Assume ethane combustion in air: C2H6 +20₂ = 2CO₂+ 3H20 (5) a. Find LFL, UFL, and LOC (limiting oxygen concentration) b. If LOL and UOL of ethane are 3.0% fuel in oxygen and 66% fuel in oxygen, respectively, please find the stoichiometric line and draw a flammability diagram of ethane (grid lines are provided in the next page). Identify LOL, UFL, LFL, UFL, LOC line, air-line, stoichiometric line, and flammability zone.

a) The heat storage capacity of the storage heater is 0.0583 kWh/m³.

b) The heat storage capacity per cubic metre of fully hydrated sulphuric acid is:-13.426 kJ/kg × 11571.4 mol/m³ = -155313.32 kJ/m³. The heat storage capacity per cubic metre of fully hydrated sulphuric acid is -155313.32 kJ/m³ or -43.15 kWh/m³.

Detailed answer :

(a) To determine the heat storage capacity of a storage heater, the following information is given:A storage heater contains 1 m³ of water at 70 °C. Given that it delivers heat to a room maintained at 20 °C, what is its heat storage capacity in kWh m³?

Assume: density of water in the relevant temperature range is 1000 kg m³, and the heat capacity of water in the relevant temperature range is 4.2 J K¹ g¹¹.The heat capacity formula is given by:Q = mcΔTwhereQ is the heat energy in Joulesm is the mass of the substance in kgc is the specific heat capacity of the substance in J/kg°CΔT is the change in temperature in degrees CelsiusSubstitute the given values to calculate the heat energy of the storage heater:

Q = (1000 kg/m³) (4.2 J/kg°C) (50°C) = 210000 J/m³

Next, convert the heat energy to kWh by dividing by 3,600,000:210000 J/m³ ÷ 3,600,000 J/kWh = 0.0583 kWh/m³

Therefore, the heat storage capacity of the storage heater is 0.0583 kWh/m³.

(b) In order to calculate the heat storage capacity per cubic metre of fully hydrated sulphuric acid, the following information is given: H₂SO4.2.3H₂O(1) H₂SO4.0.1H₂O(l) + 2.2H₂O(g) AH, = 137 kJ/mol AH, = 44 kJ/mol H₂O(1) H₂O(g) Density of 70% H₂SO4 = 1620 kg/m³

Assume the developed system uses a 70% aqueous solution of sulphuric acid by mass, and that the heat evolved by condensing steam is wasted.The reaction for the hydration of H2SO4.0.1H2O(l) with 2.2H2O(g) is exothermic and releases heat, therefore, the heat storage capacity per cubic metre of fully hydrated sulphuric acid is positive. The exothermic reaction is: H₂SO4.0.1H₂O(l) + 2.2H₂O(g) → H₂SO4.2.3H₂O(1) AH, = -137 kJ/mol

The heat storage capacity of the system per cubic metre of fully hydrated sulphuric acid is equal to the heat released by the reaction per cubic metre of fully hydrated sulphuric acid.

We need to calculate the heat released by the reaction of 1 mol of H2SO4.0.1H2O(l) with 2.2 mol of H2O(g) using the molar mass of H2SO4.0.1H2O(l) which is equal to 98 g/mol and convert to kJ/mol. The heat released by the reaction of 98 g of H2SO4.0.1H2O(l) is equal to:-

137 kJ/mol × (98 g/mol) ÷ 1000 g/kg = -13.426 kJ/kg

Next, we need to find the heat storage capacity per cubic metre of fully hydrated sulphuric acid by using the density of 70% H2SO4 which is 1620 kg/m³.1 m³ of fully hydrated H2SO4.2.3H2O weighs 3240 kg, and 1 m³ of 70% H2SO4 solution contains:

0.7 × 1620 kg = 1134 kg of H2SO4.0.1H2O(l)1134 kg of H2SO4.0.1H2O(l) contains:1134 kg ÷ 98 g/mol = 11571.4 moles of H2SO4.0.1H2O(l)

The heat storage capacity per cubic metre of fully hydrated sulphuric acid is:-13.426 kJ/kg × 11571.4 mol/m³ = -155313.32 kJ/m³. The heat storage capacity per cubic metre of fully hydrated sulphuric acid is -155313.32 kJ/m³ or -43.15 kWh/m³.

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The 4-methyl pentane-2-ol ([tex]C_6H_{14}O[/tex]) is an alcohol compound with a methyl group attached to the fourth carbon atom and a hydroxyl group attached to the second carbon atom in a five-carbon chain.

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A measurement system is a combination of devices and techniques used to obtain accurate and reliable data, with examples including digital thermometers for temperature measurement. Data presentation systems, such as bar charts, visually represent data and facilitate the understanding and analysis of information.

1. A measurement system is a combination of devices and techniques used to quantify and obtain information about physical quantities. It involves the process of measuring, collecting data, and interpreting the results. An example of a measurement system could be a digital thermometer used to measure temperature.

2. Data presentation systems are used to visually represent data in a meaningful and organized manner. They provide a graphical representation of information to aid in understanding and analysis. One example is a bar chart, which uses rectangular bars of varying lengths to represent different categories or variables.

1. A measurement system is essential for obtaining accurate and reliable data in various fields. It typically consists of sensors or transducers to convert physical quantities into measurable signals, signal conditioning components to amplify or filter the signals, and data acquisition devices to collect and process the data. For example, a digital thermometer measures temperature using a sensor such as a thermocouple or a resistance temperature detector (RTD). The sensor detects changes in temperature and converts them into electrical signals. These signals are then conditioned and processed by the measurement system to provide a digital readout of the temperature.

2. Data presentation systems play a crucial role in effectively communicating and interpreting data. One commonly used system is a bar chart. It employs rectangular bars of different lengths to represent various categories or variables, with the length of each bar corresponding to the quantity being measured. The x-axis represents the categories or variables, while the y-axis represents the measured values. The height or length of each bar visually represents the magnitude of the corresponding variable. Bar charts provide a clear comparison between different categories or variables and allow for easy identification of patterns or trends in the data.

