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Black phosphorous is a promising high mobility 2D material whose bulk form has a facecentered orthorhombic crystal structure with lattice parameters a=0.31 nm;b=0.438 nm; and c=1.05 nm. a) Determine the Bragg angles for the first three allowed reflections, assuming Cu−Kα radiation (λ=0.15405 nm) is used for the diffraction experiment. b) Determine the angle between the <111> direction and the (111) plane normal. You must show your work to receive credit.
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For the first reflection, θ = 26.74°. For the second reflection, θ = 12.67°. For the third reflection, θ = 8.16°. The angle between the <111> direction and the (111) plane normal is ≈ 25.45°.
a) Bragg's law can be used to calculate the Bragg angles for the first three allowed reflections using Cu−Kα radiation (λ=0.15405 nm) in the diffraction experiment. Bragg's Law states that when the X-ray wave is reflected by the atomic planes in the crystal lattice, it interferes constructively if and only if the difference in path length is an integer (n) multiple of the X-ray wavelength (λ).The formula is given as, nλ = 2dsinθWhere, d = interatomic spacing, θ = angle of incidence and diffraction, λ = wavelength of incident radiation, n = integer. The angle of incidence equals the angle of diffraction, and thus:θ = θ
For the first reflection, n=1, therefore, λ=2dsinθ
For the second reflection, n=2, therefore, λ=2dsinθ
For the third reflection, n=3, therefore, λ=2dsinθ
Given values: a=0.31 nm, b=0.438 nm, c=1.05 nm and Cu−Kα radiation (λ=0.15405 nm)For the (hkl) reflections, we have: dhkl = a / √(h² + k² + l²)
Substituting the given values, we get:d111 = a / √(1² + 1² + 1²)= 0.31 nm / √3 ≈ 0.18 nm
For n=1,λ = 0.15405 nm= 2d111sinθ= 2(0.18 nm)sinθsinθ = λ / 2d111= 0.15405 nm / 2(0.18 nm)= 0.4285sinθ = 0.4285θ = sin⁻¹(0.4285) = 26.74°
For n=2,λ = 0.15405 nm= 2d111sinθ= 2(0.18 nm)sinθsinθ = λ / 2d111= 0.15405 nm / 4(0.18 nm)= 0.2143sinθ = 0.2143θ = sin⁻¹(0.2143) = 12.67°
For n=3,λ = 0.15405 nm= 2d111sinθ= 2(0.18 nm)sinθsinθ = λ / 2d111= 0.15405 nm / 6(0.18 nm)= 0.1429sinθ = 0.1429θ = sin⁻¹(0.1429) = 8.16°
Therefore, the Bragg angles for the first three allowed reflections are as follows:
For the first reflection, θ = 26.74°
For the second reflection, θ = 12.67°
For the third reflection, θ = 8.16°
b) The angle between the <111> direction and the (111) plane normal is given as: tan Φ = (sin θ) / (cos θ)where, Φ is the angle between <111> and (111) plane normal and, θ is the Bragg angle calculated for the (111) reflection.
Substituting the calculated values, we get tan Φ = (sin 26.74°) / (cos 26.74°)tan Φ = 0.4915Φ = tan⁻¹(0.4915)≈ 25.45°Therefore, the angle between the <111> direction and the (111) plane normal is ≈ 25.45°.
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The number of moles of CO² which contain 8. 00g of oxygen is
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Explanation: trust me
1. (30 points total) A monochromatized ESCA instrument (equipped with an electron flood gun for charge compensation) is used to acquire data on a sample consisting of a clean platinum (Pt) plate onto which a polymer, polyethylene imine), with the repeat unit structure below, is solvent- deposited: -[CH2CH2NH]n - The binding energy (BE) for carbon in-CH2-groups (referenced to the Fermi level) is 285.0 eV. The BE for the Pt 4F7/2 line (referenced to the Fermi level) is 70.3 eV. The BE for the nitrogen 1s line (imine group) (referenced to the Fermi level) is 399.4 eV. D) For the sample with the poly(ethylene imine) deposited and the electron flood gun switched ON, the C1s speak is seen at 278 eV. What binding energy will the imine N1s peak be seen at? (calculate): Binding Energy = E) In the high resolution carbon 1s spectrum, how many peaks can be readily resolved from the peak envelope seen? (circle one) 1 2 2 3 4
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The only one peak can be seen in the high-resolution carbon 1s spectrum. Hence, the correct option is E) One peak can be readily resolved from the peak envelope seen.
