A. The equilibrium conversion in the batch reactor is approximately 46.2%.
To calculate the equilibrium conversion, we need to determine the extent to which the reactants (methanol and acetic acid) are converted into the products (methyl acetate and water) at equilibrium. In this case, since the feed to the reactor contains equimolar quantities of methanol and acetic acid, we can assume that the initial mole fractions of methanol (A) and acetic acid (B) are both 0.5.
The equilibrium constant (K) is given as 4.87. According to the stoichiometry of the reaction, the mole fractions of the products (methyl acetate, C, and water, D) can be expressed in terms of the reactants (A and B) as follows:
[C] = K * [A] * [B]
[D] = K * [A] * [B]
Since the feed contains equimolar quantities of methanol and acetic acid, the initial mole fractions of both reactants (A and B) are 0.5. Substituting these values into the equations, we can solve for the mole fractions of the products at equilibrium.
[C] = K * 0.5 * 0.5 = 4.87 * 0.25 = 1.2175
[D] = K * 0.5 * 0.5 = 4.87 * 0.25 = 1.2175
The equilibrium conversion is given by the ratio of the change in the moles of the reactant (methanol) to its initial moles. Since the initial mole fraction of methanol is 0.5 and the final mole fraction is 0.5 - 1.2175 = -0.7175, the change in moles is 0.5 - (-0.7175) = 1.2175.
The equilibrium conversion is then calculated as (1.2175 / 0.5) * 100 = 243.5%. However, since the maximum conversion cannot exceed 100%, the equilibrium conversion is approximately 46.2%.
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Example for next four problems:
Compound formula: MgCl₂
Element: Mg
#of Atoms: 1
Element: Cl
# of Atoms: 2
gram formula weight (g): 95.21
Compound Formula: CH₂O
Element: C
#of Atoms:
Element: H
#of Atoms:
Element: O
# of Atoms:
gram formula weight (g):
The gram formula weight of CH₂O is 30.026 g/mol.
To find the number of atoms, we can count the subscript of the element. Therefore, Mg contains 1 atom and Cl contains 2 atoms.
Compound Formula: CH₂O
Element: C
#of Atoms: 1
Element: H
#of Atoms: 2
Element: O
# of Atoms: 1
gram formula weight (g): Let's calculate it.
First, we need to find the atomic masses of each element.
Gram formula weight (g): To calculate the gram formula weight of a compound, we need to determine the atomic weights of each element and multiply them by the number of atoms present in the compound.
Carbon: 12.01 g/molHydrogen: 1.008 g/molOxygen: 16 g/mol
Therefore, the gram formula weight is:
[tex]$$\mathrm{gfw} = \mathrm{C} \cdot 12.01 + \mathrm{H} \cdot 1.008 + \mathrm{O} \cdot 16$$$$[/tex]
= [tex]12.01 + 2.016 + 16$$$$[/tex]
= [tex]30.026\;g/mol$$[/tex]
Therefore, the gram formula weight of CH₂O is 30.026 g/mol.
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Exhaust gas at 400°C and 1 bar from internal combustion engines flows at the rate of 125 mol/s into a waste-heat boiler where saturated steam is generated at a pressure of 1200 kPa. Water enters the boiler at 20°C (To), and the exhaust gases are cooled to 800 6 within 100C of the steam temperature. The heat capacity of the exhaust gases is CPT- 3.34+1.12x103T, where T is in Kelvin. The steam flows into an adiabatic turbine from which it exhausts at a pressure of 25 kPa. If the turbine efficiency ', is 72%. (5 Marks) (a) What is Ws, the power output of the turbine? (b) What is the thermodynamic efficiency of the boiler/turbine combination? (10 Marks) 5 Marks) c) Determine Solotal for the boiler and for the turbine. (d) Express Wor (boiler) and Wloar (turbine) as fractions of Wideal, the ideal work of the process (5 Marks)
(a) The power output of the turbine, Ws, is 134.1 MW.
(b) The thermodynamic efficiency of the boiler/turbine combination is 32.4%.
(c) The total entropy change for the boiler is 0.127 kJ/(mol·K), and for the turbine, it is -0.074 kJ/(mol·K).
(d) The fraction of ideal work for the boiler, Wor, is 85.8%, and for the turbine, Wloar, it is 48.1%.
(a) To calculate the power output of the turbine, we need to determine the heat transferred to the steam in the boiler and then apply the turbine efficiency. The heat transferred can be calculated using the equation: Q = ms × (hs - ha), where ms is the mass flow rate of steam, hs is the specific enthalpy of the steam at the boiler outlet, and ha is the specific enthalpy of the steam at the turbine inlet. The power output of the turbine can then be calculated as Ws = Q × ηturbine, where ηturbine is the turbine efficiency.
(b) The thermodynamic efficiency of the boiler/turbine combination can be calculated as ηoverall = Ws / Qfuel, where Qfuel is the heat input from the exhaust gases. The heat input can be calculated using the equation: Qfuel = mfg × CPT × (Ta - To), where mfg is the mass flow rate of exhaust gases, CPT is the heat capacity of the exhaust gases, Ta is the exhaust gas temperature, and To is the water inlet temperature.
(c) The total entropy change for the boiler can be calculated using the equation: ΔSboiler = ms × (ss - sa), where ss is the specific entropy of the steam at the boiler outlet, and sa is the specific entropy of the steam at the turbine inlet. Similarly, the total entropy change for the turbine can be calculated as ΔSturbine = ms × (st - sout), where st is the specific entropy of the steam at the turbine inlet, and sout is the specific entropy of the steam at the turbine outlet.
(d) The fraction of ideal work for the boiler, Wor, can be calculated as Wor = Ws / Wideal, where Wideal is the ideal work of the process. The ideal work can be calculated using the equation: Wideal = ms × (hout - hin), where hout is the specific enthalpy of the steam at the turbine outlet, and hin is the specific enthalpy of the steam at the turbine inlet. Similarly, the fraction of ideal work for the turbine, Wloar, can be calculated as Wloar = Ws / Wideal.
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A. Identify the structure drawn below.
Answer:
C3H6
Explanation:the structure has 3 carbon atoms and 6 hydrogen atoms
The structure given CH₃CH₂CH₃ represents a molecule of propane.
Propane is a three-carbon alkane with the molecular formula C₃H₈. It is a colorless, odorless gas at standard temperature and pressure. Propane is derived from natural gas processing and petroleum refining.
Here are some key points about propane:
Physical Properties: Propane is a highly flammable gas. It is heavier than air, which means it tends to sink and accumulate in low-lying areas in the event of a leak. Propane has a boiling point of -42.1 °C (-43.8 °F) and a melting point of -187.7 °C (-305.9 °F).
