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siniylev [52]
1 year ago
13

3. What factors affect solubility?can anyone help its due today ​

Chemistry
1 answer:
lozanna [386]1 year ago
4 0
Temperature, pressure, forces and bonds
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What are the cations and anions whose compounds are usually soluble?
HACTEHA [7]

Answer: -

Solubility of a substance depend on the balance of intermolecular forces between the solvent and solute, and the entropy change that accompanies this process.

Temperature and pressure also plays a role in solubility.

A solution having Group 1 cations like lithium, sodium, potassium etc are always soluble.

A solution having NH₄⁺ is soluble.

All salts with anion as nitrates, acetates, chlorates, and perchlorates are soluble in water.

8 0
3 years ago
What is20 grams of HNO^3
morpeh [17]

63.01284*20=1260.568

7 0
2 years ago
Read 2 more answers
A student prepares a aqueous solution of butanoic acid . Calculate the fraction of butanoic acid that is in the dissociated form
ella [17]

Answer:

15.4%

Explanation:

If Ka = 0.54 mM = 1.51x10⁻⁵

Then;

C₄H₈O₂               -------->            C₄H₇O₂⁻          +           H⁺

I                    0.54x10⁻³                             0                                0

E                   0.54x10⁻³(1-x)                      0.54x10⁻³x                0.54x10⁻³x

Recall that x is the percentage degree of dissociation

From the ICE table;

Ka = [C₄H₇O₂⁻] [ H⁺]/[C₄H₈O₂]

1.51x10⁻⁵=(0.54x10⁻³x) (0.54x10⁻³x)/ 0.54x10⁻³(1-x)  

1.51x10⁻⁵ = 0.54x10⁻³x^2/1-x

1.51x10⁻⁵(1-x) = 0.54x10⁻³x^2

1.51x10⁻⁵ - 1.51x10⁻⁵x = 0.54x10⁻³x^2

Hence;

0.54x10⁻³x^2 + 1.51x10⁻⁵x - 1.51x10⁻⁵=0

x^2 + 0.028x - 0.028 = 0

Solving the quadratic equation here;

x = 0.154 or −0.182

Ignoring the negative result, x = 0.154

Hence, fraction of butanoic acid that is in the dissociated form in this solution = 15.4%

3 0
2 years ago
The possible products of a double displacement reaction in aqueous solution are SrSO4 and NaCl. Which of these possible products
madam [21]

Answer:

SrSO4

Explanation:

According to solubility rules, we know that the sulphates of the elements of group two are insoluble in water. The solubility rules describe what chemical species are soluble in water and what species are not soluble in water.

Generally, all chlorides are soluble in water with exception of chlorides such as silver chloride. The chlorides of group one elements are usually highly soluble in water.

Since SrSO4 is a sulphate of a group two element (strontium) it will be the insoluble solid product of the double displacement reaction described in the question.

7 0
3 years ago
Use calc to determine whether it is possible to remove 99.99% Cu2 by converting it to Cu(s) in a solution mixture containing 0.1
inessss [21]

Answer:

it is possible to remove 99.99% Cu2 by converting it to Cu(s)

Explanation:

So, from the question/problem above we are given the following ionic or REDOX equations of reactions;

Cu2+ + 2e- <--------------------------------------------------------------> Cu (s) Eo= 0.339 V

Sn2+ + 2e- <---------------------------------------------------------------> Sn (s) Eo= -0.141 V

In order to convert 99.99% Cu2 into Cu(s), the equation of reaction given below is needed:

Cu²⁺ + Sn ----------------------------------------------------------------------------> Cu + Sn²⁺.

Therefore, E°[overall] = 0.339 - [-0.141] = 0.48 V.

Therefore, the change in Gibbs' free energy, ΔG° = - nFE°. Where E° = O.48V, n= 2 and F = 96500 C.

Thus, ΔG° = - 92640.

This is less than zero[0]. Therefore,  it is possible to remove 99.99% Cu2 by converting it to Cu(s) because the reaction is a spontaneous reaction.  

7 0
2 years ago
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