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3 years ago

Choose the compound that exhibits hydrogen bonding as its strongest intermolecular force.

2 answers:

4 0

Hydrogen Bonding is considered as the strongest Intermolecular Interactions between two protic compounds.

Formation of Hydrogen Bonding:
Hydrogen Bonding takes place between the Hydrogen atom bonded to most electronegative element (i.e. F, O, N) of one molecule and the most electronegative element (i.e. F, O, N) of other molecule. This is because the Hydrogen atom attains a partial positive charge which creates attractive forces with the electronegative atom with partial negative charge.
In given options Hydrogen Bonding can only exist in Methanol (CH₃-O-H). As, the hydrogen atom is directly bonded with Oxygen atom, so it will attain partial positive charge and will interact with the Oxygen atom of second Methanol molecule.

CH₃-O-H - - - - -OHCH₃ ∴ - - - - = Hydrogen Bond

Result:
Methanol (CH₃-OH) <span>exhibits hydrogen bonding as its strongest intermolecular force.</span>

mezya [45]3 years ago

4 0

CH3OH

<h2>Further Explanation </h2>

Hydrogen bonds are intermolecular attractions that occur between hydrogen atoms bound to very electronegative atoms (N, O, or F) and free electron pairs from other highly electronegative atoms. These bonds appear as N-H, O-H, and F-H bonds are very polar, where the positive partial charge on H and the negative partial charge on the electronegative atom (N, O, or F). Although stronger than most intermolecular forces, hydrogen bonds are much weaker than covalent bonds and ionic bonds. In macromolecules such as proteins and nucleic acids, this bond can occur between two parts of the same molecule and serve as an important determinant of the shape of the whole molecule.

Hydrogen bonds occur when molecules have N, O, or F atoms that have free pair electrons. The strength of hydrogen bonds is influenced by differences in electronegativity between atoms in a molecule. The greater the difference, the greater the hydrogen bonds are formed. Hydrogen bonds affect the boiling point of a compound. The greater the hydrogen bond, the higher the boiling point. However, specifically for water (H2O), two hydrogen bonds occur in each molecule. As a result, the number of hydrogen bonds is greater than hydrofluoric acid (HF) which should have the largest hydrogen bond (due to the highest electronegativity difference) so that the boiling point of water is higher than hydrofluoric acid.

Hydrogen bonds are also very influential in spectroscopy. The most obvious example is in infrared spectroscopy, where the presence of hydrogen bonds will expand stretching.

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hydrogen bonding brainly.com/question/9809034

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Grade: College

Subject: Chemistry

keywords: hydrogen, compound.

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2 years ago

Predict whether the following reactions will be exothermic or endothermic.

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Answer:

A. N₂(g) + 3H₂(g) -----> 2NH₃ exothermic

B. S(g) + O₂(g) --------> SO₂(g) exothermic

C. 2H₂O(g) --------> 2H₂(g) + O₂(g) endothermic

D. 2F(g) ---------> F₂(g) exothermic

Explanation:

The question says predict not calculate. So you have to use your chemistry knowledge, experience and intuition.

A. N₂(g) + 3H₂(g) -----> 2NH₃ is exothermic because the Haber process gives out energy

B. S(g) + O₂(g) --------> SO₂(g) is exothermic because it is a combustion. The majority, if not all, combustion give out energy.

C. 2H₂O(g) --------> 2H₂(g) + O₂(g) is endothermic because it is the reverse reaction of the combustion of hydrogen. If the reverse reaction is exothermic then the forward reaction is endothermic

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A frictionless piston cylinder device is subjected to 1.013 bar external pressure. The piston mass is 200 kg, it has an area of

Bad White [126]

Answer:

a) $T_{2} = 360.955\,K$ , $P_{2} = 138569.171\,Pa\,(1.386\,bar)$ , b) $T_{2} = 347.348\,K$ , $V_{2} = 0.14\,m^{3}$

Explanation:

a) The ideal gas is experimenting an isocoric process and the following relationship is used:

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

Final temperature is cleared from this expression:

$Q = n\cdot \bar c_{v}\cdot (T_{2}-T_{1})$

$T_{2} = T_{1} + \frac{Q}{n\cdot \bar c_{v}}$

The number of moles of the ideal gas is:

$n = \frac{P_{1}\cdot V_{1}}{R_{u}\cdot T_{1}}$

$n = \frac{\left(101,325\,Pa + \frac{(200\,kg)\cdot (9.807\,\frac{m}{s^{2}} )}{0.15\,m^{2}} \right)\cdot (0.12\,m^{3})}{(8.314\,\frac{Pa\cdot m^{3}}{mol\cdot K} )\cdot (298\,K)}$