<h2>Answer:</h2>
Arrangement of inter molecular forces from strongest to weakest.
- Hydrogen bonding
- Dipole-dipole interactions
- London dispersion forces.
<h3>Explanation:</h3>
Intermolecular forces are defined as the attractive forces between two molecules due to some polar sides of molecules. They can be between nonpolar molecules.
Hydrogen bonding is a type of dipole dipole interaction between the positive charge hydrogen ion and the slightly negative pole of a molecule. For example H---O bonding between water molecules.
Dipole dipole interactions are also attractive interactions between the slightly positive head of one molecule and the negative pole of other molecules.
But they are weaker than hydrogen bonding.
London dispersion forces are temporary interactions caused due to electronic dispersion in atoms of two molecules placed together. They are usually in nonpolar molecules like F2, I2. they are weakest interactions.
We could reduce soil erosion and recycle phosphorus from farm and human waste so that we could help make food production sustainable and prevent algae blooms. We can also do land reclamation as well to help solve this problem. With the land, we would have to design a system to where the land could be functional again in order to plant crops, trees, also to help the wildlife that was once a part of the island. Therefore if the design is done before the mining then afterward we can do the reclamation of the land which would help the people to be able to function after the mining. It would also help the future generations that come along after the previous generations. Everyone must work together in the process in order for everyone to survive. If all this is done then the people of the island would not have to import their food. The reclamation process is the most important thing that has to be designed first whether it is land, soil, water, lakes, and clay then after plant trees, vegetation, and other forms of plants to help replenish the land after the mining is done.
I hope I helped :3
Answer:
339kJ
Explanation:
Given parameters:
Mass of steam = 150g = 0.15kg
Initial temperature of steam = 100°C
Final temperature of water = 100°C
Unknown:
Quantity of heat that must be removed to condense the steam = ?
Solution:
The heat involved here is a latent heat because there is no change temperature. The process is just a phase change.
H = mL
m is the mass
L is the latent heat of vaporization = 2,260 kJ/kg
Insert the parameters and solve;
H = 0.15kg x 2,260 kJ/kg
H = 339kJ


- <u>We </u><u>have </u><u>250g </u><u>of </u><u>liquid </u><u>water </u><u>and </u><u>it </u><u>needs </u><u>to </u><u>be </u><u>cool </u><u>at </u><u>temperature </u><u>from </u><u>1</u><u>0</u><u>0</u><u>°</u><u> </u><u>C </u><u>to </u><u>0</u><u>°</u><u> </u><u>C</u>
- <u>Specific </u><u>heat </u><u>of </u><u>water </u><u>is </u><u>4</u><u>.</u><u>1</u><u>8</u><u>0</u><u>J</u><u>/</u><u>g</u><u>°</u><u>C</u>

- <u>We </u><u>have </u><u>to </u><u>find </u><u>the</u><u> </u><u>total</u><u> </u><u>number </u><u>of </u><u>joules </u><u>released</u><u>. </u>

<u>We </u><u>know </u><u>that</u><u>, </u>
Amount of heat energy = mass * specific heat * change in temperature
<u>That </u><u>is, </u>

<u>Subsitute </u><u>the </u><u>required </u><u>values </u><u>in </u><u>the </u><u>above </u><u>formula </u><u>:</u><u>-</u>




Hence, 104,500 J of heat is released to cool 250 grams of liquid water from 100° C to 0° C.

<u>We </u><u>have </u><u>to </u><u>tell </u><u>whether </u><u>the </u><u>above </u><u>process </u><u>is </u><u>endothermic </u><u>or </u><u>exothermic </u><u>:</u><u>-</u>
Here, In the above process ΔT is negative and as a result of it Q is also negative that means above process is Exothermic
- <u>Exothermic </u><u>process </u><u>:</u><u>-</u><u> </u><u>It </u><u>is </u><u>the </u><u>process </u><u>in </u><u>which </u><u>heat </u><u>is </u><u>evolved </u><u>. </u>
- <u>Endothermic </u><u>process </u><u>:</u><u>-</u><u> </u><u>It </u><u>is </u><u>the </u><u>process </u><u>in </u><u>which </u><u>heat </u><u>is </u><u>absorbed </u><u>.</u>