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The soup water would turn into soap.
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Answer: The molar enthalpy of formation for paraffin wax is -2460.5 kJ
Explanation:
The balanced chemical reaction is,

The expression for enthalpy change is,
![\Delta H=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H_f%28product%29%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H_f%28reactant%29%5D)
![\Delta H=[(n_{CO_2}\times \Delta H_{CO_2})+(n_{H_2O}\times \Delta H_{H_2O})]-[(n_{O_2}\times \Delta H_{O_2})+(n_{C_{25}H_{52}}\times \Delta H_{C_{25}H_{52}})]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B%28n_%7BCO_2%7D%5Ctimes%20%5CDelta%20H_%7BCO_2%7D%29%2B%28n_%7BH_2O%7D%5Ctimes%20%5CDelta%20H_%7BH_2O%7D%29%5D-%5B%28n_%7BO_2%7D%5Ctimes%20%5CDelta%20H_%7BO_2%7D%29%2B%28n_%7BC_%7B25%7DH_%7B52%7D%7D%5Ctimes%20%5CDelta%20H_%7BC_%7B25%7DH_%7B52%7D%7D%29%5D)
where,
n = number of moles
(as heat of formation of substances in their standard state is zero
Now put all the given values in this expression, we get
![-14800=[(25\times -393.5)+(26\times -285.5)]-[(38\times 0)+(1\times \Delta H_{C_{25}H_{52}})]](https://tex.z-dn.net/?f=-14800%3D%5B%2825%5Ctimes%20-393.5%29%2B%2826%5Ctimes%20-285.5%29%5D-%5B%2838%5Ctimes%200%29%2B%281%5Ctimes%20%5CDelta%20H_%7BC_%7B25%7DH_%7B52%7D%7D%29%5D)

Therefore, the molar enthalpy of formation for paraffin wax is -2460.5 kJ