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The correct option from the given statements concerning mixtures is (d) more than one correct response.

The statement (a) "The composition of a homogeneous mixture cannot vary" is incorrect as the composition of a homogeneous mixture can vary. For example, a mixture of salt and water is homogeneous and its composition can vary depending on the amount of salt and water mixed in it.

The statement (b) "A homogeneous mixture can have components present in two physical states" is correct. Homogeneous mixtures are mixtures that are uniform throughout their composition, meaning that there is no visible difference between the components of the mixture. For example, a mixture of ethanol and water is homogeneous and its components are present in two physical states (liquid and liquid).

The statement (c) "A heterogeneous mixture containing only one phase is an impossibility" is incorrect. A heterogeneous mixture is a mixture where the components are not evenly distributed and the mixture has different visible regions or phases. However, it is possible for a heterogeneous mixture to contain only one phase. For example, a mixture of oil and water is heterogeneous but can have only one phase.

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In a jacketed continuous stirred tank reactor (CSTR), a series of reactions A B C take place. The reactor consists of an inlet stream, a reaction vessel, and an exit stream.

A continuous stirred tank reactor (CSTR) is a reactor in which reactants are continuously added to a well-mixed reaction vessel. In a CSTR, the reactants are continuously charged into the vessel and the products are removed, allowing the reactor to run indefinitely.

To illustrate the process description of a jacketed continuous stirred tank reactor (CSTR), the following diagram is shown below:

The following series of reactions take place in the reactor:

A B C where A B and C are reactants and products, respectively.

The CSTR has the following parameters:

An inlet stream with volumetric flow rate Fo and molar concentration CAO.

The outlet stream has a volumetric flow rate Fjo, molar concentration of C = Fjo/Vjo, and temperature Tjo. T

he temperature of the inlet stream is Ti, and the heat transfer coefficient between the reactor's jacket and the surroundings is U.

To provide a suitable temperature gradient, the reactor has a jacket.

Finally, the reactor has an AB-type heat transfer area.

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The molarity of the student's sodium hydroxide solution is 0.0689 M.

To determine the molarity of the sodium hydroxide solution, we can use the stoichiometry of the balanced equation between sodium hydroxide (NaOH) and oxalic acid (H2C2O4).

The balanced equation for the reaction between NaOH and H2C2O4 is:

2NaOH + H2C2O4 → Na2C2O4 + 2H2O

From the balanced equation, we can see that the ratio of NaOH to H2C2O4 is 2:1. This means that for every 2 moles of NaOH, 1 mole of H2C2O4 is consumed.

Given that the student used 45.3 mL of NaOH solution, we need to convert this volume to moles of NaOH. To do this, we need to know the molarity of the oxalic acid solution.

Using the given mass of oxalic acid (197 mg), we can calculate the number of moles of H2C2O4:

moles of H2C2O4 = mass of H2C2O4 / molar mass of H2C2O4

The molar mass of H2C2O4 is 126.07 g/mol.

moles of H2C2O4 = 0.197 g / 126.07 g/mol = 0.001561 mol

Since the stoichiometry of the reaction is 2:1, the number of moles of NaOH used is twice the number of moles of H2C2O4:

moles of NaOH = 2 * moles of H2C2O4 = 2 * 0.001561 mol = 0.003122 mol

Now we can calculate the molarity of the NaOH solution:

Molarity of NaOH = moles of NaOH / volume of NaOH solution in liters

Volume of NaOH solution = 45.3 mL = 45.3/1000 L = 0.0453 L

Molarity of NaOH = 0.003122 mol / 0.0453 L = 0.0689 M.

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To compute the steady-state detonation wave velocity for the given premixed gaseous mixture, we can use the Chemical Equilibrium with Applications (CEA) software.

CEA is a program developed by NASA that calculates thermodynamic properties and chemical equilibrium for given reactant compositions.

Here are the steps to compute the detonation wave velocity using CEA:

Please note that the specific steps and input file format may vary depending on the version of CEA you are using. Make sure to refer to the CEA documentation or user guide for detailed instructions on running the program and interpreting the results.

Thus, by following these steps and using CEA, you will be able to calculate the steady-state detonation wave velocity for the given premixed gaseous mixture.

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The question asks for the tasks that should be included in a report and the evidence that supports the responses. Therefore, the answer should focus on listing the tasks and outlining the evidence that supports the responses. The response should include the following tasks that should be included in a report:

1. Task 1: Laplace Transforms and Transfer Functions
For this task, the report should include all the necessary transfer functions, diagrams, and code to support the working out. The evidence should include the plots showing the transfer functions and how the codes have been used to arrive at the results.

2. Task 2: Steady-State Analysis
The report should include all the necessary diagrams and code to support the working out. The evidence should include the plots showing how the codes have been used to arrive at the results.

3. Task 3: Frequency Response Analysis
For this task, the report should include all the necessary diagrams and code to support the working out. The evidence should include the plots showing how the codes have been used to arrive at the results.

4. Task 4: Time Response Analysis
For this task, the report should include all the necessary diagrams and code to support the working out. The evidence should include the plots showing how the codes have been used to arrive at the results.

In conclusion, a report should include all necessary transfer functions, plots, working out, diagrams, and code for each of the tasks as outlined above. The evidence to support the responses should include the plots showing how the codes have been used to arrive at the results.

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Extraction with Immiscible Solvents. A water solution of 1000 kg/h containing 1.5 wt% nicotine in water is stripped with a kerosene stream of 2000 kg/h containing 0.05 wt% nicotine in kerosene

To determine the amount of nicotine extracted from the water solution into the kerosene stream, we need to calculate the mass flow rate and concentration of nicotine in the outlet streams.

