A maximum of 8 electrons can share the quantum number n = 2.
Principal Quantum number has a symbol of "n". It tells you the energy level on which an electron resides. Y<span>ou need to determine exactly how many </span>orbitals<span> you have in this energy level before you can determine the number of electrons that can share the value of n.
</span>
The number of orbitals you get per energy level can be found using this formula:
<span>no. of orbitals=<span>n</span></span><span>²</span>
Each orbital can hold a maximum of two electrons, the formula would be:
<span>no. of electrons=2<span>n</span></span><span>²</span>
Using the given formulas:
<span>no. of orbitals = <span>n</span></span><span>² </span><span>= </span><span>2</span><span>² </span><span>= </span><span>4</span>
<span>no. of electrons </span><span>=</span><span>2 *</span><span> </span><span>4 </span><span>= </span><span>8 </span>
Answer:
Helium, Krypton, Neon, Argon, Radon, Oganesson and Xenon.
Explanation:
The fluid filled area between the pleural layers is called the pleural cavity.
Answer : The enthalpy of combustion per mole of 
is -2815.8 kJ/mol
Explanation :
Enthalpy change : It 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^o_{rxn}=\sum [n\times \Delta H^o_f(product)]-\sum [n\times \Delta H^o_f(reactant)]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H%5Eo_f%28product%29%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H%5Eo_f%28reactant%29%5D)
The equilibrium reaction follows:
The equation for the enthalpy change of the above reaction is:
![\Delta H^o_{rxn}=[(n_{(CO_2)}\times \Delta H^o_f_{(CO_2)})+(n_{(H_2O)}\times \Delta H^o_f_{(H_2O)})]-[(n_{(C_6H_{12}O_6)}\times \Delta H^o_f_{(C_6H_{12}O_6)})+(n_{(O_2)}\times \Delta H^o_f_{(O_2)})]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%28n_%7B%28CO_2%29%7D%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28CO_2%29%7D%29%2B%28n_%7B%28H_2O%29%7D%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28H_2O%29%7D%29%5D-%5B%28n_%7B%28C_6H_%7B12%7DO_6%29%7D%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28C_6H_%7B12%7DO_6%29%7D%29%2B%28n_%7B%28O_2%29%7D%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28O_2%29%7D%29%5D)
We are given:
Putting values in above equation, we get:
![\Delta H^o_{rxn}=[(6\times -393.5)+(6\times -285.8)]-[(1\times -1260)+(6\times 0)]=-2815.8kJ/mol](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%286%5Ctimes%20-393.5%29%2B%286%5Ctimes%20-285.8%29%5D-%5B%281%5Ctimes%20-1260%29%2B%286%5Ctimes%200%29%5D%3D-2815.8kJ%2Fmol)
Therefore, the enthalpy of combustion per mole of is -2815.8 kJ/mol
The state in which the forward reaction rate and the reverse reaction rate are equal. Th concentration of chemicals don’t change