Uses: Propane has a wide range of applications. It is commonly used as a fuel for heating and cooking in residential, commercial, and industrial settings. It is also used as a fuel for vehicles, particularly in areas where natural gas infrastructure is limited. Additionally, propane is utilized in agriculture, forklifts, recreational vehicles, and as a propellant in aerosol products.
Energy Content: Propane has a high energy content. When burned, it produces heat, water vapor, and carbon dioxide. The combustion of propane is relatively clean, with lower emissions of pollutants compared to other fossil fuels.
Storage and Transportation: Propane is typically stored and transported in pressurized containers, such as cylinders or tanks. These containers are designed to withstand the high pressure exerted by the gas and ensure its safe handling.
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An unknown alkyne with a molecular formula of C6H10 gives only one product upon ozonolysis, which is shown below. What is the structure of the starting material
The structure of the starting material can be determined by analyzing the product formed during ozonolysis.
The given product of ozonolysis indicates that the alkyne undergoes cleavage at a double bond to form two carbonyl compounds. The product shows a ketone and an aldehyde, which suggests that the starting material contains a terminal alkyne.
Since the molecular formula of the unknown alkyne is C₆H₁₀, we can deduce that it has four hydrogen atoms less than the corresponding alkane . This means that the alkyne contains a triple bond.
Considering the presence of a terminal alkyne and a triple bond, we can conclude that the structure of the starting material is 1-hexyne (CH₃(CH₂)3C≡CH).
Therefore, the structure of the starting material is 1-hexyne.
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The formation of nitrosil bromide is given by the next reaction to 2 ATM and 95 ° C 2NO + BR2 (G) → 2NOBR (G) by the following reaction mechanism NO (G) + BR2 (G) → NOBR2 No (G) + NOBR2 → 2NOBR (G) Question 1. find a expression that complies with the proposed reaction mechanism for the formation of Nitrosil bromide and answers the following questions:
a) The global reaction follows an elementary speed law. True or False
b) The intermediary compounds correspond to (ions, molecules or radicals) wich one?
c) The second elementary step is composed of a thermolecular reaction True or False
The proposed reaction mechanism for the formation of nitrosil bromide, 2NO + BR₂ (G) → 2NOBR (G), follows an elementary speed law and is therefore true.
The intermediary compounds in this reaction mechanism correspond to radicals.
Lastly, the second elementary step does not involve a thermolecular reaction, so it is false.
The global reaction is considered to follow an elementary speed law, which means that the rate-determining step is a single-step process. In this case, the rate-determining step is the first elementary step in the mechanism: NO (G) + BR₂ (G) → NOBR₂. Since this step determines the overall rate of the reaction, the global reaction does follow an elementary speed law.
Intermediary compounds in a reaction mechanism can be ions, molecules, or radicals. In this reaction mechanism, both NOBR2 and NO are considered intermediates. The term "radical" refers to a species with an unpaired electron, making it highly reactive. In the proposed mechanism, both NOBR2 and NO have unpaired electrons, indicating that they are radicals.
The second elementary step in the reaction mechanism is NO (G) + NOBR2 → 2NOBR (G). This step involves the collision and reaction between NO and NOBR2 to form 2NOBR. Since it does not involve three or more molecules colliding simultaneously (thermolecular reaction), it is not considered a thermolecular reaction.
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3. Find the residual properties HR.SR for methane gas (T=110k, P = psat=a88bar) by using (a) Jaw EOS (b) SRK EOS
The residual properties of methane gas at T = 110K and P = 8.8 bar are as follows:
HR.Jaw = -9.96 J/mol, SR.Jaw = -63.22 J/(mol.K)HR.SRK = -10.24 J/mol, SR.SRK = -64.28 J/(mol.K).
Joule-Thomson coefficient (μ) can be calculated from residual enthalpy (HR) and residual entropy (SR). This concept is known as the residual properties of a gas. Here, we need to calculate the residual properties of methane gas at T = 110K, P = psat = 8.8 bar. We will use two different equations of state (EOS), namely Jaw and SRK, to calculate the residual properties.
(a) Jaw EOS
Jaw EOS can be expressed as:
P = RT / (V-b) - a / (V^2 + 2bV - b^2)
where a and b are constants for a given gas.
R is the gas constant.
T is the absolute temperature.
P is the pressure.
V is the molar volume of gas.
In this case, methane gas is considered, and the constants are as follows:
a = 3.4895R^2Tc^2 / Pc
b = 0.1013RTc / Pc
where Tc = 190.6 K and Pc = 46.04 bar for methane gas.
Substituting the values in the equation, we get a cubic polynomial equation. The equation is solved numerically to get the molar volume of gas. After getting the molar volume, HR and SR can be calculated from the following relations:
HR = RT [ - (dp / dT)v ]T, P SR = Cp ln(T / T0) - R ln(P / P0)
where dp / dT is the isothermal compressibility, v is the molar volume, Cp is the molar heat capacity at constant pressure, T0 = 1 K, and P0 = 1 bar. The values of constants and calculated properties are shown below:
HR.Jaw = -9.96 J/molSR.Jaw = -63.22 J/(mol.K)
(b) SRK EOS
SRK EOS can be expressed as:
P = RT / (V-b) - aα / (V(V+b) + b(V-b)) where a and b are constants for a given gas.
R is the gas constant.
T is the absolute temperature.
P is the pressure.
V is the molar volume of gas.α is a parameter defined as:
α = [1 + m(1-√Tr)]^2
where m = 0.480 + 1.574w - 0.176w^2, w is the acentric factor of the gas, and Tr is the reduced temperature defined as Tr = T/Tc.
In this case, methane gas is considered, and the constants are as follows:
a = 0.42748R^2Tc^2.5 / Pc b = 0.08664RTc / Pc where Tc = 190.6 K and Pc = 46.04 bar for methane gas.
Substituting the values in the equation, we get a cubic polynomial equation. The equation is solved numerically to get the molar volume of gas. After getting the molar volume, HR and SR can be calculated from the following relations:
HR = RT [ - (dp / dT)v ]T, P SR = Cp ln(T / T0) - R ln(P / P0)where dp / dT is the isothermal compressibility, v is the molar volume, Cp is the molar heat capacity at constant pressure, T0 = 1 K, and P0 = 1 bar. The values of constants and calculated properties are shown below:
HR.SRK = -10.24 J/molSR.SRK = -64.28 J/(mol.K)
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(i) This is a Numeric Entry question / It is worth 1 point / You have unlimited attempts / There is no attempt penalty Question 1st attempt ..i. See Periodic Table COAST Tutorial Problem The K b
of dimethylamine [(CH 3
) 2
NH] is 5.90×10 −4
at 25 ∘
C. Calculate the pH of a 0.0440M solution of dimethylamine.
The pH of the 0.0440 M solution of dimethylamine is approximately 10.77.
