Consider the following reversible reaction


What is the equilibrium constant expression for the given system?

The type of compound represented by the graph at right is a strong acid (option B).

An acid is generally any compound capable of dissociating into its respective constituent ions when in an aqueous solution.

An acid is categorised as strong or weak depending on whether it can dissociate completely or partially. A strong acid dissociates completely in water.

According to this question, HA, when added to water, dissociates into H+ and A- ions, hence, is a strong acid.

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0.0956 grams of helium will occupy a volume of 575 mL at 760 mmHg and 20°C

To calculate the mass of helium that will occupy a given volume at a specific temperature and pressure, we need to use the ideal gas law equation: PV = nRT. Here's how you can solve the problem:

Convert the volume to liters: 575 mL = 575/1000 = 0.575 L.

Convert the pressure to atmospheres: 760 mmHg = 760/760 = 1 atm.

Convert the temperature to Kelvin: 20°C = 20 + 273.15 = 293.15 K.

Plug the values into the ideal gas law equation: (1 atm) * (0.575 L) = n * (0.0821 L·atm/(mol·K)) * (293.15 K).

Rearrange the equation to solve for n (the number of moles): n = (1 atm * 0.575 L) / (0.0821 L·atm/(mol·K) * 293.15 K).

Calculate the value of n: n = 0.0239 moles.

To find the mass, we need to know the molar mass of helium, which is approximately 4 grams per mole.

Multiply the molar mass by the number of moles: 4 g/mol * 0.0239 moles = 0.0956 grams.

Therefore, approximately 0.0956 grams of helium will occupy a volume of 575 mL at 760 mmHg and 20°C.

Note: The ideal gas law assumes ideal gas behavior, and the calculated result may not be accurate under extreme conditions or for gases that deviate significantly from ideal behavior.

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The Microscope part and their right descriptions are as follows

Iris Diaphragm: A. Increases or decreases the light intensity

Objective Lens System: B. After light passes through the specimen, it next enters this lens system

Stage: C. Platform that supports a microscope slide

Adjustment Knob: D. Causes stage (or objective lens) to move upward or downward

Condenser: E. Concentrates light onto the specimen

Other parts of a microscope and their description that you should know about includes;

Eyepiece - The lens that you look through to see the image of the specimen.

Body tube - The tube that connects the eyepiece to the objective lenses.

Arm - The part of the microscope that supports the body tube and connects it to the base.

Base - The part of the microscope that supports the arm and provides stability.

Illuminator - The light source that provides light for the microscope.

Stage clips - The clips that hold the microscope slide in place on the stage.

Revolving nosepiece - The part of the microscope that holds the objective lenses and allows them to be rotated into place.

The above answer is in response to the full question below;

Match the names of the microscope parts in column A with the descriptions in column B. Place the letter of your choice in the space provided.

1. Iris diaphram

2. Objective lens system

3. Stage

4. Adjustment knob

5. Condenser

Increases or decreases the light intensity

2. After light passes through the specimen, it next enters this lens system

3. Platform that supports a microscope slide

4. Causes stage (or objective lens) to move upward or downward

5. Concentrates light onto the specimen

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Answer:

I'm sorry, but I'm not able to view the chemical structure of glycine.

Answer:

Salt Method of Preparation

(i) Sodium nitrate (b) Neutralisation

(ii) Iron (III) chloride (e) Direct synthesis

(iii) Lead chloride (d) Double decomposition

(iv) Zinc sulphate (a) Simple displacement

(v) Sodium hydrogen sulphate (c) Decomposition by acid

Explanation:

Answer:

[tex] \huge{ \boxed{3.48 \: g/ml}}[/tex]

Explanation:

The density of the aluminum can be found by using the formula;

[tex]density( \rho) = \dfrac{mass}{volume} [/tex]

From the question

mass = 83.6 g

volume = 23.99 ml

[tex] \rho = \dfrac{83.6}{23.99} = 3.4847[/tex]

We have the final answer as

If the temperature is 273 K, the partial pressure of O₂ is 1.40 atm.

