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2 years ago

discuss how variations in electronegativity result in the unequal sharing of electrons in polar molecules.

1 answer:

6 0

Variations in electronegativity prompt in the unequal halves of electrons in polar molecules because when one atom is more electronegative than the other, it becomes more polar than the other.

It results in the more electronegative atom to have a slightly negative (-ve) charges, and the other atom to have partial or slightly positive(+ve) charges.

Polar molecules have unequal sharing of electrons because the atoms have unequal attraction for electrons so the sharing is unequal.

The larger the difference in electronegativity between the two atoms, the more the polar the bond.

Hydrogen bonds are involved in unequal sharing of electrons between two atoms.

To know more about variations in electronegativity in polar molecules here :

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3 years ago

At a certain temperature, 0.760 mol SO3 is placed in a 1.50 L container. 2SO3(g) = 2SO2(g) + O2(g)At equilibrium, 0.130 mol O2 i

olganol [36]

Answer:

$K_c=0.0867$

Explanation:

Moles of SO₃ = 0.760 mol

Volume = 1.50 L

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

$Molarity=\frac{0.760}{1.50\ L}$

[SO₃] = 0.5067 M

Considering the ICE table for the equilibrium as:

$\begin{matrix} & 2SO_3_{(g)} & \rightleftharpoons & 2SO_2_{(g)} & + & O_2_{(g)}\\At\ time, t = 0 & 0.5067 &&0&&0 \\ At\ time, t=t_{eq} & -2x &&2x&&x \\----------------&-----&-&-----&-&-----\\Concentration\ at\ equilibrium:- &0.5067-2x&&2x&&x\end{matrix}$

Given:

Equilibrium concentration of O₂ = 0.130 mol

Volume = 1.50 L

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

$Molarity=\frac{0.130}{1.50\ L}$

[O₂] = x = 0.0867 M

[SO₂] = 2x = 0.1733 M

[SO₃] = 0.5067-2x = 0.3334 M

The expression for the equilibrium constant is:

$K_c=\frac {[SO_2]^2[O_2]}{[SO_3]^2}$

$K_c=\frac{(0.3334)^2\times 0.0867}{(0.3334)^2}$

$K_c=0.0867$

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4 years ago

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Answer:

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