Answer:
The standard enthalpy of formation of this isomer of 
is -220.1 kJ/mol.
Explanation:
The given chemical reaction is as follows.
The expression for the entropy change for the reaction is as follows.
![\Delta H^{o}_{rxn}=[8\Delta H^{o}_{f}(CO_{2}) +9\Delta H^{o}_{f}(H_{2}O)]-[\Delta H^{o}_{f}(C_{8}H_{18})+ \frac{25}{2}\Delta H^{o}_{f}(O_{2})]](https://tex.z-dn.net/?f=%5CDelta%20H%5E%7Bo%7D_%7Brxn%7D%3D%5B8%5CDelta%20H%5E%7Bo%7D_%7Bf%7D%28CO_%7B2%7D%29%20%2B9%5CDelta%20H%5E%7Bo%7D_%7Bf%7D%28H_%7B2%7DO%29%5D-%5B%5CDelta%20H%5E%7Bo%7D_%7Bf%7D%28C_%7B8%7DH_%7B18%7D%29%2B%20%5Cfrac%7B25%7D%7B2%7D%5CDelta%20H%5E%7Bo%7D_%7Bf%7D%28O_%7B2%7D%29%5D)
Substitute the all values in the entropy change expression.
![-5104.1kJ/mol=[8(-393.5)+9(-241.8)kJ/mol]-[\Delta H^{o}_{f}(C_{8}H_{18})+ \frac{25}{2}(0)kJ/mol]](https://tex.z-dn.net/?f=-5104.1kJ%2Fmol%3D%5B8%28-393.5%29%2B9%28-241.8%29kJ%2Fmol%5D-%5B%5CDelta%20H%5E%7Bo%7D_%7Bf%7D%28C_%7B8%7DH_%7B18%7D%29%2B%20%5Cfrac%7B25%7D%7B2%7D%280%29kJ%2Fmol%5D)
Therefore, The standard enthalpy of formation of this isomer of is -220.1 kJ/mol.
Answer:HNO₃ and NO³⁻ would not function as buffer
Explanation:
The buffer solution are usually prepared by using any weak acid (which would partially dissociate) and mixing this weak acid with its own conjugate base or any weak base (which would partially dissociate) and mixing with with its conjugate acid.
A buffer solution is a solution which resists change in pH of the solution.
Since nitric acid is a very strong acid and hence neither nitric acid HNO₃ or its conjugate base NO³⁻ anionb is suitable for the preparation of buffer solution.
HCO³⁻ is a weak acid and hence it can form a buffer solution with its conjugate base CO₃²-. so they can be used to form buffer.
C₂H₅COOH is a weak acid and hence it can also form buffer solution with its conjugate base.
So only HNO₃and NO³⁻ would not be able to form buffer
So option a is the answer.
The answer is A
Mark Brainliest ☺☻
