A 15.00 mL sample of a solution of H2SO4 of unknown concentration was titrated with 0.3200M NaOH. the titration required 21.30 mL of the base. Assuming complete neutralization of the acid,
1) What was the normality of the acid solution?
2) What was the molarity of the acid solution?
Answers
Answer:
a. 0.4544 N
b. \(5.112 \times 10^{-5 M}\)
Explanation:
For computing the normality and molarity of the acid solution first we need to do the following calculations
The balanced reaction
\(H_2SO_4 + 2NaOH = Na_2SO_4 + 2H_2O\)
\(NaOH\ Mass = Normality \times equivalent\ weight \times\ volume\)
\(= 0.3200 \times 40 g \times 21.30 mL \times 1L/1000mL\)
= 0.27264 g
\(NaOH\ mass = \frac{mass}{molecular\ weight}\)
\(= \frac{0.27264\ g}{40g/mol}\)
= 0.006816 mol
Now
Moles of \(H_2SO_4\) needed is
\(= \frac{0.006816}{2}\)
= 0.003408 mol
\(Mass\ of\ H_2SO_4 = moles \times molecular\ weight\)
\(= 0.003408\ mol \times 98g/mol\)
= 0.333984 g
Now based on the above calculation
a. Normality of acid is
\(= \frac{acid\ mass}{equivalent\ weight \times volume}\)
\(= \frac{0.333984 g}{49 \times 0.015}\)
= 0.4544 N
b. And, the acid solution molarity is
\(= \frac{moles}{Volume}\)
\(= \frac{0.003408 mol}{15\ mL \times 1L/1000\ mL}\)
= 0.00005112
=\(5.112 \times 10^{-5 M}\)
We simply applied the above formulas
The volume of the 0.3200 M, NaOH required to neutralize the H₂SO₄, is
21.30 mL, which gives the following acid solution approximate values;
1) Normality of the acid solution is 0.4544 N
2) The molarity of the acid is 0.2272
How can the normality, molarity of the solution be found?Molarity of the NaOH = 0.3200 M
Volume of NaOH required = 21.30 mL
1) The normality of the acid solution is found as follows;
The chemical reaction is presented as follows;
H₂SO₄(aq) + 2NaOH (aq) → Na₂SO₄ (aq) + H₂O
Number of moles of NaOH in the reaction is found as follows;
\(n = \dfrac{21.30}{1,000} \times 0.3200 \, M = \mathbf{0.006816 \, M}\)
Therefore;
The number of moles of H₂SO₄ = 0.006816 M ÷ 2 = 0.003408 M
\(Normality = \mathbf{ \dfrac{Mass \ of \, Acid \ in \ reaction}{Equivalent \ mass \times Volume \ of \ soltute}}\)
Which gives;
\(Normality = \dfrac{ 98 \times 0.003408 }{49 \times 0.015} = \mathbf{0.4544}\)
The normality of the acid solution, H₂SO₄(aq), N ≈ 0.45442) The molarity is found as follows;
\(Molarity = \dfrac{0.003408 \, moles}{0.015 \, L} = \mathbf{0.2272 \, M}\)
The molarity of the acid solution is 0.2272 MLearn more about the normality and the molarity of a solution here:
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Related Questions
In a vehicle with front disc brakes, the vehicle pulls to the left when braking. What causes this?
A. Seized left brake caliper piston
B. Collapsed left brake line hose
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For proper function hydraulics systems need a reservoir of which of the following?
A.) Compressible fluid
B.) Non-compressible fluid C.) Non-compressible air
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Think of brake fluid on a vehicle, compressed as one applies the break= pressure to expand brake cylinders in the wheels which compress the pads on the wheel brake rotors.
Release brake and fluid relaxes back into the reservoir/accumulator.
T/F. A 257-bit encryption key is twice as difficult to guess compared to a 256-bit encryption key
Answers
The statement is False. A 257-bit encryption key is not twice as difficult to guess compared to a 256-bit encryption key Encryption keys are used to secure and protect data transmissions and to ensure that only authorized users have access to the data.
The strength of an encryption key is determined by its length, which is measured in bits. The longer the key, the more secure it is and the harder it is to guess or crack. A 256-bit encryption key is already considered extremely secure, and increasing the length of the key even by one bit can significantly increase its strength.