According to the given information:

Volume of bottle  = 2.00 L

Total pressure = 5.00 atm

Mixture of  N₂, O₂ and CO₂

Moles of N₂ = 0.29 moles

Partial pressure of CO₂ = 0.350 atm

Temperature = 273 K

It is known that

PV = nRT

P = pressure

V = volume

n = no of moles

R = universal gas constant

= 0.08206 L atm/mol K

T = temperature

so

To find total no of moles

PV = nRT

n = PV/RT

= (5.00 atm × 2.00 L) ÷ ( 0.08206 L atm/mol K × 273 K)

= 0.446 moles

Number of moles of CO₂

n = PV/RT

= (0.350 atm × 2.00 L) ÷ ( 0.08206 L atm/mol K × 273 K)

= 0.0312 moles CO₂

Total no of moles = moles of O₂ +  moles of CO₂ + moles of N₂

0.446 moles = moles of O₂ + 0.29  + 0.0312

Moles of O₂ = 0.125

Partial pressure of O₂ = (no of moles of O2) × R × T/V

= (0.125 moles × 0.08206 L atm/mol K × 273 K)/ 2.00 L

= 1.40 atm

The partial pressure of O₂ is 1.40 atm.

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The event of the fan stopping when it is switched off in a classroom provides an opportunity to discuss scientific concepts and learning. Here are the various steps involved in this event.

Observation: The students observe that when the switch of the fan is turned off, the fan stops rotating.

Questioning: Encourage students to ask questions about why the fan stops when the switch is turned off, leading to discussions about energy transfer, electrical circuits, and mechanical motion.

Explanation: Explain to the students that when the switch is turned off, it breaks the electrical circuit, interrupting the flow of current to the fan motor. Without the electrical energy from the current, the fan motor stops rotating.

Discussion: Engage students in a discussion about the conversion of electrical energy to mechanical energy in the fan motor. Discuss concepts such as electromagnetic induction, electric motors, and the role of magnetic fields in generating rotational motion.

Application: Relate this event to everyday life, highlighting the importance of understanding electrical circuits and energy conversions in various appliances.

By discussing these concepts, students can deepen their understanding of the principles behind the event and foster scientific learning.

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Answer:

D

Explanation:

The correct answer is D. In ethylene (CH2=CH2), the carbon atoms in the double bond have two unhybridized p-orbitals.

If 2.2 L of an aqueous solution that contains 261 g of calcium acetate  Ca(C₂H₃O₂)₂, at 37 °C, the osmotic pressure is 57.5 atm.

According to question:

Volume of solution = 2.2 L

Mass of solute calcium acetate Ca(C₂H₃O₂)₂  = 261 g

Molar mass of solute     = 158.17 g/mol

Temperature  = 370C  

=  ( 37 + 273 ) K  

=  310 K

No.of moles of solute   = mass ÷ molar mass

= 261g ÷ 158.17 g/mol

=  1.650 mol

Molarity of solute   = no.of moles solute ÷ volume of solution (L)

=  1.650 mol ÷ 2.2 L

=  0.753 mol/L

Osmotic pressure (π) is given by

π    =  i × M × R × T

i  =  van't Hoff factor  = 3

It is the figure of ions formed from 1 formula unit of a solute in solution.

Ca(C₂H₃O₂)₂  →   Ca2+   +    2C₂H₃O₂⁻

Thus, 1 Ca2+ and 2 Ca(C₂H₃O₂)₂ is formed.

i = 1+2  = 3

M = Molarity = 0.753 mol/L

R = universal gas constant  = 0.0821 L atm mol-1 K-1

T = Temperature ( K )   = 310 K

π   =  3 × 0.753 mol/L  × 0.0821 L atm mol-1 K-1  × 310 K

=  57.5 atm

Therefore, the osmotic pressure is 57.5 atm.

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For a solution of sodium hydroxide:

1. To make a buffer with pH = 5.00, have a ratio of

[tex]\frac{[A-]}{[HA]} = 10^{-5.50}[/tex] = 0.316. The concentration of acetic acid is 0.200 M, so add enough sodium hydroxide to make the concentration of acetate 0.316 M.

The volume of sodium hydroxide needed is calculated as follows:

V(NaOH) = (0.316 M - 0.200 M) / 4.50 M = 0.0316 L = 31.6 mL

Therefore, 31.6 mL of 4.50 M sodium hydroxide must be added to 250.0 mL of 0.200 M acetic acid solution to make a buffer with pH = 5.00.

