H2(g) +C2H4(g)→C2H6(g)
H-H +H2C =CH2→H3C-Ch3
2C -H bonds and one C-C bond are formed while enthalpy change (dH) of the reaction,
H-H: 432kJ/mol
C=C: 614kJ/mol
C-C: 413 kJ/mol
C-C: 347 kJ/mol
dH is equal to sum of the energies released during the formation of new bonds or negative sign, and sum of energies required to break old bonds or positive sign.
The bond which breaks energy is positive.
432+614 =1046kJ/mol
Formation of bond energy is negative
2(413) + 347 = 1173 kJ/mol
dH reaction is -1173 + 1046 =-127kJ/mol
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Answer:
<em>For both cases the answer is C</em>
Explanation:
We can see that the orbitals are not filled in the order of increasing energy and the Pauli exclusion principle is violated because it does not follow the correct order of the electron configuration; In the first exercise after the 2s2 orbital, the 2p2 orbital follows.
For the second exercise, you must start in order with level 1 and correctly filling each of the sublevels corresponding to each level until reaching level 7 and thus completing the desired number of electrons.
There would be 67 left because you do
The limiting reactant is chlorine (Cl2).
<u>Explanation</u>:
Limiting reactant is the amount of product formed which gets limited by the reagent without continuing it.
2 Al + 3 Cl2 ==> 2 AlCl3 represents the balanced equation.
Number of moles Al present = 34 g Al x 1 mole Al / 26.98 g
= 1.260 g moles of Al
Number of moles Cl2 present = 39 g Cl2 x 1 mole Cl2 / 35.45 g
= 1.10 g moles of Cl2
Dividing each reactant by it's coefficient in the balanced equation obtains:
1.260 moles Al / 2 = 0.63 g moles of Al
1.11 moles Cl2 / 3 = 0.36 g moles of Cl2
The reactant which produces a lesser amount of product is called as limiting reactant.
Here the Limiting reactant is Cl2.