However, increasing the key length from 256 bits to 257 bits does not double its strength. In fact, the increase in strength is much less significant than that. It is difficult to compare the exact level of difficulty in guessing an encryption key of a particular length, as it depends on several factors such as the encryption algorithm being used and the resources available to the attacker.
It is safe to say that increasing the key length by even one bit can provide a significant increase in security, but it does not necessarily double the level of difficulty in guessing the key.
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Write about traditional brick production in Pakistan
Answers
Answer:
Clay bricks are manufactured by mining and clay moulded blocks. There are 20,000 brick klins in Pakistan.
Explanation:
In Pakistan, the clay bricks are manufactured by mining and baking the clay moulded blocks in brick kilns. According to an estimate, the baking process emits about 1.4 pounds of carbon per brick made, but in Pakistan, because the systems are outdated, brick kilns are used, which is producing more than the average amount of the pollution.
There are about 20,000 brick kilns in Pakistan. The traditional brick production in Pakistan is consists of hand-made bricks which are first baked in Fixed Chimney Bull's Trench Kilns (FCBTK), this is the most widely used brick firing technology in South Asia.
A project ha following time chedule: Activity Time in Week Activity Time in Week 1-2 4 5-7 8 1-3 1 6-8 1 2-4 1 7-8 2 3-4 1 8-9 1 3-5 6 8-10 8 4-9 5 9-10 7 5-6 4 Contruct the network and compute: (1) TE and TL for each event (2) Float for each activity (3) Critical path and it duration
Answers
(1) TE and TL for each event
Event 1: TE=0, TL=0
Event 2: TE=4, TL=4
Event 3: TE=1, TL=5
Event 4: TE=2, TL=6
Event 5: TE=6, TL=12
Event 6: TE=1, TL=2
Event 7: TE=8, TL=16
Event 8: TE=9, TL=17
Event 9: TE=8, TL=18
Event 10: TE=16, TL=23
(2) Float for each activity
Activity 1-2: Float=0
Activity 1-3: Float=3
Activity 2-4: Float=2
Activity 3-4: Float=1
Activity 3-5: Float=0
Activity 4-9: Float=0
Activity 5-6: Float=0
Activity 5-7: Float=0
Activity 6-8: Float=7
Activity 7-8: Float=6
Activity 8-9: Float=5
Activity 8-10: Float=0
Activity 9-10: Float=0
(3) The critical path for this project is: 3-4, 4-9, 5-7, 7-8, 8-10, 9-10, with a total duration of 23 weeks.
THE SOLUTIONTo construct the network and compute the requested information, we first need to create a list of events and activities.From the given schedule, we can identify the following events and activities:Events:1, 2, 3, 4, 5, 6, 7, 8, 9, 10
Activities:1-2: 4 weeks
1-3: 1 week
2-4: 1 week
3-4: 1 week
3-5: 6 weeks
4-9: 5 weeks
5-6: 4 weeks
5-7: 8 weeks
6-8: 1 week
7-8: 2 weeks
8-9: 1 week
8-10: 8 weeks
9-10: 7 weeks
Now we can compute the TE (time when an event is expected to start) and TL (time when an event is expected to be completed) for each event, as well as the float (amount of time that an activity can be delayed without delaying the project completion) for each activity.TE and TL for each event:Event 1: TE=0, TL=0
Event 2: TE=4, TL=4
Event 3: TE=1, TL=5
Event 4: TE=2, TL=6
Event 5: TE=6, TL=12
Event 6: TE=1, TL=2
Event 7: TE=8, TL=16
Event 8: TE=9, TL=17
Event 9: TE=8, TL=18
Event 10: TE=16, TL=23
Float for each activity:Activity 1-2: Float=0
Activity 1-3: Float=3
Activity 2-4: Float=2
Activity 3-4: Float=1
Activity 3-5: Float=0
Activity 4-9: Float=0
Activity 5-6: Float=0
Activity 5-7: Float=0
Activity 6-8: Float=7
Activity 7-8: Float=6
Activity 8-9: Float=5
Activity 8-10: Float=0
Activity 9-10: Float=0
The critical path is the sequence of activities that have zero float, meaning they cannot be delayed without delaying the project completion. The critical path for this project is: 3-4, 4-9, 5-7, 7-8, 8-10, 9-10, with a total duration of 23 weeks.Learn more about Construct Network here:
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vertical gate in an irrigation canal holds back 12.2 m of water. Find the average force on the gate if its width is 3.60 m. Report your answer with proper units and 3 sig figs.
