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Determine Δngas for each of the following reaction:(b) 2PbO(s) + O₂(g) ⇄ 2PbO₂(s)

1 answer:

Elis [28]2 years ago

5 0

2PbO(s) + O₂(g) ⇄ 2PbO₂(s)

Then Δngas = -1

<h3>What is Δngas?</h3>

The number of moles of gas that move from the reactant side to the product side is denoted by the symbol ∆n or delta n in this equation.

Once more, n represents the growth in the number of gaseous molecules the equilibrium equation can represent. When there are exactly the same number of gaseous molecules in the system, n = 0, Kp = Kc, and both equilibrium constants are dimensionless.

<h3>Definition of equilibrium</h3>

When a chemical reaction does not completely transform all reactants into products, equilibrium occurs. Many chemical processes eventually reach a state of balance or dynamic equilibrium where both reactants and products are present.

Learn more about Equilibrium

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Draw the best Lewis structure for NH3 by filling in the bonds, lone pairs, and formal charges. (Assign bonds, lone pairs, radica

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Answer : The Lewis-dot structure of $NH_3$ is shown below.

Explanation :

Lewis-dot structure : It shows the bonding between the atoms of a molecule and it also shows the unpaired electrons present in the molecule.

In the Lewis-dot structure the valance electrons are shown by 'dot'.

The given molecule is, $NH_3$

As we know that hydrogen has '1' valence electron and nitrogen has '5' valence electrons.

Therefore, the total number of valence electrons in $NH_3$ = 5 + 3(1) = 8

According to Lewis-dot structure, there are 6 number of bonding electrons and 2 number of non-bonding electrons.

Now we have to determine the formal charge for each atom.

Formula for formal charge :

$\text{Formal charge}=\text{Valence electrons}-\text{Non-bonding electrons}-\frac{\text{Bonding electrons}}{2}$

$\text{Formal charge on N}=5-2-\frac{6}{2}=0$

$\text{Formal charge on }H_1=1-0-\frac{2}{2}=0$

$\text{Formal charge on }H_2=1-0-\frac{2}{2}=0$

$\text{Formal charge on }H_3=1-0-\frac{2}{2}=0$

Hence, the Lewis-dot structure of $NH_3$ is shown below.

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