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Bogdan
8 days ago
11

How to calculate electronegativity with 3 elements?

Chemistry
1 answer:
VMariaS [1K]8 days ago
6 0
For instance, what is the difference in electronegativity for Acetone(CH2O)? Are there two distinct values, namely 0.4 for C versus H and 1.0 for C versus O? How do you decide which one to adopt?

6 Comments

AlwaysReady1

•

Apr 3, 2016, 10:14 PM

I may not fully grasp the question, but if you’re seeking to determine a compound's electronegativity to assess its electron-attracting capability, there are various other influencing factors.

It varies depending on the compound. For example, CH2O, known as formaldehyde, has oxygen with two pairs of electrons that can be donated. Neither hydrogen nor carbon can bond further as they are already fulfilling their valence shell requirements.

Robo94

•



You're attempting to apply a concept from a binary system to a more complex one. I assume you're aiming to figure out a molecule's dipole moment. In the case of a diatomic molecule (where A is bonded to B), the potential difference can simply be determined as A minus B. For larger molecules, the calculations become much more involved.

If this inquiry is related to homework assistance, it’s a distinctly different method from what you might be accustomed to. I recommend starting with water and then expanding out from there.

Check this out: https://www.khanacademy.org/science/organic-chemistry/gen-chem-review/electronegativity-polarity/v/dipole-moment

Philosoaxolotl

•



Electronegativity pertains to single elements (or rather individual atoms) and lacks straightforward applicability to broader molecules.

What precisely are you aiming to do with this data? If you're delving into how electrons transition between molecules, the situation is more intricate—within a molecule, the more electronegative elements pull electrons from other atoms (which frequently happens in organic compounds, such as when oxygen bonds with carbon and pulls in some of its electrons). Nevertheless, this effect diminishes in lengthened molecules. The system is more complicated as molecules do not possess a single, constant electronegativity (which is more accurate for atoms); instead, they exhibit varied localized charge regions that will respond differently.

From what I gather, your question pertains to the electronegativity difference among the atoms within an acetone molecule. This indeed relies on which two atoms you are examining and won't remain constant throughout; however, the difference won't simply match the values listed in an electronegativity table due to the factors discussed earlier.

This explanation might seem a bit hazy, and I’m just an undergraduate, so please take my interpretation lightly, but I am open to clarifying further if needed.

cheeseborito

•



That statement is inaccurate.

Electronegativity represents the attraction an atom holds for the electrons in a covalent bond with another atom. Essentially, an element does not have a singular electronegativity; it fluctuates based on its bonding partners. We cannot discuss the electronegativity of an atom in isolation.

While average values are useful for practical discussions (though they may not capture the nuance), the effective electronegativity of an oxygen atom bonded to carbon will remain fairly consistent.

As far as my understanding goes, even though my definition of electronegativity may lack precision, the influence an oxygen atom has on the electrons of a carbon atom is affected by what the carbon is bonded to. For instance, the local charge around the oxygen in acetic acid will be more pronounced than that in decanoic acid.

I may have phrased the electronegativity issue poorly—what I meant was the interaction between pairs of atoms as related to one another. An oxygen will exert a consistent pull regarding a carbon atom, but the changes in local charge can differ due to the influence of surrounding atoms, making the topics we typically utilize electronegativity to clarify substantially more intricate.
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Answer:

The enthalpy of the second intermediate equation is altered by halving its value and changing the sign.

Explanation:

Let's examine both the first and second intermediate reactions alongside the overall equation concerning the examined process;

First reaction;

Ca (s) + CO₂ (g) + ½O₂ (g) → CaCO₃ (s) ΔH₁ = -812.8 kJ

Second reaction;

2Ca (s) + O₂ (g) → 2CaO (s) ΔH₂ = -1269 kJ

Thus, the overall reaction becomes;

CaO (s) + CO₂ (g) → CaCO₃ (s) ΔH =?