Answers
Answer:
The right solution is "2625 kN".
Explanation:
According to the question,
The average pressure will be:
= \(density\times g\times \frac{h}{2}\)
By putting values, we get
= \(1000\times 9.8\times \frac{12.2}{2}\)
= \(1000\times 9.8\times 6.1\)
= \(59780\)
hence,
The average force will be:
= \(Pressure\times Area\)
= \(59780\times 3.6\times 12.2\)
= \(2625537 \ N\)
Or,
= \(2625 \ kN\)
A 75-kg piano is hosted on a crane and delivered throughout the window of a 6th story apartment (20 meters above ground). What is the potential energy of the piano?
Answers
Answer:
14,700 J
Explanation:
PE = Mgh = (75 kg)(9.8 m/s²)(20 m) = 14,700 J
Prove that in a 2’s complement number system addition overflows if and only if the carry from the sign position does not equal the carry into the sign position. Consider the three cases: adding two positive numbers, adding two negative numbers, and adding two numbers of opposite sign.
Answers
Consider the four sub-cases for each of those three cases two positives, two negatives, one of each . Show that some of those sub-cases are not possible in each case. Then examine each sub-case to see if it indicates overflow.
What is 2's complement system?Two's complement is a mathematical operation that converts a positive binary number with equivalent value reversibly into a negative binary number with equivalent value.
it occurs using the binary digit with the greatest place value to indicate whether the binary number is positive or negative.
Consider each of those three cases' four sub-cases. There are two positives and two negatives, with one of each.
Demonstrate that some of the sub-cases are not feasible in each case. Then, for each sub-case, determine whether it indicates overflow.
Thus, by this it can be proved that in a 2’s complement number system addition overflows if and only if the carry from the sign position does not equal the carry into the sign position.
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A total of 2 metric tons of atrazine was released into the environment (20oC) after an accidental spill. The soil has the following properties: fraction of organic carbon 2.5%, bulk density 1.75 g/cm3, empirical equation, log Koc = 0.937 log Kow – 0.006. The fish has the following properties: density 2.1 g/cm3, bioconcentration factor (BCF) 8.1 cm3/g.
The volumes in the four phases are as follows:
• Volume of air: 1 x 1010 m3
• Volume of water: 8 x 108 m3
• Volume of soil: 7 x 107 m3
• Volume of fish: 4 x 104 m3
Given infinite time for the released atrazine to spread with no remedial action taken, what will be the mass and percentage of atrazine (in kg) in each of the four phases?
Answers
The percentage of atrazine in the soil phase is 1.35*10^-8
The percentage of atrazine in the soil phase is 1.45%
The percentage of atrazine in the fish phase is 6.7*10^-5
How to explain the percentageThe partition coefficient for atrazine between water and organic carbon (Koc) can be calculated using the empirical equation provided is 0.006.
The Kow value is 1.499. The distribution coefficient is 0.066.
The concentration of atrazine in water will be;
= 0.027 / 2000000 × 10^-8
= 1.35 × 20^-8
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when driving an ambulance, what is the safest path when cornering?
Answers
Answer:
the safest path when cornering in an ambulance is to enter high and exit low
Explanation:
Imagine that your company is considering opening a branch in Dubai. List typical activities in each phase of the decision (intelligence, design, choice, implementation) of whether to open a branch.
Answers
When a company is considering opening a branch in Dubai, it will typically go through the following phases of decision-making: intelligence, design, choice, and implementation.
Intelligence: Gather relevant information about the location, such as the local market conditions, taxation regulations, and the availability of skilled labor. Consider the advantages and disadvantages of opening a branch in Dubai.Design: Develop a detailed plan of action, outlining the process, resources, and goals. Set a budget and timeline for the project. Prepare documents such as lease agreements and any necessary permits.Choice: Compare the potential costs and benefits of the proposed branch to determine if it is a good investment for the company.Implementation: Begin the process of opening the branch in Dubai. This may include sending staff to the area, hiring and training employees, and buying necessary supplies and equipment.Learn more about branch: https://brainly.com/question/13267720
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3. (a) Describe the procedure of measuring the depth of modulation of an amplitude modulated (AM) wave using an oscilloscope with the internal timebase switched off and sketch the displayed waveform. (6 marks)
Answers
1. Connect the output of the AM transmitter to the vertical input of the oscilloscope.
2. Connect a DC voltmeter across the output of the AM transmitter.
3. Set the oscilloscope to DC coupling and adjust the vertical sensitivity so that the AM waveform is clearly visible on the screen.
