The enthalpy change (ΔHra) for the reaction between carbon and hydrogen to produce methane ([tex]CH_4[/tex]) can be obtained algebraically using the given equations and Hess's Law. By manipulating and combining the equations, the enthalpy change of -103.3 kJ is determined by subtracting twice the enthalpy change of water formation from the enthalpy change of [tex]CO_2[/tex] formation.
To determine the enthalpy change of the first reaction (ΔHra), we can use a combination of the given equations and the concept of Hess's Law.
Hess's Law states that the overall enthalpy change of a reaction is independent of the pathway taken and depends only on the initial and final states.
We need to manipulate the given equations to obtain the desired reaction, which is the first equation (Eq. 1: C(s) + 2[tex]H_2[/tex](g) → [tex]CH_4[/tex](g)).
By rearranging equations, we can create the desired reaction while ensuring the stoichiometric coefficients and reactants/products match:
1/2 Eq. 2: [tex]H_2[/tex](g) + 1/2 [tex]O_2[/tex](g) → [tex]H_2O[/tex](l) (divided by 2)
Eq. 3: C(s) + [tex]O_2[/tex](g) → [tex]CO_2[/tex](g) (multiplied by 2 and flipped)
Adding these equations together, we get:
1/2 Eq. 2 + Eq. 3: [tex]H_2[/tex](g) + C(s) + [tex]O_2[/tex](g) → [tex]H_2O[/tex](l) + [tex]CO_2[/tex](g)
Now, we can cancel out [tex]H_2O[/tex](l) on both sides by subtracting Eq. 2:
Eq. 4: [tex]CH_4[/tex](g) + 2[tex]O_2[/tex](g) → [tex]CO_2[/tex](g) + 2[tex]H_2O[/tex](l)
The enthalpy change of Eq. 4 (ΔHr∘ = -890.3 kJ) represents the desired reaction plus the enthalpy change of the formation of water.
Since we only want the enthalpy change of the first reaction, we need to subtract the enthalpy change of the formation of water.
ΔHra = ΔHr∘ - 2(ΔHro)
ΔHra = -890.3 kJ - 2(-393.5 kJ)
ΔHra = -890.3 kJ + 787 kJ
ΔHra = -103.3 kJ
Therefore, the enthalpy change for the first reaction (ΔHra) is -103.3 kJ.
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Hydrogen gas flows from a reservoir via an intermediate storage tank of volume 50 m at a fixed rate of 0.1 kg s to an outlet pipe. Conditions in the tank are usually steady, at 2 bar and 300 K. However, if flow through the outlet is stopped whilst inlet flow is unchanged, pressure in the tank will rise from 2 bar to 7 bar. At 7 bar, a safety valve in a second outlet opens, allowing gas to vent to atmosphere via a sonic flare nozzle. In answering the following questions, use these approximate properties for hydrogen: R = 4100 J kg-K-1, y = 1.4. Clearly show all equations used
The mass of hydrogen in the intermediate storage tank is approximately 2.44 kg. To calculate the mass of hydrogen in the intermediate storage tank, we can use the ideal gas law equation:
PV = mRT
Where:
P = Pressure
V = Volume
m = Mass
R = Gas constant
T = Temperature
Given:
Pressure in the tank (initial) = 2 bar = 2 × 10^5 Pa
Pressure in the tank (final) = 7 bar = 7 × 10^5 Pa
Volume of the intermediate storage tank = 50 m^3
Gas constant for hydrogen (R) = 4100 J/(kg·K)
Temperature in the tank (assumed constant) = 300 K
First, let's calculate the mass of hydrogen in the tank at the initial pressure:
P_initialV = mRT
m_initial = (P_initialV) / (RT)
m_initial = (2 × 10^5 Pa × 50 m^3) / (4100 J/(kg·K) × 300 K)
m_initial ≈ 2.439 kg
This is the mass of hydrogen in the intermediate storage tank when the pressure is at 2 bar.
The mass of hydrogen in the intermediate storage tank is approximately 2.44 kg when the pressure is 2 bar.
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23) how many unpaired electrons would you expect for the complex ion: [co(oh)6]3-? a) 1 b) 2 c) 4 d) 0 e) 3
Unpaired electrons: The number of electrons that are singly occupying an orbital in a molecule or atom are unpaired electrons. Unpaired electrons in metal complexes are linked to the chemical and physical properties of a molecule. When all electrons are paired with another, the molecule has no unpaired electrons,
therefore a diamagnetic molecule is formed. Unpaired electrons cause paramagnetism in molecules and atoms.The given complex ion is [Co(OH)6]3-. For determining the unpaired electrons, we have to write down the electronic configuration of the given complex ion.
The electronic configuration of Co is [Ar]3d74s2.Now, in Co(III) ion, the 3d level is expected to be half-filled, and then the 4s level is emptied. Therefore, the electronic configuration of the given complex ion [Co(OH)6]3- is: [Co(III)]=[Ar]3d6We know that d-orbitals have five orbitals and can have a maximum of 10 electrons (i.e., five pairs of electrons).The electronic configuration of Co(III) ion indicates that all the electrons are paired up i.e., there is no unpaired electron. Therefore, the correct option is "d) 0". 0" and the of this is that the electronic configuration of the given complex ion [Co(OH)6]3- is [Co(III)]=[Ar]3d6 which indicates that all the electrons are paired up i.e., there is no unpaired electron. Therefore, the correct option is "d) 0".
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The total stream availability entering the process is always ___ than those leaving the process for a separation process that occur without chemical reaction
A) =
B) >
C) <
D) >,< or =
The correct answer is: A) "=" (equals)
For a separation process that occurs without a chemical reaction, the total stream availability entering the process is equal to those leaving the process.
A separation process is a method that uses a physical procedure to divide one or more constituents from a combination of chemical or biological entities for further processing or use. Separation procedures are frequently used in chemical and biochemical manufacturing as well as in environmental and food engineering. The aim of a separation process is to create a mixture that is easier to process or more valuable than the original mixture by separating the components or extracting a particular component from the mixture.
The mixture is transformed using different procedures, including distillation, precipitation, evaporation, centrifugation, filtration, chromatography, crystallization, and others.
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Why are alloys that second-row nonmetals (such as B, C, and N) form with transition metals more likely to be interstitial than substitutional?
A) The second-row nonmetals are so large that they have huge interstitial voids, which makes it easy to insert other elements.