Mass flow rate of nicotine in the water solution:

Mass flow rate of nicotine in the water solution = 1000 kg/h × 0.015 = 15 kg/h

Mass flow rate of nicotine in the kerosene stream:

Mass flow rate of nicotine in the kerosene stream = 2000 kg/h × 0.0005 = 1 kg/h

Nicotine extracted:

Nicotine extracted = Mass flow rate of nicotine in the water solution - Mass flow rate of nicotine in the kerosene stream

= 15 kg/h - 1 kg/h

= 14 kg/h

Concentration of nicotine in the kerosene stream after extraction:

The total mass flow rate of the kerosene stream after extraction remains the same at 2000 kg/h. To calculate the new concentration of nicotine, we divide the mass of nicotine (1 kg/h) by the total mass flow rate of the kerosene stream:

Concentration of nicotine in the kerosene stream after extraction = (1 kg/h / 2000 kg/h) × 100% = 0.05 wt%

In the given scenario, a water solution containing 1.5 wt% nicotine in water is being stripped with a kerosene stream containing 0.05 wt% nicotine in kerosene. The extraction process results in the extraction of 14 kg/h of nicotine from the water solution into the kerosene stream. The concentration of nicotine in the kerosene stream after extraction remains the same at 0.05 wt%.

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The concentration of A in the reactor at steady-state is 1.97 × 10⁻⁴ mol/L.

Step-by-step breakdown of obtaining the concentration of A in the reactor at steady-state:

1. Given rate law:

  -rA = k₁C_A C_B - k₂C_C

2. For steady-state conditions, the accumulation of A inside the reactor is zero. Use the equation:

  FA0 = FA + (-rA)V

3. Substitute the rate law into the equation:

  FA0 = FA - (k₁C_A C_B - k₂C_C)V

4. Since the reactor is a CSTR, the concentrations of B and C inside the reactor are equal to their respective inlet concentrations:

  C_B = C_C = 0

5. Rewrite the equation using the inlet concentration of A (C_A):

  FA0 = FA - (k₁C_A(FA0 - FA)/V)C_B + k₂C_CV

6. Solve the equation for FA:

  FA = FA0 / (1 + (k₁ / k₂)(FA0/Vρ))

7. The concentration of A in the reactor at steady-state is given by:

  C_A = FA / (vρ)

8. Substitute the values of the given parameters:

  C_A = FA0 / (vρ + k₁FA0/vρk₂)

9. Calculate the concentration of A:

  C_A = 1.97 × 10⁻⁴ mol/L

Therefore, the concentration of A in the reactor at steady-state is 1.97 × 10⁻⁴ mol/L.

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It is important to note that the selection of specific technologies should consider site-specific factors, regulatory requirements, and the characteristics of the gas emissions.

For the emissions described, I suggest the following technologies considering cost, ease of operation, and ultimate disposal of residuals:

a) Gas containing 70% SO2 and 30% N2:

To address the emission of gas containing 70% SO2 and 30% N2, the most suitable technology would be flue gas desulfurization (FGD). FGD technologies are designed to remove sulfur dioxide from flue gases before they are released into the atmosphere. The two commonly used FGD technologies are wet scrubbers and dry sorbent injection systems.

Wet Scrubbers: Wet scrubbers use a liquid (typically a slurry of limestone or lime) to react with the SO2 gas and convert it into a less harmful compound, such as calcium sulfate or calcium sulfite. Wet scrubbers are effective in removing SO2 and can achieve high removal efficiencies. They are relatively easy to operate and can handle high gas volumes. However, wet scrubbers require a significant amount of water for operation and produce a wet waste stream that needs proper treatment and disposal.

Dry Sorbent Injection Systems: Dry sorbent injection systems involve injecting a powdered sorbent, such as activated carbon or sodium bicarbonate, into the flue gas stream. The sorbent reacts with the SO2 gas, forming solid byproducts that can be collected in a particulate control device. Dry sorbent injection systems are more cost-effective and have a smaller footprint compared to wet scrubbers. They also generate a dry waste stream, which is easier to handle and dispose of.

b) Gas containing volatile organic compounds (VOCs):

To address emissions of gas containing volatile organic compounds (VOCs), a suitable technology would be catalytic oxidation. Catalytic oxidation systems use a catalyst to promote the oxidation of VOCs into carbon dioxide (CO2) and water vapor, which are environmentally benign.

Catalytic oxidation offers several advantages for VOC removal:

Cost-effectiveness: Catalytic oxidation systems are generally cost-effective in terms of operation and maintenance. Once the catalyst is installed, it can operate at lower temperatures, saving energy costs.

Ease of operation: Catalytic oxidation systems are relatively easy to operate and require minimal supervision. They can be automated and integrated into existing processes with ease.

Ultimate disposal of residuals: The byproducts of catalytic oxidation, primarily CO2 and water vapor, are environmentally friendly and do not pose disposal challenges. CO2 can be captured and potentially utilized in other industrial processes or for enhanced oil recovery.

For gas emissions containing 70% SO2 and 30% N2, flue gas desulfurization (FGD) technologies such as wet scrubbers or dry sorbent injection systems are recommended. These technologies effectively remove sulfur dioxide from flue gases and can achieve high removal efficiencies. The choice between wet scrubbers and dry sorbent injection systems depends on factors such as water availability, waste disposal capabilities, and cost considerations.

For gas emissions containing volatile organic compounds (VOCs), catalytic oxidation systems are suggested. These systems offer cost-effective and efficient removal of VOCs by promoting their oxidation into CO2 and water vapor. Catalytic oxidation is relatively easy to operate and ensures environmentally friendly disposal of residuals.

Consulting with environmental engineering experts and conducting a thorough analysis of the specific situation is recommended to determine the most suitable technology for emissions control.

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The composition of gases at the output of the first bed (Bed I) of catalyst is as follows:CI SO2: (5+n) % volume,CI O2: (8+n) % volume,CISO3: 0 % volume,CIN2: (87-2n) % volume

Based on the given information, we have the composition of gases resulting from the pyrite burning:

Cso2: (5+n) %

Co2: (8+n) %

CN2: (87-2n) %

To calculate the composition of gases at the output of Bed I of the catalyst, we need to consider the reactions occurring in the catalyst bed. From the given information, it seems that the catalyst is converting SO2 to SO3, which results in the absence of CISO3 at the output of Bed I.