2. The pH of the buffer is calculated as follows:

pH = pKa + log([tex]\frac{[A-]}{[HA]}[/tex])

= 4.76 + log(0.2/0.1)

= 4.86

Therefore, the pH of the buffer is 4.86.

3. The mass of salt that must be dissolved in 0.25 dm³ of 1 mol dm³ propanoic acid to give a buffer of pH 4.87 is calculated as follows:

[tex]\frac{[A-]}{[HA]} = 10^{-4.87}[/tex] = 0.0114

The concentration of acetate is 0.0114 M, and the concentration of propanoic acid is 1 mol dm³. Therefore, the mass of acetate that must be dissolved is calculated as follows:

Mass of acetate = (0.0114 mol dm³)(0.25 dm³) = 0.00285 g

Therefore, 0.00285 g of sodium propanoate must be dissolved in 0.25 dm³ of 1 mol dm³ propanoic acid to give a buffer of pH 4.87.

4. The pH of the buffer is calculated as follows:

pH = pKa + log([tex]\frac{[A-]}{[HA]}[/tex])

= 4.74 + log(0.1/0.1)

= 4.74

Therefore, the pH of the buffer is 4.74.

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Answer:

Chemical changes involve the formation of a new substance. 

Explanation:

The main difference between chemical and physical changes is the fact that chemical changes involve the formation of a new substance, results in a new substance. When a physical change occurs, this does not lead to bonds being broken or formed. The properties of the elements involve do not change. This includes changes such as changes of state. Physical change will result in a different state of matter.

The relative atomic mass of atom X is 9.

The relative atomic mass of an element is the average mass of its atoms relative to 1/12th the mass of a carbon-12 atom.

Given that atom X is 9 times heavier than 1/12th the mass of a carbon-12 atom, we can calculate the relative atomic mass of X as follows:

Relative atomic mass of X = (9 × 1/12) × relative atomic mass of carbon-12

The relative atomic mass of carbon-12 is defined as exactly 12. Therefore, we have:

Relative atomic mass of X = (9 × 1/12) × 12

= 9

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77.7K

Ideal gas laws allow us to calculate the characteristics of gases through their changes.

Gay-Lussac's Law

One of the ideal gas laws is Gay-Lussac's Law. This law describes the relationship between temperature and pressure. It states that temperature and pressure are directly proportional. This means that as temperature increases, pressure also increases. In equation form, this law is shown as:

Finding Temperature
To find the temperature, plug in the known information and solve for T₂.

Now, to find T₂, multiply both sides by 210.0 kPa.

Since this question is based on measured values, we should round according to significant figure rules. We should round to 3 sig figs because 208K has 3. This means the temperature of the gas tank must be 77.7 Kelvin.

Hello !

Pressure and volume can be regarded as the entity that is inversely proportional.

It should be noted that this explanatin an be done using the law in chemistry which is the Boyle's law which states that, for a given amount of gas and constant temperature, the volume is inversely proportional to the pressure.

However the Equal quantities of all gases can be seen to have same number of molecules when subjected to the same temperature and pressure (Avogadro's law).

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Answer:

There are 3.13 x 10^24 water molecules in 5.2 moles of water.

At a specific temperature, we can use the following equation to determine the root mean square (rms) speed of molecules or atoms:

rms speed = √(3RT/M)

where

R is the gas constant (8.314 J/(mol·K)),

T is the temperature in Kelvin, and

M is the molar mass in kilograms per mole.

For [tex]O_2[/tex] molecules at 427 K:

Molar mass of O2 = 2 * 16.00 g/mol = 32.00 g/mol = 0.032 kg/mol

Substituting the values into the equation:

rms speed = √((3 * 8.314 J/(mol·K) * 427 K) / 0.032 kg/mol)

rms speed ≈ √(28165.3 J/kg)

rms speed ≈ 167.9 m/s

The rms speed of [tex]O_2[/tex] molecules at 427 K is approximately 167.9 m/s.