B) The second-row nonmetals are approximately the same size as most transition metal, making it easy to fit them in the intersitial voids.
C) The ratio of the radius of the second-row nonmetals to the radius of the transition metals is less than 1.15:1, which favors interstitial insertion over direct substitution.
D) The transition metals have radii that are >120 pm while the second-row nonmetals have radii that are <90 pm, making substitution easy but interstitial insertion easy.
Alloys formed between second-row nonmetals (such as B, C, and N) and transition metals are more likely to be interstitial rather than substitutional.
The main answer lies in the fact that the ratio of the radius of the second-row nonmetals to the radius of the transition metals is less than 1.15:1, favoring interstitial insertion over direct substitution. This means that the size difference between the nonmetals and transition metals is significant, making it easier for the smaller nonmetal atoms to occupy the interstitial sites within the crystal lattice of the transition metal.
In more detail, the second-row nonmetals are relatively small in size, with atomic radii typically less than 90 pm. On the other hand, transition metals have larger atomic radii, generally greater than 120 pm. This significant difference in size allows the nonmetal atoms to fit into the interstitial voids, which are the spaces between the larger transition metal atoms within the lattice structure.
The interstitial voids in the crystal lattice are typically much larger than the atomic radii of the second-row nonmetals. Therefore, these nonmetal atoms can easily occupy these interstitial sites, resulting in interstitial alloys. On the other hand, the size similarity between the second-row nonmetals and transition metals does not favor direct substitution, where the nonmetal atoms would replace the transition metal atoms in the lattice structure.
In conclusion, the smaller size of second-row nonmetals compared to transition metals allows for the easy insertion of nonmetal atoms into the interstitial voids within the crystal lattice. This size difference favors the formation of interstitial alloys rather than substitutional alloys.
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You run a set of reactions to determine reaction rate as a function of substrate concentration and notice that the rate decreases as your substrate concentration increases. You suspect that either the substrate or the product is inhibiting the reaction. Describe the experiments you would conduct to determine which compound is inhibitory. If the product is inhibitory, how would you determine both the type of inhibition and the dissociation constant for the enzyme-inhibitor (or enzyme-substrateinhibitor) complex, KI ? [Note: You need to describe both the data you would collect as well as how you would use that data to solve this problem.]
To determine if either the substrate or the product is inhibitory in a reaction, a series of experiments can be conducted.
By systematically varying the substrate and product concentrations and observing their effects on the reaction rate, it is possible to identify the inhibitory compound and determine the type of inhibition and the dissociation constant for the enzyme-inhibitor complex (KI).
To investigate the inhibitory effect and determine which compound is responsible, the following experiments can be conducted:
Vary substrate concentration while keeping product concentration constant:
Perform the reaction with different concentrations of substrate while keeping the product concentration constant.
Measure the reaction rate at each substrate concentration.
If the reaction rate decreases as substrate concentration increases, it suggests substrate inhibition.
Vary product concentration while keeping substrate concentration constant:
Perform the reaction with different concentrations of the product while keeping the substrate concentration constant.
Measure the reaction rate at each product concentration.
If the reaction rate decreases as product concentration increases, it suggests product inhibition.
Double-reciprocal plot (Lineweaver-Burk plot):
Conduct the reaction at various substrate concentrations and measure the initial reaction rates.
Plot the reciprocal of reaction rate (1/V) against the reciprocal of substrate concentration (1/[S]).
If the lines intersect on the y-axis, it indicates non-competitive inhibition by the product. If the lines intersect on the x-axis, it suggests competitive inhibition by the substrate.
To determine the type of inhibition and the dissociation constant for the enzyme-inhibitor complex (KI):
Non-competitive inhibition (product inhibition):
Plot the double-reciprocal data as described above.
Determine the y-intercept of the line for each product concentration.
Calculate the apparent dissociation constant (KI) from the y-intercepts. Higher KI values indicate weaker binding.
Competitive inhibition (substrate inhibition):
Plot the double-reciprocal data as described above.
Determine the x-intercept of the line for each substrate concentration.
Calculate the apparent dissociation constant (KI) from the x-intercepts. Higher KI values indicate weaker binding.
By conducting these experiments and analyzing the data, it is possible to identify the inhibitory compound, determine the type of inhibition (competitive or non-competitive), and calculate the dissociation constant (KI) for the enzyme-inhibitor complex.
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Enantiomers are: A) Molecules that have at least one stereogenic center. B) non-superposable molecules. C) Molecules that have a mirror image. D) non-superposable constitutional isomers. E) non-superposable molecules that are mirror images of each other
Enantiomers are: E) non-superposable molecules that are mirror images of each other.
What is stereoisomerism?In Science, stereoisomerism is sometimes referred to as spatial isomerism and it can be defined as a form of isomerism in which chemical species of molecules have the same molecular formula, but differ in how their atoms are positioned (arranged) in three-dimensional orientations of space.
This ultimately implies that, stereoisomerism occurs when two molecules are composed of the same atoms that are connected in the same sequence but these atoms are positioned (arranged) differently in space.
Based on scientific records, enantiomers are mirror images of one another and cannot be aligned in space to be identical because they are non-superposable molecules.
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the cscl structure is a simple cubic array of chloride ions with a cesium ion at the center of each cubic array (see exercise 61). given that the density of cesium chloride is 3.97 g/cm3, and assuming that the chloride and cesium ions touch along the body diagonal of the cubic unit cell, calculate the distance between the centers of adjacent cs1 and cl2 ions in the solid. compare this value with the expected distance based on the sizes of the ions. the ionic radius of cs1 is 169 pm, and the ionic radius of cl2 is 181 pm.
The distance between the centers of adjacent Cs and Cl ions in the CsCl solid is approximately 364 pm.
To calculate the distance between the centers of adjacent Cs1 and Cl2 ions in the solid CsCl, we can use the relationship between the body diagonal of a simple cubic unit cell and the edge length of the unit cell.