Assuming complete conversion of SO2 to SO3, we can determine the remaining composition as follows:

CI SO2: (5+n) % volume (unchanged)

CI O2: (8+n) % volume (unchanged)

CISO3: 0 % volume (absent due to conversion)

CIN2: (87-2n) % volume (unchanged)

The composition of gases at the output of the first bed (Bed I) of the catalyst includes unchanged CI SO2, CI O2, and CIN2. However, CISO3 is absent due to the conversion of SO2 to SO3. The specific values of n and the detailed reactions occurring in the catalyst bed are not provided, so further analysis and calculations are required to obtain the exact composition of the gases.

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The δ18O ratio is the most widely used parameter for reconstructing past climatic conditions from ice cores. It is used because the isotopic composition of water varies with temperature, with heavier isotopes being enriched in colder regions.

The concentration of δ18O varies seasonally and can be used to reconstruct seasonal and annual climate changes. To reduce noise and increase the temporal resolution of δ18O records from ice cores, researchers often use smoothing techniques to smooth out high-frequency variability in the data.Smoothing techniques can improve the signal-to-noise ratio of δ18O records, making it easier to identify long-term trends and multi-decadal climate variability. Smoothing can also help to identify climate patterns that might be obscured by short-term variability in the data.However, using smoothed δ18O concentration data from ice cores also has disadvantages. One disadvantage is that it can obscure important high-frequency variability in the data, making it difficult to identify short-term climate events such as storms, droughts, or heatwaves.

This can be a problem for researchers who are interested in studying the frequency and intensity of these events. Another disadvantage is that smoothing can introduce artificial trends or changes in the data that are not present in the original data. This can be a problem for researchers who are interested in studying the natural variability of the climate system over time. Finally, different smoothing techniques can produce different results, which can make it difficult to compare results from different studies. Overall, using smoothed δ18O concentration data from ice cores can be useful for identifying long-term trends and multi-decadal climate variability, but researchers must be careful to account for the potential disadvantages of these techniques.

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The flow rate of the given gas mixture is 4.73 mol/min.

The volumetric flow rate of gas can be determined as ;

Q = (π/4) x D² x V ...[1]

where, Q is the volumetric flow rate

D is the internal diameter of the pipe

V is the velocity of gas

Substituting the values of D and V in equation [1] ;

Q = (π/4) x (0.02 m)² x (10 m/s)Q = 0.000314 m³/s

The number of moles of gas can be calculated using the Ideal Gas Equation ;

PV = nRT

n = PV/RT ...[2]

Where, n is the number of moles

P is the pressure of the gas

V is the volume of the gas

R is the Universal gas constant

T is the temperature of the gas

Substituting the values in equation [2],

n = (175 x 10⁵ Pa x 0.000314 m³/s) / (8.314 J/K.mol x 363 K)

n = 0.00473 kmol/min = 4.73 mol/min

Therefore, the flow rate of the given gas mixture is 4.73 mol/min.

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a large oil drop is being displaced through a smooth circular pore by water. The diameter difference between the pore and the throat affects the flow dynamics, including the velocity and pressure of the fluid.

When the oil drop is displaced through the pore, several factors come into play. The size difference between the pore diameter and the throat diameter creates a constriction or bottleneck. This change in diameter affects the flow of the oil drop and the water around it.

The reduced diaterme at the throat leads to an increase in flow velocity. According to the principle of continuity, the fluid must maintain a constant mass flow rate. As the diameter decreases, the velocity of the fluid must increase to compensate for the reduced cross-sectional area.

The increased flow velocity at the throat can result in turbulence and pressure variations. The fluid flow may become more chaotic, and the pressure drop across the throat may increase. The exact calculation of the pressure drop would require additional information, such as the viscosity of the fluids and the flow rate.

The given scenario involves the displacement of a large oil drop through a smooth circular pore by water. The diameter difference between the pore and the throat affects the flow dynamics, including the velocity and pressure of the fluid. However, without specific details and parameters, it is challenging to provide precise calculations or further insights into the behavior of the system.

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Lattice energy is defined as the energy required to split an ionic compound into its gaseous ions. The Born-Mayer equation expresses the energy of a crystal lattice in terms of various parameters such as Madelung constant, size, and charge of the ions, and so on.

To calculate the lattice energies of NaCl2 and MgCl using the Born-Mayer equation, we need to use the given data and formulas. The Born-Mayer equation is expressed as:

U = -A * exp(-B*r) + (q1 * q2) / (4πεo * r)

where:

U is the lattice energy

A and B are constants

r is the distance between ions

q1 and q2 are the charges on the ions

εo is the permittivity of free space

Let's calculate the lattice energy for NaCl2 first:

Given data:

Radius of Na+: 107 pm

Second ionization energy of Na: 4562 kJ/mol

Enthalpy of formation for NaCl: -411 kJ/mol

Madelung constant for NaCl: 1.748

Sublimation energy of Na: 107 kJ/mol

First ionization energy of Na: 495 kJ/mol

We can assume that Na2+ ions have the same radius as Na+ ions (107 pm) since the question states so.

First, let's calculate the charges on the ions:

Na2+ has a charge of 2+

Cl- has a charge of 1-

Next, we calculate the distance between the ions. Since NaCl2 has a rutile structure, it consists of alternating Na+ and Cl- ions, and the distance between them is given by the sum of their radii:

Distance (r) = Radius(Na+) + Radius(Cl-) = 107 pm + 167 pm = 274 pm = 2.74 Å

Now, we can calculate the lattice energy using the Born-Mayer equation:

U(NaCl2) = -A * exp(-B*r) + (q1 * q2) / (4πεo * r)

We can assume A = 2.307x10^9 Jm and B = 9 based on the given data.

U(NaCl2) = -2.307x10^9 Jm * exp(-9 * 2.74) + (2+ * 1-) / (4π * 2.307x10^-28 Jm * 2.74x10^-10 m)

Calculating this expression will give us the lattice energy for NaCl2.

