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Korolek [52]
3 years ago
9

The intermolecular forces present in CH 3NH 2 include which of the following? I. dipole-dipole II. ion-dipole III. dispersion IV

. hydrogen bonding
Chemistry
1 answer:
astra-53 [7]3 years ago
7 0

Answer:

I. dipole-dipole

III. dispersion

IV. hydrogen bonding

Explanation:

Intermolecular forces are weak attraction force joining nonpolar and polar molecules together.

London Dispersion Forces are weak attraction force joining non-polar and polar molecules together. e.g O₂, H₂,N₂,Cl₂ and noble gases. The attractions here can be attributed to the fact that a non -polar molecule sometimes becomes polar because the constant motion of its electrons may lead to an uneven charge distribution at an instant.

Dispersion forces are the weakest of all electrical forces that act between atoms and molecules. The force is responsible for liquefaction or solidification of non-polar substances such as noble gas an halogen at low temperatures.

Dipole-Dipole Attractions are forces of attraction existing between polar molecules ( unsymmetrical molecules) i.e molecules that have permanent dipoles such as HCl, CH3NH2 . Such molecules line up such that the positive pole of one molecule attracts the negative pole of another.

Dipole - Dipole attractions are more stronger than the London dispersion forces but weaker than the attraction between full charges carried by ions in ionic crystal lattice.

Hydrogen Bonding is a dipole-dipole intermolecular attraction which occurs when hydrogen is covalently bonded to highly electronegative elements such as nitrogen, oxygen or fluorine. The highly electronegative elements have very strong affinity for electrons. Hence, they attracts the shared pair of electrons in the covalent bonds towards themselves, leaving a partial positive charge on the hydrogen atom and a partial negative charge on the electronegative atom ( nitrogen in the case of CH3NH2 ) . This attractive force is know as hydrogen bonding.

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In class, it was mentioned that at 25 °C, the pH of pure water is 7.00 becuase Kw = 1.0 × 10−14. At other temperatures, the pH v
ivanzaharov [21]

Answer:

Kw = 2.88 × 10⁻¹⁵

Explanation:

Let's consider the dissociation of water.

H₂O(l) ⇄ H⁺(aq) + OH⁻(aq)

The equilibrium constant Kw is:

Kw = [H⁺].[OH⁻]

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Answer:

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Explanation:

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