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Kay [80]
3 years ago
15

What change would shift the equilibrium system to the left?

Chemistry
2 answers:
ArbitrLikvidat [17]3 years ago
5 0

Answer: The correct option is, Adding more of gas C 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 the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

For the given equation :

A(g)+B(s)+Energy\leftrightharpoons 3C(g)

This is a type of Endothermic reaction because heat is absorbed in the reaction.

For the given options :

Option A: Adding more of gas  C to the system

If the concentration of gas C that is the product is increased, so according to the Le-Chatlier's principle, the equilibrium will shift in the direction where decrease of concentration takes place. Therefore, the equilibrium will shift in the left direction.

Option B: Heating the system

If the temperature is increased, so according to the Le-Chatlier's principle , the equilibrium will shift in the direction where increase in temperature occurs. As, this is an endothermic reaction, forward reaction will decrease the temperature. Hence, the equilibrium will shift in the right direction.

Option C: Increase the volume

If the volume of the container is increased, the pressure will decrease according to Boyle's Law. Now, according to the Le-Chatlier's principle, the equilibrium will shift in the direction where increase in pressure is taking place. As the number of moles of gas molecules is greater at the product side. So, the equilibrium will shift in the right direction.

Option D: Removing some of gas C from the system

If the concentration of gas  C that is the product is decreased, so according to the Le-Chatlier's principle, the equilibrium will shift in the direction where decrease of concentration takes place. Therefore, the equilibrium will shift in the right direction.

Hence, the correct option is, Adding more of gas C to the system.

SVEN [57.7K]3 years ago
3 0

Answer:

Adding more of gas C to the system

Explanation:

  • <em>Le Châtelier's principle</em><em> states that when there is an dynamic equilibrium, and this equilibrium is disturbed by an external factor, the equilibrium will be shifted in the direction that can cancel the effect of the external factor to reattain the equilibrium.</em>

1) Adding more of gas C to the system:

Adding more C gas will increase the concentration of the products side. So, the reaction will be shifted to the left to attain the equilibrium again.

2) Heating the system:

Heating the system will increase the concentration of the reactants side as the reaction is endothermic. so, the reaction will be shifted to the right to attain the equilibrium again.

3) Increasing the volume:

has no effect since the no. of moles of gases is the same in both reactants and products sides.

4) Removing some of gas C from the system:

Removing some of gas C from the system will decrease the concentration of the products side. So, the reaction will be shifted to the right to attain the equilibrium again.

<em>So, the right choice is: Adding more of gas C to the system.</em>

<em></em>

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Write the balanced equation for the reaction of aqueous Pb ( ClO 3 ) 2 Pb(ClO3)2 with aqueous NaI . NaI. Include phases. chemica
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<u>Answer:</u> The mass of precipitate (lead (II) iodide) that will form is 119.89 grams

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}

Molarity of NaI solution = 0.130 M

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Putting values in above equation, we get:

0.130M=\frac{\text{Moles of NaI}}{0.400L}\\\\\text{Moles of NaI}=(0.130mol/L\times 0.400L)=0.52mol

The balanced chemical equation for the reaction of lead chlorate and sodium iodide follows:

Pb(ClO_3)_2(aq.)+2NaI(aq.)\rightarrow PbI_2(s)+2NaClO_3(aq.)

The precipitate (insoluble salt) formed is lead (II) iodide

By Stoichiometry of the reaction:

2 moles of NaI produces 1 mole of lead (II) iodide

So, 0.52 moles of NaI will produce = \frac{1}{2}\times 0.52=0.26mol of lead (II) iodide

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Moles of lead (II) iodide = 0.26 moles

Molar mass of lead (II) iodide = 461.1 g/mol

Putting values in above equation, we get:

0.26mol=\frac{\text{Mass of lead (II) iodide}}{461.1g/mol}\\\\\text{Mass of lead (II) iodide}=(0.26mol\times 461.1g/mol)=119.89g

Hence, the mass of precipitate (lead (II) iodide) that will form is 119.89 grams

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matrenka [14]

Answer:

<h3>The answer is 5.24 mL</h3>

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We have the final answer as

<h3>5.24 mL</h3>

Hope this helps you

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