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3 years ago

In which of the following reactions does a decrease in the volume of the reaction vessel at constant

Chemistry

1 answer:

Black_prince [1.1K]3 years ago

6 0

Answer:

The correct option is: A) 2H₂(g) + O₂(g) → 2H₂O(g)

Explanation:

According to the Le Chatelier's principle, change in the volume of the reaction system causes equilibrium to shift in the direction that reduces the effect of the volume change.

When the volume decreases then the pressure of the reaction vessel increases, then the equilibrium shifts towards the reaction side that produces less number of moles of gas.

A) 2H₂(g) + O₂(g) → 2H₂O(g)

The number of moles of reactant is 3 and number of moles of product is 2.

Therefore, when volume decreases, the equilibrium shifts towards the product side, thereby favoring the formation of products.

B) NO₂(g) + CO(g) → NO(g) + CO₂(g)

The number of moles of reactant and product both is 2.

Therefore, when the volume decreases, the equilibrium does not shift in any direction.

C) H₂(g) + I₂(g) → 2HI(g)

The number of moles of reactant and product both is 2.

Therefore, when the volume decreases, the equilibrium does not shift in any direction.

D) 2O₃(g) → 3O₂(g)

The number of moles of reactant is 2 and number of moles of product is 3.

Therefore, when volume decreases, the equilibrium shifts towards the reactant side, thereby favoring the formation of reactants.

E) MgCO₃(s) → MgO(s) + CO₂(g)

The number of moles of reactant is 1 and number of moles of product is 2.

Therefore, when volume decreases, the equilibrium shifts towards the reactant side, thereby favoring the formation of reactants.

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Water 2270g is heated until it just begins to boil. If the water absorbs 5.51 ×10^5 J of heat in the process, what was the initi

kolbaska11 [484]

Answer:

42.01 °C

Explanation:

Step 1: explain the problem

We have to find the initial temperature, when a certain amount of heat raises this sample of water to its boiling point ( 100 °C)

⇒this amount of heat = 5.51 x 10Δ^5 J

We will use the formule : Q = mcΔ T

with Q = heat transfer ( J)

with m = mass of the substance (g)

with c = specific heat ( J/g °C)

with Δ T = temperature change ( in °C or K)

Water has a specific heat of 4.186 J/g °C

Step 2 : Calculate the initial temperature

We have to rearrange the formule first:

Δ T = Q / mc

In this case we have :

Δ T = 5.51 * 10^5 J / 2270 g * 4.186 J/g °C = 57.99

⇒ The final temperature of the water is the boiling point (100 °C) and the change of temperature is 57.99. This means that the boiling point is 57.99 °C higher than the initial temperature.

This means : Δ T = Tboiling point - Tinitial

Δ T = 57.99 °C = 100 °C - Tinitial

Tinitial = 100 °C - 57.99 °C = 42.01 °C

The initial temperature of the water is 42.01 °C

6 0

4 years ago

Kinetic energy and potential energy are often considered to be forms of mechanical energy. List three other forms of energy and

mafiozo [28]

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3 years ago

3. Suppose you wanted to design an experiment to test the composition of a mixture that includes sodium phenoxide (NaC6H5O). You

Liula [17]

Answer:

21.5mL of a 0.100M HCl are required

Explanation:

The sodium phenoxide reacts with HCl to produce phenol and NaCl in a 1:1 reaction.

To solve this question we need to find the moles of sodium phenoxide. These moles = Moles of HCl required to reach equivalence point and, with the concentration, we can find the needed volume as follows:

Mass NaC6H5O:

1.000g * 25% = 0.250g NaC6H5O

Moles NaC6H5O -116.09g/mol-

0.250g NaC6H5O * (1mol/116.09g) = 2.154x10⁻³ moles = Moles of HCl required

Volume 0.100M HCl:

2.154x10⁻³ moles HCl * (1L/0.100mol) = 0.0215L =

<h3>21.5mL of a 0.100M HCl are required</h3>

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3 years ago

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Ket [755]

Answer:

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Explanation:

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