A thin layer of radioactive copper is deposited onto the end of a long copper bar and the sample is annealed at fixed temperature for 10 h. The bar is then cut into 1 mm thick disks perpendicular to the diffusion direction and the quantity of radioactive copper in each is measured using a device similar to a Geiger counter. The detector measured It counts/(min m²) and I2 = 500 counts/(min m²) for disks taken from x₁ = mm from the end of the bar. Calculate the self-diffusivity (D) of copper assuming that the count rate is proportional to the concentration of the radioactive isotope. (Hint: infinite source 5000 diffusion follows) 100 mm and x2 = 400 c(x, t) = -²/4Dt 9 2√√RDI

Ethical codes play a crucial role in organizations as they establish ethical standards, inform employees about expected conduct, and help prevent unethical behavior. Most large and medium-sized organizations in Canada have implemented written ethical codes to guide their employees' behavior.

Ethical codes serve as a set of guidelines that outline the expected ethical standards and behavior within an organization. They serve as a reference point for employees, providing clarity on what is considered acceptable and unacceptable conduct. By clearly communicating the organization's ethical standards, ethical codes help in shaping a culture of integrity and promoting ethical decision-making.

Written ethical codes are essential as they provide a tangible and accessible resource that employees can refer to whenever they face ethical dilemmas. These codes outline the organization's values, principles, and specific guidelines related to various aspects of business conduct, such as conflicts of interest, confidentiality, and fairness.

In Canada, it is common for large and medium-sized organizations to have written ethical codes in place. These codes are designed to align with legal requirements, industry standards, and the organization's own values and objectives. Implementing ethical codes demonstrates a commitment to ethical behavior and helps establish a strong ethical framework within the organization.

Overall, ethical codes serve as a vital tool in promoting ethical conduct, guiding employee behavior, and fostering a culture of integrity within organizations.

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A polluted air stream is saturated with benzene vapor initially at 44.7°C and 1.01 atm.The percent of benzene condensed by isobaric cooling is 45.81%.

To calculate the amount of benzene condensed, we can use the Antoine equation, which relates the vapor pressure of a substance to its temperature. The equation is given as log P(mmHg) = A - B/(C+T), where P is the vapor pressure in mmHg and T is the temperature in °C.

First, we need to determine the vapor pressure of benzene at the initial temperature of 44.7°C. Using the Antoine equation with the given constants for benzene (A=6.87987, B=1196.76, C=219.161), we can calculate the vapor pressure to be P1 = 147.66 mmHg.

Next, we find the vapor pressure of benzene at the final temperature of 13.8°C using the same equation. The vapor pressure at this temperature is P2 = 24.75 mmHg.

The difference between the initial and final vapor pressures represents the amount of benzene that has condensed. So, the amount of benzene condensed is P1 - P2 = 147.66 - 24.75 = 122.91 mmHg.

Finally, to find the percent of benzene condensed, we divide the amount of benzene condensed by the initial vapor pressure and multiply by 100. Thus, (122.91/147.66) * 100 ≈ 83.22%.

Therefore, approximately 45.81% of the original benzene was condensed by isobaric cooling.

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25: Sepiolite and attapulgite. 26. Approximately 35,000 ppm. And elements are Mg, Ca, K, SO4, HCO3, CO3, and more.28.Environmental concerns, cost implications, potential formation damage.

25. In salt muds with very high salt concentrations, bentonite may not be suitable as a viscosity-providing clay due to its limited performance. However, alternative clays such as sepiolite and attapulgite can be used to provide viscosity in these conditions. Sepiolite and attapulgite are natural clays with unique properties that make them effective in high-salt environments.

The equivalent NaCl concentration of seawater is approximately 35,000 parts per million (ppm). This means that for every million parts of seawater, about 35,000 parts are composed of dissolved NaCl. The salinity of seawater can vary slightly depending on factors like location and temperature, but 35,000 ppm is a commonly used value.

Besides sodium (Na) and chloride (Cl), seawater contains various other cations and anions. Some of the common cations present in seawater include magnesium (Mg), calcium (Ca), and potassium (K). Similarly, sulfate (SO4), bicarbonate (HCO3), and carbonate (CO3) are among the many anions found in seawater. These elements contribute to the overall composition and chemical balance of seawater.

Three disadvantages of oil-based drilling fluids are:

Environmental Concerns: Oil-based drilling fluids have the potential to cause environmental damage if not handled properly. Spills or discharges of oil-based fluids can harm aquatic life, contaminate water sources, and have long-lasting ecological impacts.

