Answer:
N- 1s2 2s2 2p3
Mg- 1s2 2s2 2p6 3s2
O- 1s2 2s2 2p4
F- 1s2 2s2 2p5
Al-1s2 2s2 2p6 3s2 3p1
Explanation:
Order of decreasing atomic radius
Mg,Al, N,O,F
Order of increasing ionization energy
Mg,Al, N,O,F
Reason:
Atomic radius decreases with increase in nonmetallic character. Looking at the electronic configurations, as effective nuclear charge increases, the atom becomes smaller and the attractive force between the nucleus and the outermost electrons increases. Hence, the radius of the atom decreases and ionization energy increases. Note that the addition of more orbital electrons implies addition of more nuclear charge since the both must exactly balance for the atom to remain electrically neutral. The more the electrons in the outermost shell, the higher the first ionization energy.
<u>Answer:</u> The heat of hydrogenation of the reaction is coming out to be 234.2 kJ.
<u>Explanation:</u>
Enthalpy change 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_{rxn}=\sum [n\times \Delta H_{(product)}]-\sum [n\times \Delta H_{(reactant)}]](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%3D%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H_%7B%28product%29%7D%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H_%7B%28reactant%29%7D%5D)
For the given chemical reaction:

The equation for the enthalpy change of the above reaction is:
![\Delta H_{rxn}=[(1\times \Delta H_{(C_4H_{10})})]-[(1\times \Delta H_{(C_4H_6)})+(2\times \Delta H_{(H_2)})]](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%3D%5B%281%5Ctimes%20%5CDelta%20H_%7B%28C_4H_%7B10%7D%29%7D%29%5D-%5B%281%5Ctimes%20%5CDelta%20H_%7B%28C_4H_6%29%7D%29%2B%282%5Ctimes%20%5CDelta%20H_%7B%28H_2%29%7D%29%5D)
We are given:

Putting values in above equation, we get:
![\Delta H_{rxn}=[(1\times (-2877.6))]-[(1\times (-2540.2))+(2\times (-285.8))]\\\\\Delta H_{rxn}=234.2J](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%3D%5B%281%5Ctimes%20%28-2877.6%29%29%5D-%5B%281%5Ctimes%20%28-2540.2%29%29%2B%282%5Ctimes%20%28-285.8%29%29%5D%5C%5C%5C%5C%5CDelta%20H_%7Brxn%7D%3D234.2J)
Hence, the heat of hydrogenation of the reaction is coming out to be 234.2 kJ.
Answer:
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Explanation:
hloride,which reactant is in excess?how many moles of aluminium chloride can be produced during the reaction.hloride,which reactant is in excess?how many moles of aluсиськижопыссиськасуксиськипись loride can be produced during the reaction.hloride,which reactant is in excess?how many moles of aluminium chloride can be produced during the reaction.hloride,whichйух reactant is in excess?how many moles of aluminium chloride can be produced during the reaction.