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Artist 52 [7]
2 years ago
9

1. For each of the molecules below, determine the electron geometry, molecule geometry, and bond

Chemistry
1 answer:
Alexxx [7]2 years ago
8 0

Answer:

CCl4- tetrahedral bond angle 109°

PF3 - trigonal pyramidal bond angles less than 109°

OF2- Bent with bond angle much less than 109°

I3 - linear with bond angles = 180°

A molecule with two double bonds and no lone pairs - linear molecule with bond angle =180°

Explanation:

Valence shell electron-pair repulsion theory (VSEPR theory) helps us to predict the molecular shape, including bond angles around a central atom, of a molecule by examination of the number of bonds and lone electron pairs in its Lewis structure. The VSEPR model assumes that electron pairs in the valence shell of a central atom will adopt an arrangement which tends to minimize repulsions between these electron pairs by maximizing the distance between them. The electrons in the valence shell of a central atom are either bonding pairs of electrons, located primarily between bonded atoms, or lone pairs. The electrostatic repulsion of these electrons is reduced when the various regions of high electron density assume positions as far apart from each other as possible.

Lone pairs and multiple bonds are known to cause more repulsion than single bonds and bond pairs. Hence the presence of lone pairs or multiple bonds tend to distort the molecular geometry geometry away from that predicted on the basis of VSEPR theory. For instance CCl4 is tetrahedral with no lone pair and four regions of electron density around the central atom. This is the expected geometry. However OF2 also has four regions of electron density but has a bent structure. The molecule has four regions of electron density but two of them are lone pairs causing more repulsion. Hence the observed bond angle is less than 109°.

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Answer:

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Explanation:

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3 years ago
Can somebody help me please? it’s urgent
Anettt [7]

Answers:

See attached table

Explanation:

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Answer:

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Calculate the pH of a solution in which one normal adult dose of aspirin (640 mg ) is dissolved in 10 ounces of water. Express y
d1i1m1o1n [39]

The pH of the solution in which one normal adult dose aspirin is dissolved is :  2.7

Given data :

mass of aspirin = 640 mg = 0.640 g

volume of water = 10 ounces = 0.295735 L

molar mass of aspirin = 180.16 g/mol

moles of aspirin = mass / molar mass = 0.00355 mol

<h3>Determine the pH of the solution </h3>

First step : <u>calculate the concentration of aspirin</u>

= moles of Aspirin / volume of water

= 0.00355 / 0.295735

= 0.012 M

Given that pKa of Aspirin = 3.5

pKa = -logKa

therefore ; Ka = 10^{-3.5} = 3.162 * 10^{-4}

From the Ice table

3.162 * 10^{-4} = \frac{x + H^+}{[aspirin]}  = \frac{x^{2} }{0.012-x}

given that the value of Ka is small we will ignore -x

x² = 3.162 * 10^{-4} * 0.012

x = 1.948 * 10^{-3}  

Therefore

[ H⁺ ] = 1.948 * 10^{-3}

given that

pH = - Log [ H⁺ ]

     = - ( -3 + log 1.948 )

     = 2.71 ≈ 2.7

Hence we can conclude that The pH of the solution in which one normal adult dose aspirin is dissolved is :  2.7

Learn more about Aspirin : brainly.com/question/2070753

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Answer:

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