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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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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Exerase Question 2: [15] Batch Reactor and CSTR
The saponification of ethyl acetate CH3COOC₂H₂ + NaOH CH3COONa + C₂H5OH is an elementary and irreversible reaction. In a laboratory experiment, a well-stirred batch reactor was charged with an aqueous solution containing NaOH and ethyl acetate, both of initial concentrations equal to 0.1 mol dm1. It was found that when the reactor was operated at 300 °K, the ethyl acetate conversion was 18% after 15 minutes. It is desired to produce 10 mol min-¹ of C₂H5OH in a CSTR via the above reaction. For this, 5 dm³/min of ethyl acetate is to be fed to the reactor together with an equal volumetric flowrate of NaOH and the streams are to be mixed immediately before entering the CSTR. If 90% conversion of ethyl acetate is to be achieved, determine the necessary reactor volume - assume that the operating temperature of the CSTR is 310 °K and that the activation energy for the reaction is 82 000 cal/mol.
iven data:Aqueous solution contains NaOH and ethyl acetate,Initial concentration of NaOH and ethyl acetate=0.1 MConversion of ethyl acetate=18%Operating temperature of reactor (T1)=300 KDesired product=C2H5OHProduction rate=10 mol/minVolumetric flow rate of ethyl acetate (V1)= 5 dm³/minVolumetric flow rate of NaOH (V2)= 5 dm³/minOperating temperature of CSTR (T2)= 310 KActivation energy(Ea)= 82,000 cal/molTo find:
The necessary reactor volume in the CSTR.90% conversion of ethyl acetate is to be achieved.Reaction: CH3COOC₂H₂ + NaOH → CH3COONa + C₂H5OHThe conversion of ethyl acetate is given as: X = 0.18Let's write the mole balance for the CSTR,Total flow rate = F = V1 + V2 = 10 dm³/minLet's consider the conversion of ethyl acetate as X.FEAc = FA0 - XFA0 = 0.1 mol/dm³We need to determine the concentration of NaOH in the reactor as it's not given in the problem. We know that NaOH and Ethyl acetate are reacting in a 1:1 molar ratio. Hence the concentration of NaOH required for this reaction can be determined as:CAOH = CEAc = 0.1 mol/dm³Let's calculate the rate constant using Arrhenius equation:k = A e(-Ea/RT)where A = pre-exponential factor Ea = activation energy = 82000 cal/molR = gas constant = 1.987 cal/mol K (considering temperature in Kelvin)T = operating temperature = 300 KSubstituting the values in the above equation,k1 = 1.31 x 10⁹ dm³/mol minNow, the volume of the CSTR can be determined using the formula for a CSTR.V = F X / (-rA)CA = CEAc (1-X)where, rA = -k CA CBV = (V1 + V2)X / (-rA)CAV = 5 X 0.9 / (1.31 10⁹ 0.1 (1-0.9))V = 190.8 dm³The necessary reactor volume in the CSTR is 190.8 dm³.About ReactorA chemical reactor is a vessel where a chemical reaction takes place. The design of this reactor depends on many variables that can be studied in chemical engineering.
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Copper has a density of 8.96 g/cm³. What is the mass of 17.4 L of copper? Mass = ….. g
A load of asphalt weighs 38,600 lbs and occupies a volume of 8720 L. What is the density of this asphalt in g/L? ….. g/L
The mass of 17.4 L of copper is 155.90 g. The density of the asphalt is 4.42 g/L.
To find the mass of 17.4 L of copper, we can use the formula Mass = Density x Volume. Given that the density of copper is 8.96 g/cm³, we need to convert the volume from liters to cubic centimeters (cm³) to ensure the units match. One liter is equal to 1000 cm³, so the volume of 17.4 L is 17,400 cm³. Plugging these values into the formula, we get Mass = 8.96 g/cm³ x 17,400 cm³ = 155,904 g. Rounding to two decimal places, the mass of 17.4 L of copper is 155.90 g.
Step 2: Copper has a specific density of 8.96 g/cm³, which means that for every cubic centimeter of copper, it weighs 8.96 grams. In order to find the mass of a given volume, we can use the formula Mass = Density x Volume. However, it is important to ensure that the units are consistent. In this case, the given volume is in liters, while the density is in grams per cubic centimeter. To address this, we need to convert the volume from liters to cubic centimeters. Since 1 liter is equal to 1000 cm³, we can convert 17.4 liters to cubic centimeters by multiplying it by 1000, resulting in 17,400 cm³.
By substituting the values into the formula, we have Mass = 8.96 g/cm³ x 17,400 cm³ = 155,904 g. Rounding the answer to two decimal places, we find that the mass of 17.4 L of copper is 155.90 g.