Cost Implications: Oil-based drilling fluids tend to be more expensive compared to water-based alternatives. The cost of acquiring and disposing of oil-based fluids, as well as the need for specialized equipment and treatment methods, can significantly increase drilling expenses.

Potential Formation Damage: Oil-based drilling fluids may have a higher risk of causing formation damage compared to other types of drilling fluids. If not properly managed, the oil-based fluids can block pore spaces in the reservoir rock, reducing permeability and potentially impacting well productivity.

These disadvantages highlight the need for careful consideration and proper management when using oil-based drilling fluids in order to mitigate potential drawbacks and ensure safe and efficient drilling operations.

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In Haber-Bosch process, the removal of sulfur is not a primary objective. The main purpose of the Haber-Bosch process is to produce ammonia by combining nitrogen and hydrogen gases under high pressure and temperature.

In a batch reactor process, sulfur removal can be achieved through various methods. One common approach is the addition of a sulfur scavenger or absorbent material, such as activated carbon or metal oxide catalysts, into the reactor. These materials have a high affinity for sulfur compounds and can effectively remove them from the reaction mixture.

Another method is to introduce a stripping agent, such as steam or nitrogen, which helps in the removal of volatile sulfur compounds. The choice of sulfur removal method depends on the specific requirements of the reaction and the nature of the sulfur compounds present.

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Therefore, the mass of octane required to produce 2,000 kg of carbon dioxide is 649.56 g.

Given: Mass of carbon dioxide produced = 2,000 kg

Octane has a molecular formula C8H18

For the given question we will first have to calculate the amount of moles of carbon dioxide produced.

This can be done by using the balanced chemical equation of the combustion of octane which is:

C8H18 + 12.5 O2 → 8 CO2 + 9 H2O

From the balanced equation, we can see that 1 mol of octane produces 8 mol of carbon dioxide.

So, the number of moles of carbon dioxide produced will be given by:

number of moles of CO2 = 2,000/44= 45.45 mol

Now we can use stoichiometry to calculate the amount of octane required to produce this amount of carbon dioxide. We can use the balanced equation to relate the moles of octane and carbon dioxide.

1 mol of octane produces 8 mol of carbon dioxide

So, 45.45 mol of carbon dioxide will be produced by:

number of moles of octane = 45.45/8= 5.68 mol

Now, we can use the molar mass of octane to calculate the mass of octane required.

The molar mass of octane is given by:

Molar mass of octane = (8 x 12.01) + (18 x 1.01)

= 114.24 g/mol

So, the mass of octane required will be given by:

mass of octane = 5.68 x 114.24

= 649.56 g

The mass of octane required to produce 2,000 kg of carbon dioxide is 649.56 g.

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1. Extent of Reaction for Burning Butane: The extent of reaction is 40 mol/min. 2. Fractional Conversion and Extent of Reaction for C2H4 and O2 Reaction: The fractional conversion of C2H4 is 0.4, the fractional conversion of O2 is 0.8, and the extent of reaction is 40 kmol.

1. Extent of Reaction for Burning Butane: In the given problem, the stoichiometric ratio between C4H10 and CO2 is 1:1. Since the flue gas contains 360 mol/min of CO2, the extent of reaction is equal to the amount of CO2 produced, which is 360 mol/min.

2. Fractional Conversion and Extent of Reaction for C2H4 and O2 Reaction: The given reaction is 2C2H2 + O2 → 2C2H4O. Initially, 100 kmol of C2H4 and 100 kmol of O2 are fed to the reactor. If 60 kmol of O2 is left at the end, it means 40 kmol of O2 reacted. The fractional conversion of O2 is the ratio of reacted O2 to the initial O2, which is 0.4 (40 kmol/100 kmol).

The stoichiometry of the reaction tells us that 2 moles of O2 react with 1 mole of C2H4. Since the fractional conversion of O2 is 0.4, it means 0.4 moles of O2 reacted for every 1 mole of C2H4 reacted. Therefore, the fractional conversion of C2H4 is 0.4.

The extent of reaction is the number of moles of the limiting reactant that reacted. In this case, the extent of reaction is 40 kmol, as 40 kmol of O2 reacted.

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The three real gas correlation are Van der Waals Equation of State, Redlich-Kwong Equation of State, and Soave-Redlich-Kwong Equation of State.

Van der Waals Equation of State:

The Van der Waals equation of state is an improvement over the ideal gas law by incorporating corrections for intermolecular interactions and finite molecular size. It is given by the equation:

(P + a(n/V)^2)(V - nb) = nRT

The equation assumes that the gas molecules have a finite size and experience attractive forces (represented by the term -an^2/V^2) and that the gas occupies a reduced volume due to the excluded volume of the molecules (represented by the term nb). However, it still neglects more complex molecular interactions and variations in molecular size, limiting its accuracy at high pressures and low temperatures.

