Answer:
Experiments to determine mechanisms involve looking at indirect evidence to help support or disprove a proposed mechanism.
Most intermediates are not typically isolated to determine reaction mechanisms.
Carbocations are very reactive and are typically not isolated for analysis.
Scientists can prove that a specific mechanism exists.
Evidence of intermediates sometimes can be seen using techniques such as nuclear magnetic resonance spectroscopy
Explanation:
The study of reaction mechanism and chemical kinetics often form the main thrust of study in organic, inorganic and physical chemistry.
We often want to know the actual processes involved in the conversion of one specie to another. Unfortunately, this information may have to be obtained indirectly by certain chemical reactions or by the use of new instrumental methods such as nuclear magnetic resonance spectroscopy.
Many organic reactions have carbocation intermediates. These carbocations are relatively short-lived and are transient intermediates which are rarely isolated unless they are isolated in a molecular cage using a macromolecule or in superacids.
By intensive study, scientists can proof or disprove the authenticity of any proposed mechanism.
We must know that a transition state has partial bonds. It is often an extremely short-lived specie which cannot be isolated.
Answer : The energy of the photon emitted is, -12.1 eV
Explanation :
First we have to calculate the 
orbit of hydrogen atom.
Formula used :
where,
= energy of 
orbit
n = number of orbit
Z = atomic number of hydrogen atom = 1
Energy of n = 1 in an hydrogen atom:
Energy of n = 2 in an hydrogen atom:
Energy change transition from n = 1 to n = 3 occurs.
Let energy change be E.
The negative sign indicates that energy of the photon emitted.
Thus, the energy of the photon emitted is, -12.1 eV
Answer: +178.3 kJ
Explanation:
The chemical equation follows:
The equation for the enthalpy change of the above reaction is:
![\Delta H^o_{rxn}=[(1\times \Delta H^o_f_{(CaO(s))})+(1\times \Delta H^0f_{CO_2}]-[(1\times \Delta H^o_f_{(CaCO_3(s))})]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%281%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28CaO%28s%29%29%7D%29%2B%281%5Ctimes%20%5CDelta%20H%5E0f_%7BCO_2%7D%5D-%5B%281%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28CaCO_3%28s%29%29%7D%29%5D)
We are given:
Putting values in above equation, we get:
![\Delta H^o_{rxn}=[(1\times (-635.1))+(1\times (-393.5))]-[(1\times (-1206.9))]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%281%5Ctimes%20%28-635.1%29%29%2B%281%5Ctimes%20%28-393.5%29%29%5D-%5B%281%5Ctimes%20%28-1206.9%29%29%5D)
The DH°rxn for the decomposition of calcium carbonate to calcium oxide and carbon dioxide is +178.3 kJ
The result when the OH- ions go as additional acid is added is that t<span>hey remain in solution but change concentration. </span>This is because when you add more acid to hydroxide, it will change its concentration due to the acidity added.