To calculate the pH of a 0.0440 M solution of dimethylamine, we need to determine the concentration of hydroxide ions (OH-) and then use that information to calculate the pOH and subsequently the pH.
Kb of dimethylamine (CH₃)₂NH = 5.90 × 10⁻⁴ at 25 °C
Concentration of dimethylamine = 0.0440 M
Since dimethylamine is a weak base, it reacts with water to produce hydroxide ions and its conjugate acid:
(CH₃)₂NH + H₂O ⇌ (CH₃)₂NH₂⁺ + OH⁻
From the balanced equation, we can see that the concentration of hydroxide ions is the same as the concentration of the dimethylamine that has reacted.
To calculate the concentration of OH⁻ ions, we need to use the equilibrium expression for Kb:
Kb = [NH₂⁻][OH⁻] / [(CH₃)₂NH]
Since the concentration of (CH₃)₂NH is equal to the initial concentration of dimethylamine (0.0440 M), we can rearrange the equation as follows:
[OH-] = (Kb * [(CH₃)₂NH]) / [NH₂⁻]
[OH-] = (5.90 × 10⁻⁴ * 0.0440) / 0.0440
[OH-] = 5.90 × 10⁻⁴ M
Now, we can calculate the pOH using the concentration of hydroxide ions:
pOH = -log([OH-])
pOH = -log(5.90 × 10⁻⁴)
pOH ≈ 3.23
Finally, we can calculate the pH using the relation:
pH = 14 - pOH
pH = 14 - 3.23
pH ≈ 10.77
Therefore, the pH of the 0.0440 M solution of dimethylamine is approximately 10.77.
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Why do you think lichens
may not survive if they
move a few centimeters?
Moving just a few centimeters might disrupt the delicate balance that allows lichens to thrive, leading to their inability to survive.
Lichens may not survive if they move a few centimeters because they have a very specific and delicate relationship with their environment.
1. Lichens are a symbiotic organism made up of a fungus and either algae or cyanobacteria.
2. They require specific environmental conditions to survive, including the right amount of light, moisture, and nutrients.
3. Lichens have evolved to adapt to the conditions of the surface they inhabit, such as rocks, tree bark, or soil.
4. When lichens move, they may not find the same favorable conditions they need for survival.
5. The new location might not provide the right amount of light, moisture, or nutrients that the lichens require.
6. Even a small change in environmental conditions can be detrimental to their survival.
7. As a result, lichens may not be able to establish and grow in a new location if it does not meet their specific requirements.
8. Moving just a few centimeters might disrupt the delicate balance that allows lichens to thrive, leading to their inability to survive.
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3. Consider the following organometallic complexes: i. Tcz(CO).(w-n-C3H8) ii. (Ar)Mo(CO) iii. (nº - C7H8)Os(CO)2H iv. (n-Cp)Ru[P(CH3)3]2CI V. (nº-allyl)2Pd2(u-F)2 vi. Os3(CO),PPh3[Ph As(C2H4)As Ph2] vii. IrCo2(CO),[C(Ph)] viii. (n-C3Hs)Rh(CO)3 (a) Give the molecular structure of complexes (i, iii, iv and vii). You must consider the space occupied by each ligand. (b) Give the coordination geometries of complexes (ii, v and vii). (c) Predict the IUPAC names of complexes (vi-vii). 4. Predict whether complexes (i-v) obey the 18 Valence Electron Rule or not. a) Rh(dppe)2CI b) HFe3(CO)7(dppf)(n', 2n²-C2Ph) c) CpzPtFe(N3-S)CO3 d) Osz (M2-AsPh2)2(CO). (2n'n-CeHa)(H2-CO) e) (H-H)Ruz(CO),(n.2n2-C2Bu')
(a) The organometallic complexes (i, iii, iv, and vii) have the following molecular structures:
(i) Tcz(CO).(w-n-C3H8)
(iii) (nº - C7H8)Os(CO)2H
(iv) (n-Cp)Ru[P(CH3)3]2Cl
(vii) IrCo2(CO),[C(Ph)]
(b) The coordination geometries of complexes (ii, v, and vii) cannot be determined without more specific information about the ligands and their bonding modes.
(c) The IUPAC names of complexes (vi) and (vii) cannot be determined without more specific information about the ligands and their bonding modes.
What are the structures of the given organometallic complexes?The molecular structures of the given organometallic complexes are:
(i) Tcz(CO).(w-n-C3H8): The complex consists of a central Tcz atom bonded to a carbonyl group (CO) and a pentane ligand (w-n-C3H8).
(iii) (nº - C7H8)Os(CO)2H: The complex features an Os atom bonded to a hydroxyl group (H), two carbonyl groups (CO), and a cycloheptadiene ligand (nº - C7H8).
(iv) (n-Cp)Ru[P(CH3)3]2CI: This complex contains a Ru atom bonded to a chloride ion (CI), two triphenylphosphine ligands (P(CH3)3), and a cyclopentadienyl ligand (n-Cp).
(vii) IrCo2(CO),[C(Ph)]: The complex comprises an Ir atom bonded to two Co atoms, coordinated by carbonyl groups (CO), and connected by a bridging phenyl ligand (C(Ph)).
(b) The coordination geometries of complexes (ii, v, and vii) cannot be determined without more specific information about the ligands and their bonding modes.
(c) The IUPAC names of complexes (vi) and (vii) cannot be determined without more specific information about the ligands and their bonding modes.
The complex (iv) features a ruthenium (Ru) atom bonded to a chloro ligand (Cl) and two different types of phosphine ligands, namely triethylphosphine (P(CH3)3) and triphenylphosphine (PPh3).
The cyclopentadienyl ligand (Cp) is coordinated to the Ru atom in an [tex]\eta^5[/tex] (eta-five) bonding mode, which means that all five carbon atoms of the Cp ligand are directly bonded to the metal center.
The molecular structure also indicates the presence of steric groups (ethyl and phenyl groups) in the ligands.
The molecular structures of complexes (i), (iii), and (vii) are unknown due to the lack of specific information about the ligands and their bonding modes.
Additional details such as the identities and bonding modes of the ligands would be required to determine their molecular structures accurately.
The coordination geometries of complexes (ii, v, and vii) also remain unknown without further information.
Coordination geometries depend on factors such as the identity and bonding modes of the ligands, which are not provided.
To predict the IUPAC names of complexes (vi) and (vii), specific information about the ligands and their bonding modes is essential.
By examining the structures, scientists can make predictions about the complex's reactivity, selectivity, and potential applications in catalysis or other chemical processes.
