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3 years ago

5

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

2 answers:

Kipish [7]3 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]3 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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<u>Answer:</u> The number of $OH^-$ ions dissociated are $8.57\times 10^{11}$

<u>Explanation:</u>

We are given:

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Putting values in above equation, we get:

$1.17\times 10^{-12}M=\frac{\text{Moles of }OH^-}{1.243L}\\\\\text{Moles of }OH^-=(1.17\times 10^{-12}mol/L\times 1.243L)=1.424\times 10^{-12}mol$

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