of the gas

(V_{1}) represents the initial volume of the gas

(V_{2}) represents the final volume of the gas

(T_{1}) represents the initial temperature of the gas

(T_{2}) represents the final temperature of the gas

So;

\frac{150 * 1.75}{210 * 1.30} = \frac{260}{T_{2} }

(T_{2}) =260 k

<u><em>Therefore the final temperature of the gas is 260 k</em></u>

<u><em></em></u>

3 0
2 years ago
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