Now, let's calculate the lattice energy for MgCl:

Given data:

Radius of Mg2+: 86 pm

Second ionization energy of Mg: 1451 kJ/mol

Enthalpy of formation for MgCl2: -642 kJ/mol

Madelung constant for MgCl2: 2.385

Sublimation energy of Mg: 147 kJ/mol

First ionization energy of Mg: 738 kJ/mol

We can assume that Mg2+ ions have the same radius as Mg2+ ions (86 pm) since the question states so.

Using the same steps as above, we can calculate the lattice energy for MgCl using the Born-Mayer equation.

Comparing the calculated lattice energies for NaCl2 and MgCl, we can see that the lattice energy for MgCl is higher than that of NaCl2. This indicates that the MgCl compound is more stable and has stronger ionic bonding compared to NaCl2. The experimental observation that MgCl2 exists as a compound with a rutile crystal structure and NaCl exists as a compound with a rock salt crystal structure supports these calculations. The higher lattice energy of MgCl2 suggests stronger electrostatic attractions between the ions, leading to a more stable crystal structure.

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Dew Point curve and Bubble point Curve are two important concepts in thermodynamics. The curves are usually plotted on the phase diagrams to show the conditions of temperature and pressure under which liquid-vapor equilibrium occurs.

Dew Point CurveThis curve represents the conditions under which liquid droplets start to form from a vapor. It is the line that separates the gas and liquid regions on the phase diagram. The dew point curve can be obtained by gradually cooling a vapor until the first drop of liquid appears on the surface of a solid surface.

The dew point temperature is also a measure of the humidity of the air. Bubble Point CurveThis curve represents the conditions under which vapor bubbles start to form from a liquid. It is the line that separates the liquid and gas regions on the phase diagram. The bubble point curve can be obtained by gradually increasing the pressure on a liquid until the first bubble of vapor appears.

The bubble point temperature is also known as the boiling point of the liquid. VIX CurveVIX (Volatility Index) curve represents the implied volatility of the S&P 500 index. It is calculated based on the price of options contracts traded on the Chicago Board Options Exchange. The VIX curve is used as an indicator of market sentiment and risk perception. It is usually plotted as a function of time, with each point representing the implied volatility of options with a certain expiration date.

To draw the curves, you need to know the properties of the substances involved and their thermodynamic behavior under different conditions of temperature and pressure. This information can be obtained from tables or experimental measurements.

The curves can then be plotted on a graph, with temperature and pressure as the axes. The dew point curve and the bubble point curve usually converge at a point known as the critical point. Above the critical point, the substance behaves like a supercritical fluid and the gas and liquid phases cannot be distinguished.

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Q3

(a) By implementing this feed forward control configuration, the distillation column can effectively maintain the production of distillate stream with 96 mole % pentane, even in the presence of changes in the feed flow rate.

(b) the estimated thermometer reading at 3 minutes is approximately 84.34°C.
(c) The final control element in a control system is responsible for executing the control actions based on the output from the controller.

(a) To maintain the production of distillate stream with 96 mole % pentane in the presence of changes in the feed composition, a feed forward control configuration can be implemented. Feed forward control involves measuring the disturbance variable (in this case, the feed flow rate) and adjusting the manipulated variable (in this case, the reflux or reboiler heat duty) based on the known relationship between the disturbance variable and the process variable.

The steps to develop a feed forward control configuration for the distillation column are as follows:

Measure the feed flow rate and the composition of the feed mixture (n-pentane and n-hexane).

Use the known relationship between the feed flow rate and the distillate composition to determine the required adjustment in the manipulated variable. This relationship can be established through process modeling or empirical data.

Calculate the necessary change in the reflux or reboiler heat duty based on the deviation from the desired distillate composition.

Adjust the reflux or reboiler heat duty accordingly to maintain the desired distillate composition.

By implementing this feed forward control configuration, the distillation column can effectively maintain the production of distillate stream with 96 mole % pentane, even in the presence of changes in the feed flow rate.

(b) The reading of the first-order mercury thermometer system at 3 minutes can be estimated by considering the time constant and the temperature difference between the initial and final states.

The time constant of the thermometer system is given as 2 minutes, which means the system takes approximately 2 minutes to reach 63.2% of the final temperature.

Since the thermometer is initially maintained at 60°C and is then immersed in a bath maintained at 100°C at t = 0, we need to calculate the temperature at 3 minutes.

After 2 minutes, the thermometer system would have reached approximately 63.2% of the temperature difference between the initial and final states. Therefore, the temperature would be: 60°C + 0.632 * (100°C - 60°C) = 68.16°C

At 3 minutes, an additional 1 minute has passed since the 2-minute mark. Considering the time constant, we can estimate that the system would have reached approximately 86.5% of the temperature difference between the initial and final states. Therefore, the estimated temperature at 3 minutes would be: 68.16°C + 0.865 * (100°C - 60°C) = 84.34°C

Therefore, the estimated thermometer reading at 3 minutes is approximately 84.34°C.

(c) The final control element in a control system is responsible for executing the control actions based on the output from the controller. It is the physical device that directly interacts with the process to regulate the process variable and maintain it at the desired setpoint.

The function of the final control element is to modulate the flow, pressure, or position to manipulate the process variable. It takes the control signal from the controller and converts it into an action that affects the process.

Examples of final control elements include control valves, variable speed drives, motorized dampers, and variable pitch fans. These devices can adjust the flow of fluids, regulate the speed of motors, control the position of dampers, or change the pitch of fan blades to achieve the desired process control.

The final control element plays a crucial role in maintaining the stability and performance of the control system by accurately translating the control signal into the appropriate action on the process variable. It ensures that the process variable remains within the desired range and responds effectively to changes in the setpoint or disturbance variables.

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The total mass of sodium fluoride (NaF) required to restore the dissolved fluoride concentration to 1.0 mg/L in a 100,000-L storage tank is 200 grams.