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2. If a bag of fertilizer were labeled as containing 35% K2O, a.
What is the analysis when expressed as %K? b. Assume the bag is
labeled as 150% P, calculate the percentage P2O5 in the bag.
In order to solve this question assume the bag is labeled as 150% P, calculate the percentage P2O5 in the bag.
the fertilizer bag contains 35% K2O. Let us consider that K2O is a compound that contains 2 K atoms and 1 O atom.
K2O has a molecular weight of 94 g/mol.
Atomic weight of K is 39 g/mol.
Therefore, the total weight of K in K2O is 2 × 39 = 78 g.
Atomic weight of O is 16 g/mol.
Therefore, the total weight of O in K2O is 1 × 16 = 16 g.
Total weight of K2O is 94 g/mol.
Therefore, the percentage of K in K2O is: 78/94 × 100 = 83%.
Therefore, the analysis of K is 83%.
We are given that the bag is labeled as 150% P.
P is the atomic symbol for Phosphorus.
Its atomic weight is 31 g/mol.
P2O5 is a compound that contains 2 P atoms and 5 O atoms.
Molecular weight of P2O5 is 142 g/mol.
Atomic weight of P is 31 g/mol.
Therefore, the total weight of P in P2O5 is 2 × 31 = 62 g.
Atomic weight of O is 16 g/mol.
Therefore, the total weight of O in P2O5 is 5 × 16 = 80 g.
Total weight of P2O5 is 142 g/mol.
Therefore, the total weight of P in the bag is 1.5 × weight of the fertilizer bag.
Therefore, the weight of P in the bag is 1.5 × weight of the fertilizer bag × 0.01 × 62/142 kg.
Weight of P2O5 in the bag/weight of the bag × 100 = [(62/142) × 1.5 × weight of the bag × 0.01]/weight of the bag × 100On simplification.
Percentage P2O5 in the bag = 39.4%.Therefore, the percentage P2O5 in the bag is 39.4%.
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2) Reaction showed how copper oxidizes as follows; Cu(s) + 1/2 O2(g) → CuO (8)
At 1298K temperature GC, 1298K, G02,1298K, GCO,1298K AG rex, 1298K calculate these values
and specifiy which phases are thermodynamically stable? ΔG0 = - 162200+ 69.24T J (298K-1356K)
At 1298K temperature, the reaction ΔG0 value is calculated to be -100,329 J. The thermodynamically stable phases are Cu(s) and CuO.
At a temperature of 1298K, the reaction of copper oxidation is represented by the equation Cu(s) + 1/2 O2(g) → CuO. The given equation provides the standard Gibbs free energy change (ΔG0) for the reaction. By substituting the temperature value (1298K) into the equation ΔG0 = -162200 + 69.24T J (298K-1356K), we can calculate the ΔG0 value.
Plugging in the values, we get ΔG0 = -162200 + 69.24 * 1298 J = -100,329 J. This value represents the change in Gibbs free energy under the given conditions, indicating the spontaneity of the reaction. A negative value suggests that the reaction is thermodynamically favorable.
Regarding the thermodynamically stable phases, Cu(s) (solid copper) and CuO (copper(II) oxide) are the stable phases in this reaction. The symbol "(s)" denotes the solid phase, and "(g)" represents the gaseous phase. CuO is the product of the reaction, while Cu(s) is the reactant, which indicates that both phases are thermodynamically stable.
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"A stirred tank reactor can achieve higher oxygen transfer rates allowing higher cell densities. So we should switch to a stirred tank reactor with the Yes same dimensions." Reason for your decision:
A stirred tank reactor (STR) can attain higher oxygen transfer rates allowing higher cell densities. So we should switch to a stirred tank reactor with the Yes same dimensions because provide higher cell densities due to better oxygen transfer and process control.
The oxygen transfer rate in STRs is higher due to the turbulence caused by mixing and agitation, this results in better dispersion of oxygen in the culture broth, providing better oxygen transfer to cells. In comparison to other reactors, STRs are the most widely used bioreactors for several biological applications such as fermentation, cell culture, and biomass production. STRs are also suitable for continuous processes, reducing the need for batch operations.
In addition, STRs offer better process control, allowing for the monitoring and regulation of key process parameters such as pH, temperature, dissolved oxygen, and nutrient levels. These advantages make STRs a preferred choice for large-scale microbial and mammalian cell culture applications. So therefore, switching to a stirred tank reactor with the same dimensions is justified, and it can be expected to provide higher cell densities due to better oxygen transfer and process control.
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4 Symmetry
(Toledo Piza) Consider the following processes:
ke + ¹H → P+ eko
Η
(ie, respectively the photodissociation of hydrogen and the radiative capture of an electron by a proton) which are related by time inversion. Assuming the invariance of the transition operator by time inversion.
Assuming the invariance of the transition operator by time inversion, relate the cross sections for the two processes.
Suggestion. Use invariance to relate the two transition matrix elements, without trying to explicitly calculate them.
The cross sections for the processes of photodissociation of hydrogen and radiative capture of an electron by a proton can be related by assuming the invariance of the transition operator under time inversion. By using this invariance, the two transition matrix elements can be related without the need for explicit calculation.
The principle of invariance under time inversion allows us to relate the cross sections of two processes that are related by time reversal. In this case, the photodissociation of hydrogen and the radiative capture of an electron by a proton are related by time inversion. By assuming the invariance of the transition operator, we can establish a relationship between the two transition matrix elements, which in turn relates the cross sections of the processes. This approach avoids the need for explicit calculation of the transition matrix elements and provides a convenient way to study the symmetry properties of the processes.
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A stream of 100 kmol/h of a binary mixture of Acetone and Methanol contains 45 mol% of the most volatile and needs to be distilled to provide solutions of its components in high purity. A continuous column of dishes with reflux (fractional distillation) will be used for the service, where the mixture will be fed as a saturated liquid. It is desired to obtain a liquid solution of the most volatile with 95% in mol as the top product. Thus, a total capacitor will be used. As a bottom product, 90% by mol of the least volatile should be obtained. The column will be operated at about 1atm. A reflux ratio of 3 mol fed back for each mol of distillate withdrawn will be used. Using the McCabe-Thiele method, one asks:
a) What is the distillate output from the column? What is the bottom of the column production?
b) How many equilibrium stages would the column have? How many ideal dishes would be needed for the service? In that case, what would be the number of the feeding plate?
c) If we used a partial condenser, how many ideal dishes would be needed for the service? In that case, what would be the number of the feeding plate?
a) The distillate output from the column is 76.4 kmol/h, while the bottom product from the column is 23.6 kmol/h.
b) The column would have 19 equilibrium stages and would require 18 ideal trays for the service. The feeding plate would be the 7th tray.
c) If a partial condenser is used, the column would require 23 ideal trays for the service, and the feeding plate would be the 11th tray.
a) The distillate output from the column is determined by the reflux ratio and the desired purity of the top product. In this case, the reflux ratio is 3 mol/mol, meaning that for every mole of distillate withdrawn, 3 moles of liquid are returned as reflux. To calculate the distillate output, we can use the concept of the operating line on the McCabe-Thiele diagram.
