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Ierofanga [76]
2 years ago
11

Describe the three steps of solution formation. be sure to include the details regarding the dissolving of the substance

Chemistry
2 answers:
Elza [17]2 years ago
8 0

The three steps involve;

Step 1: Separation/expansion of the solute particles

Step 2: Separation/expansion of the solvent particles

Step 3; Combining the solute and solvent particles


The first two steps are usually endothermic. Step 3, nonetheless, can be either exothermic or endothermic and is significant in determining whether the dissolving process will be endothermic or exothermic.


IceJOKER [234]2 years ago
8 0

Answer:

The dissolution process is carried out in three completely independent stages:

1. Separation of solvent molecules

2. Separation of solute molecules.

3. Mixture of solvent and solute molecules.

Explanation:

A solution occurs when the solute and solvent molecules meet by intermolecular attractions, which should not be interpreted as chemical bonds. When a substance dissolves in another substance, the solute particles spread over the solvent. These particles break into positions occupied by solvent molecules. Its ease of replacement is dependent on the strength of three interactions: between solvent and solvent, between solute and solute, and between solvent and solute.

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umka21 [38]

Hi!


The correct options would be:

1. Cathode - <em>reduction</em>

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Wires have a universal role of being a pathway for the transport of electrons in circuit. This role is also the same in the wires involved in an electrochemical cells where they are used to transport electrons from the anodic half cell, and this electron transport results in the generation of electricity in the internal circuit of the electrochemical cell.


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4 0
2 years ago
A sample of gas contains 0.1500 mol of HCl(g) and 7.500×10-2 mol of Br2(g) and occupies a volume of 9.63 L. The following reacti
Furkat [3]

Answer:

9.63 L.

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

2HCl(g) + Br_2(g)\rightarrow 2HBr(g) + Cl_2(g)

So the consumed amounts of hydrochloric acid and bromine are the same to the beginning based on:

n_{Br_2}^{consumed}=0.1500molHCl*\frac{1molBr_2}{2molHCl}=0.075molBr_2

In such a way, the yielded moles of hydrobromic acid and chlorine are:

n_{HBr}=0.1500molHCl*\frac{2molHBr}{2molHCl}=0.1500molHBr \\n_{Cl_2}=0.1500molHCl*\frac{1molCl_2}{2molHCl}=0.075molCl_2

Thus, the volume of the sample, after the reaction is the same as no change in the total moles is evidenced, that is 9.63L.

Best regards.

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