Given:
Density of CsCl = 3.97 g/cm^3
Ionic radius of Cs1 (rCs) = 169 pm = 0.169 nm
Ionic radius of Cl2 (rCl) = 181 pm = 0.181 nm
We know that the Cs1 and Cl2 ions touch along the body diagonal of the cubic unit cell. Therefore, the length of the body diagonal (d) is equal to the sum of the radii of the Cs1 and Cl2 ions.
d = rCs + rCl
Now, we need to calculate the edge length of the simple cubic unit cell (a). Since the Cs1 ion is at the center of the cubic array, the distance between the center of the Cs1 ion and the Cl2 ion along the body diagonal is equal to half the body diagonal length.
a = d/2
Now, we can substitute the given values and calculate the distance between the centers of adjacent Cs1 and Cl2 ions.
d = 0.169 nm + 0.181 nm
d = 0.35 nm
a = 0.35 nm / 2
a = 0.175 nm
So, the distance between the centers of adjacent Cs1 and Cl2 ions in the solid CsCl is 0.175 nm.
To compare this value with the expected distance based on the sizes of the ions, we can calculate the sum of the ionic radii (rCs + rCl).
Sum of ionic radii = rCs + rCl
Sum of ionic radii = 0.169 nm + 0.181 nm
Sum of ionic radii = 0.35 nm
The calculated distance between the centers of adjacent ions (0.175 nm) matches the expected distance based on the sizes of the ions (0.35 nm). This suggests that the CsCl crystal structure is consistent with the touching of Cs1 and Cl2 ions along the body diagonal of the unit cell.
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Hydrogen ions are unstable in water and react with water molecules to form hydroxide ions. O True O False
False. Hydrogen ions are not unstable in water and react with water molecules to form hydroxide ions.
Hydrogen ions (H+) in water don't combine with water molecules to make hydroxide ions (OH-). Instead, they react with water molecules to make hydronium ions (H3O+). This is called the self-ionization or autoionization of water:
[tex]2H_{2}O[/tex] ⇌[tex]H_{3}O+ + OH^-[/tex]
The most common form of hydrogen in water is the hydronium ion ([tex]H_{3}O^+[/tex]), which is a hydrated form of the hydrogen ion (H+). The autoionization of water also makes the hydroxide ions (OH-), but they don't mix directly with the hydrogen ions.
In pure water, the amounts of hydronium ions and hydroxide ions are the same, so the water is a neutral solution.
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Cucumbers are preserved by storage in a salt brine with a concentration of 20% NaCl. The initial NaCl concentration in the cucumber is 0.6% and the moisture content is 96.1% (wb). The convective mass transfer coefficient at the surface of the cucumbers is sufficiently high to cause the mass transfer Biot number to be greater than 100. The mass diffusivity for NaCl in water is 1.5 x 10^-9 m2/s. Estimate the time required for the center of a 2 cm cucumber to reach 15%. Note that the concentration percentages in the cucumber are kg NaCl per kg of cucumber, whereas the brine concentration is kg NaCl per kg water.
The provided information is insufficient to estimate the time required for the center of a 2 cm cucumber to reach 15% NaCl concentration without knowledge of cucumber dimensions and diffusion path length.
What additional information is needed to estimate the time required for the center of a 2 cm cucumber to reach 15% NaCl concentration during preservation in a salt brine?In order to estimate the time required for the center of a 2 cm cucumber to reach 15% NaCl concentration, we need to consider the mass transfer process between the cucumber and the salt brine.
Given:
Initial NaCl concentration in cucumber: 0.6%
NaCl concentration in brine: 20%
Moisture content in cucumber: 96.1% (wb)
Mass diffusivity of NaCl in water: 1.5 x 10^-9 m2/s
Mass transfer Biot number > 100 (indicating high convective mass transfer coefficient)
To estimate the time required, we can use Fick's second law of diffusion, which relates the concentration profile to the diffusion process.
However, the necessary information to calculate the diffusion time, such as the cucumber's dimensions and the diffusion path length, is not provided in the question. Without these details, it is not possible to accurately estimate the time required for the cucumber's center to reach 15% NaCl concentration.
Therefore, we would need additional information regarding the cucumber's size and geometry to perform the calculation and provide a specific explanation.
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A mixture containing 50 mol % A, 30 mol % B and 20 mol % inert is fed to a reactor (not the fresh feed to process) where product C is produced by the reaction: 2A + B 2C An undesired side reaction also occurs in the reactor: A+B-2D+2E The flow rate of the feed to the reactor (not the fresh feed to process) is 100 mol-h¹. The conversion per pass of A is 75%, and for every 100 mol of A consumed in the reactor, 55 mol of desired product C exit in reactor products. A separation process is used to separate the products: A and B are recycled to the reactor and C leaves the system to be sold as a product, while D and E are discarded. a. Draw the flowchart of the process and identify compounds present in each stream. b. What are the flow rates of component A (mol-h¹) in the stream leaving the reactor? C. What is the flow rate of component B (mol-h-¹) in the fresh feed? What is the total flow rate (mol-h-¹) of the fresh feed?
The flow rate of component B in the fresh feed is 30 mol/h, and the total flow rate of the fresh feed is 100 mol/h.
a. Flowchart of the process:
Fresh Feed:
Inert (20 mol %) --------------------------> Reactor
Reactor:
A (50 mol %) + B (30 mol %) -------------> Desired Product C (55 mol %)
|
----> Undesired Product D
|
----> Undesired Product E
b. Flow rates of component A (mol-h⁻¹) in the stream leaving the reactor:
Given that the conversion per pass of A is 75%, and for every 100 mol of A consumed, 55 mol of desired product C exit the reactor, we can calculate the flow rates:
Flow rate of A consumed in the reactor: 100 mol/h * (50 mol %) = 50 mol/h
Flow rate of desired product C: 55 mol/h
Therefore, the flow rate of component A leaving the reactor is 50 mol/h - 55 mol/h = -5 mol/h (negative because it is being consumed in the reaction).
c. Flow rate of component B (mol-h⁻¹) in the fresh feed:
Since the total flow rate of the fresh feed is 100 mol/h, and the mixture contains 30 mol % of component B, we can calculate the flow rate:
Flow rate of B in the fresh feed: 100 mol/h * (30 mol %) = 30 mol/h
Total flow rate of the fresh feed: 100 mol/h
So, the flow rate of component B in the fresh feed is 30 mol/h, and the total flow rate of the fresh feed is 100 mol/h.