D) The binding energy for the imine N1s peak is 514.1 eV.
E) One peak can be readily resolved from the peak envelope seen.
Explanation: When the electron flood gun is turned on, the excess energy given to electrons to neutralize the surface charge is absorbed by the sample which leads to inelastic scattering.
Thus, if the electron flood gun is turned on, then the binding energy of C1s would shift by 7 eV to lower energy and become 278 eV. So, the binding energy for the N1s peak of imine can be calculated as:
Binding Energy of N1s peak = (Measured binding energy of C1s peak) + (Binding energy difference of C1s and N1s) = 278 eV + (399.4 eV - 285.0 eV) = 514.4 eVHigh-resolution carbon 1s spectrum
The carbon atoms present in the carbon-carbon (C-C) single bond of poly(ethylene imine) have a binding energy of 285.0 eV.
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The safety hierarchy is essential for every plant and engineered device. In the BPCS (basic process control system) layer for highly exothermic reaction, we better be sure that temperature T stays within allowed range. The measure we protect against an error in the temperature sensor (reading too low) causing a dangerously high temperature could be ___________________________________________________. The failure position of a control valve is selected to yield the safest condition in the process, so for the reactor with exothermic reaction we should select "fail open" valve, as shown in following figure, by considering the reason that ________________________________________________________.
In the SIS (safety interlock system to stop/start equipment), the reason why we do not use the same sensor that used in BPCS is that _____________________________________________________. In relief system, the goal is usually to achieve reasonable pressure (prevent high pressure or prevent low pressure), the capacity should be for the "worst case" scenario, the action is automatic (it does not require a person), and it is entirely self-contained (no external power required), in which the reason why it needs not electricity is that _______________________________________________.
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In the BPCS (basic process control system) layer for a highly exothermic reaction, we better be sure that the temperature T stays within the allowed range. The measure we protect against an error in the temperature sensor (reading too low) causing a dangerously high temperature could be to install a second temperature sensor that can detect any erroneous reading from the first sensor. This will alert the BPCS system and result in appropriate actions. The failure position of a control valve is selected to yield the safest condition in the process, so for the reactor with exothermic reaction, we should select "fail-open" valve, which will open the valve during a failure, to prevent the reaction from building pressure. This will avoid any catastrophic situation such as a sudden explosion.
In the SIS (safety interlock system to stop/start equipment), the reason why we do not use the same sensor that is used in BPCS is that if there is an issue with the primary sensor, then the secondary sensor, which is in SIS, will not give the same reading as the primary. This will activate the SIS system and result in appropriate action to maintain the safety of the process. In relief system, the goal is usually to achieve reasonable pressure (prevent high pressure or prevent low pressure). The capacity should be for the "worst-case" scenario, the action is automatic (it does not require a person), and it is entirely self-contained (no external power required).
The reason why it needs no electricity is that in case of an emergency like a power cut, the relief valve still must function. Therefore, it has to be self-contained to operate in the absence of any external power.
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MATLAB. A company aims to produce a lead-zinc-tin of 30% lead, 30% zinc, 40% tin alloy at minimal cost. The problem is to blend a new alloy from nine other purchased alloys with different unit costs as follows 30 alloy supplier 1 2 3 4 5 6 7 8 9 lead 10 10 10 40 60 30 30 50 20 zinc 10 30 50 30 30 40 20 40 30 tin 80 60 10 10 40 30 50 10 50 price/unit weight 4.1 4.3 5.8 6.0 7.6 7.5 7.3 6.9 7.3 To construct the model for optimization, consider the following:
1. the quantity of alloy is to be optimized per unit weight
2. the 30–30–40 lead–zinc–tin blend can be framed as having a unit weight, i.e., 0.3 + 0.3 + 0.4 = 1 unit weight
3. since there are 9 alloys to be acquired, it means there are 9 quantities to be optimized.
4. there are 4 constraints to the optimization problem:
(a) the sum of alloys must be kept to the unit weight
(b) the sum of alloys for lead must be kept to its composition.
