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vova2212 [387]
3 years ago
6

statement to analyze and classify as science protoscience pseudoscience or non-science the statement i classified is

Chemistry
1 answer:
saw5 [17]3 years ago
5 0

Answer:question?

Explanation:

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Identify a type of strong intermolecular force that exists between water molecules, but does not exist between carbon dioxide mo
timurjin [86]

Intermolecular forces of attraction hold the molecules together. These forces determine the physical properties of substances like melting and boiling points. There are five types of intermolecular forces: Hydrogen bonding, dipole-dipole interactions, ionic interactions, ion-dipole interactions and dispersion forces.

Hydrogen bonding is a stronger force of attraction between hydrogen atom and an electronegative atom (F, N, and O). So, water molecules exhibit hydrogen bonding.

In carbon dioxide molecules, although each C=O is polar the molecule as a whole will be non polar due to symmetry. Therefore, the only intermolecular forces in CO2 will be dispersion forces.

Hence, Hydrogen bonding exists between water molecules but not carbon dioxide molecules.

7 0
3 years ago
What is a hydrogen bond?
marin [14]

Explanation: Hydrogen bonds are the strongest one of the intermolecular forces. A hydrogen bond is a bond between hydrogen in one molecule or the other ones are fluorine and nitrogen. So it's between a hydrogen in one molecule and an electronegative atom in another molecule. So they always involve hydrogen.

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3 years ago
Use the Henderson-Hasselbalch equation, eq. (3), to calculate the pH expected for a buffer solution prepared from this acid and
notka56 [123]

Answer:

pH=4.56

Explanation:

Hello there!

In this case, given the Henderson-Hasselbach equation, it is possible for us to compute the pH by firstly computing the concentration of the acid and the conjugate base; for this purpose we assume that the volume of the total solution is 0.025 L and the molar mass of the sodium base is 234 - 1 + 23 = 256 g/mol as one H is replaced by the Na:

n_{acid}=\frac{0.2g}{234g/mol}=0.000855mol\\\\n_{base}= \frac{0.2g}{256g/mol}=0.000781mol

And the concentrations are:

[acid]=0.000855mol/0.025L=0.0342M

[base]=0.000781mol/0.025L=0.0312M

Then, considering that the Ka of this acid is 2.5x10⁻⁵, we obtain for the pH:

pH=-log(2.5x10^{-5})+log(\frac{0.0312M}{0.0342M} )\\\\pH=4.60-0.04\\\\pH=4.56

Best regards!

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Even though the neighborhood shown by the green star was not next to any of these fires, how was the neighborhood affected by th
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Im pretty sure the answers b

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Complete one rotation. 

Hope this helps.
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