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2 years ago

5

Identify which of the following molecules can exhibit hydrogen bonding as a pure liquid.

2 answers:

Kipish [7]2 years ago

7 0

N2H4

H3CO2H

C2H2O4

This should be the right answer, let me know if its not so i can know for future.

stira [4]2 years ago

6 0

The following molecules can exhibit hydrogen bonding as a pure liquid is NH3 (amonia), HC3 -- O -- OH (methanol), CH3CO2H (acetic acid)

<h3>Further explanation </h3>

Hydrogen bonding is directly connected to Nitrogen, Oxygen, and Fluoride and it (N, O, F) represents the hydrogen bonding. Whereas molecule is the smallest particle in a chemical element or compound that has the chemical properties of that element or compound

Identify which of the following molecules can exhibit hydrogen bonding as a pure liquid.

NH3 (amonia) : Yes, because ammonia has 3 bonds, each between Nitrogen and a Hydrogen.
HC3 -- O -- OH (methanol) : Yes, because oxygen and hydrogen in the hydroxyl group that branching off of the carbon share a hydrogen bond.
CH2F2 (difluoromethane) : No, because the carbon bonds with two hydrogens and two fluorines, but hydrogen and fluorine never bond with each other.
CF4 (tetrafluoromethane) : No, because the carbon bonds with four fluorine atoms, hence no hydrogen involved.
HI (hydroiodic acid) : No, because F, O, N involved.
CH3CO2H (acetic acid) : Yes, because hydrogen bonds with oxygen directly in the hydroxyl substituent group that forms off of the second carbon
C7H6O (benzaldehyde): No, because this is a benzene ring with oxygen and hydrogen branching off of one of the carbons.

<h3>Learn more</h3>

Learn more about hydrogen bonding brainly.com/question/1814257
Learn more about molecules brainly.com/question/1462404
Learn more about benzaldehyde brainly.com/question/12631576

<h3>Answer details</h3>

Grade: 9

Subject: chemistry

Chapter: hydrogen bonding

Keywords: hydrogen bonding, molecules, amonia, tetrafluoromethane, benzaldehyde

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(a) Moles of ammonium chloride = 0.243 moles

(b) $Molarity_{ammonium\ chloride}=0.824\ M$

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Explanation:

(a) Mass of ammonium chloride = 13.0 g

Molar mass of ammonium chloride = 53.491 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{13.0\ g}{53.491\ g/mol}$

<u>Moles of ammonium chloride = 0.243 moles</u>

(b) Moles of ammonium chloride = 0.243 moles

Volume = 295 mL = 0.295 L ( 1 mL = 0.001 L)

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

$Molarity_{ammonium\ chloride}=\frac{0.243}{0.295}$

$Molarity_{ammonium\ chloride}=0.824\ M$

(c) Moles of ammonium chloride = 0.0500 moles

Volume = ?

Molarity = 0.824 M

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

$0.824\ M=\frac{0.0500}{Volume}$

<u>Volume = 0.05 / 0.824 L = 0.06068 L = 60.68 mL</u>

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