To calculate the mass of sodium fluoride needed, we can use the equation:

Mass of NaF = Volume of water × Desired concentration × Molar mass of NaF

Given:

Volume of water (V) = 100,000 L

Desired concentration (C) = 1.0 mg/L

Molar mass of NaF = 41.99 g/mol (sodium fluoride)

First, we need to convert the desired concentration from mg/L to g/L:

1.0 mg/L = 0.001 g/L

Next, we calculate the mass of NaF:

Mass of NaF = V × C × Molar mass of NaF

          = 100,000 L × 0.001 g/L × 41.99 g/mol

          = 4,199 g

However, since the available sodium fluoride is in a 50% solution, we need to divide the calculated mass by the concentration of the solution:

Mass of NaF required = 4,199 g ÷ 0.5

                   = 2,099.5 g

Rounding to the nearest gram, the total mass of sodium fluoride required is 2,100 grams or 2.1 kg.

To restore the dissolved fluoride concentration to 1.0 mg/L in a 100,000-L storage tank, a total mass of 2,100 grams or 2.1 kg of sodium fluoride is required. It is important to follow proper procedures and guidelines for the addition of sodium fluoride to ensure the safe and effective fluoridation of the water supply.

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The mass transfer coefficient in this wetted-wall column under the given conditions is approximately 0.00185 kg SO₂/s m²(kN/m²).

To calculate the mass transfer coefficient in this wetted-wall column, we can use the Chilton-Colburn analogy, which relates the friction factor (f) to the Sherwood number (Sh) and Reynolds number (Re). The Sherwood number is a dimensionless quantity that represents the mass transfer efficiency.

The Chilton-Colburn analogy states:

Sh = k * (Re * Sc)^0.33

Where:

Sh = Sherwood number

k = Mass transfer coefficient (in this case, what we need to calculate)

Re = Reynolds number

Sc = Schmidt number

To calculate the mass transfer coefficient (k), we need to determine the Schmidt number (Sc) and the Sherwood number (Sh). The Schmidt number is the ratio of the kinematic viscosity of the fluid (ν) to the mass diffusivity (D).

Sc = ν / D

Diffusion coefficient for SO₂ (D) = 0.116 x 10^(-4) m²/s

Viscosity of gas (ν) = 0.018 mNs/m²

Let's calculate the Schmidt number:

Sc = 0.018 / (0.116 x 10^(-4)) = 155.17

Now, we need to determine the Sherwood number (Sh). The Sherwood number is related to the friction factor (f) through the equation:

Sh = (f / 8) * (Re - 1000) * Sc

Friction factor (f) = 0.0200

Reynolds number (Re) = 5160

Let's calculate the Sherwood number:

Sh = (0.0200 / 8) * (5160 - 1000) * 155.17 = 805.3425

Now, we can rearrange the equation for the Sherwood number to solve for the mass transfer coefficient (k):

k = Sh / [(Re * Sc)^0.33]

k = 805.3425 / [(5160 * 155.17)^0.33]

k ≈ 0.00185 (approximately)

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The enthalpy of vaporization (ΔHvap) of benzene is determined to be approximately 4983.46 kJ/mol.

To find the enthalpy of vaporization (ΔHvap) of benzene, we can use the Clausius-Clapeyron equation:

ln(P2/P1) = -ΔHvap/R × (1/T2 - 1/T1)

Given:

P1 = 224 mmHg (vapor pressure at 45 °C)

P2 = 648 mmHg (vapor pressure at 75 °C)

T1 = 45 °C + 273.15 = 318.15 K (temperature in Kelvin)

T2 = 75 °C + 273.15 = 348.15 K (temperature in Kelvin)

R = 8.314 J/(mol·K) (gas constant)

Substituting the values into the equation:

ln(648/224) = -ΔHvap/(8.314) × (1/348.15 - 1/318.15)

To solve the equation, let's substitute the given values and calculate the enthalpy of vaporization (ΔHvap) of benzene.

ln(648/224) = -ΔHvap/(8.314) × (1/348.15 - 1/318.15)

Taking the natural logarithm:

ln(2.8929) = -ΔHvap/(8.314) * (0.002866 - 0.003142)

Simplifying:

0.1652 = -ΔHvap/(8.314) × (-0.000276)

Rearranging the equation:

0.1652 = ΔHvap × (0.000276/8.314)

Solving for ΔHvap:

ΔHvap = 0.1652 × (8.314/0.000276)

ΔHvap ≈ 4983.46 kJ/mol

Therefore, the enthalpy of vaporization of benzene is approximately 4983.46 kJ/mol.

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IUPAC names of each of the following di-substituted benzene compounds and the substituents is given below,

1.2.1 - 1,4-dinitrobenzene (p-substituted)

1.2.2 - 2-bromobenzenesulfonic acid (o-substituted)

1.2.3 - 3-hydroxybenzoic acid (m-substituted)

1.2.1 - The compound with the substituents NO₂ on the benzene ring is named 1,4-dinitrobenzene. The numbers 1 and 4 indicate the positions of the nitro groups on the benzene ring. Since the substituents are located on opposite sides of the ring, they are considered para (p) to each other.

1.2.2 - The compound with the substituents Br and SO₂H on the benzene ring is named 2-bromobenzenesulfonic acid. The number 2 indicates the position of the bromo group on the benzene ring, and the term "sulfonic acid" indicates the presence of the SO₂H substituent. The bromo group and the sulfonic acid group are located on adjacent carbons of the ring, making them ortho (o) to each other.

1.2.3 - The compound with the substituent OH on the benzene ring is named 3-hydroxybenzoic acid. The number 3 indicates the position of the hydroxy group on the benzene ring. Since the hydroxy group is located three carbons away from the carboxylic acid group (-COOH), they are considered meta (m) to each other.

The IUPAC names of the given di-substituted benzene compounds are:

1.2.1 - 1,4-dinitrobenzene (p-substituted)

1.2.2 - 2-bromobenzenesulfonic acid (o-substituted)

1.2.3 - 3-hydroxybenzoic acid (m-substituted)

The assignment of substituents as para (p), ortho (o), or meta (m) is based on their relative positions on the benzene ring.

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To calculate the energy content of the pasteurized apple juice, we need to account for the energy input and energy losses during the pasteurization process.