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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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4. (20 points total) An electrically conducting sample is placed in an XPS spectrometer. The sample is irradiated with x-rays from an Al Ka source (1486 eV). The kinetic energy of electrons emitted from one particular orbital as measured within the spectrometer is 500 eV. The work function of the spectrometer is 4 eV. The work function of the sample is 3 eV. What is the binding energy of the electron?
The electron's binding energy is 993 eV.
XPS is an analytical tool that employs high-intensity X-rays to identify the chemical state of surface elements. An XPS spectrum displays the energies of detected electrons; a broad peak is generated by every electron orbital, with the binding energy on the x-axis and the signal intensity on the y-axis.
Binding energy is the energy required to separate an electron from its atom and is determined by the chemical environment. The higher the atomic number of the atom's core, the stronger the binding energy of the electrons to the atom's nucleus.
The potential energy required to eject an electron from the metal's Fermi level is referred to as the work function, and it is represented by Φ. The energy required to detach an electron from its atomic orbital is referred to as the binding energy, which is denoted by BE.
The binding energy (BE) can be calculated using the following formula:
BE = hν - Φ - KE
where h is Planck's constant, ν is the frequency of incident radiation, KE is the kinetic energy of the photoelectron, and Φ is the work function.
According to the problem given, the work function of the spectrometer is 4 eV, while that of the sample is 3 eV. KE of electron is 500 eV. Therefore, putting all the given values in the above formula we get,
BE = hν - Φ - KEBE = (6.626x10⁻³⁴ J s)(2.418x10¹⁷ s⁻¹) - (3+4) eV - 500 eV
BE = 993 eV
Therefore, the electron's binding energy is 993 eV.
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540 kg/h of sliced fresh potato (82.11% moisture, the balance is solids) is fed to a forced convection dryer. The air used for drying enters at 68°C, 1 atm, and 14.5% relative humidity. The potatoes exit at only 2.18% moisture content. If the exiting air leaves t 86.9% humidity at the same inlet temperature and pressure, what is the mass flow rate of the outlet air? Type your answer as a whole number rounded off to the units digit.
The mass flow rate of the outlet air is 12 kg/h
In the given scenario, 540 kg/h of sliced fresh potato with 82.11% moisture is fed into a forced convection dryer. The objective is to reduce the moisture content of the potatoes to 2.18%. The air used for drying enters the dryer at 68°C, 1 atm, and 14.5% relative humidity. It is required to determine the mass flow rate of the outlet air, which leaves the dryer at 86.9% humidity, under the same inlet temperature and pressure conditions.
To solve this problem, we can use the concept of mass balance. The mass flow rate of the outlet air can be calculated by subtracting the mass of the dried potatoes from the mass of the fresh potatoes. The moisture content in the dried potatoes can be determined by multiplying the mass flow rate of the potatoes with their respective moisture content.
First, we calculate the mass of dried potatoes:
Mass of dried potatoes = Mass flow rate of potatoes × (1 - moisture content of dried potatoes)
Mass of dried potatoes = 540 kg/h × (1 - 0.0218) = 528.42 kg/h
Next, we can calculate the mass flow rate of the outlet air by subtracting the mass of dried potatoes from the mass flow rate of the fresh potatoes:
Mass flow rate of outlet air = Mass flow rate of fresh potatoes - Mass of dried potatoes
Mass flow rate of outlet air = 540 kg/h - 528.42 kg/h = 11.58 kg/h
Rounded off to the units digit, the mass flow rate of the outlet air is 12 kg/h.
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Using a logarithmic concentration diagram, determine the pH of a solution containing 10-2 M acetic acid and 2 x 10-2 M sodium acetate.
The pH of this solution is approximately 4.74, indicating it is slightly acidic. The presence of sodium acetate, a salt of acetic acid, acts as a buffer and helps maintain the pH of the solution.
The pH of a solution containing[tex]10^-2[/tex] M acetic acid and 2 x[tex]10^-2[/tex] M sodium acetate can be determined using a logarithmic concentration diagram.
To determine the pH of the solution, we need to consider the dissociation of acetic acid and the hydrolysis of sodium acetate. Acetic acid (CH3COOH) is a weak acid that partially dissociates in water, releasing hydrogen ions (H+) and acetate ions (CH3COO-).
The dissociation of acetic acid can be represented as follows:
CH3COOH ⇌ H+ + CH3COO-
The equilibrium constant for this dissociation is known as the acid dissociation constant (Ka). The pKa value of acetic acid is approximately 4.74. The pKa is the negative logarithm of the Ka value.
In the given solution, we have both acetic acid and sodium acetate. Sodium acetate (CH3COONa) is a salt that dissociates completely in water, releasing sodium ions (Na+) and acetate ions (CH3COO-). The acetate ions from sodium acetate can react with any additional H+ ions present in the solution through hydrolysis, which helps maintain the pH.
