Answer:
No. No new substance is formed during Physical change.
<u>Answer:</u> The heat of hydrogenation of the reaction is coming out to be 234.2 kJ.
<u>Explanation:</u>
Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as 
The equation used to calculate enthalpy change is of a reaction is:
![\Delta H_{rxn}=\sum [n\times \Delta H_{(product)}]-\sum [n\times \Delta H_{(reactant)}]](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%3D%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H_%7B%28product%29%7D%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H_%7B%28reactant%29%7D%5D)
For the given chemical reaction:

The equation for the enthalpy change of the above reaction is:
![\Delta H_{rxn}=[(1\times \Delta H_{(C_4H_{10})})]-[(1\times \Delta H_{(C_4H_6)})+(2\times \Delta H_{(H_2)})]](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%3D%5B%281%5Ctimes%20%5CDelta%20H_%7B%28C_4H_%7B10%7D%29%7D%29%5D-%5B%281%5Ctimes%20%5CDelta%20H_%7B%28C_4H_6%29%7D%29%2B%282%5Ctimes%20%5CDelta%20H_%7B%28H_2%29%7D%29%5D)
We are given:

Putting values in above equation, we get:
![\Delta H_{rxn}=[(1\times (-2877.6))]-[(1\times (-2540.2))+(2\times (-285.8))]\\\\\Delta H_{rxn}=234.2J](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%3D%5B%281%5Ctimes%20%28-2877.6%29%29%5D-%5B%281%5Ctimes%20%28-2540.2%29%29%2B%282%5Ctimes%20%28-285.8%29%29%5D%5C%5C%5C%5C%5CDelta%20H_%7Brxn%7D%3D234.2J)
Hence, the heat of hydrogenation of the reaction is coming out to be 234.2 kJ.
Answer:
Relating to orbit or orbits
Explanation:
Answer:
they would react to make aluminum oxide and nitrogen
Explanation:
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