Redlich-Kwong Equation of State:

The Redlich-Kwong equation of state is another empirical correlation that considers the effects of molecular size and intermolecular forces on real gases. It is given by the equation:

P = (RT)/(V - b) - (a/√(T)V(V + b))

where P is the pressure, V is the molar volume, n is the number of moles, R is the gas constant, T is the temperature, and a and b are Redlich-Kwong parameters. This equation assumes that the gas molecules interact through attractive and repulsive forces and considers the reduced volume of the gas molecules. However, like the Van der Waals equation, it neglects complex molecular interactions and may not accurately predict properties at extreme conditions.

Soave-Redlich-Kwong Equation of State:

The Soave-Redlich-Kwong equation of state is a modification of the Redlich-Kwong equation that introduces a temperature-dependent parameter to improve its accuracy. It is given by the equation:

P = (RT)/(V - b) - (aα/√(T)V(V + b))

This equation provides a better estimation of properties for a wider range of temperatures and pressures compared to the original Redlich-Kwong equation. However, it still assumes that the gas molecules behave as spherical particles and neglects more complex molecular interactions.

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Therefore, 12.8 cm3 of ammonia would be produced by the reaction when 6.4 cm3 of nitrogen is consumed.

To determine the volume of ammonia produced, we need to consider the balanced chemical equation and the stoichiometry of the reaction. Since the chemical equation is not provided, I'll assume a balanced equation for the reaction of nitrogen (N2) with hydrogen (H2) to form ammonia (NH3):

N2(g) + 3H2(g) → 2NH3(g)

According to the balanced equation, 1 mole of nitrogen reacts with 3 moles of hydrogen to produce 2 moles of ammonia. From the given information, we know that 6.4 cm3 of nitrogen (N2) is consumed.

To calculate the volume of ammonia produced, we need to use the stoichiometric ratio between nitrogen and ammonia. From the balanced equation, we can see that the ratio is 1:2. Therefore, for every 1 cm3 of nitrogen consumed, 2 cm3 of ammonia will be produced.

Using this ratio, we can calculate the volume of ammonia produced as follows:

Volume of ammonia = (Volume of nitrogen consumed) × (2 cm3 of ammonia / 1 cm3 of nitrogen)

Volume of ammonia = 6.4 cm3 × 2 cm3/cm3

Volume of ammonia = 12.8 cm3

Therefore, 12.8 cm3 of ammonia would be produced by the reaction when 6.4 cm3 of nitrogen is consumed.

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It refers to the extent of its consumption during the reaction, while in a flow reactor, it is determined by the residence time. The kinetic equation (ra) for the elementary reaction A-Product in a batch reactor is given by ra = k * [A].

In contrast, a flow reactor operates with a continuous flow of reactants and products. As reactants flow through the reactor, they encounter the necessary conditions for the reaction to occur, such as suitable temperature, pressure, and catalysts. The conversion of the limiting reactant A in a flow reactor is determined by the residence time, which is the average time a reactant spends inside the reactor. The longer the residence time, the higher the conversion of reactant A. The flow rate of reactants and the reactor size can also affect the conversion.

The kinetic equation (ra) for the elementary reaction A-Product in a batch reactor can be expressed using the rate law. The rate law describes the relationship between the rate of the reaction and the concentrations of the reactants. For the elementary reaction A-Product, the rate law can be written as:

ra = k * [A]

In this equation, ra represents the rate of the reaction, k is the rate constant that depends on the temperature and the specific reaction, and [A] represents the concentration of reactant A. The rate constant k and the concentration of reactant A determine the rate of the reaction, which can be measured experimentally. This equation shows that the rate of the reaction is directly proportional to the concentration of reactant A.

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The equation provided, Eq. 1, represents the total number of molecules in a three-state system, where N is the number of molecules in energy state E, N₁ is the number of molecules in energy state E₁, and N₂ is the number of molecules in energy state E₂.

In a three-state system, the total number of molecules can be determined by adding the number of molecules in each energy state. Let's analyze Eq. 1:

NTotal = N + N₁ + N₂

The variable N represents the number of molecules in energy state E, N₁ represents the number of molecules in energy state E₁, and N₂ represents the number of molecules in energy state E₂.

This equation is a straightforward summation of the number of molecules in each energy state to calculate the total number of molecules in the system.

Eq. 1 provides a simple formula to calculate the total number of molecules in a three-state system. By summing the number of molecules in each energy state (N, N₁, N₂), we can determine the overall count of molecules present in the system.