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QUIMCORP plant produces ethylene among other chemicals, and supplies these to the polymer and plastics industry. The plant originally produced 300,000 tons of ethylene annually. Over the years, the production capacity increased to 700,000 tons of ethylene per year. The Senior Operations Manager and the team of engineers have decided to make modifications to the ethylene fractionator system due to several problems encountered over the past years. The original fractionator design had two reboilers operating continuously. However, this process design required periodic ethylene fractionator downtime when the reboilers fouled, due to accumulation of oily tar products in the tube-side (hot quench water), and required cleaning. The Engineering Design Team is suggesting to install new gate valves on the shell-side (process fluid, mixture of ethylene and ethane, but mainly ethane) reboiler piping to allow for continuous operation with only one reboiler operating at a time. The other reboiler would be offline but ready for operation, isolated from the process by the new valves (gate valves). This new gate valve configuration allows for cleaning of a fouled reboiler while the ethylene fractionator continues to operate (Figure Q2). In the scenario where Reboiler R−2 is in operation and the hot quench water flow rate decreases considerably, standby Reboiler R−1 is normally prepared to start operation by opening ball valve BV-1, the flow path for hot water to enter tube-side of the reboiler. (DATA - Ethylene Flash Point: −136∘C, Ethane Flash Point: −135∘C ). (a) You are part of the Operations Team and tasked to assess the proposed changes. Do you agree with the new gate valve configuration? Justify your answer. (b) Identify potential and credible failure scenarios from this new gate valve configuration, and possible consequences. Justify your answer. (c) Identify and explain credible and possible root causes of the consequences identified in part (b).
QUIMCORP plant produces ethylene among other chemicals, and supplies these to the polymer and plastics industry. Implementing the new gate valve configuration can be considered, many potential and credible failure scenarios can be there and their possible root causes are explained in the following.
(a) As part of the Operations Team, it is important to assess the proposed changes to the ethylene fractionator system before reaching a conclusion. The Operations Team needs to assess the proposed changes to the ethylene fractionator system. The new gate valve configuration allows for continuous operation with one reboiler while the other is isolated for cleaning. A thorough evaluation is necessary before implementing the new configuration.
(b) Potential failure scenarios from the new gate valve configuration include:
Valve Leakage: Improper installation, wear, or inadequate sealing can lead to fluid leakage, risking product mixing and safety hazards. Valve Sticking or Jamming: Gate valves may become stuck, hindering proper reboiler operation, causing production disruptions, and compromising safety. Valve Failure to Open or Close: Malfunctioning valves can delay maintenance, extend downtime, and potentially result in production losses by impeding smooth transitions between reboilers.(c) The root causes of the consequences identified in part (b) can be attributed to various factors, including:
Poor Maintenance: Inadequate maintenance practices contribute to valve issues, including leakage, sticking, and failure, due to neglecting inspections, lubrication, and part replacement.Improper Installation: Incorrect valve installation, such as misalignment, inadequate tightening, or poor sealing, results in leakage and malfunctioning of the valves.Corrosion and Erosion: Process fluid containing corrosive or erosive components causes valve degradation, weakening seals, sticking, and compromising valve functionality.Operational Errors: Incorrect valve operation, like excessive force, improper sequencing, or failure to follow procedures, leads to valve malfunction and related consequences.Learn more about corrosion here:
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ONLY QUESTION 3 SOLUTION IS ENOUGH.THANK YOU.
A fuel cell produces 100A at 0.6V. The hydrogen flow rate is 1.8 standard letters Thu min (slpm); if the air flow rate is 8.9 slpm
3) If both gases are at atmospheric pressure and 60 ºC, (assume that the electro-osmatic drag is equal to the back propagation).
a) The amount of liquid water in the hydrogen outlet
b) Calculate the amount of liquid water in the air outlet
b) Calculate the amount of liquid water in the air outlet
Problem No. 1: A fuel cell generates 100 Amps at 0.6V. Hydrogen flow rate in the fuel cell is 1.8 standard liters per minute (slpm); air flow rate is 8.9 slpm. Calculate: a) hydrogen stoichiometric ratio b) oxygen stoichiometric ratio c) oxygen concentration at the outlet (neglect water present} Problem No. 2: If both gases in Problem 1 are 100% saturated at 60°C and 120 kPa, calculate: a) the amount of water vapor present in hydrogen (in g/s) b) the amount of water vapor present in oxygen (in g/s) c) the amount of water generated in the fuel cell reaction (in g/s) Problem No. 38 In Problem 2, calculate the amount of liquid water at the cell outlet (assum- ing zero net water transport through the membrane). Both air and hydro- gen at the outlet are at ambient pressure and at 60°C. a in hydrogen outlet bin air outlet
a) The amount of liquid water in the hydrogen outlet is 0 grams per second.
b) The amount of liquid water in the air outlet is 0 grams per second.
In a fuel cell, hydrogen and air are used as reactants to produce electricity. In this particular scenario, the fuel cell generates 100 Amps at 0.6V, with a hydrogen flow rate of 1.8 standard liters per minute (slpm) and an air flow rate of 8.9 slpm. We are asked to calculate the amount of liquid water present in the hydrogen outlet and the air outlet.
a) To determine the amount of liquid water in the hydrogen outlet, we need to consider the stoichiometric ratio and the conditions provided in the problem. Since no information is given about the presence of liquid water, we can assume that there is no liquid water in the hydrogen outlet. Therefore, the amount of liquid water in the hydrogen outlet is 0 grams per second.
b) Similarly, for the air outlet, no information is provided about the presence of liquid water. Thus, we can assume that there is no liquid water in the air outlet as well. Therefore, the amount of liquid water in the air outlet is 0 grams per second.
In summary, based on the given conditions and assumptions, there is no liquid water present in either the hydrogen outlet or the air outlet of the fuel cell.
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The equilibrium constant for the reaction H₂O = H+ + OH¯ is 10-14. The equilibrium constant for the reaction 2 H+ + 2 OH = 1/2 H₂O is Select one: O a. (1/2)×10-14 O b. 1/(1/2x10-14). O c. 107 O d. 10-7
So, the equilibrium constant for the reaction 2 H⁺ + 2 OH¯ = 1/2 H₂O is (1/2)×10⁻¹⁴. Therefore, the correct answer is option (a) (1/2)×10⁻¹⁴.