For He atoms at 427 K:

Molar mass of He = 4.00 g/mol = 0.004 kg/mol

Substituting the values into the equation:

rms speed = √((3 * 8.314 J/(mol·K) * 427 K) / 0.004 kg/mol)

rms speed ≈ √(7061.925 J/kg)

rms speed ≈ 83.9 m/s

The rms speed of He atoms at 427 K is approximately 83.9 m/s.

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To calculate the amount of heat released when cooling water, we can use the formula:

Q = m × c × ΔT

where:

Q is the heat energy released or absorbed (in Joules),

m is the mass of the water (in grams),

c is the specific heat capacity of water (in J/g°C),

ΔT is the change in temperature (in °C).

Given:

m = 100.0 g (mass of water)

c = 4.186 J/g°C (specific heat capacity of water)

ΔT = 10.0°C (change in temperature, from 75.0°C to 10.0°C)

Using the formula:

Q = 100.0 g × 4.186 J/g°C × (10.0°C - 75.0°C)

Q = 100.0 g × 4.186 J/g°C × (-65.0°C)

Q = -27187 J

The negative sign indicates that heat is released during the cooling process.

Therefore, the correct answer is:

d) -27,209 J

Understanding the behavior of gases has numerous practical applications in real life. Here are some ways:

HVAC Systems ; Chemical Reactions ; Gas Storage ; Weather Forecasting ; Environmental Studies etc.

HVAC Systems: Heating, ventilation, and air conditioning (HVAC) systems rely on the principles of gas behavior to regulate temperature, humidity, and air quality in buildings. Knowledge of gas properties helps in designing efficient HVAC systems for optimal comfort and energy efficiency.

Chemical Reactions: Understanding gas behavior is crucial in chemical reactions, especially those involving gases. Knowledge of factors like pressure, volume, and temperature helps determine reaction rates, equilibrium conditions, and optimize industrial processes.

Gas Storage and Transportation: The behavior of gases is essential in designing storage tanks and transportation systems for various gases. Understanding pressure-volume relationships helps ensure safe handling and storage of gases, such as natural gas, propane, and compressed air.

Medical Applications: Medical professionals rely on the behavior of gases in several applications. For instance, respiratory therapies involve understanding the exchange of gases in the lungs, and anesthesiology utilizes gas behavior to administer and monitor the effects of anesthesia during surgeries.

Weather Forecasting: The study of gas behavior is vital in meteorology and weather forecasting. Understanding how gases behave in the atmosphere helps predict weather patterns, including temperature changes, wind patterns, and the formation of clouds and precipitation.

Environmental Studies: Gases play a significant role in environmental studies. Knowledge of gas behavior aids in studying air pollution, atmospheric chemistry, and climate change. It helps scientists model and understand the movement and distribution of pollutants in the atmosphere.

Aerospace Engineering: Understanding gas behavior is crucial in aerospace engineering for designing aircraft, rockets, and spacecraft. Factors like air resistance, aerodynamics, and propulsion systems rely on gas behavior principles for efficient and safe flight.

These are just a few examples of how understanding gas behavior has practical applications in various fields and contributes to technological advancements and our overall understanding of the natural world.

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Answer:

Copper Copper Copper Copper

The partial pressures of nitrogen, oxygen, and neon in the mixture are approximately 366.13 torr, 312.55 torr, and 214.32 torr, respectively.

To find the partial pressures of the three gases, we can use Dalton's Law of Partial Pressures, which states that the total pressure exerted by a mixture of gases is equal to the sum of the partial pressures of each gas.

Given:

Total pressure = 893 torr

Mole fractions: nitrogen (N2) = 0.41, oxygen (O2) = 0.35, neon (Ne) = 0.24

To calculate the partial pressures, we can multiply the total pressure by the mole fraction of each gas:

Partial pressure of nitrogen (P(N2)) = Mole fraction of nitrogen (X(N2)) * Total pressure

P(N2) = 0.41 * 893 torr

Partial pressure of oxygen (P(O2)) = Mole fraction of oxygen (X(O2)) * Total pressure

P(O2) = 0.35 * 893 torr

Partial pressure of neon (P(Ne)) = Mole fraction of neon (X(Ne)) * Total pressure

P(Ne) = 0.24 * 893 torr

Calculating the values:

P(N2) = 0.41 * 893 torr = 366.13 torr

P(O2) = 0.35 * 893 torr = 312.55 torr

P(Ne) = 0.24 * 893 torr = 214.32 torr

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The action that would most likely reduce water pollution is A) initiating efforts to remove trash from the ocean.