Using a logarithmic concentration diagram, we can determine that the pH of the solution containing [tex]10^-2[/tex] M acetic acid and 2 x [tex]10^-2[/tex] M sodium acetate is approximately 4.74, which is slightly acidic.
The presence of sodium acetate acts as a buffer, helping to resist changes in pH by absorbing excess H+ ions or releasing additional H+ ions as needed to maintain the pH within a certain range.
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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 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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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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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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Find the probability that a ground-state electron will be found within 0.0010 nm of the wall in an infinitely deep potential well of width 0.20 nm.
The probability of finding a ground-state electron within 0.0010 nm of the wall in an infinitely deep potential well of width 0.20 nm is 0.012.
The probability of finding a ground-state electron within 0.0010 nm of the wall in an infinitely deep potential well of width 0.20 nm can be determined using the wave function of the electron.
The wave function is given by the equation : ψn(x) = (2/L)^1/2 sin(nπx/L) where
L is the width of the potential well
n is the quantum number
x is the position of the electron within the well
The probability of finding the electron within a given distance from the wall can be found by integrating the wave function over that distance.
To find the probability of finding the electron within 0.0010 nm of the wall, we need to integrate the wave function over the range 0 to 0.0010 nm :
Probability = ∫[ψn(x)]^2 dx from 0 to 0.0010 nm
Probability = ∫[(2/L)^1/2 sin(nπx/L)]^2 dx from 0 to 0.0010 nm
= (2/L) ∫sin^2(nπx/L) dx from 0 to 0.0010 nm
Probability = (2/L) [L/2 - (L/2) cos(2nπx/L)] from 0 to 0.0010 nm
Probability = 1 - cos(2nπx/L)
So, the probability of finding a ground-state electron within 0.0010 nm of the wall in an infinitely deep potential well of width 0.20 nm is given by :
Probability = 1 - cos(2πx/L)
Probability = 1 - cos[(2π)(0.0010 nm)/(0.20 nm)] = 0.012
Thus, the probability of finding a ground-state electron within 0.0010 nm of the wall in an infinitely deep potential well of width 0.20 nm is 0.012
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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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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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Which isotope of helium is more tightly bound, 72H or 52H? (Atomic mass of 7He = 7.027991 u and atomic mass of 5He = 5.012057 u) OA. 5₂H OB.72H C. Both isotopes are equally bound. D. Not enough information.
Option b-A The isotope ⁷₂H (7He) is more tightly bound than ⁵₂H (5He).
The stability of an isotope depends on its binding energy, which represents the amount of energy required to break apart the nucleus into its constituent particles. Higher binding energy indicates greater stability and tighter binding of nucleons within the nucleus.
To determine which isotope is more tightly bound, we compare their binding energies. The binding energy is related to the mass defect, which is the difference between the sum of the masses of the individual nucleons and the actual mass of the nucleus.
In this case, the atomic mass of ⁷₂H (7He) is 7.027991 u, and the atomic mass of ⁵₂H (5He) is 5.012057 u. The greater the mass defect, the more tightly bound the nucleus. Since the mass defect of ⁷₂H (7He) is greater than that of ⁵₂H (5He), it implies that ⁷₂H (7He) has a higher binding energy and is more tightly bound.
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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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3. (a) (b) Answer ALL parts. Zeolites can be utilised as shape selective catalysts by exploiting the various shapes and dimensions of zeolite pores and channels. Discuss three types of shape selective catalysis, illustrating your answer with suitable examples and diagrams in each case. [18 marks] Silicon-germanium is an important material that is used in the fabrication of transistors. The phase diagram for the silicon-germanium system is given below. A sample of composition 20 wt% Si is made up and held at 1150 °C. Based on this information, answer the following questions: (1) What is the composition of any solid phase present? (ii) What is the composition of any liquid phase present? (iii) How much of each phase is present? (iv) What are the melting points of pure silicon and pure germanium? 1500 1400 Liquid (L) 1300 Temperature (°C) a+L 1200 LLLLLL 1100 Solid(a) 1000 900 0 Ge 20 40 60 80 100 Composition (wt% Si) Si [16 marks) Q3 continues overleaf- (c) Using a suitable diagram, write a note on the rate of diffusion as it relates to solid-state reactions. [6 marks)
Shape-selective catalysis is a type of catalysis in which the reactive molecules are restricted to move along a certain path and within a certain shape by the catalytic surface.
Three types of shape-selective catalysis are there, and they are as follows:
1. Intraparticle: The reaction molecules can only reach the active sites on the exterior surface of the particle.
E.g., the decomposition of isopropyl alcohol to acetone over an activated carbon catalyst.
2. Intermolecular: The reaction molecules can only approach the active sites when they are present in a particular orientation or conformation.