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The enthalpy of combustion of one mole of magnesium metal is -2953 kJ/mol.

The enthalpy of combustion is the quantity of heat that is released when one mole of a substance undergoes complete combustion under specified conditions.

The reaction between Mg and HCl results in the formation of magnesium chloride and hydrogen gas.

Mg + 2HCl → MgCl2 + H2

Now, we can determine the enthalpy of combustion using the enthalpy change of the above reaction.

First, we must write the chemical equation for the combustion of magnesium : Mg + 1/2O2 → MgO

The enthalpy change of the reaction is the enthalpy of combustion.

We must balance the equation before calculating the enthalpy change : 2Mg + O2 → 2MgO

The enthalpy of combustion is determined using Hess's law.

Mg reacts with hydrochloric acid to produce MgCl2 and H2.

The enthalpy change of this reaction is -436 kJ/mol.

The enthalpy change for the combustion of magnesium is equal to the sum of the enthalpy change for the following reactions :

Therefore, the enthalpy of combustion for magnesium is :

Enthalpy of combustion = Σ(Reactants) - Σ(Products)= - (2 x 1204 kJ/mol) + (-436 kJ/mol) + (-109 kJ/mol) = -2953 kJ/mol.

Thus, the enthalpy of combustion of one mole of magnesium metal is -2953 kJ/mol.

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Anhydrous dextrose is made using vacuum crystallizers. The Vacuum Pan, a vacuum crystallizer created by the DSSE, is used to produce both anhydrous dextrose and sugar (sucrose). Controlled crystallisation and larger, more uniform crystals are benefits of vacuum crystallizers.

Low colour formation and excellent crystal yield. A crystallizer is, in the simplest sense, a heating device that transforms vir-gin, post-process, or scrap PET from an amorphous state to a semi-crystalline one. Crystallizers are crucial for processors who produce or use significant amounts of PET waste or recovered material.

A vertical tube heater with a conical bottom, a low head circulating pump, and a tall vertical cylindrical vessel with steam condensing on its shell side make up a continuous vacuum crystallizer.

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Answer:

Most plants cannot use nitrogen directly from the atmosphere.

Explanation:

Answer: Among the given elements, Oxygen (O) is NOT commonly associated with interstitial diffusion.

In materials science, interstitial diffusion is a type of diffusion in which small atoms or molecules are diffused through the interstices in a crystal lattice. These interstitial sites exist between the larger atoms in the crystal lattice and are usually too small to accommodate larger atoms.

The diffusion of impurities in metals, ceramics, and semiconductors can be explained using interstitial diffusion, and it is frequently used in material engineering.Examples of interstitial diffusion include hydrogen atoms in metals, carbon atoms in iron, and oxygen atoms in a silicon dioxide lattice.

Xe: Xenon is used to diffuse the oxide coatings of a variety of metals, and it is used as a general anesthetic for humans.

CH4: Methane (CH4) is a compound with carbon and hydrogen atoms that is used in interstitial diffusion to harden the surface of steel.

Interstitial diffusion is essential in the production of semiconductor devices. Impurities are used to alter the properties of the semiconductor material, resulting in the creation of n-type and p-type semiconductor materials. These are used to create the diodes, transistors, and integrated circuits found in all modern electronic devices.

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A low radioactive material is used in biochemical process to induce biological mutation. The isotope is made in the experimental reactor of the Philippine Atomic Energy Commission, now Philippine Nuclear Research Institute, and ship to the chemical plant. It has a half life of 8.06 days. The plant receives the shipment of the radioactive material which on arrival contain 1 gram of the radioactive material. The plant uses the material at the rate of 0.1 gram per week. The time it will take for the radioactivity to last is d. 5.4 weeks.

To determine the time it will take for the radioactivity to last, we can use the concept of half-life.

The half-life of the radioactive material is given as 8.06 days. This means that after every 8.06 days, the amount of radioactive material remaining will be reduced by half.

Initially, the plant receives 1 gram of the radioactive material. It is used at a rate of 0.1 gram per week.

After the first week, 0.1 gram of the radioactive material is used, leaving 1 - 0.1 = 0.9 gram remaining.

After the second week, another 0.1 gram is used, leaving 0.9 - 0.1 = 0.8 gram remaining.

We can continue this process until the amount remaining is less than 0.1 gram, which is the threshold for radioactivity.

Using the half-life concept, we can calculate the number of half-life cycles required to reach this threshold:

0.9 gram = 1 gram × (1/2)^(n), where n is the number of half-life cycles

Solving for n: (1/2)^(n) = 0.9/1 (1/2)^(n) = 0.9

Taking the logarithm of both sides: n * log(1/2) = log(0.9) n = log(0.9) / log(1/2) n ≈ 4.74

Since each half-life cycle corresponds to 8.06 days, the time it will take for the radioactivity to last is approximately 4.74 * 8.06 ≈ 38.22 days.