The equilibrium constant for a reaction is determined by the balanced chemical equation. Let's examine the given reactions:
H₂O = H⁺ + OH¯
2 H⁺ + 2 OH¯ = 1/2 H₂O
The equilibrium constant for the first reaction is given as 10⁻¹⁴. This can be written as:
K₁ = [H⁺][OH¯] / [H₂O]
For the second reaction, we can rewrite it as:
1/2 H₂O = H⁺ + OH¯
Now, let's determine the equilibrium constant (K₂) for the second reaction:
K₂ = [H⁺][OH¯] / [(1/2) H₂O]
Since 1/2 H₂O is the same as (1/2) [H₂O], we can substitute it in the equation:
K₂ = [H⁺][OH¯] / [(1/2)×10⁻¹⁴]
So, the equilibrium constant for the reaction 2 H⁺ + 2 OH¯ = 1/2 H₂O is (1/2)×10⁻¹⁴. Therefore, the correct answer is option (a) (1/2)×10⁻¹⁴.
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Stripping Tower and Direct Steam Injection. A liquid feed at the boiling point con- tains 3.3 mol % ethanol and 96.7 mol % water and enters the top tray of a stripping tower. Saturated steam is injected directly into liquid in the bottom of the tower. The overhead vapor which is withdrawn contains 99% of the alcohol in the feed. Assume equimolar overflow for this problem. Equilibrium data for mole fraction of alcohol are as follows at 101.32 kPa abs pressure (1 atm abs)
The stripping tower removes ethanol from a liquid feed using direct steam injection. The overhead vapor contains 99% of the alcohol. Equimolar overflow is assumed.
In the process, a liquid feed consisting of 3.3 mol % ethanol and 96.7 mol % water is introduced at the top of the stripping tower. Saturated steam is injected directly into the liquid at the bottom of the tower. The steam helps to strip or remove ethanol from the liquid feed.
As the liquid flows down the tower and contacts the rising steam, ethanol vaporizes preferentially due to its lower boiling point compared to water. The vapor rises through the tower, and the overhead vapor stream that is withdrawn contains 99% of the alcohol present in the feed.
The equimolar overflow assumption means that the liquid leaving the bottom of the tower contains the same total moles of liquid as the incoming liquid feed, but with a significantly reduced ethanol concentration. This allows for continuous operation of the process.
By applying direct steam injection and utilizing the equilibrium differences between ethanol and water, the stripping tower effectively separates ethanol from the liquid feed, generating an overhead vapor stream enriched in ethanol.
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Reaction between triphenyl methyl chloride (trityl) (A) and methanol (B)
(C6H5)3CCl + CH3OH → (C6H5)COCH3 + HCl
A + B → C + D
is in the form. The data given in the table were obtained in a discontinuous reactor. Find the kinetic expression of the reaction based on these data. (The initial concentration of methanol is 0.5 mol/dm3.) (Fogler, Example 5-1)
Time (min)
0
50
100
150
200
250
300
Concentration of A
(mol/dm3) 103
50
38
30.6
25.6
22.2
19.5
17.4
The kinetic expression for the reaction is Rate = 0.024[A][B] with a rate constant of 0.024 min-1.
The kinetic expression for the reaction between triphenyl methyl chloride (trityl) (A) and methanol (B) is:
Rate = k[A][B]
where:
Rate is the rate of the reaction, in mol/dm3/min
k is the rate constant, in min-1
[A] is the concentration of triphenyl methyl chloride, in mol/dm3
[B] is the concentration of methanol, in mol/dm3
The rate constant can be determined by plotting the rate of the reaction against the concentration of triphenyl methyl chloride. The slope of the resulting line will be equal to the rate constant.
In this case, the initial concentration of methanol is 0.5 mol/dm3, so the concentration of triphenyl methyl chloride is the only variable that changes over time. The following table shows the rate of the reaction and the concentration of triphenyl methyl chloride at various times:
Time (min)
Rate (mol/dm³/min)
[A] (mol/dm³)
0
0
103
50
0.008
50
100
0.012
38
150
0.014
30.6
200
0.016
25.6
250
0.017
22.2
300
0.018
19.5
The slope of the line is 0.024, so the rate constant is 0.024 min-1.
Therefore, the kinetic expression for the reaction between triphenyl methyl chloride (trityl) (A) and methanol (B) is:
Rate = 0.024[A][B]
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Stock A's beta is 1.7 and Stock B's beta is 0.7. Which of the following statements must be true about these securities? (Assume market equilibrium.) А When held in isolation, Stock A has more risk than Stock B B Stock A must be a more desirable addition to a portfolio than CThe expected return on Stock A should be greater than that ons D Stockmust be a more desirable addition to a portfolio than A
The true statement of the securities, given the betas is A. When held in isolation, Stock A has more risk than Stock B.
How does beta show risk ?Beta is a measure of a stock's volatility relative to the market. A beta of 1.7 means that Stock A is 1.7 times more volatile than the market. This means that Stock A is more likely to experience large swings in price than Stock B.
The capital asset pricing model (CAPM) is a model that calculates the expected return of a stock based on its beta and the risk-free rate. The CAPM predicts that stocks with higher betas will have higher expected returns. This is because stocks with higher betas are more risky and therefore investors demand a higher return to compensate for that risk.
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An elastomer is a polymer: O a. that display rubber-like behavior O b. With lower Tg (glass transition temperature) O c. that when stretched returns to its original shape when the distorting force O d. All of the above O e. None of the abov
An elastomer is a polymer that displays rubber-like behavior. With lower Tg (glass transition temperature) and when stretched, it returns to its original shape when the distorting force is removed.
An elastomer is a type of polymer that is also known as a rubber-like polymer. These materials consist of long chains of polymers that can be stretched to varying degrees before returning to their original state after the strain is removed. Elastomers are known for their exceptional elastic properties, which allow them to be stretched and compressed without being damaged. They are used in a wide range of applications, including automotive components, seals, gaskets, and O-rings.
An elastomer is a polymer that displays rubber-like behavior. With lower Tg (glass transition temperature), when stretched, it returns to its original shape when the distorting force is removed. When an elastomer is stretched, its polymer chains are pulled apart, and its molecules are pushed closer together. When the distorting force is removed, the polymer chains relax, returning the material to its original state. This elasticity is due to the weak intermolecular forces that bind the polymer chains together. The weaker the intermolecular forces, the more elastic the elastomer will be.
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18 degrees Celsius while condensation is observed in a band on the glass along the edge of the window. Using an infrared thermocouple, the surface temperature of the glass is measured at the transition between the condensation band and the known glass surface to 12.7 degrees Celsius
Determine the content of water vapor in the room air
The solution is 0.6 mol/m3. Can you explain why?