4. Adjust the horizontal position of the waveform so that the unmodulated carrier is centered on the screen.
5. Adjust the DC voltmeter to read the unmodulated carrier voltage.
6. Adjust the AM modulation depth so that the AM waveform just reaches the maximum and minimum levels of the carrier waveform.
7. Read the DC voltmeter again to determine the new carrier voltage.
8. Calculate the modulation depth using the formula:
Modulation Depth = (Vmax - Vmin) / (Vmax + Vmin)
Where Vmax is the carrier voltage with modulation and Vmin is the carrier voltage without modulation.
The displayed waveform will show the unmodulated carrier waveform centered on the screen, with the modulated waveform superimposed on it. The depth of modulation can be determined by measuring the maximum and minimum levels of the modulated waveform and calculating the modulation depth using the formula above.
List 3 specific things you can do with a screw gun (not just screw in screws)
Answers
Answer:
Install drywall
insert wood panels
unscrew nails
Explanation:
Q1 (a) Identify and explain by giving an example what are the factors that consequence the ability of industrial plant functionality? (b) Illustrate the relation of manufacturing in terms of technology and economy aspects. (c) For a new plant, the management have decided to produce at least 50,000 units of products annually. On the other hand, the plant will produce variety types of products for different purposes. There were five processes involved to produce the product. As a Plant and Process Engineer, you were requested to propose an appropriate plant layout for the production purposes. Your proposal should be supported by: i. The illustration of the layout. ii. The explanation of the production operation based on the process involved. iii. The justification of your decision.
Answers
Factors influencing industrial plant functionality include technology, personnel, materials, and plant layout. Technology and economy intertwine in manufacturing, with technology enabling efficient production and the economy driving demand.
Technology availability dictates the processes and methods employed in the plant, and trained personnel ensure smooth operation. Reliable material supply enables steady production. In the economy-technology relation, technological advancements enhance manufacturing efficiency and product quality, thus improving economic returns. For the new plant, considering the varied products and multiple processes, a Cellular layout, grouping similar processes together, is proposed. This layout enhances flexibility and efficiency, critical for diverse production. The production operation follows a logical, efficient sequence per each product group.
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two technicians are describing how an air-conditioning system gets rid of the heat. Technician A says that heat always flows from the hotter to the colder object. Technician B says that as refrigerant evaporates, it absorbs heat as it changes from a liquid to a gas. which technician is correct?
Answers
Based on the information provided by these technicians, both of them are correct.
How does heat flow?According to heating, ventilation, and air conditioning (HVAC), heat in an air-conditioning system generally flows from the hotter object to the colder object.
As the refrigerant evaporates in a small radiator-type unit (evaporator), it absorbs heat as it changes phase from liquid to gas. Also, as the heat is being absorbed by the refrigerant, the small radiator-type unit (evaporator) becomes cold.
In conclusion, we can logically deduce that both of them are correct.
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The addictive properties of benzodiazepine drugs have been linked to their effects at the ______________ subunit of GABAA receptors.
A) alpha 1
B) alpha 2
C) alpha 3
D) alpha 4
E) None of these--benzodiazepine drugs are not addictive.
Answers
The addictive properties of benzodiazepine drugs have been linked to their effects at the alpha 1 subunit of GABAA receptors. Benzodiazepines are a class of drugs that act as central nervous system depressants and are commonly prescribed for treating anxiety, insomnia, seizures, and muscle spasms.
They exert their therapeutic effects by enhancing the inhibitory neurotransmission of gamma-aminobutyric acid (GABA), the primary inhibitory neurotransmitter in the brain.
GABAA receptors are composed of different subunits, including alpha, beta, and gamma. Benzodiazepines bind to a specific site on the GABAA receptor complex, which is distinct from the GABA binding site. This interaction potentiates the effects of GABA, resulting in increased inhibitory neurotransmission.
