Answer: In a chemical equation the reactants are located on the left side and the products are located on the right side.
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
Answer:
I think the primary producers start the flow of energy
Answer:
Electronegativity, symbol χ, measures the tendency of an atom to attract a shared pair of electrons (or electron density).
An atom's electronegativity is affected by both its atomic number and the distance at which its valence electrons reside from the charged nucleus.
The higher the associated electronegativity, the more an atom or a substituent group attracts electrons.
Answer:
The rock will sink.
Explanation:
If the object has a higher density then the liquid it is being placed in it will sink.