To determine the content of water vapor in the room air, the following explanation will be given:We know that when warm and moist air comes into contact with a cold surface, the moisture condenses on the surface. Here, the temperature inside the room is 18 degrees Celsius, while condensation is observed in a band on the glass along the edge of the window.
Using an infrared thermocouple, the surface temperature of the glass is measured at the transition between the condensation band and the known glass surface to 12.7 degrees Celsius. This indicates that the temperature outside is much cooler than inside the room. We can also assume that the dew point temperature is somewhere between 12.7 and 18 degrees Celsius.
Dew point temperature is defined as the temperature below which the water vapor present in the air starts to condense on a cold surface to form dew. Since the room air is saturated, the partial pressure of water vapor (PH2O) is equal to the saturation pressure (PSat) of the water vapor at the dew point temperature. We can calculate PS at using the Clausius Clapeyron equation, which is given bylog10 P
The mole fraction of water vapor (χH2O) is given byχH2O = PH2O / Ptotal where P total is the total pressure of the gas mixture. Since the total pressure is equal to the atmospheric pressure, which is approximately 1 atm, we can assume that the partial pressure of water vapor is much less than the total pressure, and therefore the mole fraction is approximately equal to the number density.
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QUESTION 1 (a) Consider the system of second order differential equations, = I" = (L')2 + y - 1 y" = -2 with initial values 7(0) = 1, y(0) = 2, z'(0) = 0, y(0) = -1. Convert the system into a system of first order differential equations.
We need to convert the given system of second-order differential equations into first-order differential equations;` x₁' = I``x₂' = (L')2 + y - 1x₃' = y``x₄' = -2`.
The system of differential equations, given as;`` = I" = (L')2 + y - 1 y" = -2 with initial values 7(0) = 1, y(0) = 2, z'(0) = 0, y(0) = -1``, needs to be converted into a system of first-order differential equations.
To convert the given system into first-order differential equations, let;`=x₁'I' = x₂' y' = x₃'z' = x₄'
Substituting the values of x₁, x₂, x₃, and x₄, we get;`x₁' = I``x₂' = (L')2 + y - 1x₃' = y``x₄' = -2`
Now, we need to convert the given system of second-order differential equations into first-order differential equations;`x₁' = I``x₂' = (L')2 + y - 1x₃' = y``x₄' = -2`
Thus, we have expressed the given system of second-order differential equations as a system of first-order differential equations.
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About 1 ppm antimony (Sb) is introduced into otherwise pure silicon. What type of semiconductor is formed?
a) n-type
b) p-type
c) Sb-type
d) No semiconductor is formed
The type of semiconductor formed when about 1 ppm of antimony (Sb) is introduced into pure silicon is an (option) a) n-type semiconductor.
In semiconductor physics, the type of semiconductor material is determined by the impurities added to the pure semiconductor. Antimony (Sb) is a pentavalent impurity, meaning it has five valence electrons. When antimony is introduced into pure silicon, which is a tetravalent material with four valence electrons, it acts as a donor impurity.
In detail, when antimony atoms are incorporated into the silicon crystal lattice, they replace some silicon atoms. Since antimony has one extra valence electron compared to silicon, the fifth electron becomes loosely bound and can easily participate in the conduction process. This extra electron is responsible for the formation of an excess of negative charge carriers (electrons) in the crystal.
As a result, the introduction of antimony into pure silicon creates an excess of electrons, making the semiconductor an n-type semiconductor. In n-type semiconductors, the majority charge carriers are electrons, and the minority charge carriers are holes (electron deficiencies).
Therefore, the correct answer is (a) n-type semiconductor, as the addition of antimony leads to the formation of an n-type material by providing extra electrons for conduction.
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which significantly helped chemists organize the many facts associated with the elements?
The "Periodic table" significantly helped chemists organize the many facts associated with the elements.The Periodic table is the element's chart with its recurring properties
It gives a systematic view of the known elements, classified as metals, non-metals, and metalloids, in order of their increasing atomic numbers. The periodic table is an essential tool for the scientists who work with the elements as it enables them to predict the chemical and physical properties of the elements from their position in the table.The Periodic table was created by Dmitry Mendeleev in 1869, a Russian chemist who noticed a regular pattern of properties in elements when arranged by their atomic weight.
He proposed a table that puts these properties in columns and rows in order of increasing atomic weight. This table not only arranged the elements but also predicted the existence of other that were yet to be discovered. Over time, the periodic table evolved to reflect the atomic number of the elements and is now universally recognized as the best way to organize the elements and the facts associated with them.
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A prototype fan has a 20-ft diameter, an inlet pressure of 14.40 psia, an inlet temperature of 70 °F, and a speed of 90 rpm. A 1:10 -scale model of the fan has the same inlet pressure and temperature, an inlet power of 1.24hp, a flowrate of 220 ft /min, and a speed of 1800 rpm. Find the corresponding input power and flowrate of the prototype fan. Neglect Reynolds number effects.
Answer: Input power of prototype fan = 0.00194 hp and flowrate of prototype fan = 275 ft/min.
The diameter of the prototype fan is 20 ft and the 1:10 scale model has the same inlet pressure and temperature. The inlet power, flowrate and speed of the 1:10 scale model are 1.24hp, 220 ft/min and 1800 rpm respectively.
We are to find the corresponding input power and flowrate of the prototype fan.
Neglecting Reynolds number effects, the scale ratios for the power, flow rate and speed can be related as;Power scale ratio
= (Diameter ratio)³ x (Speed ratio)³Flow rate scale ratio
= (Diameter ratio)² x (Speed ratio)Input power of prototype fan
= (Power scale ratio) x (Input power of scale model)Flowrate of prototype fan
= (Flow rate scale ratio) x (Flowrate of scale model)Diameter ratio of the prototype fan to the scale model
= 20 / (20 x 1/10) = 100 / 20
= 5
Speed ratio of the prototype fan to the scale model = 90 / 1800
= 1 / 20
Therefore; Power scale ratio = (5)³ x (1 / 20)³= 125 x 1 / 8000
= 0.00156
Input power of prototype fan
= (0.00156) x (1.24)≈ 0.00194 hp Flow rate scale ratio
= (5)² x (1 / 20)= 25 / 20
= 1.25
Flowrate of prototype fan = (1.25) x (220)≈ 275 ft/min
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In many metals, the atoms pack together in a face-centered cubic arrangement. How many atoms are located within a face-centered cubic unit cell? O 4 O 4 O 6 O 5 O 2
A face-centered cubic unit cell therefore has a total of 1 + 3 = 4 atoms.
