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Topic: Formation of ice + hexagonal unit cell  (Read 1719 times)

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Offline Rutherford

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Formation of ice + hexagonal unit cell
« on: May 01, 2013, 10:22:20 AM »
1. During the formation of ice out of liquid water:
A) energy is set free
B) the surrounding cools down
C) the pressure in the surrounding decreases
D) the mass of water increases
E) the volume of the total portion of ice and water
decreases.

I thought that A and C are correct, but C isn't. The molecules will harder leave the solid than the liquid thus the pressure should decrease. Why it isn't the case here?


2. Lithium cobalt oxide has the formula: LiCoO2. I just can't derive by looking at the unit cell (attached). I counted:
nO=2·1/2(the top and bottom atoms)+12·1/6(the atoms at the edge)+6(the atoms inside)=9
nCo=6·1/3(the atoms at the edges)+1(the atom at the center)=3
nLi=6
So the formula should be Li2CoO3. Where am I wrong?

Online Corribus

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Re: Formation of ice + hexagonal unit cell
« Reply #1 on: May 01, 2013, 10:31:05 AM »
1. During the formation of ice out of liquid water:
A) energy is set free
B) the surrounding cools down
C) the pressure in the surrounding decreases
D) the mass of water increases
E) the volume of the total portion of ice and water
decreases.

I thought that A and C are correct, but C isn't. The molecules will harder leave the solid than the liquid thus the pressure should decrease. Why it isn't the case here?
Does ice have a larger or smaller volume than liquid water?  Now suppose you put a quantity of water in a sealed container and freeze it.  What happens to the volume above the water?  Does it shrink or expand? 

So what happens to the pressure?  (Assuming solubility of the atmosphere in the ice is the same as the water.)
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

Offline Rutherford

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Re: Formation of ice + hexagonal unit cell
« Reply #2 on: May 01, 2013, 10:54:16 AM »
The volume of ice is bigger. Understood what you meant. I didn't think of other particles in the surroundings. Thanks.

The second question left.

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