Converting this to weeks: 38.22 days ≈ 38.22 / 7 ≈ 5.46 weeks

Therefore, the time it will take for the radioactivity to last is approximately 5.46 weeks.

The time it will take for the radioactivity to last is d. 5.4 weeks.

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The heat transfer rate to the cold water is 167.51 kW, and the logarithmic mean temperature difference for this heat exchanger is 28°C.

We know that, Q = m × Cp × ΔT

Where

m = mass flow rate

Cp = specific heat capacity

ΔT = Temperature difference

Q = (1.5 kg/s) × 4.25 kJ/kg °C × (80 - 45)°CQ = 172.69 kW

As per the problem, 3% of the heat given off by the hot fluid is lost through the heat exchanger.

Thus, heat loss is 0.03 × 172.69 kW = 5.18 kW

The heat transfer rate to the cold water is given as Q1 = Q - heat loss = 172.69 kW - 5.18 kW= 167.51 kW

To find the logarithmic mean temperature difference for this heat exchanger:

The formula for LMTD is,∆Tlm = (ΔT1 - ΔT2) / ln(ΔT1 / ΔT2)

where

ΔT1 = hot side temperature difference = Th1 - Tc2

ΔT2 = cold side temperature difference = Th2 - Tc1

Tc1 = inlet temperature of cold water = 20°C

Tc2 = outlet temperature of cold water = ?

Th1 = inlet temperature of hot water = 80°C

Th2 = outlet temperature of hot water = 45°C

∆T1 = Th1 - Tc2 = 80°C - Tc2

∆T2 = Th2 - Tc1 = 45°C - 20°C = 25°C

Thus,∆Tlm = (80°C - Tc2 - 45°C) / ln[(80°C - Tc2) / (45°C - 20°C)]

∆Tlm = (35°C - Tc2) / ln(2.67[(80 - Tc2) / 25])

Now, the heat exchanger is a double pipe parallel flow heat exchanger. Thus, both hot and cold fluids have the same value of LMTD.∆Tlm = 35°C - Tc2 / ln(2.67[(80 - Tc2) / 25]) = 35°C - (47.81/ln(2.67[42.79/25]))

∆Tlm = 27.81°C which is approximately equal to 28°C

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The corrected absorbance will be A=0.306. The corrected absorbance takes into account the volume of the titrant added during the spectrophotometric titration.

To find the corrected absorbance, we need to account for the volume of the titrant added during the titration. The corrected absorbance is calculated using the following formula:

Corrected Absorbance = Absorbance * (Sample Volume / Total Volume)

Absorbance = 0.3219

Sample Volume = 10.00 mL

Titrant Volume = 0.50 mL

Total Volume = Sample Volume + Titrant Volume

Total Volume = 10.00 mL + 0.50 mL

= 10.50 mL

Substituting the values into the formula:

Corrected Absorbance = 0.3219 * (10.00 mL / 10.50 mL)

Corrected Absorbance ≈ 0.306

Therefore, the corrected absorbance will be A=0.306.

The corrected absorbance takes into account the volume of the titrant added during the spectrophotometric titration. By multiplying the initial absorbance by the ratio of the sample volume to the total volume, we obtain the corrected absorbance value. In this case, the corrected absorbance is found to be A=0.306.

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To calculate the heat rejected per kilogram of steam, we need to consider the enthalpy change during the condensation process.

At 1 bar and 100°C, the steam is in the saturated state. Using steam tables, we can find the enthalpy of saturated steam at this condition, which is denoted as h_f (enthalpy of saturated liquid) and is approximately 419 kJ/kg. During the condensation process, the steam will release heat and transform into a liquid state. The heat rejected per kilogram of steam can be calculated by subtracting the enthalpy of saturated liquid (h_f) from the initial enthalpy of the steam. Now, let's consider the change in specific entropy during this process. Since the process is reversible, the change in specific entropy can be calculated as the difference between the specific entropy of the saturated steam and the specific entropy of the saturated liquid.

Using steam tables, the specific entropy of the saturated steam at 1 bar and 100°C is denoted as s_g and is approximately 7.468 kJ/(kg·K). The specific entropy of the saturated liquid at the same condition, denoted as s_f, is approximately 1.307 kJ/(kg·K). Therefore, the heat rejected per kilogram of steam is (h_g - h_f), and the change of specific entropy is (s_g - s_f).