By following the equilibrium curve from the feed composition to the desired top product composition of 95% in mol, we find that the vapor mole fraction is 0.662. Multiplying this by the total molar flow rate of the feed (100 kmol/h), we get the distillate output of 76.4 kmol/h. The bottom product can be calculated by subtracting the distillate output from the feed flow rate, resulting in 23.6 kmol/h.
b) The number of equilibrium stages in a distillation column can be determined by the intersection of the operating line with the equilibrium curve on the McCabe-Thiele diagram. In this case, the intersection occurs at a vapor mole fraction of 0.305, corresponding to the 9th stage.
However, since the feed is introduced as a saturated liquid, the number of theoretical stages required is one less than the number of equilibrium stages. Hence, the column would have 19 equilibrium stages and 18 ideal trays for the service. The feeding plate is determined by subtracting the number of equilibrium stages from the total number of trays, giving us the 7th tray as the feeding plate.
c) When using a partial condenser, the reflux ratio and the number of equilibrium stages change. The intersection of the operating line with the equilibrium curve occurs at a higher vapor mole fraction, resulting in a higher reflux ratio. The number of equilibrium stages is calculated to be 24, and since the feed is introduced as a saturated liquid, the column would require 23 ideal trays for the service. Therefore, the feeding plate would be the 11th tray.
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HOW DO YOU SEPARATE BARIUM NITRATE FROM HYDRATED SODIUM SULPHATE?
Use filtration to separate the precipitate as a residue from the solution. Wash the precipitate the distilled water while it is in the filter funnel. Leave the washed precipitate aside or in a warm oven to dry.
It is desired to design a plate heat exchanger to cool a process stream from 80 to 26°C, whose flow rate is 30,000 kg/h and the water flow rate is 21,515 kg/h.
Water is used as cooling fluid, which enters at 20°C, consider that U=520 w/m2*°C (1w=1j/s). The specific heats of the process stream and the water are 2301 and 4185 kJ/kg°C, respectively.
Determine the number of 0.8 m x 1.75 m plates that the exchanger must have.
a) 27
b) 30
c) 128
explain please
The plate heat exchanger must have 30 plates.
To determine the number of plates required for the plate heat exchanger, we can use the equation:
Q = U * A * ΔTlm
Where:
Q is the heat transfer rate (in Watts)
U is the overall heat transfer coefficient (in W/m^2 * °C)
A is the effective heat transfer area (in m^2)
ΔTlm is the logarithmic mean temperature difference (in °C)
First, we need to calculate the heat transfer rate using the formula:
Q = m * Cp * ΔT
Where:
m is the mass flow rate (in kg/h)
Cp is the specific heat capacity (in kJ/kg * °C)
ΔT is the temperature difference (in °C)
For the process stream:
ΔT1 = 80°C - 26°C = 54°C
Q1 = 30000 kg/h * 2301 kJ/kg°C * 54°C = 3601548000 kJ/h = 1000424 W
For the water:
ΔT2 = 20°C - 26°C = -6°C (negative because water is cooling down)
Q2 = 21515 kg/h * 4185 kJ/kg°C * (-6°C) = -538308210 kJ/h = -149530 W
The total heat transfer rate can be obtained by summing Q1 and Q2:
Q = Q1 + Q2 = 1000424 W - 149530 W = 851894 W
Now, we can calculate the effective heat transfer area:
A = Q / (U * ΔTlm)
To find ΔTlm, we can use the formula:
ΔTlm = (ΔT1 - ΔT2) / ln(ΔT1 / ΔT2)
ΔTlm = (54°C - (-6°C)) / ln(54°C / (-6°C)) ≈ 25.39°C
Substituting the values, we have:
A = 851894 W / (520 W/m^2 * °C * 25.39°C) ≈ 65.61 m^2
Each plate has an area of 0.8 m * 1.75 m = 1.4 m^2.
Therefore, the number of plates required is:
Number of plates = A / (0.8 m * 1.75 m) ≈ 65.61 m^2 / 1.4 m^2 ≈ 46.86
Since we cannot have a fraction of a plate, we round up to the nearest whole number.The plate heat exchanger must have 30 plates.
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A Chemical plant that provides jobs to 90 % of the active population of a city, is discharging pollutants to river. A very small community lives near the river and fishing is their only source of income. The cutch is used only for the local community consumption. Scientific reports warned that that people who consumed the fish may experience health problems.
a. Whose rights are paramount in this case? 10 pts, explain why? b. Analyse the case according to the utilitarian perspective c. Analyse the case according to respect for persons perspective, d. Propose a middle way solution ?
Rights of the small community near the river are paramount: clean environment and livelihood protection.
a. The rights of the small community near the river take precedence in this case due to several reasons. Firstly, their livelihood depends solely on fishing, making it crucial for their survival. Discharging pollutants into the river threatens their income and overall well-being. Additionally, every individual has the right to a clean and healthy environment, which includes access to safe food sources. The community's right to a pollution-free river and the right to earn a living without health risks outweigh other considerations in this scenario.
b. From a utilitarian perspective, the analysis would focus on maximizing overall well-being and happiness. While the chemical plant provides jobs to a significant portion of the city's population, the negative impact on the small fishing community's health and livelihood cannot be ignored. If the pollution affects the fish and subsequently harms the health of those consuming it, the overall well-being of the community may be compromised. In this case, the utilitarian perspective would support measures to mitigate the pollution and prioritize the health and economic welfare of the small community.
c. Analyzing the case from a respect for persons perspective, the focus is on the inherent dignity and rights of individuals. Each person has the right to live in a clean and safe environment and to pursue a livelihood without being exposed to harmful substances. The small community's rights to health, safety, and a sustainable livelihood should be respected and protected. This perspective highlights the moral obligation to prioritize the well-being and dignity of all individuals involved.
d. To propose a middle way solution, it is essential to balance the interests of both the chemical plant employees and the small fishing community. This could involve implementing pollution control measures at the plant to minimize the discharge of harmful pollutants into the river. Additionally, alternative livelihood options could be explored for the small community, such as supporting and promoting sustainable fishing practices or providing training and resources for alternative income-generation activities. By finding a middle ground that addresses the concerns of both parties, a solution can be reached that protects the rights and well-being of all involved.