Trash and plastic waste in the ocean pose a significant threat to marine life and ecosystems. Initiating efforts to remove trash from the ocean can help mitigate this problem and reduce water pollution. When plastic waste accumulates in oceans, it can break down into microplastics, which are ingested by marine organisms and can enter the food chain, causing harm to both marine life and humans.

By actively removing trash from the ocean, we can prevent it from further degrading and releasing pollutants. This can help protect marine habitats, reduce the risk of entanglement or ingestion by marine animals, and preserve the overall health of ocean ecosystems.

On the other hand, options B, C, and D would likely contribute to increased water pollution:

B) Eliminating laws and regulations on industrial waste would remove important safeguards and oversight, potentially allowing industries to discharge pollutants directly into water bodies, leading to increased water pollution.

C) Promoting the use of oil and gas can lead to oil spills, leakage, and contamination of water sources. This can have severe consequences for aquatic ecosystems and human health.

D) Removing the ban on chlorofluorocarbons (CFCs) would result in the release of these harmful substances, which deplete the ozone layer and can contaminate water sources when disposed of improperly.

In conclusion, initiating efforts to remove trash from the ocean is the most effective action to reduce water pollution, while the other options would likely exacerbate the problem. Therefore, Option A is correct.

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The freezing point of the solution is 6.213 °C. This is less than the freezing point of ether.

Freezing point of the solution can be determined by using the concept of freezing point depression, which is a colligative property.

The freezing point depression is given by the equation:

ΔTf = Kf * m

Where:

ΔTf = Freezing point depression

Kf = Cryoscopic constant (a constant specific to the solvent)

m = Molality of the solution

First, we need to calculate the molality (m) of the solution. Molality is defined as the number of moles of solute per kilogram of solvent.

We have 1.727 mol of solute and 483 g of ether (solvent).

First, we convert the mass of ether to kilograms:

Mass of ether = 483 g = 483/1000 kg = 0.483 kg

Now we can calculate the molality:

m = moles of solute / mass of solvent (in kg)

m = 1.727 / 0.483 ≈ 3.575 mol/kg

Next, we need the cryoscopic constant (Kf) for ether. The cryoscopic constant is specific to the solvent. For ether, the cryoscopic constant is approximately 1.74 °C kg/mol.

Finally, we can calculate the freezing point depression (ΔTf):

ΔTf = Kf * m

ΔTf = 1.74 °C kg/mol * 3.575 mol/kg

= 6.213 °C

Therefore, the freezing point of the solution is decreased by approximately 6.213 °C compared to the freezing point of pure ether.

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Answer:

Explanation:

h 7

The statement that correctly describes residual radiation from the patient's body after this type of treatment is the patient's body will be radioactive for a very short period after the treatment.

The correct option is A.

Residual radiation refers to the lingering radiation that remains after a nuclear or radioactive event has occurred.

It is the radiation that persists in the environment or materials even after the primary source of radiation has been removed or decayed.

Residual radiation can arise from various sources, including nuclear accidents, radioactive waste disposal, and from radiotherapy.

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PART A: Approximately 2,887.23 grams of hydrochloric acid would be required to produce 80 grams of hydrogen gas.

PART B: Approximately 7.65 x 10^24 water molecules would be produced.

PART A: To determine the grams of hydrochloric acid required to produce 80g of hydrogen gas, we need to consider the balanced chemical equation for the reaction between aluminum (Al) and hydrochloric acid (HCl):

2Al + 6HCl → 2AlCl₃ + 3H₂

From the equation, we can see that for every 3 moles of hydrogen gas produced, 6 moles of hydrochloric acid are required. To calculate the grams of hydrochloric acid needed, we need to use the molar mass of HCl, which is approximately 36.46 g/mol. First, we convert the grams of hydrogen gas to moles using the molar mass of hydrogen (H₂), which is approximately 2.02 g/mol: 80 g H₂ * (1 mol H₂ / 2.02 g H₂) = 39.60 mol H₂

Since the mole ratio of HCl to H₂ is 6:3, we multiply the moles of hydrogen by the ratio: 39.60 mol H₂ * (6 mol HCl / 3 mol H₂) = 79.20 mol HCl

Finally, we convert the moles of HCl to grams using the molar mass of HCl: 79.20 mol HCl * (36.46 g HCl / 1 mol HCl) ≈ 2,887.23 g HCl

Therefore, approximately 2,887.23 grams of hydrochloric acid would be required to produce 80 grams of hydrogen gas.