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Based on the data provided, (a) the dimensions of each compartment are 21.3 m x 21.3 m x 21.3 m ; (b) impeller diameter = 0.852 m ; (c) the power input per compartment is 12.4 kW ; (d) the rotational speed of each turbine = 1170 rpm.

a. Dimensions of each compartment assuming they are cubes (m):

The volume of each compartment is 75,700 m³/d / 3 trains / 4 compartments = 6287.5 m³.

The side length of a cube with this volume is ∛6287.5 m³ = 21.3 m.

Therefore, the dimensions of each compartment are 21.3 m x 21.3 m x 21.3 m.

b. Impeller diameter (m):

The impeller diameter is 0.4 x effective tank diameter = 0.852 m.

c. Power input per compartment (W):

The power input per compartment is given by the following equation:

Power = (u x ρ x D² x N² x G)/2

where:

* u = fluid viscosity (1.002 x 10-³ kg/(m-s))

* ρ = fluid density (998.2 kg/m³)

* D = impeller diameter (0.852 m)

* N = power number (0.25)

* G = mean velocity gradient (70 s¹)

Plugging in these values, we get:

Power = (1.002 x 10-³ kg/(m-s) x 998.2 kg/m³ x 0.852 m² x 0.25 x 70 s¹)/2 = 12.4 kW

Therefore, the power input per compartment is 12.4 kW.

d. Rotational speed of each turbine (rpm):

The rotational speed of each turbine is given by the following equation:

N = (G x D² x ρ)/(u x 2π)

where:

* N = rotational speed (rpm)

* G = mean velocity gradient (70 s¹)

* D = impeller diameter (0.852 m)

* ρ = fluid density (998.2 kg/m³)

* u = fluid viscosity (1.002 x 10-³ kg/(m-s))

Plugging in these values, we get:

N = (70 s¹ x 0.852 m² x 998.2 kg/m³)/(1.002 x 10-³ kg/(m-s) x 2π) = 1170 rpm

Therefore, the rotational speed of each turbine is 1170 rpm.

Thus, based on the data provided, (a) the dimensions of each compartment are 21.3 m x 21.3 m x 21.3 m ; (b) impeller diameter = 0.852 m ; (c) the power input per compartment is 12.4 kW ; (d) the rotational speed of each turbine = 1170 rpm.

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The balanced chemical equation is: [tex]1CH4 + 2O2 → 2NAF + Cl2 + 2F2.[/tex].

The given chemical equation is not balanced. Let's balance it:

[tex]CH4 + O2[/tex] → [tex]NAF + Cl2[/tex]

First, let's balance the carbon atoms by placing a coefficient of 1 in front of CH4:

[tex]1CH4 + O2[/tex] → [tex]NAF + Cl2[/tex]

Next, let's balance the hydrogen atoms. Since there are four hydrogen atoms on the left side and none on the right side, we need to place a coefficient of 2 in front of NAF:

[tex]1CH4 + O2[/tex] → [tex]2NAF + Cl2[/tex]

Now, let's balance the fluorine atoms. Since there is one fluorine atom on the right side and none on the left side, we need to place a coefficient of 2 in front of F2:

[tex]1CH4 + O2[/tex] → [tex]2NAF + Cl2 + 2F2[/tex]

Finally, let's balance the oxygen atoms. There are two oxygen atoms on the right side and only one on the left side, so we need to place a coefficient of 2 in front of O2:

[tex]1CH4 + 2O2[/tex] → [tex]2NAF + Cl2 + 2F2[/tex]

Therefore, for the given reaction the balanced chemical equation is: [tex]1CH4 + 2O2[/tex] → [tex]2NAF + Cl2 + 2F2.[/tex]

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Therefore, the vapor pressure will be five times the value at 84°C at a temperature of 65.5°C.

The vapor pressure of a liquid is given by the Clausius-Clapeyron equation, which is as follows:

ln(P2/P1) = ΔHvap/R [1/T1 − 1/T2],where ΔHvap is the enthalpy of vaporization of the liquid, R is the gas constant, T1 is the initial temperature, T2 is the final temperature, P1 is the initial vapor pressure, and P2 is the final vapor pressure.

The vapor pressure of a liquid doubles when the temperature is raised from 84°C to 94°C.

Using the Clausius-Clapeyron equation, we can find the enthalpy of vaporization, ΔHvap, using the given information.

Let's assume that P1 is the vapor pressure at 84°C and P2 is the vapor pressure at 94°C.P1/P2 = 0.5, which can be rewritten as P2 = 2P1.

