Look at the diagram. Which shows the correct arrangement of electrons in a chlorine molecule?
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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.

0.6 is the equilibrium constant for the given reaction.

To calculate the equilibrium constant (K) for the given reaction, we need to use the molar concentrations of the reactants and products at equilibrium. The equilibrium constant expression is given by:

[tex]K= [H_{2}O]^{2} / ([H_{2}^{2} * [O_{2}])[/tex]

Given the moles of H2, O2, and H2O in the 8.1 L container, we can convert them to molar concentrations by dividing the number of moles by the volume:

[H2] = 1.83 moles / 8.1 L

[O2] = 2.33 moles / 8.1 L

[H2O] = 3.95 moles / 8.1 L

Substituting these values into the equilibrium constant expression, we have:

K = [tex](3.95 / 8.1)^{2}[/tex] / ([tex](1.83 / 8.1)^{2}[/tex] * (2.33 / 8.1))

Evaluating this expression and rounding to one decimal place, we find the equilibrium constant to be:

K ≈ 0.6

Therefore, the equilibrium constant for the given reaction is approximately 0.6.

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

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

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

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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answer and explanation :)

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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Answer: If a given sample of metal has a mass of 2.68 g and a volume of 1.03 cm³, the density of the metal will be 2.6019417476 g/cm³.

Explanation:

To find out the density of any object we must have known values of mass of the object and volume of the object.

Mass- Mass is the amount of matter present in any object or particle. The S.I. unit of mass is the kilogram.

Volume- Volume is defined as the amount of space occupied by an object or particle. The measuring unit of volume is cubic meter (m³)- for larger volumes and cubic centimeters (ccm³) and cubic millimeters (cmm³) for smaller volumes.

Density- Density is the measurement that compares the mass of an object with its volume. The S.I. unit of density is kilogram per cubic meter (kg /m³) and the C.G.S unit is gram per cubic centimeter ( g/ ccm³). Density is denoted by rho (ρ).

The density of an object can be calculated by the following formula:

                    Density (ρ) = mass (m)/ volume (v)

In the given question, the mass of the object is 2.68 g. i.e. m = 2.68 g and the volume of the given sample is 1.03 cm³  i.e. v = 1.03 cm³.

Hence, by using the above formula and putting the values of mass and volume, we can calculate the density of the sample as below-

                Density = mass (m)/ volume (v)

                               = 2.68 /  1.03

                               = 2.6019417476 g/cm³

Therefore, for the given sample of metal that has a mass of 2.68 g and volume of 1.03 cm³ will have a density of 2.6019417476 g/cm³.

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Using the equilibrium constant expression and given concentrations of SO₂ and O₂, the approximate concentration of SO₃ at equilibrium is 0.436 M, rounded to 2 significant figures.

The equilibrium constant expression for the given reaction is:

Kc = [SO₃]² / ([SO₂]² * [O₂])

Given that the equilibrium constant (Kc) is 2.5 x 10³, [SO₂] = 0.0651 M, and [O₂] = 0.114 M, we can substitute these values into the equilibrium constant expression:

2.5 x 10³ = [SO₃]² / (0.0651² * 0.114)

To solve for [SO₃], we can rearrange the equation:

[SO₃]² = (2.5 x 10³) * (0.0651² * 0.114)

[SO₃]² = 0.1902643

Taking the square root of both sides:

[SO₃] = √(0.1902643)

[SO₃] ≈ 0.436 M

Therefore, at equilibrium, the concentration of SO₃ is approximately 0.436 M, rounded to 2 significant figures.

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The equilibrium constant for the reaction as it has been shown is [tex]5.7 * 10^{18}[/tex]

The quantitative expression of the size of a chemical process at equilibrium is the equilibrium constant, abbreviated as K. It links the reactant and product concentrations (or partial pressures) in a chemical process and gives details on the make-up of the equilibrium mixture. It offers crucial details regarding the proportions of reactants and products.

We know that;

ΔG = -RTlnKp

Thus we have that;

Kp =[tex]e^-[/tex](ΔG/RT)

Kp = [tex]e^-[/tex](-107000 /8.314 * 298)

=[tex]5.7 * 10^{18}[/tex]

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