Answer: The enthalpy of formation of 
is -396 kJ/mol
Explanation:
Calculating the enthalpy of formation of
The chemical equation for the combustion of propane follows:
The equation for the enthalpy change of the above reaction is:
![\Delta H^o_{rxn}=[(2\times \Delta H^o_f_{(SO_3(g))})]-[(2\times \Delta H^o_f_{(SO_2(g))})+(1\times \Delta H^o_f_{(O_2(g))})]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%282%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28SO_3%28g%29%29%7D%29%5D-%5B%282%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28SO_2%28g%29%29%7D%29%2B%281%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28O_2%28g%29%29%7D%29%5D)
We are given:
Putting values in above equation, we get:
![-198=[(2\times \Delta H^o_f_{(SO_3(g))})]-[(2\times \Delta -297)+(1\times (0))]\\\\\Delta H^o_f_{(SO_3(g))}=-396kJ/mol](https://tex.z-dn.net/?f=-198%3D%5B%282%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28SO_3%28g%29%29%7D%29%5D-%5B%282%5Ctimes%20%5CDelta%20-297%29%2B%281%5Ctimes%20%280%29%29%5D%5C%5C%5C%5C%5CDelta%20H%5Eo_f_%7B%28SO_3%28g%29%29%7D%3D-396kJ%2Fmol)
The enthalpy of formation of is -396 kJ/mol
Answer:
HNO₃ (aq) —> H⁺ (aq) + NO₃¯ (aq)
Explanation:
From the question given above
HNO₃ + H₂O —> ?
Nitric acid, HNO₃ reacts with water, H₂O to form aqueous solution of nitric acid as illustrated below:
HNO₃ + H₂O —> HNO₃ (aq)
Nitric acid is a strong acid and, so will ionised completely when dissolved in water. This is illustrated below:
HNO₃ (aq) —> H⁺ (aq) + NO₃¯ (aq)
No, they do not. Carbon dioxide has a linear geometry because the lone pair and bond pair repulsion cancels out; however, water has a bent structure because only the oxygen atom possesses a lone pair which brings the bonding electron pairs closer.
Hi!
I'm not entirely sure about this so I'm sorry if I'm wrong but I think it would be helium.
Again in not entirely sure but i hope this helped you, i hope you have a great day, afternoon, or night!