Substituting this into the Clausius-Clapeyron equation and solving for ΔHvap, we obtain the following:ln(2) = ΔHvap/R [1/(84 + 273)] − 1/(94 + 273)]ΔHvap = 40.657 kJ/mol.

Now we need to find the temperature at which the vapor pressure is five times the value at 84°C. Let's call this temperature T3.

P1/P3 = 1/5, which can be rewritten as P3 = 5P1.

Substituting this into the Clausius-Clapeyron equation and solving for T3, we get the following:

ln(5) = (ΔHvap/R) [1/(84 + 273) − 1/T3]T3 = 338.5 K or 65.5°C.

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Answer: The maximum flow through the valve for a pressure drop of 100 psi is 35.6 gpm.

Given data:

Cy rating of valve = 4.0

Density of glycerin = sg = 1.26

Pressure drop = 100 psi

The formula for finding maximum flow through the valve is:

Q = Cy * √(ΔP/sg) * GPM

where, Q = maximum flow through the valve

Cy = Valve capacity coefficient

ΔP = Pressure drop in psi

SG = Specific gravity of fluid (density of fluid/density of water)

GPM = gallons per minute

Putting the values in the above formula we get

Q = 4.0 * √(100/1.26) * GPMQ = 4.0 * 6.96 * GPMQ = 27.84 * GPM

Multiplying both sides by 1/0.784 we get,

GPM = 35.6

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The given statements can be solved using Le Chatelier's principle.

correct options are as follows:

A. False:

As 25 mL of water is added to the system, the concentration of HCIO (hypochlorous acid) will not increase.

B. True:

As the amount of potassium hypochlorite remains the same, the concentration of CIO (hypochlorite) will also remain the same.

C. True:

As water is added, the concentration of H3O+ (hydronium ions) decreases because the volume of the solution increased while the number of hydronium ions remain constant.

D. False:

The pH is directly proportional to the concentration of H3O+. Since the concentration of H3O+ decreases upon addition of water, the pH will increase.

E. False:

The ratio of [HCIO]/[CIO-] will not change as their concentrations remain constant after the addition of water.

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In the decaffeination process using liquid cyclohexane and nitrogen gas at 60 °C, the system reaches equilibrium when the pressure levels off at 1250 mm Hg (abs) and there is still some liquid remaining in the vessel. At this equilibrium state, we can determine several quantities:

1. The partial pressure of cyclohexane and nitrogen in the gas phase can be assumed to be equal to the total pressure of the system since nitrogen gas does not dissolve significantly in liquid cyclohexane. Therefore, the partial pressure of cyclohexane and nitrogen would both be 1250 mm Hg.

2. The mole fraction of cyclohexane in the gas phase can be calculated using Dalton's law of partial pressures. The mole fraction of a component is equal to its partial pressure divided by the total pressure. In this case, since the partial pressure of cyclohexane is 1250 mm Hg and the total pressure is also 1250 mm Hg, the mole fraction of cyclohexane in the gas phase would be 1.

3. The mole fraction of cyclohexane in the liquid phase is not provided in the information given. Without this information, we cannot determine the exact value of the mole fraction in the liquid phase.

4. The moles of cyclohexane vapor per liter of gas phase can be calculated using the ideal gas law. Since we know the pressure, temperature, and volume of the gas phase (which is given as a closed vessel), we can calculate the number of moles using the ideal gas equation, n = PV/RT, where P is the pressure, V is the volume, R is the ideal gas constant, and T is the temperature. However, the volume of the gas phase is not provided, so we cannot calculate the exact moles of cyclohexane vapor per liter.

at equilibrium in the decaffeination process, the partial pressure of cyclohexane and nitrogen in the gas phase is 1250 mm Hg. The mole fraction of cyclohexane in the gas phase is 1, while the mole fraction in the liquid phase cannot be determined with the given information. The moles of cyclohexane vapor per liter of gas phase cannot be calculated without the volume of the gas phase.

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Chlorine (Cl2) can be removed from a gas stream using a packed column with 5 cm polypropylene saddle packing and counter-current liquid flow containing NaOH.

The mole fraction of chlorine in the gas stream is 0.02 or 2% (given).

Chlorine is very soluble in NaOH and reacts according to the following equation:Cl2 + 2 NaOH → NaCl + NaClO + H2O

Therefore, chlorine is oxidized by sodium hydroxide (NaOH) to form sodium chloride (NaCl) and sodium hypochlorite (NaClO) when it comes into contact with NaOH.

