Answer:
ΔS° = 180.5 J/mol.K
Explanation:
Let's consider the following reaction.
4 NH₃(g) + 5 O₂(g) → 4 NO(g) + 6 H₂O(g)
The standard molar entropy of the reaction (ΔS°) can be calculated using the following expression.
ΔS° = ∑np × S°p - ∑nr × S°r
where,
ni are the moles of reactants and products
S°i are the standard molar entropies of reactants and products
ΔS° = 4 mol × S°(NO(g)) + 6 × S°(H₂O(g)) - 4 mol × S°(NH₃(g)) - 5 mol × S°(O₂(g))
ΔS° = 4 mol × 210.8 J/K.mol + 6 × 188.8 j/K.mol - 4 mol × 192.5 J/K.mol - 5 mol × 205.1 J/K.mol
ΔS° = 180.5 J/K
This is the change in the entropy per mole of reaction.
A visual representation of covalent bonding which represents the valence shell electrons in the molecule is said to be a Lewis structure. The lines represents the shared electron pairs and dots represents the electrons that are not involved in the bonding i.e lone pairs.
Number of valence electrons in each atom:
For Carbon,
= 4
For Hydrogen,
= 1
For Nitrogen,
= 5
The Lewis structure of
is shown in the attached image.
The formula of calculating formula charge =
-(1)
where, F.C is formal charge, V.E is number of valence electrons, N.E is number of non-bonding electrons and B.E is number of bonding electrons.
Now, calculating the formal charge:
For
on left side:

For
:

For
on right side:

The formula charge of each atom other than hydrogen is shown in the attached image.
Answer:
It comes from the chemical reaction of The wax and the gas because the wax is made up of carbon-based chemicals so the gas and carbon make the candle light =)
Explanation:
No, Heat moves from concentrations of high to low.