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stepan [7]
1 year ago
9

Which forces are intramolecular and which intermolecular?

Chemistry
1 answer:
sergiy2304 [10]1 year ago
7 0

There must be an intramolecular force. The oxygen atoms are produced as a result of the breakdown of oxygen molecules. Intramolecular force is necessary to stop the oxygen (O2) in the air from changing into the O atom.

Which force causes attraction between O2 molecules?

The result is the London dispersion force, a fleeting attractive attraction, which is created when the electrons in two neighboring atoms occupy positions that temporarily cause the atoms to form dipoles. This interaction is commonly described by the phrase "induced dipole-induced dipole attraction".

What is the difference between intramolecular forces and intermolecular forces which type is stronger?

In general, intramolecular forces are greater than intermolecular forces. Ion-dipole interaction exerts the strongest intermolecular force, followed by hydrogen bonds, dipole-dipole interaction, and London dispersion. Examples. Hydrogen bonding forces, London dispersion forces, and dipole-dipole forces are the three different kinds of intermolecular interactions. The three different kinds of intramolecular forces are metal bonds, ionic bonds, and covalent bonds.

Learn more about intramolecular forces: brainly.com/question/28170469

#SPJ4

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What is the amount of moles in 67 g of C​
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Explanation:

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What is the mass, in grams, of 50.0L of N2 at STP?
vodka [1.7K]

Hey there!:

Molar mass N₂ = 28.0134 g/mol

28.0134 g ------------------- 22.4 L (at STP )

mass  N₂ -------------------- 50.0 L

mass N₂ = 50.0 x 28.0134 / 22.4

mass N₂ = 1400.67 / 22.4

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Hope this helps!

6 0
2 years ago
Calculate the following quantity: molarity of a solution prepared by diluting 45.45 mL of 0.0404 M ammonium sulfate to 550.00 mL
dybincka [34]

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M_2=3.34x10^{-3}M

Explanation:

Hello!

In this case, since a dilution process implies that the moles of the solute remain the same before and after the addition of diluting water, we can write:

M_1V_1=M_2V_2

Thus, since we know the volume and concentration of the initial sample, we compute the resulting concentration as shown below:

M_2=\frac{M_2V_2}{V_1} =\frac{45.45mL*0.0404M}{550.00mL}\\\\M_2=3.34x10^{-3}M

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