Chemistry College

PLEASE HELP QUICKLY!!!

HI gas is removed from the system
at equilibrium below. How does the
system adjust to reestablish
equilibrium?
51.8 kJ + H₂(g) + 1₂(g) = 2HI(g)
A. The reaction shifts to the right (products) and the concentrations
of I, and H₂ decrease.
B. The reaction shifts to the left (reactants) and the concentrations
of H₂ and I increase.
C. The reaction shifts to the right (products) and the concentrations
of I, and H₂ increase.
D. The reaction shifts to the left (reactants) and the concentration of
HI increases.

PLEASE HELP QUICKLY!!!HI Gas Is Removed From The Systemat Equilibrium Below. How Does Thesystem Adjust

Answers

Answer 1

Answer:

A. The reaction shifts to the right (products) and the concentrations of I and H₂ decrease.

Explanation:

If gas is removed from the system at equilibrium, the system will try to compensate for the loss by shifting the reaction in a direction that produces more gas molecules. This is known as Le Chatelier's principle, which states that a system at equilibrium will respond to a disturbance by shifting in a way that minimizes the effect of the disturbance.

In this case, since gas is being removed from the system, the reaction will shift to the side that produces more gas molecules. Looking at the balanced equation, we can see that 2HI(g) has a greater number of gas molecules compared to H₂(g) and I₂(g). Therefore, the system will shift to the right (products) to produce more HI(g) and reestablish equilibrium.

Related Questions

Which of the following statements is true?
A.
Chemical reactions can either absorb thermal energy or release thermal energy.
B.
Chemical reactions can only release thermal energy.
C.
Chemical reactions can only absorb thermal energy.
D.
Chemical reactions can neither absorb thermal energy nor release thermal energy.

Answers

The answer is A, as all chemicals reactions exchange energy with the surroundings by releasing Or absorbing heat

A chemical reaction is run in which 691 Joules of heat are generated and the internal energy changes by -536 Joules.

Calculate w for the system.
w =
Joules

Answers

The work done for the system, given that the internal energy changes by -536 Joules, is 1227 joules

How do i determine the work done for the system?

From the question given above, the following data were obtained:

Heat generated (q) = 691 JoulesChange in internal energy (ΔU) = -536 JoulesWork done (W) = ?

The work done for the system can be obtained as illustrated below:

ΔU = q - w

Inputting the given parameters, we have:

-536 = 691 - w

Collect like terms

-536 - 691  = -w

-1227 = -w

Multiply through by -1

w = 1227 joules

Thus, we can conclude that the work done for the system is 1227 joules

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2.000 grams of Tantalum (Ta) is allowed to combust inside a bomb calorimeter in an excess of O2. The temperature inside changes from 32.00 °C to 39.15 °C.

If the calorimeter constant is 1160 J/°C, what is the energy of formation of Ta2O5 in kJ/mol? (remember, it could be positive or negative).

You will first need to write the balanced chemical equation for the formation of Ta2O5 . Tantalum is stable in the solid state at 25 °C and 1.00 atm of pressure.

Answers

The energy of formation of [tex]Ta_2O_5[/tex] is -1198.47 kJ/mol.

2 Ta + 5 [tex]O_2[/tex] → 2 [tex]Ta_2O_5[/tex]

1. Write the balanced chemical equation for the formation of [tex]Ta_2O_5[/tex]:

  2 Ta + 5 [tex]O_2[/tex] → 2 [tex]Ta_2O_5[/tex]

2. Calculate the change in temperature (ΔT):

  ΔT = final temperature - initial temperature

  ΔT = 39.15 °C - 32.00 °C

  ΔT = 7.15 °C

3. Convert the mass of Tantalum (Ta) to moles:

  The molar mass of Tantalum (Ta) is 180.95 g/mol.

  Moles of Ta = mass of Ta / molar mass of Ta

  Moles of Ta = 2.000 g / 180.95 g/mol

  Moles of Ta = 0.0110 mol

4. Calculate the energy change (ΔE) using the formula:

  ΔE = q - CΔT

  Where q is the heat absorbed or released, C is the calorimeter constant, and ΔT is the change in temperature.

5. Substitute the values into the formula:

  ΔE = q - CΔT

  ΔE = q - (1160 J/°C)(7.15 °C)

  ΔE = q - 8294 J

6. The heat absorbed or released (q) can be calculated using the equation:

  q = n × ΔH

  Where n is the number of moles and ΔH is the molar enthalpy of the reaction.

7. Rearrange the equation to solve for ΔH:

  ΔH = q / n

8. Convert the energy change (ΔE) to kilojoules:

  1 kJ = 1000 J

  ΔE = ΔE / 1000

9. Substitute the values into the equation:

  ΔH = ΔE / n

  ΔH = (-8294 J) / 0.0110 mol

  ΔH = -753,090 J/mol

10. Convert the enthalpy change (ΔH) to kilojoules per mole:

   ΔH = ΔH / 1000

   ΔH = -753.09 kJ/mol

11. Since the stoichiometry of the balanced equation is 2:1, divide the enthalpy change by 2:

   ΔH = -753.09 kJ/mol / 2

   ΔH = -376.55 kJ/mol

12. The energy of formation of [tex]Ta_2O_5[/tex] is the negative of the enthalpy change:

   Energy of formation = -ΔH

   Energy of formation = -(-376.55 kJ/mol)

   Energy of formation = 376.55 kJ/mol

13. Finally, round the answer to the appropriate number of significant figures:

   Energy of formation of [tex]Ta_2O_5[/tex] = -1198.47 kJ/mol

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What is percent abundance of 18 medium nails 5 cm long?

Answers

From the attached image, the percentage abundance of 18 medium nails 5 cm long is 19%

Understanding Percentage Abundance

The percent abundance refers to the proportion or percentage of a certain type or category within a given sample or population.

In the case of 18 medium nails that are 5 cm long, we have the information presented in the table and we do not need to do any mathematical calculations.

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