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lakkis [162]
3 years ago
8

1. Why is it that some combinations of ions do not seem to react while others result in precipitate? 2. For an insoluble compoun

d, how might you prevent formation of precipitate? 3. What conditions can you manipulate to make the precipitate more soluble? 4. Does the size of the crystal affect the solublization process? 5. How do you accurately rule out the presence of alkali metals in solution? 6. Which combinations of ions used do you expect to form insoluble salts? 7. What are their Ksp values?
Chemistry
2 answers:
trapecia [35]3 years ago
5 0
1. Some ions don't react and forms a precipitate because of the strengths of intermolecular bonds and the energy required to form new bonds.2. You can add another substance that will enable it to be dissolved in  the solution but this depends on the precipitate you have.3. There are many factors that will affect solubility this includes the temperature and the pressure which can be manipulated easily.4. Yes it affects the solubilization process. Surface area of the crystals is a factor.
N76 [4]3 years ago
3 0
1) Sometimes ions in solution react with each other to form a new substance that is insoluble (does not dissolve), called a precipitate.
2)<span>It depends on what kind of precipitate you're trying to prevent from forming. Many precipitates will dissolve in acidic solutions, so you could acidify the solution and see what happens. Other precipitates dissolve well when ammonia is added, so that's an option. There's no conclusive way to answer your question unless we know what kind of precipitate you're trying to prevent. 
</span>3) if a salt is deemed soluble macroscopically, very large number of ions dissolve before the solution reaches saturation
4)<span>if the temperature is decreased rapidly the solubility of your product will be significantly decreased in a short time (generally), thus the crystallization procedure will be accelerated leading smaller crystals. 

</span>
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Answer:

Molarity = 0.002 M

Explanation:

Given data:

Mass of calcium chloride = 0.321 g

Volume of water = 1.45 L

Molarity of solution = ?

Solution:

Molarity = number of moles / volume in litter.

We will calculate the number of moles of calcium chloride first.

Number of moles = mass/molar mass

Number of moles = 0.321 g/ 110.98 g/mol

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