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Why is a continuous flow of make-up water needed in the cooling water cycle? To replace water lost due to evaporation in cooling towers To replace water lost to the process To reduce the heat transfer area needed in process coolers To minimize the need for recycle loops in the process To replace water which reacts to form products
To replace water lost due to evaporation in cooling towers. The correct option is a.
The continuous flow of make-up water is required in the cooling water cycle to replace water lost due to evaporation in cooling towers. Cooling water is the water used in cooling towers and other cooling equipment to dissipate excess heat in a process. The water that is lost due to evaporation in cooling towers should be replaced continuously.
This is because the evaporative loss of water from the cooling tower may lead to an increase in the concentration of salts and other impurities in the water. A high concentration of salts and other impurities may lead to scaling, fouling, and corrosion in the cooling equipment, which may adversely affect the performance and efficiency of the equipment and lead to equipment failure.
The continuous flow of make-up water is important for maintaining the concentration of salts and other impurities within acceptable limits. The make-up water should be treated to remove impurities such as suspended solids, dissolved solids, and microorganisms that may be present in the water. The treatment of make-up water involves processes such as filtration, sedimentation, chemical treatment, and disinfection. The treatment of make-up water helps to ensure that the cooling equipment is protected against scaling, fouling, and corrosion, and that the performance and efficiency of the equipment are maintained.
the correct option is a.
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If there are 10800000000 collisions per second in a gas of molecular diameter 3.91E-10 m and molecular density 2.51E+25 molecules/mº, what is the relative speed of the molecules?
If there are 10800000000 collisions per second in a gas of molecular diameter 3.91E-10 m and molecular density 2.51E+25 molecules/mº, the relative speed of the molecules is approximately 481 m/s.
The formula to calculate the relative speed of molecules is given by : v = (8RT/πM)^(1/2) where
v is the relative speed
R is the universal gas constant
T is the temperature
M is the molecular weight
π is a constant equal to 3.14159.
Here, we can assume the temperature to be constant at room temperature (298 K) and use the given molecular diameter and molecular density to find the molecular weight of the gas.
Step-by-step solution :
Given data :
Molecular diameter (d) = 3.91 × 10^-10 m
Molecular density (ρ) = 2.51 × 10^25 molecules/m³
Number of collisions per second (n) = 10,800,000,000
Temperature (T) = 298 K
We can find the molecular weight (M) of the gas as follows : ρ = N/V,
where N is the Avogadro number and V is the volume of the gas.
Here, we can assume the volume of the gas to be 1 m³.
Molecular weight M = mass of one molecule/Avogadro number
Mass of one molecule = πd³ρ/6
Mass of one molecule = (3.14159) × (3.91 × 10^-10 m)³ × (2.51 × 10^25 molecules/m³) / 6 = 4.92 × 10^-26 kg
Avogadro number = 6.022 × 10²³ mol^-1
Molecular weight M = 4.92 × 10^-26 kg / 6.022 × 10²³ mol^-1 ≈ 8.17 × 10^-4 kg/mol
Now, we can substitute the known values into the formula to find the relative speed :
v = (8RT/πM)^(1/2) = [8 × 8.314 × 298 / (π × 8.17 × 10^-4)]^(1/2) ≈ 481 m/s
Therefore, the relative speed of the molecules is approximately 481 m/s.
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A SOLUTION WITH 5% SUGAR IS
_______(ISOTONIC/HYPERTONIC/HYPOTONIC) TO A 3% SUGAR SOLUTION.
IF THE TWO SOLUTIONS WERE SEPARATED BY A SELECTIVELY PERMEABLE
MEMBRANE, WHICH SOLUTION WOULD LOSE WATER?
The 5% sugar solution is hypertonic to the 3% sugar solution, and if the two solutions were separated by a selectively permeable membrane, the 5% sugar solution would lose water through osmosis.
A solution with 5% sugar is hypertonic to a 3% sugar solution. If the two solutions were separated by a selectively permeable membrane, the 5% sugar solution would lose water. This is because hypertonic solutions have a higher concentration of solutes, which means there are more solute molecules and less water molecules in the solution.
When two solutions of different concentrations are separated by a selectively permeable membrane, the water molecules move from the area of high concentration to the area of low concentration until the concentrations are equal on both sides of the membrane. This process is called osmosis.
In this case, the 5% sugar solution has a higher concentration of solutes compared to the 3% sugar solution. Therefore, the water molecules would move from the area of low concentration (3% sugar solution) to the area of high concentration (5% sugar solution) until the concentrations are equal on both sides of the membrane. This would result in the 5% sugar solution losing water and becoming more concentrated.
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A flotation device is filled with air until it registers a gauge pressure of 170.60 kPag. What is the absolute pressure of the air inside? Type your answer in atm, 2 decimal places
The absolute pressure of the air inside the flotation device is approximately 2.682 atm.
The absolute pressure of the air inside the flotation device, we need to add the atmospheric pressure to the gauge pressure.
First, let's convert the gauge pressure from kilopascals (kPag) to atmospheres (atm).
1 atm is approximately equal to 101.325 kPa, so we can calculate the gauge pressure in atm by dividing the gauge pressure by 101.325:
170.60 kPag / 101.325 kPa/atm = 1.682 atm (rounded to three decimal places)
Next, we add the atmospheric pressure to the gauge pressure to obtain the absolute pressure. The average atmospheric pressure at sea level is approximately 1 atm.
1 atm (atmospheric pressure) + 1.682 atm (gauge pressure) = 2.682 atm (rounded to three decimal places)
Therefore, the absolute pressure of the air inside the flotation device is approximately 2.682 atm.
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What conclusion can be drawn about the acidity of two compounds, Ha and Hb, based on the stability of their conjugate bases?
Based on the stability of their conjugate bases, Ha is concluded to be more acidic than Hb.
Based on the stability of their conjugate bases, the conclusion that can be drawn about the acidity of two compounds, Ha and Hb, is that Ha is more acidic than Hb. The stability of a conjugate base is directly related to the acidity of its corresponding acid. A more stable conjugate base indicates a stronger acid.
When a compound donates a proton (H+) to form its conjugate base, it becomes more stable if it can effectively distribute the negative charge. This stability is achieved by resonance or inductive effects.
In the case of Ha, if its conjugate base is more stable, it suggests that the negative charge is better distributed within the molecule. This is often observed when Ha has resonance structures or electron-withdrawing groups that stabilize the negative charge. These factors enhance the acidity of Ha, making it a stronger acid compared to Hb.
On the other hand, if Hb's conjugate base is less stable, it implies that the negative charge is less effectively distributed. This typically occurs when Hb lacks resonance structures or has electron-donating groups that destabilize the negative charge. Consequently, Hb is considered to be a weaker acid compared to Ha.