According to Hess's law, which states that the total heat change in a reaction is equal to the sum of the heat changes for each step, we cannot simply sum the enthalpies for this overall reaction. Instead, we obtain the overall enthalpy by halving the second intermediate reaction's enthalpy and changing its sign before adding, as illustrated below;

Enthalpy of Intermediate reaction 1 + ½(-Enthalpy of Intermediate reaction 2) = Enthalpy of Overall reaction

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A water tank can hold 1 m3 of water. When it’s empty, how much liters is needed to refill it?
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A chamber with a fixed volume is shown above. The temperature of the gas inside the chamber before heating is 25.2 C and it’s pr
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Answer:

Explanation:

Given data:

Initial temperature T₁ = 25.2°C = 298.2K

Initial pressure P₁ = 0.6atm

Final temperature = 72.4°C = 345.4K

What we need to find:

Final pressure = ?

To determine this, we apply a modified version of the combined gas law with constant volume. This simplifies our calculations to:

\frac{P_{1} }{T_{1} }   = \frac{P_{2} }{T_{2} }

Here, P and T signify pressure and temperatures, 1 refers to initial and 2 to final temperatures.

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Answer:

Chemists observe phenomena on a macroscopic level which informs their understanding of microscopic aspects.

Explanation:

Many critical chemical insights arise from macroscopic observations because most scientific instruments currently cannot directly evidence microscopic events. Data gathered from these larger-scale observations can yield valuable insights into the nature of specific microscopic interactions.

This is particularly true in atomic structure studies. The majority of evidence that contributed to our understanding of atomic structure was obtained from macroscopic observations and subsequently provided crucial information regarding the atom's microscopic configuration.

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1) reacting hydrochloric acid with nickel:

Balanced molecular equation: Ni(s) + 2HCl(aq) → NiCl₂(aq) + H₂(g).

Ionic equation: Ni(s) + 2H⁺(aq) + 2Cl⁻(aq) → Ni²⁺(aq) + 2Cl⁻(aq) + H₂(g).

Net ionic equation: Ni(s) + 2H⁺(aq) → Ni²⁺(aq) + H₂(g).

In this reaction, nickel undergoes oxidation, changing from an oxidation state of 0 to +2, while hydrogen is reduced from +1 to 0 (H₂).

2) reacting sulfuric acid with iron:

Balanced molecular equation: Fe(s) + H₂SO₄(aq) → FeSO₄(aq) + H₂(g).

Ionic equation: Fe(s) + 2H⁺(aq) + SO₄²⁻(aq) → Fe²⁺(aq) + SO₄²⁻(aq) + H₂(g).

Net ionic equation: Fe(s) + 2H⁺(aq) → Fe²⁺(aq) + H₂(g).

In this scenario, iron is oxidized from an oxidation state of 0 to +2, while hydrogen experiences reduction from +1 to 0 (H₂).

3) hydrobromic acid reacting with magnesium:

Balanced molecular equation: Mg(s) + 2HBr(aq) → MgBr₂(aq) + H₂(g).

Ionic equation: Mg(s) + 2H⁺(aq) + 2Br⁻(aq) → Mg²⁺(aq) + 2Br⁻(aq) + H₂(g).

Net ionic equation: Mg(s) + 2H⁺(aq) → Mg²⁺(aq) + H₂(g).

This reaction sees magnesium oxidized from 0 to +2, and hydrogen reduced from +1 to 0 (H₂).

4) acetic acid reacting with zinc:

Balanced molecular equation: Zn(s) + 2CH₃COOH(aq) → (CH₃COO)₂Zn(aq) + H₂(g).

Ionic equation: Zn(s) + 2H⁺(aq) + 2CH₃COO⁻(aq) → Zn²⁺(aq) + 2CH₃COO⁻(aq) + H₂(g).

Net ionic equation: Zn(s) + 2H⁺(aq) → Zn²⁺(aq) + H₂(g).

Here, zinc gets oxidized from 0 to +2 (Zn²⁺), while hydrogen is reduced from +1 to 0 (H₂).

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