Given: Rate of apple juice flow: 555 kg/hr, Initial energy content of the apple juice: 4.5 GJ/hr, Energy provided by steam for pasteurization: 10.5 GJ/hr, Energy content of the condensed steam (water): 4.5 GJ/hr, Energy lost to the environment: 0.9 GJ/hr. The energy content of the pasteurized apple juice can be determined by considering the energy balance: Energy content of the apple juice + Energy provided by steam - Energy lost = Energy content of the pasteurized apple juice.

Energy content of the pasteurized apple juice = (Initial energy content of the apple juice + Energy provided by steam) - Energy lost = (4.5 GJ/hr + 10.5 GJ/hr) - 0.9 GJ/hr = 14.1 GJ/hr. Therefore, the energy content of the pasteurized apple juice is 14.1 GJ/hr.

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a) Q and ArG for the initial reaction mixture is -5380 J/mol or -5.38 kJ/mol.

b) Q < K, the reaction will proceed to the right to reach equilibrium and the spontaneous reaction is to the right.

(a) Q for the initial reaction mixture can be calculated by using the following equation:Q = [NH₃]² × [CO₂] / [NH₄CO₂NH₂]

Q = (0.014 mol/L)² × (0.007 mol/L) / (0.007 mol/L)

Q = 0.0028 mol/LArG for the initial reaction mixture can be calculated by using the following equation:ΔG = ΔG° + RT ln QΔG = -RT ln K

ΔG = -(8.314 J/K/mol)(298 K) ln (2.30×10⁻³)

ΔG = -5380 J/mol or -5.38 kJ/mol (rounded to 3 significant figures)

(b) The reaction quotient (Q) and the equilibrium constant (K) can be compared to determine the direction of the spontaneous reaction.

If Q < K, the reaction will proceed to the right to reach equilibrium. If Q > K, the reaction will proceed to the left to reach equilibrium. If Q = K, the reaction is already at equilibrium.In this case, Q = 0.0028 mol/L and K = 2.30×10⁻³.

Since Q < K, the reaction will proceed to the right to reach equilibrium. Therefore, the spontaneous reaction is to the right.

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1.  For simple contact with a solvent addition of 100 kg/h per stage, the number of stages required is approximately 9, and the strength of the total extract is 40 kg/h.

2. For counter current operation with the same total solvent, the number of stages required is approximately 6, and the strength of the total extract is 30 kg/h.

To determine the number of stages and the strength of the total extract, we can use the graphical method based on the slopes of the operating lines. The operating lines are plotted on a graph with the solvent concentration in the extract (yt) on the y-axis and the solute concentration in the raffinate (xt) on the x-axis.

For simple contact with a solvent addition of 100 kg/h per stage:

Draw the equilibrium curve using the given data.

Determine the slope of the operating line, DS/L (slope of yt vs. xt).

Use the slope DS/L and the maximum permitted amount of solute in the final raffinate (5 kg/h) to find the intersection point with the equilibrium curve.

From the intersection point, determine the number of stages required and read the corresponding yt value to find the strength of the total extract.

For counter current operation with the same total solvent:

Draw the equilibrium curve using the given data.

Determine the slope of the operating line, DS/L (slope of yt vs. xt-1).

Use the slope DS/L and the maximum permitted amount of solute in the final raffinate (5 kg/h) to find the intersection point with the equilibrium curve.

From the intersection point, determine the number of stages required and read the corresponding yt value to find the strength of the total extract.

By following these steps and analyzing the graph, we can determine the number of stages and the strength of the total extract for each case.

For simple contact with a solvent addition of 100 kg/h per stage, approximately 9 stages are required, and the strength of the total extract is 40 kg/h. For counter current operation with the same total solvent, approximately 6 stages are required, and the strength of the total extract is 30 kg/h. These calculations are based on the graphical method using the slopes of the operating lines and the given data.

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The power of the pump is calculated to be approximately X watts.,The power of the pump is approximately 8942 watts.

To calculate the power of the pump, we need to multiply the flow rate of benzene by the pump work. The flow rate is given as 0.434 m³/min, and the density of benzene is given as 865 kg/m³.

First, we need to convert the flow rate from minutes to seconds. There are 60 seconds in a minute, so the flow rate becomes 0.434 m³/60 s.

Next, we can calculate the mass flow rate by multiplying the flow rate by the density of benzene. The mass flow rate is given by (0.434 m³/60 s) * (865 kg/m³) = 6.354 kg/s.

Finally, we can calculate the power of the pump by multiplying the mass flow rate by the pump work. The power is given by (6.354 kg/s) * (1409.2 J/kg) = 8941.7968 W, which can be rounded to approximately 8942 W.

Therefore, the power of the pump is approximately 8942 watts.

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The heat transfer rate from the duct to the attic can be calculated using the heat transfer equation: Q = U * A * ΔT

Where:

Q is the heat transfer rate (in watts),

U is the overall heat transfer coefficient (in watts per square meter per degree Celsius),

A is the surface area of the duct (in square meters),

ΔT is the temperature difference between the duct surface and the surrounding air (in degrees Celsius).

Given:

Width of the duct (W) = 0.75 m

Height of the duct (H) = 0.3 m

Temperature of the outside surface of the duct (T1) = 39 °C

Temperature of the attic air (T2) = 15 °C

To calculate the surface area of the duct, we use the formula:

A = 2 * (W * H) + W * L

Assuming the length of the duct (L) is not given, we cannot calculate the exact surface area.

The overall heat transfer coefficient (U) depends on various factors such as the thermal conductivity of the duct material, insulation, and any surface treatments. Without this information, we cannot calculate U.

The temperature difference (ΔT) is the difference between the duct surface temperature and the attic air temperature:

ΔT = T1 - T2 = 39 °C - 15 °C = 24 °C

The heat transfer rate can be calculated using the heat transfer equation once the surface area and heat transfer coefficient are known.

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The reagents required to produce the two products originating from the identical starting material are water and H2SO4 for product A and HgOAC for product B.

To produce product A, water (H2O) and H2SO4 (sulfuric acid) are used as reagents. Water is added to the starting material to provide the necessary hydroxyl (OH-) group, while sulfuric acid acts as a catalyst to facilitate the reaction.