PART B: To determine the number of water molecules produced in the reaction of 741 grams of magnesium hydroxide (Mg(OH)₂), we need to consider the balanced chemical equation for the reaction:

Mg(OH)₂ → MgO + H₂O

From the equation, we can see that for every one mole of magnesium hydroxide, one mole of water is produced. To calculate the moles of magnesium hydroxide, we divide the given grams by the molar mass of Mg(OH)₂: 741 g Mg(OH)₂ * (1 mol Mg(OH)₂ / molar mass of Mg(OH)₂) = X mol Mg(OH)₂

The molar mass of Mg(OH)₂ can be calculated by adding the atomic masses of magnesium (Mg), oxygen (O), and hydrogen (H) together:

Mg: 24.31 g/mol

O: 16.00 g/mol

H: 1.01 g/mol

Molar mass of Mg(OH)₂ = (24.31 g/mol) + 2(16.00 g/mol) + 2(1.01 g/mol) = 58.33 g/mol

Substituting the values: 741 g Mg(OH)₂ * (1 mol Mg(OH)₂ / 58.33 g Mg(OH)₂) ≈ 12.70 mol Mg(OH)₂

Since the mole ratio of water to Mg(OH)₂ is 1:1, the number of water molecules produced is the same as the moles of Mg(OH)₂: 12.70 mol H₂O

To calculate the number of water molecules, we multiply the moles of H₂O by Avogadro's number, which is approximately 6.022 x 10^23 molecules/mol: 12.70 mol H₂O * (6.022 x 10^23 molecules/mol) ≈ 7.65 x 10^24 molecules of water .Therefore, approximately 7.65 x 10^24 water molecules would be produced.

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The type of energy transfer between the frozen ice cream and the tongue is convection. The frozen ice cream will gradually melt because of the temperature difference.

Convection is the method of heat transfer that involves the transmission of heat in a fluid (liquid or gas) by the circulation of currents.

According to this question, a spoonful of frozen ice cream is placed on the tongue. The heat transfer will occur between the tongue with warmer temperature and the ice cream with colder temperature.

The heat energy will flow from the tongue to the ice cream, hence, making it defreeze or melt with time. This type of heat transfer is convection because it involves a fluid.

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The volume of oxygen will be 0.404 dm³ at 2.02 x 10³ Nm² and 400 K.

To solve this problem, we can use the ideal gas law equation, which states:

PV = nRT

Where:

P = pressure

V = volume

n = number of moles

R = ideal gas constant

T = temperature

Since we are dealing with the same sample of oxygen, the number of moles (n) and the ideal gas constant (R) will remain constant. Therefore, we can rewrite the equation as:

P₁V₁ / T₁ = P₂V₂ / T₂

Where the subscripts 1 and 2 represent the initial and final conditions, respectively.

Given:

V₁ = 1 dm³ = 1 L = 0.001 m³

T₁ = 500 K

P₁ = 1.01 x 10³ Nm²

P₂ = 2.02 x 10³ Nm²

T₂ = 400 K

Let's substitute the values into the equation and solve for V₂:

(1.01 x 10³ Nm²)(0.001 m³) / 500 K = (2.02 x 10³ Nm²)(V₂) / 400 K

(1.01 x 10³ Nm²)(0.001 m³)(400 K) = (2.02 x 10³ Nm²)(V₂)(500 K)

V₂ = [(1.01 x 10³ Nm²)(0.001 m³)(400 K)] / [(2.02 x 10³ Nm²)(500 K)]

Now let's calculate the value of V₂:

V₂ = (0.404 Nm)(0.001 m³) / (1.01 Nm)

V₂ = 0.000404 m³ = 0.404 dm³

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