E.g., the hydrolysis of ethyl acetate over zeolites.
3. Intramolecular: The reaction molecules are large and can only reach the active sites if they are present in a certain orientation or conformation.
E.g., disproportionation of ethylbenzene over zeolites.
(b) 1. Solid phase present: The composition of the solid phase present can be found by reading the vertical line of 20 wt% Si from the solid phase boundary of the phase diagram. It tells us that the solid present at 1150 °C is silicon.
2. Liquid phase present: The composition of the liquid phase present can be found by reading the vertical line of 20 wt% Si from the liquid phase boundary of the phase diagram. It tells us that the liquid present at 1150 °C is a eutectic mixture of silicon and germanium.
3. Quantity of each phase present: The phase rule states that P + F = C + 2.
P = 2 (solid and liquid phases) C = 2 (composition of the solid and liquid phases) F = 0 (no degrees of freedom at a particular temperature and pressure) . Therefore, the system is invariant, implying that only one combination of the two phases can co-exist at a certain temperature and pressure.
4. The melting point of pure silicon and pure germanium is 1410 °C and 938 °C, respectively.
(c) Diffusion in Solid-State Reactions: When reactants are in a solid-state, they need to diffuse into and around the solid to come into contact with each other. The rate of diffusion can be increased by increasing the surface area and temperature. A simple schematic diagram of the rate of diffusion as it relates to solid-state reactions is shown below:
Where:
ΔC/dx: Concentration gradient
D: Diffusion coefficient
A: Surface area
C: Concentration
T: Temperature
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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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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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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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Discuss using diagrams how porosity and particle size affect a well's ability to provide enough quantities of water.
Porosity and particle size both play an important role in the amount of water that a well can provide.
The porosity of a rock refers to the amount of pore space it has, which is the space between the grains. Larger pore space means that more water can be stored. In contrast, smaller pore spaces limit the amount of water that can be stored. Particle size, on the other hand, affects the ability of water to move through the rock. Larger particles mean larger pore spaces, which in turn, means that more water can be stored. Smaller particles mean smaller pore spaces, which limit the amount of water that can be stored.
Wells that have larger pore spaces and larger particle sizes can store more water and therefore have the potential to provide larger quantities of water. Conversely, wells that have smaller pore spaces and smaller particle sizes can only store limited amounts of water. Porosity and particle size are important to consider when constructing wells since they affect the amount of water that can be drawn from a well. The diagrams below show how porosity and particle size affect the ability of a well to provide enough quantities of water. A diagram showing how porosity affects a well's ability to provide enough quantities of water. A diagram showing how particle size affects a well's ability to provide enough quantities of water.
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Question 3 (7 points out of 20) The first order gas phase reaction: A 2B with k -0.3 mole/(kg-catalyst min*atmtakes place in an isothermal packed bed reactor. The feed, which is 75% in A and 25% inert, enters the reactor at 400 K and total pressure of 10 atm with the total flow rate of 40 mole/min. If there is no pressure drop along the length of the packed bed reactor, calculate the weight of catalyst needed to produce 36 mole/min of product B.
Step 1: The weight of catalyst needed to produce 36 mole/min of product B is -120 kg.
To calculate the weight of catalyst needed, we need to consider the stoichiometry of the reaction and the molar flow rates. The given reaction is A 2B, which means that for every 2 moles of A reacted, we obtain 1 mole of B.
Given that the feed contains 75% A and 25% inert gas, we can calculate the molar flow rates of A and inert gas. The total molar flow rate is given as 40 mole/min, so the molar flow rate of A would be 0.75 * 40 = 30 mole/min, and the molar flow rate of the inert gas would be 0.25 * 40 = 10 mole/min.
Since the reaction is first-order and takes place in a packed bed reactor with no pressure drop, the rate constant (k) is -0.3 mole/(kg-catalyst min*atm). We can use this information to calculate the weight of catalyst needed.
The rate equation for the reaction can be written as r = k * P_A, where r is the reaction rate, k is the rate constant, and P_A is the partial pressure of A. In this case, P_A can be calculated as (molar flow rate of A) / (total flow rate) * (total pressure). So, P_A = (30 mole/min) / (40 mole/min) * (10 atm) = 7.5 atm.
Now, we can use the rate equation to solve for the weight of catalyst. r = k * P_A can be rearranged as r / k = P_A. Since we want to produce 36 mole/min of product B, the reaction rate would be 36 mole/min. Plugging in these values, we get 36 mole/min / -0.3 mole/(kg-catalyst min*atm) = 7.5 atm.
Simplifying the equation, we find that the weight of catalyst needed (X) is X = 36 mole/min / (-0.3 mole/(kg-catalyst min*atm)) = -120 kg.
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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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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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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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"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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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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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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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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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.