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Illusion [34]
1 year ago
8

How the molecule’s properties affect the physical properties of a molecule

Chemistry
1 answer:
Lilit [14]1 year ago
8 0

The intermolecular forces, such as hydrogen bonds or van der Waals attractions, which draw one molecule to its neighbors, govern a substance's physical properties. Due to the relatively weak intermolecular forces of attraction, molecular substances typically take the form of gases, liquids, or low melting point solids.

<h3>How do the intermolecular forces affect physical properties?</h3>

The forces that bind two molecules together are known as intermolecular forces. Intermolecular forces have an impact on physical properties. Strong and weak forces both exist; the stronger the force, the more energy is needed to separate the molecules from one another. As intermolecular forces increase melting, boiling, and freezing points rise.

The following intermolecular forces are listed in order of strength:

  • Van der Waals dispersion forces
  • Van der Waals dipole-dipole interactions
  • Hydrogen bonding
  • Ionic bonds

It would take very little energy to separate two molecules if they are connected by van der Waals dispersion forces. On the other hand, it requires a lot more energy to separate two molecules that are joined together by ionic bonds.

To know more about molecules refer to: brainly.com/question/1819972

#SPJ1

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What happens to a glass of sugar solution when sugar is added to it?
galina1969 [7]

Answer:

The molarity of the solution increases.

Explanation:

Molarity is the measure of the concentration of the solute in the solution. In this case, the solvent is the sugar solution and the solute is the sugar.

If sugar is ADDED to the already sugary solution, then there would be more sugar. Therefore, the sugar (solute) would increase in number.

This means that the answer is the third choice: The molarity of the solution increases.

The answer would not be the first or second choice because there isn't anything in the question that implies water. It just says sugar solution.

The answer is not the last choice because the sugar concentration does not decrease after you have added more sugar to it. It increases.

7 0
3 years ago
Read 2 more answers
I NEED HELP PLEASE,THANKS!
Yakvenalex [24]

Answer:

Covalent bonding

Explanation:

In covalent bonding, the electrons are shared to fill the octet rule (8 electrons in valence shell). CCl4 tends to do covalent bonding because the the 4 valence electrons are Carbon are shared with the Chlorine atoms so that each chlorine atom has a full octet and chlorine shares its electrons to fill the octet of carbon.

Also, since carbon and chlorine are both non-metal, non-metal things exhibit covalent bonding thus this is covalent bonding as well. Ionic boding is for metal and non metal pair where electrons are transferrred, in our case, electrons are shared, they are not transferred.

6 0
3 years ago
Consider the formation of [Ni(en)3]2+ from [Ni(H2O)6]2+. The stepwise ΔG∘ values at 298 K are ΔG∘1 for first step=−42.9 kJ⋅mol−1
timurjin [86]

Answer:

kf = 1.16 x 10¹⁸

Explanation:

Step 1: [Ni(H₂O)₆]²⁺  + 1en → [Ni(H₂O)₄(en)]²⁺  ΔG°1 = -42.9 kJmol⁻¹

Step 2: [Ni(H₂O)₄(en)]²⁺  + 1en → [Ni(H₂O)₂(en)₂]²⁺  ΔG°2 = -35.8 kJmol⁻¹

Step 3: [Ni(H₂O)₂(en)₂]²⁺ + 1en →  [Ni(en)₃]²⁺  ΔG°3 = -24.3 kJmol⁻¹

________________________________________________________

Overall reaction: [Ni(H₂O)₆]²⁺  + 3en → [Ni(en)₃]²⁺  ΔG°r

ΔG°r = ΔG°1 + ΔG°2 + ΔG°3

ΔG°r = -42.9 - 35.8 - 24.3

ΔG°r = -103.0 kJmol⁻¹

ΔG°r = -RTlnKf

-103,000 Jmol⁻¹ =  - 8.31 J.K⁻¹mol⁻¹ x 298 K x lnKf

kf = e ^(-103,000/-8.31x298)

kf = e ^41.59

kf = 1.16 x 10¹⁸

7 0
3 years ago
An atom whose valence electrons conform to the octet rule is:
Readme [11.4K]

Answer: D. less likely to form any bond

Explanation:

3 0
3 years ago
If the freezing point of an aqueous 0.10 m glucose solution is −x°c, what is the approximate freezing point of a 0.10 m nacl sol
maks197457 [2]
Answer is: the approximate freezing point of a 0.10 m NaCl solution is -2x°C.
V<span>an't Hoff factor (i) for NaCl solution is approximately 2.
</span>Van't Hoff factor (i) for glucose solution is 1.<span>
Change in freezing point from pure solvent to solution: ΔT = i · Kf · m.
Kf - molal freezing-point depression constant for water is 1,86°C/m.
m -  molality, moles of solute per kilogram of solvent.
</span>Kf and molality for this two solutions are the same, but Van't Hoff factor for sodium chloride is twice bigger, so freezing point is twice bigger.
8 0
3 years ago
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