The correct answer is 3) 2CO2(g) ⇄ 2CO(g) + O2(g)
this is the correct one because it is a decomposition reaction and all the number of atoms is equal on both sides.
there are 2 C atoms on both sides.
and 4 O atoms on both sides.
and 1) the atoms numbers are equal on both sides but not correct as it not a
correct number as it has 1/2 O2.
and 2) CO2(g) ⇆ CO(g) + O2
the number of O atoms is not equal on both sides of the equation.
we have 2 O atoms on the left side and 3 O atoms on the right side.
so, this not a balanced equation.
4) also not correct 2CO(g) + O2 ⇆ 2CO2
as it is not a decomposition reaction and the 2CO & O2 are as reactants not products.
so the correct answer is 3) 2CO2(g) ⇆ 2CO(g) + O2(g)
Answer:
For n=3 and l=1=p
It is 3p-orbital.
Magnetic quantum number m
l
have values from -l to +l and total of 2l+1 values.
Forl=1, m
l
values are:
m
l
=−1,0,1 for l=1; total m
l
values =3= Number of orbitals
Each orbital can occupy maximum of two electron
Number of electrons =2×3=6
Thus 6 electrons will show same quantum number values of n=3 and l=1.
Number of elements with last electron in 3p orbitals = 6
The answer is elements gain electrons. Oxidation reduction is elements lose electrons. And oxygen is added/lost can be a type of oxidation/reduction reaction.
The percent yield of the reaction : 89.14%
<h3>Further explanation</h3>
Reaction of Ammonia and Oxygen in a lab :
<em>4 NH₃ (g) + 5 O₂ (g) ⇒ 4 NO(g)+ 6 H₂O(g)</em>
mass NH₃ = 80 g
mol NH₃ (MW=17 g/mol):

mass O₂ = 120 g
mol O₂(MW=32 g/mol) :

Mol ratio of reactants(to find limiting reatants) :

mol of H₂O based on O₂ as limiting reactants :
mol H₂O :

mass H₂O :
4.5 x 18 g/mol = 81 g
The percent yield :
