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faust18 [17]
3 years ago
6

What is the definition of a common ion as it applies to Le Chatelier’s principle?

Chemistry
2 answers:
navik [9.2K]3 years ago
6 0
The definition of a common ion as it applies to Le Chatelier's principle is an ion that is present in an equilibrium system and a compound added to the system. This is the case since in Le Chatelier's principle, it is based on an equilibrium system and where the reaction shifts to the left or to the right; towards the products or the reactants side.  
Juliette [100K]3 years ago
6 0

Answer: D) an ion that is present in an equilibrium system and a compound added to the system

Explanation: Any change in the equilibrium is studied on the basis of Le-Chatelier's principle.

This principle states that if there is any change in a equilibrium reaction, the equilibrium will shift in a direction so as to minimize the effect.

Thus when a common ion is introduced to an equilibrium reaction, the equilibrium will shift in a direction where the concentration of common ion is decreasing.

For example: In a equilibrium reaction for dissociation of acetic acid:

CH_3COOH\rightleftharpoons CH_3COO^-+H^+

If HCl is introduced wth H^+ as common ion.

HCl\rightarrow H^++Cl^-

The equilibrium will shift in backward direction where the concentration of H^+ is decreasing.

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How many moles of a gas would occupy 22.4 Liters at 273 K and 1 atm?<br><br><br> Read bottom comment
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Question 3 0
Tanya [424]

Answer:

Option D. KBr < KCl < NaCl

Explanation:

We'll begin by calculating the number of mole of each sample.

This can be obtained as follow:

For NaCl:

Mass = 1 g

Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol

Mole of NaCl =?

Mole = mass /Molar mass

Mole of NaCl = 1/58.5

Mole of NaCl = 0.0171 mole

For Kbr:

Mass = 1 g

Molar mass of KBr = 39 + 80 = 119 g/mol

Mole of KBr =?

Mole = mass /Molar mass

Mole of KBr = 1/119

Mole of KBr = 0.0084 mole

For KCl:

Mass = 1 g

Molar mass of KCl = 39 + 35.5 = 74.5 g/mol

Mole of KCl =?

Mole = mass /Molar mass

Mole of KCl = 1/74.5

Mole of KCl = 0.0134 mole

Summary

Sample >>>>>>>> Number of mole

NaCl >>>>>>>>>> 0.0171

KBr >>>>>>>>>>> 0.0084

KCl >>>>>>>>>>> 0.0134

Arranging the number of mole of the sampl in increasing order, we have:

KBr < KCl < NaCl

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