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goldfiish [28.3K]
4 years ago
6

Identify the types of intermolecular forces found for each of the liquids and relate these to the shape of the drop as seen from

the side. Include the terms cohesive and adhesive forces properly in your explanations.
What forces (dipole-dipole, hydrogen bonds, or dispersion forces) are present in each sample?
a) C12H26 (oil)
b) H2O (water)
Chemistry
1 answer:
pav-90 [236]4 years ago
8 0

Explanation:

Oil molecules are non-polar, and they can't form hydrogen bonds. Dispersion forces are present in C12H26 (oil).

H20 (water) are polar, has hydrogen bonds, it also has dipole-induced dipole and London dispersion forces.

The difference between them is that adhesion refers to the clinging of unlike molecules and cohesion refers to the clinging of like molecules.

In C12H26 (oil)  the adhesive forces are stronger than the cohesive forces as a result, oil molecules tend to stick to the walls of the container.

In  H2O (water) the cohesive forces is greater than the adhesive forces, as a result water molecules tend to stick together.

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Grace [21]

Answer:

electrons

Explanation:

because it rotates the nucleus around it

3 0
4 years ago
What are the electron configurations of an electrically neutral atom of sodium, electrically neutral atom of neon, and electrica
MA_775_DIABLO [31]
Na - 1s^2, 2s^2, 2p^6, 3s^1

Ne - 1s^2, 2s^2, 2p^6

O - 1s^2, 2s^2, 2p^4
4 0
4 years ago
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A 19.0 L helium tank is pressurized to 26.0 atm. When connected to this tank, a balloon will inflate because the pressure inside
vesna_86 [32]

Answer:

The new volume of the balloon when the pressure equalised with the pressure of the atmosphere = 494 L.

The balloon expands by am additional 475 L.

Explanation:

Assuming Helium behaves like an ideal gas and temperature is constant.

According to Boyle's law for ideal gases, at constant temperature,

P₁V₁ = P₂V₂

P₁ = 26 atm

V₁ = 19.0 L

P₂ = 1 atm (the balloon is said to expand till the pressure matches the pressure of the atmpsphere; and the pressure of the atmosphere is 1 atm)

V₂ = ?

P₁V₁ = P₂V₂

(26 × 19) = 1 × V₂

V₂ = 494 L (it is assumed the balloon never bursts)

The new volume of the balloon when the pressure equalised with the pressure of the atmosphere = 494 L.

The balloon expands by am additional 475 L.

Hope this Helps!!!

6 0
4 years ago
A room is a cube that measures 4 meters in each dimension. If the temperature changes from 10 C to 30 C, and the pressure decrea
Katen [24]

Answer:

More air will enter the room since the volume of the room increases.

Explanation:

We'll begin by calculating the initial volume (V₁) of the room. This can be obtained as follow

Length (L) of room = 4 m

Initial volume (V₁) =..?

V₁ = L³

V₁ = 4³

V₁ = 64 m³

Next, we shall determine if there is an increase in the volume of the room or not. This can be obtained as follow:

Data obtained from the question include:

Initial volume (V₁) = 64 m³

Initial temperature (T₁) = 10 °C + 273 = 283 K

Final temperature (T₂) = 30 °C + 373 = 303 K

Initial pressure (P₁) = 760 mmHg

Final pressure (P₂) = 718 mmHg

Final volume (V₂) =?

P₁V₁/T₁ = P₂V₂/T₂

760 × 64 / 283 = 718 × V₂ / 303

48640 / 283 = 718 × V₂ / 303

Cross multiply

283 × 718 × V₂ = 48640 × 303

203194 × V₂ = 14737920

Divide both side by 203194

V₂ = 14737920 / 203194

V₂ = 72.53 m³

SUMMARY

Initial volume (V₁) of room = 64 m³

Final volume (V₂) of room = 72.53 m³

From the calculations made above, we can see that the volume of the room increases from 64 m³ to 72.53 m³. Thus more air will enter the room since the volume of the room increases.

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