(c) the sum of alloys for zinc must be kept to its composition.
(d) the sum of alloys for tin must be kept to its composition.
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MATLAB can be used to optimize the production of a lead-zinc-tin alloy that contains 30% lead, 30% zinc, and 40% tin at the least expense by blending nine different alloys with various unit costs as shown below:
A lead-zinc-tin alloy of 30% lead, 30% zinc, and 40% tin can be formulated as having a unit weight, i.e., 0.3 + 0.3 + 0.4 = 1 unit weight. The aim is to blend a new alloy from nine purchased alloys with different unit costs, with the quantity of alloy to be optimized per unit weight.
Here are the four constraints of the optimization problem:
(a) The sum of alloys must be kept to the unit weight.
(b) The sum of alloys for lead must be kept to its composition.
(c) The sum of alloys for zinc must be kept to its composition.
(d) The sum of alloys for tin must be kept to its composition.
Mathematically, let Ai be the quantity of the ith purchased alloy to be used per unit weight of the lead-zinc-tin alloy. Then, the cost of blending the new alloy will be:
Cost per unit weight = 4.1A1 + 4.3A2 + 5.8A3 + 6.0A4 + 7.6A5 + 7.5A6 + 7.3A7 + 6.9A8 + 7.3A9
Subject to the following constraints:
(i) The total sum of the alloys is equal to 1. This can be represented mathematically as shown below:
A1 + A2 + A3 + A4 + A5 + A6 + A7 + A8 + A9 = 1
(ii) The total sum of the lead alloy should be equal to 0.3. This can be represented mathematically as shown below:
0.1A1 + 0.1A2 + 0.1A3 + 0.4A4 + 0.6A5 + 0.3A6 + 0.3A7 + 0.5A8 + 0.2A9 = 0.3
(iii) The total sum of the zinc alloy should be equal to 0.3. This can be represented mathematically as shown below:
0.1A1 + 0.3A2 + 0.5A3 + 0.3A4 + 0.3A5 + 0.4A6 + 0.2A7 + 0.4A8 + 0.3A9 = 0.3
(iv) The total sum of the tin alloy should be equal to 0.4. This can be represented mathematically as shown below:
0.8A1 + 0.6A2 + 0.1A3 + 0.1A4 + 0.4A5 + 0.3A6 + 0.5A7 + 0.1A8 + 0.5A9 = 0.4
The optimization problem can then be solved using MATLAB to obtain the optimal values of A1, A2, A3, A4, A5, A6, A7, A8, and A9 that will result in the least cost of producing the required alloy.
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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
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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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6. If I took a 10 mL sample from 2 litres of a 100 mM solution of NaCl (sodium chloride or common table salt), what would be the concentration of NaCl in my 10 mL sample?
Give an example of when you would record experimental data in a table and explain why this is more appropriate than listing or describing the results.
8. Name 2 common functions that you would use on your calculator (not the simple operator’s addition, subtraction, division, and multiplication).
9. If you saw the scientific term 560 nm, what topic do you think might being discussed? Explain why you think this.
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The concentration of NaCl in the 10 mL sample would be 2000 mM. Two common functions on a calculator are exponentiation and square root. The term "560 nm" likely relates to the wavelength or color of light in a scientific context.
To calculate the concentration of NaCl in the 10 mL sample taken from a 100 mM (millimolar) solution, we can use the formula:
[tex]C_1V_1 = C_2V_2[/tex]
Where:
Rearranging the formula, we have:
[tex]C_2 = (C_1V_1) / V_2[/tex]
Substituting the given values:
[tex]C_2[/tex] = (100 mM * 2 liters) / 10 mL
Now we need to convert the volume units to the same measurement. Since 1 liter is equal to 1000 mL, we can convert the volume of the solution to milliliters:
[tex]C_2[/tex] = (100 mM * 2000 mL) / 10 mL
[tex]C_2[/tex] = 20,000 mM / 10 mL
[tex]C_2[/tex] = 2000 mM
Therefore, the concentration of NaCl in the 10 mL sample would be 2000 mM.
