Dissolving Solids: Can They Be Reformed?

  • Thread starter Thread starter erka15
  • Start date Start date
  • Tags Tags
    Solids
Click For Summary
Dissolving solids involves understanding solubility and equilibrium, where a solid can either dissolve or remain undissolved. The solubility product constant (Ksp) quantifies the equilibrium concentrations of the dissolved ions. Factors such as temperature and concentration influence whether a solid will dissolve or reform. The discussion highlights the importance of Ksp in determining solubility behavior. Overall, the topic revolves around the reversible nature of dissolving solids and their equilibrium states.
erka15
Messages
3
Reaction score
0
I would type the problem but it's really long. I just need to know what the two different results are for dissolving solids because I completely forgot. I think it has to do with if you can bring them back or not ?
 
Physics news on Phys.org
Are you asking about diffusion or solubility? It seems like you know what you are searching for better than we do, try google or wikipedia.
 
I'm talking about solubilty. I've tried to look it up but i can't find any thing.
 
Well every substance has a "Ksp" called the solubility which is the product of the concentrations of the products at equilibrium. Most substances will dissolve then go back to solid and redissolve, and based on different factors like temp, conc. of reactants, etc. they will reach a state of equilibrium, the values for the products at this point are the ksp.

For example

Take the Reaction
aX -> Ca+ + Dx-

[a+]^C[x-]^D = Ksp(solubility product) for compound aX

Does that shed any light on this question? I think this is probably posted in the wrong area.
 
Last edited:
Ok, well, I think the wikipedia link helped a little bit. Thanks any way.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 10 ·
Replies
10
Views
2K
Replies
15
Views
2K
Replies
18
Views
6K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
4
Views
928
Replies
2
Views
2K
Replies
21
Views
2K
Replies
8
Views
14K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K