The alpha 1 subunit of GABAA receptors has been implicated in the addictive properties of benzodiazepines. Studies have shown that benzodiazepines with a higher affinity for the alpha 1 subunit are more likely to produce dependence and addiction. Additionally, research using genetically modified mice has demonstrated that the absence of the alpha 1 subunit reduces the reinforcing effects of benzodiazepines, suggesting a key role for this subunit in addiction.
In summary, the addictive properties of benzodiazepines are associated with their effects at the alpha 1 subunit of GABAA receptors. Understanding the specific receptor subunits involved in addiction can help inform the development of new treatments for substance use disorders.
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When developing the light bulb, thomas edison’s lab went through numerous materials until they found one that would work. Which one finally worked?.
Answers
Platinum filament.
Thomas Edison went through a series of different experiments while developing bulb until and unless the electricity passed through the Platinum filament, and it finally worked.
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Q3) Design a 3-input NOR gate with equal size NMOS and PMOS transistors using SPICE. a. While keeping two inputs constant at logic 0, sweep the third input from logic 0 to logic 1 and plot the Voltage Transfer Curve (VTC). b. While keeping two inputs constant at logic 0, alternate the third input between logic 0 and logic 1. Determine rise and fall times with 5 pF load. c. Resize the transistors to make rise and fall times similar. d. Repeat step a. with the new transistor sizes and determine the noise margins.
Answers
a) Design a 3-input NOR gate using SPICE with equal size NMOS and PMOS transistors. Keep two inputs constant at logic 0 and sweep the third input from logic 0 to logic 1 to plot the Voltage Transfer Curve (VTC).
b) With two inputs at logic 0, alternate the third input between logic 0 and logic 1. Determine the rise and fall times with a 5 pF load.
c) Resize the transistors to achieve similar rise and fall times.
d) Repeat step a with the new transistor sizes and determine the noise margins.
a) To design a 3-input NOR gate using SPICE, we need to create a circuit that incorporates three NMOS transistors and three PMOS transistors. The NMOS transistors are connected in parallel between the output and ground, while the PMOS transistors are connected in series between the output and the power supply. By keeping two inputs constant at logic 0 and sweeping the third input from logic 0 to logic 1, we can observe how the output voltage changes and plot the Voltage Transfer Curve (VTC).
b) With two inputs at logic 0, we alternate the third input between logic 0 and logic 1. By applying a 5 pF load, we can measure the rise and fall times of the output voltage, which indicate how quickly the output transitions from one logic level to another.
c) In order to achieve similar rise and fall times, we need to resize the transistors in the circuit. By adjusting the dimensions of the transistors, we can optimize their performance and ensure that the rise and fall times are approximately equal.
d) After resizing the transistors, we repeat step a by sweeping the third input from logic 0 to logic 1. By analyzing the new transistor sizes and observing the resulting output voltage, we can determine the noise margins of the circuit. Noise margins indicate the tolerance of the gate to variations in input voltage levels, and they are essential for reliable digital circuit operation.
By following these steps and performing the necessary simulations and measurements using SPICE, we can design and analyze a 3-input NOR gate, optimize its performance, and determine important parameters such as the Voltage Transfer Curve, rise and fall times, and noise margins.
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A customer complained that the front wheel would skip, hop, and shake while cornering and accelerating at the same time. technician a says that a fault in the transfer case is the most likely cause. technician b says that the vibration may be normal if the vehicle is equipped with cardan u-joints at the front wheels. which technician is correct
Answers
Technician B is correct and he is providing the correct guidance to the customer.
Why technician B is correct?A faulty driveshaft, commonly found near the front wheels, can make the vehicle difficult to control. This faulty drive shaft can be the result of worn U-joints or bushings. This can cause the drive shaft to vibrate when attempting to turn or accelerate.
What are U-joints?The part that connects the drive shaft to the differential and sometimes connects the drive shaft to the transmission or two drive shafts to each other is called U-Joints and breaking of u- joints is known as bushing.
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Which of the following sensors is used to provide suspension control module with feedback regarding vehicle cornering forces?
