Answer:
C. The potential energy change for a chemical reaction.
Explanation:
The reaction coordinate q illustrates, graphically, the energy changes during exothermic and endothermic reactions. This graphical representation of the energy changes in the course of a chemical reaction is known as reaction coordinates. A reaction coordinate is a graphical sequence of steps by which the reaction progresses from reactants through activated complexes to products. Reaction coordinates explain how far a reaction has proceeded towards the products or from the reactants.
From the images attached below, we can see the reaction coordinates in the reaction profiles.
Answer:
This is an example of convection
Answer:
The standard enthalpy change for the reaction at 
is -2043.999kJ
Explanation:
Standard enthalpy change (
) for the given reaction is expressed as:
![\Delta H_{rxn}^{0}=[3mol\times \Delta H_{f}^{0}(CO_{2})_{g}]+[4mol\times \Delta H_{f}^{0}(H_{2}O)_{g}]-[1mol\times \Delta H_{f}^{0}(C_{3}H_{8})_{g}]-[5mol\times \Delta H_{f}^{0}(O_{2})_{g}]](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%5E%7B0%7D%3D%5B3mol%5Ctimes%20%5CDelta%20H_%7Bf%7D%5E%7B0%7D%28CO_%7B2%7D%29_%7Bg%7D%5D%2B%5B4mol%5Ctimes%20%5CDelta%20H_%7Bf%7D%5E%7B0%7D%28H_%7B2%7DO%29_%7Bg%7D%5D-%5B1mol%5Ctimes%20%5CDelta%20H_%7Bf%7D%5E%7B0%7D%28C_%7B3%7DH_%7B8%7D%29_%7Bg%7D%5D-%5B5mol%5Ctimes%20%5CDelta%20H_%7Bf%7D%5E%7B0%7D%28O_%7B2%7D%29_%7Bg%7D%5D)
Where 
refers standard enthalpy of formation
Plug in all the given values from literature in the above equation:
![\Delta H_{rxn}^{0}=[3mol\times (-393.509kJ/mol)]+[4mol\times (-241.818kJ/mol)]-[1mol\times (-103.8kJ/mol)]-[5mol\times (0kJ/mol)]=-2043.999kJ](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%5E%7B0%7D%3D%5B3mol%5Ctimes%20%28-393.509kJ%2Fmol%29%5D%2B%5B4mol%5Ctimes%20%28-241.818kJ%2Fmol%29%5D-%5B1mol%5Ctimes%20%28-103.8kJ%2Fmol%29%5D-%5B5mol%5Ctimes%20%280kJ%2Fmol%29%5D%3D-2043.999kJ)
Alpha partical is a He nucleus. When decaying alpha particle mass is reduced by 4 and atomic number is reduced by 2.
The actual element which has 102 protons is No (Nobelium).
Since it has 167 neutrons, the mass = protons + neutrons = 102 + 167 = 269
after an alpha decay, the new element formed has 100 protons which is Fm ( Fermium)
the alpha decaying equation is,
₁₀₂²⁶⁹No → ₁₀₀²⁶⁵Fm + ₂⁴α + heat
the total mass and the atomic number( numbe rof protons) must be equal in both sides.