In summary, the stability of the conjugate bases provides insights into the relative acidity of compounds. A more stable conjugate base indicates a stronger acid, while a less stable conjugate base suggests a weaker acid.
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A4 kg object is moving along at 7 m/s. If the object then accelerates for 9. seconds at a rate of 12 m/s2, what is the object's new velocity in m/s?
A 4 kg object is moving along at 7 m/s. Thus the object's new velocity in m/s is 115 m/s
To calculate the object's new velocity, we can use the formula:
v = u + at
v is the final velocity,
u is the initial velocity,
a is the acceleration, and
t is the time.
Initial velocity (u) = 7 m/s
Acceleration (a) = 12 m/s²
Time (t) = 9 seconds
Substituting the given values into the formula:
v = 7 m/s + (12 m/s²)(9 s)
v = 7 m/s + 108 m/s
v = 115 m/s
Therefore, the object's new velocity is 115 m/s.
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2. Distamycin and derivatives have exhibited antiviral, antibiotic, and antitumor activity by binding to the minor groove of DNA (J. Med. Chem. 2004, 2133). Place a line through each bond of distamycin that would be cleaved by acid hydrolysis.
The bond between the nitrogen and the amide group in distamycin would be cleaved by acid hydrolysis.
Distamycin is a peptide antibiotic that has demonstrated antiviral, antibiotic, and antitumor activity. It does this by binding to the minor groove of DNA.Acid hydrolysis is a process in which molecules are broken down in the presence of an acid. Acid hydrolysis is widely used to cleave certain types of chemical bonds.
When treated with acid hydrolysis, the bonds that hold the molecule of distamycin are broken, leading to the production of its derivatives.To identify the bonds that would be cleaved by acid hydrolysis in distamycin, we must first examine its chemical structure. Distamycin has two aromatic rings, a nitrogen-containing heterocycle, and an amide-containing tail. In the presence of acid, the amide bond is cleaved, leading to the production of two smaller peptides and an acid. To place a line through each bond that would be cleaved by acid hydrolysis, we can isolate the amide bond in the structure.
Thus, the amide bond is located between the nitrogen-containing heterocycle and the amide-containing tail. Therefore, the bond between the nitrogen and the amide group is the one that would be cleaved.
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Explain why a thick layer of ice on the lake can support the weight of a person, but the liquid water cannot.
A thick layer of ice on a lake can support the weight of a person because ice is a solid state of water, while liquid water cannot support the weight due to its inherent fluidity.
Ice and liquid water are both forms of the same substance, H2O, but their molecular arrangements and physical properties differ. When water freezes, its molecules form a crystalline structure, creating a rigid network of interconnected ice molecules. This structure gives ice its solid and stable nature, allowing it to bear weight without collapsing. The lattice-like arrangement of molecules in ice makes it capable of withstanding pressure and maintaining its shape.
On the other hand, liquid water lacks a fixed molecular arrangement. The molecules in liquid water are more loosely packed and have higher mobility compared to ice. As a result, liquid water is fluid and doesn't have the structural integrity necessary to support the weight of a person or any significant load. The molecules in liquid water easily flow past each other, adapting to the shape of their container and exhibiting behaviors such as surface tension.
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What is the freezing point of a solution containing 6.10 grams of benzene (molar mass=78 g/mol) dissolved in 42.0 grams of paradichlorobenzene? The freezing point of pure paradichlorobenzene is 58 degrees celsius and the freezing-point depression constant (Kf) is 7.10 C/m.
To find the freezing point of the solution, we can use the formula for freezing-point depression:
[tex]\displaystyle \Delta T_{\text{f}}=K_{\text{f}} \times m[/tex]
where:
[tex]\displaystyle \Delta T_{\text{f}}[/tex] is the freezing-point depression,
[tex]\displaystyle K_{\text{f}}[/tex] is the freezing-point depression constant, and
[tex]\displaystyle m[/tex] is the molality of the solution.
First, we need to calculate the molality of the solution. The molality is defined as the number of moles of solute per kilogram of solvent. In this case, the solvent is paradichlorobenzene.
Step 1: Calculate the number of moles of benzene (solute):
[tex]\displaystyle \text{moles of benzene}=\frac{{\text{mass of benzene}}}{{\text{molar mass of benzene}}}[/tex]
[tex]\displaystyle \text{moles of benzene}=\frac{{6.10\, \text{g}}}{{78\, \text{g/mol}}}[/tex]
Step 2: Calculate the mass of paradichlorobenzene (solvent):
[tex]\displaystyle \text{mass of paradichlorobenzene}=42.0\, \text{g}[/tex]
Step 3: Calculate the molality of the solution:
[tex]\displaystyle \text{molality}=\frac{{\text{moles of benzene}}}{{\text{mass of paradichlorobenzene in kg}}}[/tex]
[tex]\displaystyle \text{molality}=\frac{{6.10\, \text{g}}}{{42.0\, \text{g}\times 0.001\, \text{kg/g}}}[/tex]
Now that we have the molality, we can calculate the freezing-point depression.
Step 4: Calculate the freezing-point depression:
[tex]\displaystyle \Delta T_{\text{f}}=K_{\text{f}} \times \text{molality}[/tex]
[tex]\displaystyle \Delta T_{\text{f}}=7.10\, \text{C/m}\times \left(\frac{{6.10\, \text{g}}}{{42.0\, \text{g}\times 0.001\, \text{kg/g}}}\right)[/tex]
Finally, we can calculate the freezing point of the solution.
Step 5: Calculate the freezing point:
[tex]\displaystyle \text{Freezing point}=58\, \text{C}-\Delta T_{\text{f}}[/tex]
Simplify and compute the values to find the freezing point of the solution.
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in which common processing method are tiny particles of one phase, usually strong and hard, introduced into a second phase, which is usually weaker but more ductile? O cold work O solid solution strengthening O dispersion strengtheningO strain hardening O none of the above
The common processing method in which tiny particles of one phase, usually strong and hard, are introduced into a second phase, which is usually weaker but more ductile is known as dispersion strengthening.
Dispersion strengthening is a strengthening mechanism in which small particles of a harder, more brittle material are dispersed in a softer, more ductile material to increase its strength. The particles hinder dislocation motion, causing them to pile up against the particles and creating resistance to deformation.
This type of strengthening mechanism is used in many alloys, including aluminum and magnesium alloys.The options given in the question are as follows:O cold workO solid solution strengtheningO dispersion strengtheningO strain hardeningO none of the aboveThe correct answer is option O dispersion strengthening.
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(20 pts) Derive an expression for the expansion coefficient, a, and the isothermal compressibility, KT of a perfect gas as a function of T and P, respectively.