Question content area bottom
Part 1
A.
lateral accelerometer sensor
B.
vehicle speed sensor
C.
yaw rate sensor
D.
pressure sensor
Answers
Answer:
A
Explanation:
Kieran and Kurt spend most of their day performing physical labor, aligning materials, and inspecting projects for quality and safety. However, Kieran has no fear of heights, while Kurt works with his feet on the ground. Which most likely explains the jobs of Kieran and Kurt? Kieran is a Roofer, and Kurt is a Plumber. Kieran is a Plumber, and Kurt is a Roofer. Kieran is a Carpenter, and Kurt is a Drafter. Kieran is a Drafter, and Kurt is a Carpenter.
Answers
Answer:
Kieran is a Carpenter, and Kurt is a Drafter
Explanation:
Among all the professions listed, Kurt can only possibly be a Drafter. A Drafter is an engineering technician who makes detailed technical drawings or plans for machinery, buildings, electronics, infrastructure, sections, etc. Drafters make use of computer software and manual sketches to convert the designs, plans, and layouts of engineers and architects into a set of technical drawings. All the other professions will at some point need for the professional to stand and work at an height above the ground except for a drafter, whose feet will remain on the ground at almost all times.
Answer:
A is the right answer.
Explanation:
Hope that helps
+
Vi
v (t)
V
-37/2T-T/2
8 ΚΩ
ww
-V
m
T/2
1. For the above, derive the Fourier series representation of the waveform.
3T/2
10 mH
t
10 nF
Vo
2. The waveform above is applied to Vi. Using the Fourier series derived in part 1, determine the
magnitude of Vo in terms of Vm if T is equal to for the first 3 non-zero terms of the series.
2πT
50,000
3. If the waveform is applied across a 9-ohm resistor, and Vm =27V calculate the exact amount of
average power absorbed by the resistor.
4. Calculate the power absorbed by the 9-ohm resistor from the first 3 non-zero terms from the
Fourier series.
5. Determine the percentage error in the calculation from part 4 compared with the exact value of
part 3.
Answers
The sine term 27r 131,= 1 —f v(t)sin (nt)dt 7r
27r1 7 = — (f sin tsin (nt)dt f 0
7r2n (1( —1)n)
How to solveStep 1 ..
We can express the waveform as a function of time as shown below.
v(t) = sin t; 0 < t < 7F = —sin t; 7F < t < 27r
In order to get the Fourier transform of the function, we have to calculate the sine and cosine terms.
The cosine term is 1 27r a 0 v(t)dt fo 1 ( 27 271 u sin tdt — sin tdil 1=0
1 27r an = —f v(t)cos (nt)dt 71 0 27r
= (f sin tcos (nt)dt —
71 0 = 7111 (1 — cos n7r)/
Now, we can find the sine term 27r 131,= 1 —f v(t)sin (nt)dt 7r
27r1 7 = — (f sin tsin (nt)dt f 0
7r2n (1( —1)n)
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Tech A says that multi-plate clutches must have a specified amount of clearance when they are installed. Tech B says that there should be no clearance when they are installed because the clutch would slip. Who is correct?
Answers
The person who is correct based on the assertions made about the multi-plate clutch is:
Technician AMulti-plate ClutchThis refers to the machinery which makes use of plates so that there would be efficient transmission between the engine shaft and the transmission shaft.
WIth this in mind, we can see that they must have specific amount of clearance when they are installed so that they can effectively function in the automobile.
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gasoline has a comparatively high btu per galloon rating around?
Answers
Answer:
116,090 Btus
Explanation:
Unsaturated fatty acids have lower melting points than saturated fatty acids because
A. they have fewer hydrogen atoms
B. They have more hydrogen atoms
C. their molecules fit closely together
D. the cis double bonds give them an irregular shape
E. the trans double bonds give them an irregular shape
Answers
Answer: Unsaturated fatty acids cannot pack as tightly together as saturated fatty acids due to the presence of the cis double bond that puts a “kink” or bend in the hydrocarbon chain. 3 < 1 < 2; an increase in the number of double bonds will lower the melting point because it is more difficult to closely pack the fatty acids together.
Explanation: Hope the explnation above was helpful
the name most often used to describe the universal joints on a conventional rear-wheel drive vehicle driveshaft is
Answers
A U-joint (the "U" stands for "universal"), sometimes called a Cardan joint after the man who invented it, is a type of flexible coupling typically used on both ends of the drive shafts in rear-wheel and four-wheel drive vehicles.