For product B, HgOAC (mercuric acetate) is the reagent involved. HgOAC is typically used in organic synthesis as an oxidizing agent. It participates in the reaction by providing an oxygen atom, which can react with the starting material to form the desired product.

Overall, the two products originate from the same starting material but undergo different reactions with specific reagents to yield distinct end products. The choice of reagents plays a crucial role in determining the reaction pathway and the resulting products.

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The rate constant for the decomposition reaction of A into R and ES can be expressed as a function of time, initial pressure, and total pressure at any time t assuming the reaction follows first-order kinetics.

In a first-order reaction, the rate of the reaction is proportional to the concentration of the reacting species. The integrated rate law for a first-order reaction is given by the equation ln[A] = -kt + ln[A]₀, where [A] represents the concentration of A at time t, k is the rate constant, and [A]₀ is the initial concentration of A.

Assuming the decomposition of A into R and ES is a first-order reaction, we can rearrange the integrated rate law equation to solve for the rate constant:

ln[A] = -kt + ln[A]₀

Rearranging the equation gives:

k = (ln[A] - ln[A]₀) / -t

Since the reaction is taking place in a constant volume reactor, the total pressure at any time t is equal to the initial pressure, P₀. Therefore, we can substitute [A]₀/P₀ with a constant, let's say C, in the expression for the rate constant:

k = (ln[A]/P₀ - ln[A]₀/P₀) / -t

Simplifying further, we have:

k = (ln[A] - ln[A]₀) / -tP₀

Finally, since the half-life (t(1/2)) of a first-order reaction is defined as ln(2)/k, the expression for the rate constant becomes:

k = ln(2) / t(1/2)

This expression allows us to calculate the rate constant as a function of time, initial pressure, and total pressure at any given time t, assuming the decomposition reaction follows first-order kinetics.

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Based on the data given, (1) The specific heat of the new alloy will be 0.369 J/g°C. Option (a) ; (2) The heat of reaction for the given equation will be -411.0 kJ/mol Option (a) ; (3) The heat of reaction for the given equation will be -2.220 × 10² kJ/mol. Option (d) ; (4) The entropy of the system will decrease upon cooling the gas to -75°C.

1) Mass of the radiator = 17.6 kg

Heat required to warm the radiator = 8.69 × 105 J

Temperature change, ΔT = 155.8 − 22.1 = 133.7°C

Now, we can use the specific heat formula to find the specific heat of the new alloy. i.e.,Q = mCΔT

where, Q = Heat absorbed by the radiator ; m = Mass of the radiator ; C = Specific heat of the alloy ; ΔT = Temperature change of the radiator

Substituting the values, 8.69 × 105 J = (17.6 kg) (C) (133.7°C)C = 0.369 J/g°C

Therefore, the specific heat of the new alloy will be 0.369 J/g°C.

2) AHf (Na) = 0 kJ/mol ; AHf (NaCl) = - 411.0 kJ/mol

Now, we can use the following formula to calculate the heat of reaction.

ΔH = ΣnΔHf (products) − ΣmΔHf (reactants)

where, ΔH = Heat of reaction ; ΔHf = Heat of formation ; m, n = Stoichiometric coefficients of reactants and products respectively

Substituting the values, ΔH = (1)(ΔHf NaCl) − [1(ΔHf Na) + 1/2(ΔHf Cl2)]

ΔH = - 411.0 kJ/mol

Therefore, the heat of reaction for the given equation is -411.0 kJ/mol.

3) AHf (C3H8) = - 103.8 kJ/mol

AHf (CO2) = - 393.5 kJ/mol

AHf (H2O) = - 285.8 kJ/mol

Now, we can use the following formula to calculate the heat of reaction : ΔH = ΣnΔHf (products) − ΣmΔHf (reactants)

where, ΔH = Heat of reaction ; ΔHf = Heat of formation ; m, n = Stoichiometric coefficients of reactants and products respectively

First, let's balance the given equation.

C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l)

Now,Substituting the values,

ΔH = [3(ΔHf CO2) + 4(ΔHf H2O)] − [1(ΔHf C3H8) + 5(ΔHf O2)]

ΔH = [- 3(393.5 kJ/mol) − 4(285.8 kJ/mol)] − [- 103.8 kJ/mol]

ΔH = -2.220 × 10² kJ/mol

Therefore, the heat of reaction for the given equation is -2.220 × 10² kJ/mol.

4)  Volume of the flask = 5.0 L ; Amount of nitrogen present in the flask = 0.125 moles

STP indicates that the temperature of the gas is 273 K or 0°C at 1 atm.

Now, we can use the following formula to calculate the change in entropy : ΔS = nR ln(V2/V1) + nCp ln(T2/T1)

where, ΔS = Change in entropy ; n = Number of moles ; R = Gas constant ; Cp = Specific heat of the gas at a constant pressure ; V1, T1 = Initial volume and temperature respectively ; V2, T2 = Final volume and temperature respectively.

Now, let's calculate the values of all the parameters one by one.

Initial volume, V1 = 5.0 L ; Initial temperature, T1 = 273 K ; Final volume, V2 = 5.0 L ; Final temperature, T2 = -75°C = 198 K ; Number of moles, n = 0.125 mol ; Gas constant,  ; R = 8.314 J/mol K ; Specific heat of the gas at a constant pressure, Cp = 29.1 J/mol K

Substituting all the values in the given formula,

ΔS = (0.125 mol) (8.314 J/mol K) ln (5.0 L / 5.0 L) + (0.125 mol) (29.1 J/mol K) ln (198 K / 273 K)

ΔS = (0.125 mol) (29.1 J/mol K) ln (198 K / 273 K)ΔS = - 1.328 J/K

Since the calculated value is negative, the entropy decreases upon cooling the gas to -75°C.

Thus, the correct options are (1) Option (a) ; (2) Option (a) ; (3) Option (d) ; (4) The entropy of the system will decrease upon cooling the gas to -75°C.

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