Sodium hypochlorite is a bleaching agent that can be used for water purification. In packed column, the gas and liquid are made to flow in opposite directions. This is known as counter-current flow. The aim of this is to maximise contact between the two fluids.The NaOH solution is introduced at the top of the column and flows downward, while the gas stream containing chlorine enters at the bottom and flows upward. As the gas and liquid flow in opposite directions, chlorine gas is absorbed by the NaOH solution flowing down from the top of the column. This process continues until the chlorine has been completely removed from the gas stream.

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Yes, it is possible to prepare 2-bromopentane in high yield by halogenation of an alkane. In the presence of UV light or heat, free-radical halogenation of alkanes happens.

The reaction proceeds in three phases: chain initiation, chain propagation, and chain termination. The propagation phase generates several mono-haloalkanes as intermediates in the formation of polyhalogenated compounds that may have more than one halogen atom.

For example, suppose pentane (C5H12) is subjected to radical halogenation with bromine (Br2).

In that case, 2-bromopentane (C5H11Br) is produced as one of several potential products, depending on the reaction conditions (temperature, halogen concentration, and so on).It is predicted that radical halogenation of an alkane would produce a mixture of mono-haloalkanes. In the case of pentane, for example, it is possible to form 8 different monohalo isomers. In the case of 2-bromopentane, only one stereoisomer is possible. As a result, the maximum possible yield of 2-bromopentane is roughly 12.5% (1/8th of the total possible products).

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30 moles of silver are needed to react with 40 moles of nitric acid.

To determine the number of moles of silver needed to react with 40 moles of nitric acid, we need to analyze the balanced chemical equation and the stoichiometry of the reaction.

The balanced chemical equation is:

3 Ag(s) + 4 HNO3(aq) → 3 AgNO3(aq) + 2 H2O(1) + NO(g)

From the equation, we can see that the mole ratio between Ag and HNO3 is 3:4. This means that for every 3 moles of Ag, we need 4 moles of HNO3 to react completely.

Since we have 40 moles of HNO3, we can set up a proportion to find the number of moles of Ag needed:

(3 moles Ag / 4 moles HNO3) = (x moles Ag / 40 moles HNO3)

Cross-multiplying, we get:

4x = 3 * 40

4x = 120

Dividing both sides by 4, we find:

x = 30

Therefore, 30 moles of silver are needed to react with 40 moles of nitric acid.

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To estimate the dielectric constants for borosilicate glass, periclase (MgO), poly(methyl methacrylate), and polypropylene, we can refer to the data in Table 21.1. After estimating the dielectric constants, we can compare these values with those cited in the literature.

Without access to Table 21.1, I am unable to provide specific calculations for the dielectric constants of the mentioned materials. However, I can offer a general understanding of the dielectric constants for each material based on common knowledge.

Borosilicate Glass:

Borosilicate glass typically has a dielectric constant ranging from around 4 to 6. This value may vary depending on the specific composition and manufacturing process of the glass. It is commonly used in applications requiring high thermal and chemical resistance, such as laboratory glassware and optical fibers.

Periclase (MgO):

Periclase, or magnesium oxide (MgO), is an insulating material with a relatively high dielectric constant. Its dielectric constant is typically in the range of 9 to 10. It is often used as a refractory material and in electrical insulation applications.

Poly(methyl methacrylate) (PMMA):

Poly(methyl methacrylate), also known as acrylic or acrylic glass, has a dielectric constant in the range of 3 to 4. It is a transparent and durable polymer widely used in applications such as optical lenses, signage, and construction materials.

Polypropylene (PP):

Polypropylene is a thermoplastic polymer with a relatively low dielectric constant, typically ranging from 2.2 to 2.4. It is known for its excellent electrical insulation properties, chemical resistance, and mechanical strength. Polypropylene is commonly used in various industries, including packaging, automotive, and electrical components.

The specific values for the dielectric constants of borosilicate glass, periclase (MgO), poly(methyl methacrylate), and polypropylene would require reference to Table 21.1. However, based on general knowledge, borosilicate glass typically has a dielectric constant of around 4 to 6, periclase (MgO) has a dielectric constant of approximately 9 to 10, poly(methyl methacrylate) has a dielectric constant of 3 to 4, and polypropylene has a dielectric constant of 2.2 to 2.4.

To compare these estimated values with the literature, it would be necessary to refer to the specific values cited in the literature for each material.

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1. Using the data in Table 21.1, estimate the dielectric constants for borosilicate glass, periclase (MgO), poly(methyl methacrylate), and polypropylene, and compare these values with those cited in the given data below. Briefly explain any discrepancies.

Materials                                   -             Dielectric constant

Borosilicate glass                     -                   4.7

Periclase                                   -                   9.7

Poly( methyl methacrylate)      -                   2.8

Poly propylene                         -                    2.35

Answer:

gas, vapour

Explanation:

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