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Arturiano [62]
3 years ago
6

HELP HELP HELP!! Explain the various factors that impact the solubility of substances in water.

Chemistry
2 answers:
astraxan [27]3 years ago
8 0

Answer:

1) Temperature

When salts like KNO3, NANO3 and KCl are added in water the test tube ecomes cold, which means the heat is being absorbed. Solubility usually increases with increase in temperature for such solutes.

On the other hand, when salts like Li2SO4 and Ce2(SO4)3 are dissolved in water, test tube turns warm which indicates the heat is released. In such cases the solubility of salt decreases with increase in temperature.

In some cases,the heat is neither absorbed nor releasedwhen dissolved in water. When salts like NaCl is added in water, the temperature remains the same, and hence temperature has minimal effect on solubility.

2) Pressure

The solubility of solid and liquids aren't affected by pressure but the solubility of gas in a liquid is directly propotional to the pressure of gas. This is also explained through Henry's Law. i.e.

m ∝ P or m = KP

where "m" is the amount of gas dissolved.

Hope that helps, it might be a little too detailed and lenthy, but it explains everything you asked.

Agata [3.3K]3 years ago
4 0

Explanation:

There are two direct factors that affect solubility: temperature and pressure. Temperature affects the solubility of both solids and gases, but pressure only affects the solubility of gases.

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How do the prefixes of the –saccharide words relate to the structure of the sugar?
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The prefixes tell you how many of each there are.

Explanation:

So for this example, mono would mean there is one sugar, di meaning 2 sugars and so on. Hope this helped!

Here is a table:

mono- 1    hexa-6

di-2           hepta- 7

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(a) what product or products would you expect to obtain by the addition of bromine to cis-cinnamic acid, assuming that it reacts
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3 years ago
If you have two pistons in a system, which exerts the larger force?
Alik [6]

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How much energy would it take to heat a section of the copper tubing that weighs about 665.0 g, from 15.71 ∘C to 27.09 ∘C ? Copp
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Answer:

2914 J

Explanation:

Step 1: Given data

  • Mass of the copper tubing (m): 665.0 g
  • Initial temperature: 15.71 °C
  • Final temperature: 27.09 °C
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Step 2: Calculate the temperature change

ΔT = 27.09 °C - 15.71 °C = 11.38 °C

Step 3: Calculate the energy required (Q)

We will use the following expression.

Q = c × m × ΔT

Q = 0.3850 J/g.°C × 665.0 g × 11.38 °C

Q = 2914 J

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