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pav-90 [236]
4 years ago
13

25 POINTS!!! HELP FAST

Chemistry
2 answers:
FromTheMoon [43]4 years ago
8 0

im not sure wow super hard question

DIA [1.3K]4 years ago
6 0

Can you please send me whole question picture to better assist you.

Thank you

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When a 3.80 g sample of C8H18(l) is burned in a bomb calorimeter, the temperature of the calorimeter rises by 27.3 oC. The heat
Fudgin [204]

<u>Answer:</u> The enthalpy of the reaction is -5112.5 kJ/mol

<u>Explanation:</u>

To calculate the heat absorbed by the calorimeter, we use the equation:

q=c\Delta T

where,

q = heat absorbed

c = heat capacity of calorimeter = 6.18 kJ/°C

\Delta T = change in temperature = 27.3°C

Putting values in above equation, we get:

q=6.18kJ/^oC\times 27.3^oC=168.714kJ

Heat absorbed by the calorimeter will be equal to the heat released by the reaction.

<u>Sign convention of heat:</u>

When heat is absorbed, the sign of heat is taken to be positive and when heat is released, the sign of heat is taken to be negative.

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Given mass of octane = 3.80 g

Molar mass of octane = 114 g/mol

Putting values in above equation, we get:

\text{Moles of octane}=\frac{3.80g}{114g/mol}=0.033mol

To calculate the enthalpy change of the reaction, we use the equation:

\Delta E=\frac{q}{n}

where,

q = amount of heat released = -168.714 kJ

n = number of moles = 0.033 moles

\Delta E = enthalpy change of the reaction

Putting values in above equation, we get:

\Delta E=\frac{-168.714kJ}{0.033mol}=-5112.5kJ/mol

Hence, the enthalpy of the reaction is -5112.5 kJ/mol

4 0
3 years ago
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