Answer: The enthalpy of combustion, per mole, of butane is -2657.4 kJ
Explanation:
The balanced chemical reaction is,
The expression for enthalpy change is,
Putting the values we get :
2 moles of butane releases heat = 5314.8 kJ
1 mole of butane release heat = ![\frac{5314.8}{2}\times 1=2657.4kJ](https://tex.z-dn.net/?f=%5Cfrac%7B5314.8%7D%7B2%7D%5Ctimes%201%3D2657.4kJ)
Thus enthalpy of combustion per mole of butane is -2657.4 kJ
Answer:
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Explanation:
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The second and first one but if it isn’t 2 choices then 1