Answer:
Cloud condensation nuclei or CCNs are small particles typically 0.2 µm, or 1/100 the size of a cloud droplet on which water vapor condenses. Water requires a non-gaseous surface to make the transition from a vapour to a liquid; this process is called condensation.
Answer:
The correct option is: c. phospholipid
Explanation:
Phospholipids, also known as glycerophospholipids, are the derivatives of fatty acids which is a major structural component of the cell membranes.
Phospholipid is the class of lipids that is composed of a <u>glycerol molecule that forms ester bonds with the two long-chain fatty acids and one phosphate group.</u>
<u>Therefore, Molecule A is a </u><u>phospholipid.</u>
Given :
Mass of oxygen containing carbon monoxide (CO) is 2.666 gram .
To Find :
How many grams of carbon (C) would be present in carbon monoxide (CO) that contains 2.666 grams of oxygen (O) .
Solution :
By law of constant composition , a given chemical compound always contains its component elements in fixed ratio (by mass) and does not depend on its source and method of preparation.
So , volume of solution does not matter .
Moles of oxygen , 
.
Now , molecule of CO contains 1 mole of C .
So , moles of C is also 0.167 mole .
Mass of carbon , 
.
Therefore , mass of carbon is 2 grams .
Hence , this is the required solution .
Displacement is the distance and direction of an object's change in position from the starting.
Hence option B is correct.
Hope this helps!
<span>294400 cal
The heating of the water will have 3 phases
1. Melting of the ice, the temperature will remain constant at 0 degrees C
2. Heating of water to boiling, the temperature will rise
3. Boiling of water, temperature will remain constant at 100 degrees C
So, let's see how many cal are needed for each phase.
We start with 320 g of ice and 100 g of liquid, both at 0 degrees C. We can ignore the liquid and focus on the ice only. To convert from the solid to the liquid, we need to add the heat of fusion for each gram. So multiply the amount of ice we have by the heat of fusion.
80 cal/g * 320 g = 25600 cal
Now we have 320 g of ice that's been melted into water and the 100 g of water we started with, resulting in 320 + 100 = 420 g of water at 0 degrees C. We need to heat that water to 100 degrees C
420 * 100 = 42000 cal
Finally, we have 420 g of water at the boiling point. We now need to pump in an additional 540 cal/g to boil it all away.
420 g * 540 cal/g = 226800 cal
So the total number of cal used is
25600 cal + 42000 cal + 226800 cal = 294400 cal</span>