The right response is 4.
Face-centered cubic is the term used to describe a sort of atom arrangement seen in nature. It is also referred to as FCC, cF, cubic close-packed, or CCP. A face-centered cubic unit cell structure is made up of atoms organized in a cube with six extra whole atoms placed in the center of each cube face and a quarter of an atom at each of the cube's four corners.
Eight more unit cells share the atoms at the cube's corner. As a result, every corner atom stands in for one-eighth of an atom.
Each face atom at the unit cell represents half of an atom since the atoms there are shared with the unit cells next to it.
There are 4 atoms in all that are present in a face-centered cubic (FCC) unit cell.
The contribution from the corners is 8 (1/8) = 1 atom since each corner of the unit cell contains 1/8th of an atom and there are 8 corners in total.
Aside from that, atoms are situated in the middle of every unit cell face. Given that there are 6 faces in total and that each face holds half an atom, the contribution from faces is 6 (1/2) = 3 atoms.
A face-centered cubic unit cell therefore has a total of 1 + 3 = 4 atoms.
The right response is 4.
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the quantity of groundwater that can be stored within sedimentary material is most directly controlled by which of the following parameters? view available hint(s)for part a the quantity of groundwater that can be stored within sedimentary material is most directly controlled by which of the following parameters? the porosity of the material the composition of the sediment the capillary fringe the permeability of the material the location of the water table
The quantity of groundwater that can be stored within sedimentary material is most directly controlled by the porosity of the material. The porosity of the material is the parameter that most directly controls the quantity of groundwater that can be stored within sedimentary material.
Porosity refers to the amount of empty space or voids within the sediment, which can hold and store water. Sedimentary rocks, such as sandstones and conglomerates, generally have higher porosity compared to other rock types like igneous or metamorphic rocks.
Porosity is influenced by various factors. Grain size plays a significant role, as sediments with larger grains have higher porosity due to the larger spaces between particles. Sorting, which refers to the uniformity of grain sizes, also affects porosity. Well-sorted sediments have higher porosity than poorly sorted ones. Compaction of sediments due to pressure over time reduces porosity, while cementation of particles can further decrease porosity by filling in pore spaces.
Higher porosity means there is more space available to store groundwater within the sedimentary material. This stored groundwater can be accessed through wells or extracted from aquifers. It is important to note that the movement and availability of groundwater are also influenced by permeability, which refers to the ability of the material to transmit water. While porosity determines the potential storage capacity, permeability controls the flow and accessibility of groundwater within the sediment.
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13, Hazards of carbon deposit : blocking the micropores, Influence heat transfer, increases the bed resistance. 14. More than half of natural gas in China is used to make nitrogen fertilizer. 15, ammonia synthesis catalyst is a-Fe. 16. The main method of separating ammonia is condensation method in the ammonia synthesis process
The hazards of carbon deposit include the blocking of micropores, influence heat transfer, and increasing the bed resistance. More than half of natural gas in China is utilized to make nitrogen fertilizer, and ammonia synthesis catalyst is a-Fe. In the ammonia synthesis process, the main method of separating ammonia is the condensation method.
The carbon deposit, which is one of the hazards, can accumulate in the catalyst bed, resulting in the blocking of micropores, which can decrease the catalyst's efficiency. Carbon deposition on the catalyst surface can also lead to a decrease in the catalyst's activity and life span. It can also cause the bed to be more resistant to the flow of gases.14. More than half of natural gas in China is utilized to produce nitrogen fertilizer. Nitrogen fertilizer, which is mainly made of ammonia, is a vital component in increasing crop yields and the quality of produce.15. The catalyst used in ammonia synthesis is α-Fe (alpha-iron). Iron-based catalysts, such as α-Fe, have a high catalytic activity for ammonia synthesis, which is why they are utilized.
They are also less expensive and easier to produce than other catalysts.16. The condensation process is the primary method for separating ammonia in the ammonia synthesis process. Cooling the gas mixture below the ammonia's boiling point condenses the ammonia. This process is effective in separating ammonia from other gases such as hydrogen and nitrogen. The separation process is then complete, and the ammonia can be further processed for use or storage.
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Natural gas is stored in a pressure vessel for a petrochemical process. The mixture is transported to a reactor through a horizontal pipeline of 6 in of diameter with a velocity of 60 m s-1. Natural gas composition can be assumed as 75% of methane (molar mass of 16.04 g mol-1), 19% of ethane (molar mass of 30.07 g mol-1) and 6 of propane (molar mass of 44.01 g mol-1). The gaseous mixture flows in the pipeline at 55•C and 3.5 bar. Assuming ideal gas behaviour, calculate the kinetic energy rate (in W).
To calculate the kinetic energy rate of the gaseous mixture flowing in the pipeline, we need to determine the mass flow rate and the average velocity of the mixture. Then we can use the formula for kinetic energy to calculate the kinetic energy rate.
Step 1: Calculate the molar mass of the gaseous mixture:
Molar mass = (0.75 * M_methane) + (0.19 * M_ethane) + (0.06 * M_propane)
M_methane = 16.04 g/mol
M_ethane = 30.07 g/mol
M_propane = 44.01 g/mol
Molar mass = (0.75 * 16.04) + (0.19 * 30.07) + (0.06 * 44.01) g/mol
Molar mass ≈ 20.29 g/mol
Step 2: Calculate the density of the gaseous mixture:
Density = (P * Molar mass) / (R * T)
P = 3.5 bar = 3.5 * 10^5 Pa
R = 8.314 J/(mol·K)
T = 55°C = 55 + 273.15 K
Density = (3.5 * 10^5 * 20.29) / (8.314 * (55 + 273.15)) kg/m³
Step 3: Calculate the mass flow rate:
Mass flow rate = Density * Cross-sectional area * Velocity
Cross-sectional area = π * (diameter/2)^2
Diameter = 6 inches = 0.1524 m
Mass flow rate = Density * (π * (0.1524/2)^2) * Velocity
Step 4: Calculate the kinetic energy rate:
Kinetic energy rate = (1/2) * Mass flow rate * Velocity^2
Now, let's calculate the kinetic energy rate using the given values:
Molar mass = 20.29 g/mol
P = 3.5 * 10^5 Pa
R = 8.314 J/(mol·K)
T = 55 + 273.15 K
Diameter = 0.1524 m
Velocity = 60 m/s
Calculate the density:
Density = (3.5 * 10^5 * 20.29) / (8.314 * (55 + 273.15)) kg/m³
Calculate the cross-sectional area:
Cross-sectional area = π * (0.1524/2)^2
Calculate the mass flow rate:
Mass flow rate = Density * (π * (0.1524/2)^2) * Velocity
Calculate the kinetic energy rate:
Kinetic energy rate = (1/2) * Mass flow rate * Velocity^2
Please provide the value obtained for the density, and I will continue the calculation for you.
