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2 years ago

When the compound BaCl2 forms , what happens to the Ba and Cl ions

Chemistry

2 answers:

Korvikt [17]2 years ago

6 0

Ba stays as Ba+2 and Cl stays as Cl-

Anna007 [38]2 years ago

4 0

Answer : The correct answer is Ba²⁺ and Cl⁻ ions .

This can be explained using Solubility concept .

Solubility is defined as property referring to ability of a substance , SOLUTE , to dissolves in a SOLVENT . It is measured at maximum amount of solute that dissolves in solvent as equilibrium .

Solubility of any compound can be checked using Solubility Rule ( Image ).

BaCl₂ is salt of Chloride . Since the solubility rules says that salts if chlorides are soluble , Hence BaCl₂ is also soluble .

SO when BaCl₂ forms in aqueous solution , it again dissociates to forms ions since it is soluble in aqueous solution . It produced one Ba²⁺ ion and two ions of Cl⁻ .

The BaCl₂ dissociates as follows :

$BaCl_2 (aq) \rightarrow Ba^2^+ (aq) + 2 Cl^- (aq)$

Hence even BaCl₂ forms but it remain as dissociated ionic form as Ba²⁺ and Cl⁻ ions.

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Consider the following reaction where Kc = 1.80×10-2 at 698 K:

Klio2033 [76]

Answer:

The system is not in equilibrium and the reaction must run in the forward direction to reach equilibrium.

Explanation:

The reaction quotient Qc is a measure of the relative amount of products and reagents present in a reaction at any given time, which is calculated in a reaction that may not yet have reached equilibrium.

For the reversible reaction aA + bB⇔ cC + dD, where a, b, c and d are the stoichiometric coefficients of the balanced equation, Qc is calculated by:

$Qc=\frac{[C]^{c}*[D]^{d} } {[A]^{a}*[B]^{b}}$

In this case:

$Qc=\frac{[H_{2} ]*[I_{2} ] } {[HI]^{2}}$

Since molarity is the concentration of a solution expressed in the number of moles dissolved per liter of solution, you have:

$[H_{2} ]=\frac{2.09*10^{-2} moles}{1 Liter}$ =2.09*10⁻² $\frac{moles}{liter}$

$[I_{2} ]=\frac{4.14*10^{-2} moles}{1 Liter}$ =4.14*10⁻² $\frac{moles}{liter}$

$[I_{2} ]=\frac{0.280 moles}{1 Liter}$ = 0.280 $\frac{moles}{liter}$

So,

$Qc=\frac{2.09*10^{-2} *4.14*10^{-2} } {0.280^{2} }$

Qc= 0.011

Comparing Qc with Kc allows to find out the status and evolution of the system:

If the reaction quotient is equal to the equilibrium constant, Qc = Kc, the system has reached chemical equilibrium.

If the reaction quotient is greater than the equilibrium constant, Qc> Kc, the system is not in equilibrium. In this case the direct reaction predominates and there will be more product present than what is obtained at equilibrium. Therefore, this product is used to promote the reverse reaction and reach equilibrium. The system will then evolve to the left to increase the reagent concentration.

If the reaction quotient is less than the equilibrium constant, Qc <Kc, the system is not in equilibrium. The concentration of the reagents is higher than it would be at equilibrium, so the direct reaction predominates. Thus, the system will evolve to the right to increase the concentration of products.

Being Qc=0.011 and Kc=1.80⁻²=0.018, then Qc<Kc. <u><em>The system is not in equilibrium and the reaction must run in the forward direction to reach equilibrium.</em></u>

8 0

3 years ago

4. The volume of a sample of a gas at STP is 200.0 ml. If the pressure is increased to 4.00 atmospheres (temperature constant),

Elina [12.6K]

Answer : The value of new volume is, 50.0 mL

Explanation :

Boyle's Law : It is defined as the pressure of the gas is inversely proportional to the volume of the gas at constant temperature and number of moles.

$P\propto \frac{1}{V}$