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3 years ago

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

Chemistry

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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The pK1, pK2, and pKR of the amino acid lysine are 2.2, 9.1, and 10.5, respectively. The pK1, pK2, and pKR of the amino acid arg

almond37 [142]

Answer:

pH 9,8 is likely to work best for this separation

Explanation:

Ion exchange chromatography is a chemical process where molecules are separated by affinity to an ion exchange resin. To separate different aminoacids you must use the isoelectric point (That is the pH where the aminoacid will be in its neutral form).

For lysine, PI is:

$pH = \frac{1}{2} (9,1+10,5) =$ 9,8

For arginine:

$pH = \frac{1}{2} (9,0+12,5) =$ 10,75

At pH = 9,8 lysine will be in its neutral form and will not be retain in the column but arginine will be in +1 charge being retained by the ion exchange resin.

Thus, <em>pH 9,8 is likely to work best for this separation</em>

I hope it helps!

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3 years ago

How many moles of carbon, hydrogen, and oxygen are present in a 100-g sample of ascorbic acid?

Y_Kistochka [10]

There are:

3.41 moles of C

4.54 moles of H

3.40 moles of O.

Why?

To solve the problem, the first thing that we need to do is to write the chemical formula of the ascorbic acid.

$C_{6}H_{8}O_{6}$

Now, we know that there are 100 grams of the compound, so, the masses of each element will represent the percent in the compound.

We have that:

$C_{6}=12.0107g*6=72.08g\\\\H_{8}=1.008g*8=8.064g\\\\O_{6}=15.999g*6=95.994g\\\\C_{6}H_{8}O_{6}=72.08g+8.064g+95.994g=176.138g$

To know the percent of each element, we need to to the following:

$C=\frac{72.08g}{176.138g}*100=0.409*100=40.92(percent)\\\\H=\frac{8.064g}{176.138g}*100=4.58(percent)\\\\O=\frac{95.994}{176.138g}*100=54.49(percent)$

So, we know that for the 100 grams of the compound, there are:

40.92 grams of C

4.58 grams of H

54.49 grams of O

We know the molecular masses of each element:

$C=12.0107\frac{g}{mol}\\\\H=1.008\frac{g}{mol}\\\\O=15.999\frac{g}{mol}{mol}$

Now, to calculate the number of moles of each element, we need to divide the mass of each element by the molecular mass of each element:

$C=\frac{40.92g}{12.010\frac{g}{mol}}=3.41mol\\\\H=\frac{4.58g}{1.008\frac{g}{mol}}=4.54mol\\\\O=\frac{54.49g}{15.999\frac{g}{mol}}=3.40mol$

Hence, we have that there are 3.41 moles of C, 4.54 moles of H, and 3.40 moles of O.

Have a nice day!

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3 years ago

Calculate the volume of a 0.5M solution containing 20g of NaOH

Tju [1.3M]

Answer:

Explanation:

First, let us calculate the number of mole present in 20g of NaOH. This is illustrated below:

Mass = 20g

Molar Mass of NaOH = 23 + 16 + 1 = 40g/mol

Number of mole =?

Number of mole = Mass /Molar Mass

Number of mole of NaOH = 20/40 = 0.5mol

From the question given, we obtained the following data:

Molarity = 0.5M

Mole = 0.5mole

Volume =?

Molarity = mole /Volume

Volume = mole /Molarity

Volume = 0.5/0.5

Volume = 1L

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2H(+) + SO4(2-) + Ca(2+) + 2I(-) -> CaSO4(s) + 2H(+) + 2I(-)
The signs in brackets are the subscripts for the charge of the ion. This is the complete ionic equation. The net ionic equation is:
Ca(2+) + SO4(2-) -> CaSO4

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