Two common functions that you would use on a calculator, other than the basic arithmetic operations (addition, subtraction, multiplication, and division), are:
a) Exponentiation: This function allows you to calculate a number raised to a specific power. It is commonly denoted by the "^" symbol. For example, if you want to calculate 2 raised to the power of 3, you would enter "[tex]2^3[/tex]" into the calculator, which would give you the result of 8.
b) Square root: This function enables you to find the square root of a number. It is often represented by the "√" symbol. For instance, if you want to calculate the square root of 9, you would enter "√9" into the calculator, which would yield the result of 3.
These functions are frequently used in various mathematical calculations and scientific applications.
When encountering the scientific term "560 nm," it is likely that the topic being discussed is related to the electromagnetic spectrum and wavelengths of light. The term "nm" stands for nanometers, which is a unit of measurement used to express the length of electromagnetic waves, including visible light.
The wavelength of light in the visible spectrum ranges from approximately 400 nm (violet) to 700 nm (red). The value of 560 nm falls within this range and corresponds to yellow-green light. This range of wavelengths is often discussed in various scientific fields, such as physics, optics, and biology when studying the properties of light, color perception, or interactions between light and matter.
Overall, seeing the term "560 nm" suggests a focus on the wavelength or color of light in a scientific context.
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2. The experienced analyst who normally conducts these analyses fell ill and will be unable to analyze the urine samples for the drug in time for the sporting event. In order for the laboratory manager to assign a new analyst to the task, a "blind sample" experiment was done. a. The results for the blind sample experiment for the determination of Methylhexaneamine in a urine sample are shown in Table 1 below. Table 1: Results of blind sample analysis. Response factor (F) Analyst results Internal Standard Concentration 0.25 ug/ml 0.35 mg/ml Signals 522 463 Sample Analysis ? 1.05 ug/ml 15 ml 10 ml Original concentration Volume added to sample Total Volume Signals 25 ml 400 418 i. Provide justification why an internal standard was used in this analysis instead of a spike or external standard? ii. Determine the response factor (F) of the analysis. iii. Calculate the concentration of the internal standard in the analyzed sample. iv. Calculate the concentration of Methylhexaneamine in the analyzed sample. v. Determine the concentration of Methylhexaneamine in the original sample. b. Explain how the results from the blind sample analysis can be used to determine if the new analyst should be allowed to conduct the drug analysis of the athletes' urine samples. c. Urine is considered to be a biological sample. Outline a procedure for safe handling and disposal of the sample once the analysis is completed.
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a.i) Justification of why an internal standard was used in this analysis instead of a spike or external standard:
An internal standard was used in this analysis instead of a spike or external standard because an internal standard is a compound that is similar to the analyte but is not present in the original sample. The use of an internal standard in analysis corrects the variation in response between sample runs that can occur with the use of an external standard. This means that the variation in the amount of analyte in the sample will be corrected for, resulting in a more accurate result.
ii) Response factor (F) of the analysis can be calculated using the following formula:
F = (concentration of internal standard in sample) / (peak area of internal standard)
iii) Concentration of the internal standard in the analyzed sample can be calculated using the following formula:
Concentration of internal standard in sample = (peak area of internal standard) × (concentration of internal standard in original sample) / (peak area of internal standard in original sample)
iv) Concentration of Methylhexaneamine in the analyzed sample can be calculated using the following formula:
Concentration of Methylhexaneamine in sample = (peak area of Methylhexaneamine) × (concentration of internal standard in original sample) / (peak area of internal standard)
v) Concentration of Methylhexaneamine in the original sample can be calculated using the following formula:
Concentration of Methylhexaneamine in the original sample = (concentration of Methylhexaneamine in the sample) × (total volume) / (volume of sample) = (concentration of Methylhexaneamine in the sample) × (25 ml) / (15 ml) = 1.67 × (concentration of Methylhexaneamine in the sample)
b. The results from the blind sample analysis can be used to determine if the new analyst should be allowed to conduct the drug analysis of the athletes' urine samples. The new analyst should be allowed to conduct the analysis if their results are similar to the results of the blind sample analysis. If their results are significantly different, this could indicate that there is a problem with their technique or the equipment they are using, and they should not be allowed to conduct the analysis of the athletes' urine samples.
c. Procedure for safe handling and disposal of the sample once the analysis is completed:
i) Label the sample container with the sample name, date, and analyst's name.
ii) Store the sample container in a refrigerator at 4°C until it is ready to be analyzed.
iii) Once the analysis is complete, dispose of the sample container according to the laboratory's waste management protocols. The laboratory should have protocols in place for the safe disposal of biological samples. These protocols may include autoclaving, chemical treatment, or incineration.