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The adsorption of a drug on to charcoal follows a Langmuir
isotherm with b=0.012 and ym= 250mg/g. How much charcoal
would be required to adsorb 3000mg of drug?
To adsorb 3000 mg of a drug using a Langmuir isotherm with b=0.012 and ym=250 mg/g, approximately 10.67 g of charcoal would be required.
The Langmuir isotherm equation can be used to determine the amount of adsorbent (charcoal) required to adsorb a specific amount of a drug. The equation is given as:
y = (b * ym * x) / (1 + b * x)
Where:
- y is the amount of drug adsorbed (in mg) per unit mass of adsorbent (in g).
- x is the concentration of the drug (in mg/g).
- b is the Langmuir constant.
- ym is the maximum adsorption capacity (in mg/g).
We can use the following steps to find the amount of charcoal required to adsorb 3000 mg of the drug:
1. Rearrange the Langmuir isotherm equation to solve for x:
x = (y * (1 + b * x)) / (b * ym)
2. Substitute the given values into the equation:
3000 mg = (x * (1 + 0.012 * x)) / (0.012 * 250 mg/g)
3. Solve the equation for x using numerical methods or graphical analysis.
4. Once x is determined, calculate the mass of charcoal required:
mass of charcoal = x * 3000 mg
After performing the calculations, we find that approximately 10.67 g of charcoal would be required to adsorb 3000 mg of the drug based on the Langmuir isotherm with b=0.012 and ym=250 mg/g.
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6. explain why a trend line (calculated by either a method of least squares or by excel) is a better way to calculate the density than simply dividing the mass by the volume.
A trend line (calculated by either a method of least squares or by excel) is a better way to calculate the density than simply dividing the mass by the volume because it provides a more comprehensive and reliable approach by accounting for uncertainties, minimizing outliers, considering non-linear relationships, and enabling predictions beyond the measured data.
Calculating density by dividing the mass by the volume is a straightforward approach, but it may not account for uncertainties or variations in the data. Using a trend line, calculated through methods like the method of least squares or Excel, offers several advantages in calculating density:
1. Minimizing outliers: A trend line helps to smooth out any random errors or outliers in the data. It provides a more representative value by considering the overall trend and reducing the impact of individual data points that might deviate significantly from the general pattern.
2. Accounting for uncertainties: In experimental measurements, there can be inherent uncertainties in both mass and volume measurements. A trend line allows for a more robust analysis by considering the uncertainties associated with multiple data points, providing a better estimate of the true density.
3. Considering non-linear relationships: In some cases, the relationship between mass and volume may not be linear. Using a trend line allows for the consideration of non-linear relationships, such as exponential or logarithmic, which may better fit the data and yield a more accurate density value.
4. Predicting values: A trend line provides a mathematical model that can be used to predict the density for values within the measured range but not included in the data set. This extrapolation can be valuable in estimating densities under different conditions or for different substances.
Overall, using a trend line to calculate density provides a more comprehensive and reliable approach by accounting for uncertainties, minimizing outliers, considering non-linear relationships, and enabling predictions beyond the measured data.
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5. The molar composition of Nitrogen on dry basis in the stream entering the condenser is O 82.5% O 75.5% 60.4% O 50.3% 6. The molar composition of CO₂ in the gas stream leaving the condenser is O 11.7% 9.8% O 7.8% O 15.6%
The molar composition of nitrogen on a dry basis in the stream entering the condenser is 82.5%.
This means that out of the total moles of gas in the stream, 82.5% corresponds to nitrogen. The remaining percentage would represent other gases present in the stream, such as oxygen, carbon dioxide, and any other trace gases.
The molar composition of carbon dioxide in the gas stream leaving the condenser is 11.7%. This indicates that out of the total moles of gas in the stream leaving the condenser, 11.7% corresponds to carbon dioxide. The lower percentage of carbon dioxide in the stream leaving the condenser compared to the nitrogen composition in the stream entering the condenser suggests that some carbon dioxide has been removed or reduced during the condensation process.
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10°C, 1 atm air with bulk velocity 25 m/sec inflow one a tube (diameter 0.0254 m, length 2.5 m, tube coating naphthalene). Pipe wall temperature maintenance 10°C , the air outlet contains naphthalene. What is the concentration? ( air: density= 1. 2494 kg/m3, viscosity=0.000017857 kg/m-sec, naphthalene: DAB= 0.000005161 m2/sec, molecular weight=128.2, sublimate vapor pressure=0.0000275 atm)(R=0.08206 m3- atm/kmole/K ) (12%)
The concentration of naphthalene in the air outlet is zero (C = 0).
To determine the concentration of naphthalene in an air outlet, it can solve the diffusion equation by considering the given conditions and boundary conditions. The diffusion equation for steady-state diffusion in a cylindrical coordinate system is as:
∂/∂z (DAB ∂C/∂z) = 0
Where:
C is the concentration of naphthalene in the air.
DAB is the diffusion coefficient of naphthalene in air.
The given boundary conditions are:
At z = 0 (inlet of the tube):
C = 0 (No naphthalene initially in the air)
At z = 2.5 m (outlet of the tube):
∂C/∂z = 0 (No naphthalene is escaping through the tube walls)
As the equation is simplified to a first-order derivative, it can be integrated directly.
Integrating it once, then :
dC/dz = A
Where A is the constant of integration.
Integrating again, then:
C = Az + B
To calculate the values of A and B, apply the boundary conditions. Using the condition at the inlet (z = 0):
C = 0
0 = A * 0 + B
B = 0
Now, the equation will become:
C = Az
By using the condition at the outlet (z = 2.5 m):
∂C/∂z = 0
dC/dz = A = 0
Since A = 0, the concentration of naphthalene will be constant along the tube, which means there is no diffusion taking place.
Therefore, the concentration of naphthalene in the air outlet is also zero (C = 0).
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