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3 years ago

In an aqueous solution of lithium chloride LiCl, what ions or molecules are present in solution

1 answer:

7 0

In an aqueous solution of lithium chloride (LiCl), the ions present in solution are lithium cations (Li+) and chloride anions (Cl-), and the water molecules (H2O).

Lithium chloride is an ionic salt, and therefore, fully dissociates or ionizes upon dissolution in water. From the chemical formula of lithium chloride, we can say that 1 lithium ion (positively-charged) only needs 1 chloride ion (negatively charged) to form the salt. Thus, upon dissociation of 1 molecule of LiCl, only 1 lithium ion, and 1 chloride ion form.

Water is a molecule formed by covalent bonds and generally, does not dissociate to form ions. However, if we consider the very minor contribution of the dissociation of water, hydroxide (OH-) ions and hydronium (H3O+) ions will also be present. However for water, very few molecules, only 1 in 550 million, are dissociated at any given time. Thus, we can consider the presence of the OH- and H3O+ ions to be negligible and just identify the molecular undissociated form of water to be present.

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A solution at 25 degrees Celsius has a pH of 4.48. What is the pOH of this solution?

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3 years ago

A mixture containing initially 3.90 mole of NO(g) and 0.88 mole of CO2(g) was allowed to react in a flask of volume 1.00 L at a

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Answer:

Equilibrium constant Kc for the reaction will be 1.722

Explanation:

O2(g)+NO(g)→CO(g)+ NO2(g)

0.88 3.9 --- ---

0.88x 3.9-x x x

GIVEN:

0.88X-X= 0.11

⇒ X=0.77

CO2(g)+NO(g) → CO(g) + NO2(g)

0.88 3.9 --- ---

0.88-x 3.9-x x x

= 3.13 0.77 0.77

=0.11

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4 years ago

An equilibrium mixture of PCl₅(g), PCl₃(g), and Cl₂(g) has partial pressures of 217.0 Torr, 13.2 Torr, and 13.2 Torr, respective

Papessa [141]

Answer:

PCl₅ = 223.4 torr

PCl₃ = 6.8 torr

Cl₂ = 26.4 torr

Explanation:

For gas substances, the equilibrium constant can be calcultaed based on the partial pressures (Kp). For a generic reaction:

aA(g) + bB(g) ⇄ cC(g) + dD(g)

$Kp = \frac{(pC)^c*(pD)^d}{(pA)^a*(pB)^b}$ , where pX is the partial pressure of X.

The reaction with the gas mixture given is:

PCl₅(g) ⇄ PCl₃(g) + Cl₂(g)

Kp = [(pPCl₃)*(pCl₂)]/(pPCl₅)

Kp = [13.2*13.2]/217

Kp = 0.803

When more Cl₂ is added, for Le Chatêlier's principle, the equilibrium will shift for the left, more PCl₅ will be formed, and the equilibrium will be reestablished.

The initial total pressure was 243.4 torr, so if it jumps to 263.0 torr, it was added 19.6 torr of Cl₂, so the partial pressure of Cl₂ is 32.8 torr. For the reaction:

PCl₅(g) ⇄ PCl₃(g) + Cl₂(g)

217.0 13.2 32.8 Initial

+x -x -x Reacts (stoichiometry is 1:1:1)

217 + x 13.2-x 32.8-x Equilibrium

So, the equilibrium constant must be:

$Kp = \frac{(13.2-x)*(32.8-x)}{217+x}$

0.803 = (432.96 - 46x + x²)/(217 + x)

432.96 - 46x + x² = 174.251 + 0.803x

x² - 46.803x + 258.71 = 0

By Bhaskara's equation:

Δ = (46.803)² - 4*1*258.71

Δ = 1,155.68

x =[-(- 46.803) ±√1,155.68]/2

x' = (46.803 + 33.99)/2

x' = 40.40

x'' = (46.803 - 33.99)/2

x'' = 6.40

x < 32.8, so x = 6.40

The new partial pressures are:

PCl₅ = 217.0 + 6.40 = 223.4 torr

PCl₃ = 13.2 - 6.40 = 6.8 torr

Cl₂ = 32.8 - 6.40 = 26.4 torr

6 0

3 years ago