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How many liters of oxygen will be required to react with .56 liters of sulfur dioxide?
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Oxygen of 0.28 liters will be required to react with 0.56 liters of sulfur dioxide.
To determine the number of liters of oxygen required to react with sulfur dioxide, we need to examine the balanced chemical equation for the reaction between sulfur dioxide ([tex]SO_2[/tex]) and oxygen ([tex]O_2[/tex]).
The balanced equation is:
2 [tex]SO_2[/tex]+ O2 → 2 [tex]SO_3[/tex]
From the equation, we can see that 2 moles of sulfur dioxide react with 1 mole of oxygen to produce 2 moles of sulfur trioxide.
We can use the concept of stoichiometry to calculate the volume of oxygen required. Since the ratio between the volumes of gases in a reaction is the same as the ratio between their coefficients in the balanced equation, we can set up a proportion to solve for the volume of oxygen.
The given volume of sulfur dioxide is 0.56 liters, and we need to find the volume of oxygen. Using the proportion:
(0.56 L [tex]SO_2[/tex]) / (2 L [tex]SO_2[/tex]) = (x L [tex]O_2[/tex]) / (1 L [tex]O_2[/tex]2)
Simplifying the proportion, we have:
0.56 L [tex]SO_2[/tex]= 2x L [tex]O_2[/tex]
Dividing both sides by 2:
0.56 L [tex]SO_2[/tex]/ 2 = x L [tex]O_2[/tex]
x = 0.28 L [tex]O_2[/tex]
Therefore, 0.28 liters of oxygen will be required to react with 0.56 liters of sulfur dioxide.
It's important to note that this calculation assumes that the gases are at the same temperature and pressure and that the reaction goes to completion. Additionally, the volumes of gases are typically expressed in terms of molar volumes at standard temperature and pressure (STP), which is 22.4 liters/mol.
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Production of Renewable Ammonia In recent years, significant interest has been paid to developing fuel and chemicals from renewable feedstocks, In this regard, you are requested to design a plant to produce 150 000 metric tons per annum of Ammonia (at least 99.5 wt. %). The hydrogen to nitrogen feed ratio is 3:1. The feed also contains 0.5 % argon. The feed is available at 40°C and 20 atm. The plant should operate for 330 days in a year, in order to allow for shutdown and maintenance. The plant is to be built in Nelson Mandela Bay. In this assessment, you need to assess the feasibility of such a process by conducting a conceptual design, that covers the following topics: 1.1. Design basis 1.2. Literature Survey 1.3. Process Description 1.4. Preliminary block flow diagram (BFD) and process flow diagram (PFD) 1.4.1. Block diagram of the entire process 1.4.2. Process flow diagram for ammonia synthesis 1.5. Preliminary major equipment list
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It's important to note that this is a preliminary list, and a detailed engineering study would be required to finalize the equipment selection and sizing based on specific process conditions and requirements.
Based on the provided information, here is a preliminary major equipment list for the plant designed to produce 150,000 metric tons per annum of ammonia:
Feedstock Preparation:
Feedstock Heat Exchanger
Feedstock Filters
Reforming Section:
Primary Reformer
Secondary Reformer
Waste Heat Boiler
Steam Drum
High-Temperature Shift Converter
Low-Temperature Shift Converter
CO2 Removal Unit
Synthesis Loop:
Ammonia Synthesis Converter
Methanation Converter
Separation and Purification:
Ammonia Separator
Ammonia Purification Column
Methane Separator
Methane Purification Column
Compression and Storage:
Ammonia Compressors
Ammonia Storage Tanks
Nitrogen Compressors
Utilities:
Steam Generation Unit
Cooling Tower
Air Compressors
Power Generation Unit
Safety Systems:
Safety Relief Valves
Emergency Shutdown System
Fire Protection Equipment
It's important to note that this is a preliminary list, and a detailed engineering study would be required to finalize the equipment selection and sizing based on specific process conditions and requirements. Additionally, the list does not include all auxiliary equipment and instrumentation required for the plant's operation.
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