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Likurg_2 [28]
3 years ago
11

Describing the Effects of Surfactar

Chemistry
2 answers:
krok68 [10]3 years ago
5 0

Answer:

The forces between the water molecules are called as the hydrogen bonding, while the forces of attraction between the ethanol molecules are dipole-dipole forces. Along, with the alignment or arrangement of the surface molecules makes it very clear to know more about the characteristics of the water and ethanol molecules, and the way they behave when any number of insects walk across these surfaces. So, the water characteristics makes it more easy to understand that the insects will find it more easy to walk on the surface of the water medium just because of the capacity of the molecules to with hold the amount of insects weight in a given parameter. And thus water surface is suitable for an insect to walk on rather then walking on the surface of ethanol.So, its all about the surface tension of the water medium that provides us with an option for the insect to travel on it surface.

<u>Water and ethanol two different mediums:</u>

The two different mediums represents a more different form or nature to a insect when it comes to letting it to walk over its surface. As there are different forms of bonding inside the molecules of the two different mediums, the ethanol has the dipole-dipole forces between its molecules making it much weaker to support for lifting the total weight of the insect present on its surface.

Svetlanka [38]3 years ago
3 0

Explanation:

An insect would have an easier time walking on the surface of water than on the surface of ethanol. Water's stronger intermolecular forces lead to higher surface tension. Higher surface tension allows water to support the insect. I hope this helps.

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A 10.0 g sample of propane, C3H8, was combusted in a constant-volume bomb calorimeter. The total heat capacity of the bomb calor
Liono4ka [1.6K]

Answer:

83ºC

Explanation:

A bomb calorimeter is an instrument used to measure the heat that release or absorb a particular reaction.

The reaction of combustion of propane is:

C₃H₈ +  5O₂ → 3 CO₂ + 4 H₂O ΔH = -2222kJ/mol

<em>1 mole of propane release 2222kJ</em>

10.0g of propane (Molar mass: 44.1g/mol).

10.0g ₓ (1mol/ 44.1g) = <em>0.227 moles of C₃H₈</em>

If 1 mole of propane release 2222kJ, 0.227moles will release (Release because molar heat is < 0):

0.227 moles of C₃H₈ ₓ (2222kJ / mol) = 504kJ.

Our calorimeter has a constant of 8.0kJ/ºC, that means if there are released 8.0kJ, the bomb calorimeter will increase its temperature in 1ºC. As there are released 504kJ:

504kJ ₓ (1ºC / 8.0kJ) = 63ºC will increase the temperature in the bomb calorimeter.

As initial temperature was 20ºC, final temperature will be:

<h2>83ºC</h2>
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4 years ago
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