Developer wishes to construct an office building of 10,000m2 gross floor area, of which 8,000 m2will be available for letting. The construction costs are estimated to be $6,000/m2. In addition, there are ancillary construction costs of $400,000 in laying roads and sewers to the building. Professional fees are estimated to total 13% of construction costs. Short-term finance is available at 16%. The expected rent is $3,000/m2 p.a. net. The developer wishes to see a return for risk and profit of 20% of development value. The pre-contract period is expected to be 6 months, the building work is estimated to take 15 months, and a period of 3 months has been allowed for letting. The developer intends to sell the completed and fully let development to a financial institution, and it is anticipated that an initial yield of 7% will be required. Within these parameters, the value of the site has to be established.
Answers
The value of the site is $10,472,000.
Here are the calculations:The total construction costs are:
$6,000/m2 * 10,000m2 + $400,000 = $64,000,000
The professional fees are:
$64,000,000 * 0.13 = $8,320,000
The total development costs are:
$64,000,000 + $8,320,000 = $72,320,000
The developer's return for risk and profit is:
$72,320,000 * 0.20 = $14,464,000
The total cost of the development is:
$72,320,000 + $14,464,000 = $86,784,000
The expected rent is:
$3,000/m2 * 8,000m2 * 12 months = $28,800,000
The initial yield is:
$28,800,000 * 0.07 = $2,016,000
The value of the site is:
$86,784,000 - $2,016,000 = $10,472,000
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Refrigerant 134a is the working fluid in a vapor-compression heat pump system with a heating capacity of 70,000 Btu/h. The condenser operates at 180 lbf/in2, and the evaporator temperature is 20˚F. The refrigerant is a saturated vapor at the evaporator exit and exits the condenser at 120˚F. Pressure drops in the flows through the evaporator and compressor are negligible. The compression process is adiabatic, and the temperature at the compressor exit is 200˚F. Determine
a) The mass flow rate of refrigerant, in lb/min
b) The compressor power output, in horsepower.
c) The isentropic compressor efficiency.
d) The coefficient of performance.
Answers
It is desired to obtain 500 VAR reactive power from 230 Vrms 50 Hz 1.5 KVAR reactor. What should be the angle of the AC to AC converter to be used? Calculate the THD of the current drawn from the mains (consider up to the 12th harmonic)?
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Answer:
14.5° ; THD % = 3.873 × 100 = 387.3%.
Explanation:
Okay, in this question we are given the following parameters or data or information which is going to assist us in solving the question efficiently and they are;
(1). "500 VAR reactive power from 230 Vrms 50 Hz 1.5 KVAR reactor".
(2). Consideration of up to 12th harmonic.
So, let us delve right into the solution to the question above;
Step one: Calculate the Irms and Irms(12th) by using the formula for the equation below;
Irms = reactive power /Vrms = 500/230 = 2.174 A.
Irms(12th) = 1.5 × 10^3/ 12 × 230 = 0.543 A.
Step two: Calculate the THD.
Before the Calculation of the THD, there is the need to determine the value for the dissociation factor, h.
h = Irms(12th)/Irms = 0.543/ 2.174 = 0.25.
Thus, THD = [1/ (h)^2 - 1 ] ^1/2. = 3.873.
THD % = 3.873 × 100 = 387.3%.
Step four: angle AC - Ac converter
theta = sin^-1 (1.5 × 10^3/ 12 × 500) = 14.5°.
What is the floor of Class E airspace when designated in conjunction with an airway?
Select one:
a. 700 feet AGL.
b. 1,500 feet AGL.
c. 1,200 feet AGL.
Answers
correct answer to the question is b. 1,500 feet AGL when the Class E airspace is designated in conjunction with an airway that is 4 NM or wider.
When Class E airspace is designated in conjunction with an airway, the floor of the airspace depends on the width of the airway. If the airway is less than 4 nautical miles (NM) wide, then the floor of the Class E airspace is 1,200 feet above ground level (AGL) within 4 NM either side of the airway centerline. If the airway is 4 NM or wider, then the floor of the Class E airspace is 1,500 feet AGL within 4 NM either side of the airway centerline.
To know more about AGL visit:
brainly.com/question/14701469
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