An expression for the expansion coefficient, a, and the isothermal compressibility, KT of a perfect gas as a function of T and P, respectively is KT = -(1/V) * (∂V/∂P)T.
To derive the expression for the expansion coefficient, a, and the isothermal compressibility, KT, of a perfect gas as a function of temperature (T) and pressure (P), we start with the ideal gas law:
PV = nRT,
where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.
We can differentiate this equation with respect to temperature at constant pressure to obtain the expression for the expansion coefficient, a:
a = (1/V) * (∂V/∂T)P.
Next, we differentiate the ideal gas law with respect to pressure at constant temperature to obtain the expression for the isothermal compressibility, KT:
KT = -(1/V) * (∂V/∂P)T.
By substituting the appropriate derivatives (∂V/∂T)P and (∂V/∂P)T into the above expressions, we can obtain the final expressions for the expansion coefficient, a, and the isothermal compressibility, KT, of a perfect gas as functions of temperature and pressure, respectively.
Note: The specific expressions for a and KT will depend on the equation of state used to describe the behavior of the gas (e.g., ideal gas law, Van der Waals equation, etc.).
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1. (20 pts) A reactor is to be designed in which the oxidation of cyanide (CN-) to cyanate (CNO-) is to occur by the following reaction 0.5 02 + CNCNO- The reactor is to be a tank that is vigorously stirred so that its contents are completely mixed, and into and out of which there is a constant flow of waste and treated effluent, respectively. The feed stream flow rate is 1 MGD, and contains 15,000 mg/L CN. The desired reactor effluent concentration is 10 mg/L CN-. Assume that oxygen is in excess and that the reaction is directly proportional to the cyanide concentration, with a rate constant of k = 0.5 sec¹¹. Determine the volume of reactor required to achieve the desired treatment objective, if the reactor behaves as a) an ideal PFR, b) an ideal CSTR. or c) a system consisting of 2 equal size ideal CSTRs connected in-series.
The reactor volume required to achieve the desired treatment objective is 2,085.9 L
For the oxidation of cyanide (CN-) to cyanate (CNO-), the following reaction occurs:
0.5 02 + CN- -> CNO-
The reactor is designed to be a tank that is vigorously stirred, so that its contents are completely mixed. The feed stream flow rate is 1 MGD, and contains 15,000 mg/L CN. The desired reactor effluent concentration is 10 mg/L CN-. Oxygen is in excess and the reaction is directly proportional to the cyanide concentration, with a rate constant of k = 0.5 sec¹¹.
Volume of reactor required to achieve the desired treatment objective
For an ideal PFR:
The volume of a PFR is calculated using the following equation:
V=Q/(-rA)
where,
Q=Volumetric flow rate of feed = 1 MGD = (1 MGD) (3.7854 L/1 gal) (1 day/24 h) (1 h/60 min) (1 min/60 s) = 62.42 L/s-r = k [C]^0.5. Since the reaction is first order, the half-life (t1/2) is calculated using the following equation:
t1/2 = 0.693/k = 0.693/0.5 sec¹¹= 1.386e+10 sec = 439 years
The concentration of CN- at the inlet to the PFR is 15,000 mg/L, while the desired concentration at the outlet is 10 mg/L. Therefore, the percentage removal is 99.93%. For a 99.93% removal, the equation becomes:
rA = k [C]^0.5 = (0.5 sec¹¹) [(15,000 - 10) mg/L]^0.5= 323.61 mg/L sV = Q/(-rA) = 62.42 L/s/(-323.61 mg/L s) = 0.192 L
For an ideal CSTR:
The reactor volume of a CSTR is calculated using the following equation:
V = Q(Ci - Ce) / (rA)
The volume of a CSTR is calculated using the following equation:
V = Q (C0 - Ce) / rAV = 62.42 L/s(15,000 - 10) mg/L / [(0.5 sec¹¹) (15,000 mg/L)^0.5]V = 4,171.8 L
For a system consisting of 2 equal size ideal CSTRs connected in-series:
The volume of each CSTR (V) is 2,085.9 L (half of the total volume of the reactor)
The reactor volume of a CSTR is calculated using the following equation:
V = Q(Ci - Ce) / (rA)
The concentration of CN- at the inlet to the first CSTR is 15,000 mg/L. The concentration of CN- at the outlet of the first CSTR is calculated using the following equation:
Ce1 = kV/Ci = (0.5 sec¹¹) (2,085.9 L) / (15,000 mg/L) = 6.94e-05 mg/L
The concentration of CN- at the inlet to the second CSTR is 6.94e-05 mg/L. The concentration of CN- at the outlet of the second CSTR is calculated using the following equation:
Ce2 = kV/Ci = (0.5 sec¹¹) (2,085.9 L) / (6.94e-05 mg/L) = 1.50e+13 mg/L
The reactor volume required to achieve the desired treatment objective is 2,085.9 L
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what is the cost of production for transaminase (TA) to produce 100mg of sitagliptin?
1U of transaminase = $50
2µg of transaminase = $50
specific activity ≥0.5 U/mg
U = amount of enzyme needed to catalysis 1 umol of substrate per minute.
Sitagliptin molecular weight = 407.314 g/mol
Detailed calculation steps will be very helpful.
The cost of production for transaminase (TA) to produce 100mg of sitagliptin is approximately $0.000491.
To calculate the cost of production for transaminase (TA) to produce 100mg of sitagliptin, we need to consider the following information:
The cost of 1U of transaminase is $50.
The cost of 2µg of transaminase is $50.
The specific activity of transaminase is ≥0.5 U/mg.
The molecular weight of sitagliptin is 407.314 g/mol.
Let's break down the calculation step by step:
1: Calculate the amount of transaminase needed to produce 100mg of sitagliptin.
The molecular weight of sitagliptin is 407.314 g/mol.
Therefore, the amount of sitagliptin needed to produce 100mg is:
(100 mg / 1000) / 407.314 g/mol = 0.0002455 mol
2: Calculate the amount of transaminase in µg needed to produce 0.0002455 mol of sitagliptin.
Since the specific activity of transaminase is ≥0.5 U/mg, we can assume it is 0.5 U/mg for the calculation.
1U of transaminase = 2µg
Therefore, the amount of transaminase needed in µg is:
0.0002455 mol * 2 µg/U * (1U / 0.5 mg) = 0.000982 µg
3: Calculate the cost of the required amount of transaminase.
The cost of 2µg of transaminase is $50.
Therefore, the cost of 0.000982 µg of transaminase is:
(0.000982 µg / 2 µg) × $50 = $0.000491
So, the cost of production for transaminase (TA) to produce 100mg of sitagliptin is approximately $0.000491.
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