MAKE A WELCOME NOTE AS THE PROJECT MANAGER. (NOT MORE THAN 20 WORDS.) Case scenario: Taking the role as Project Manager for Biz Wiz Project Pte Ltd, you have been approached by Margaret House, CEO for Bounce Fitness to undertake a project to develop a café for Bounce Fitness in Sydney.
1. MAKE A WELCOME NOTE AS THE PROJECT MANAGER. (NOT MORE THAN 20 WORDS.) Case scenario: Taking the role as Project Manager for Biz Wiz Project Pte Ltd, you have been approached by Margaret House, CEO for Bounce Fitness to undertake a project to develop a café for Bounce Fitness in Sydney.
Dear Margaret House and Bounce Fitness team, I am thrilled to lead the project in developing a café for Bounce Fitness in Sydney.
2. what is 18 groups in the PTE?
Answer:
In the context of PTE (Pearson Test of English) Academic, "18 groups" is not a term or concept that I am familiar with.
The PTE Academic exam assesses the English language proficiency of non-native English speakers. It measures a candidate's ability to communicate effectively in academic settings and includes four sections: Speaking & Writing, Reading, Listening, and Writing. Each section consists of several tasks that evaluate various language skills.
If you could provide more context or information about what specifically you are referring to with "18 groups," I may be able to provide a more accurate answer.
3. 1. BELCKNIG 2. HPO PTES 5. CLEOS PTES PE scrambled letters po yan.. nonsense = report and 1 star
Answer:
1. Blecking
2. Hop Step
3. Close Step
Explanation:
Hope it helps :)
4. Affiliated computer services of india pvt ltd
please specify question.
5. Combustion of 2.016 g glucose C6H12O6 at 298 K in a bomb calorimeter with heat capacity 9.55 kJ/K gives a temperature rise of 3.282 K. Which of the following is the energy of combustion of glucose a. -15 kJ/mol B.2798 kJ/mol C.15 kJ/mol d. -2798 kJ/mol e.None of the above
Answer:
D
Explanation:
by finding heat capacity and then molar heat capacity using data given
6. The machine does work of 49 kJ as it realeses 107 kJ of heat. How much work in KJ must be done on the machine to maintain the internal heat at 30 kJ?
Answer:
The first law of thermodynamics states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system:
ΔU = Q - W
where ΔU is the change in internal energy, Q is the heat added, and W is the work done.
In this problem, the machine releases 107 kJ of heat, so Q = -107 kJ (negative because heat is released). The work done by the machine is 49 kJ, so W = 49 kJ. We want to find the work that must be done on the machine to maintain the internal heat at 30 kJ, so ΔU = 30 kJ - (-107 kJ) = 137 kJ.
Therefore, we can rearrange the equation to solve for the work done:
W = Q - ΔU
W = -107 kJ - 137 kJ
W = -244 kJ
The negative sign indicates that work must be done on the machine to maintain the internal heat at 30 kJ, and the magnitude of the work is 244 kJ. Therefore, 244 kJ of work must be done on the machine to maintain the internal heat at 30 kJ.
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Brain!iest, Thank you
7. Calculate the internal energy of the system that absorbs 50 kJ of heat and 65 kJ work done by the system. A. + 15 J B. -15 kJ C. +20 kJ D. -20 kJ
SOLUTION:
Step 1: Determine the signs for heat and work.
Since heat is absorbed, then
q = +50 kJ
Since the work is done by the system, then
w = -65 kJ
Step 2: Calculate the change in internal energy.
[tex]\begin{aligned} \Delta E & = q + w \\ & = \text{+50 kJ} + (-\text{65 kJ}) \\ & = \boxed{-\text{15 kJ}} \end{aligned}[/tex]
Hence, the answer is B. -15 kJ.
[tex]\\[/tex]
#CarryOnLearning
8. 5. What is the value of of G when H = 25.7 kj and S = 108.3J/K or 0.1083 kj/K?a. -6.6 kjb.-38 kjc. 20 kjd. 6.6 kj
SOLUTION:
Assume that the temperature is 298 K.
[tex]\begin{aligned} \Delta G & = \Delta H - T\Delta S \\ & = \text{25.7 kJ} - (\text{298 K})(\text{0.1083 kJ/K}) \\ & = \text{25.7 kJ} - \text{32.3 kJ} \\ & = \boxed{-\text{6.6 kJ}} \end{aligned}[/tex]
Hence, the answer is a. -6.6 kJ.
[tex]\\[/tex]
#CarryOnLearning
9. Advanced electronics company ltd history brief
Answer:
Advanced Electronics Company (AEC) is a limited liability company in Saudi Arabia that was established in 1988 under the Economic Offset Program as per the directives of the government of Saudi Arabia. AEC is a regional organization specializing in the field of modern electronics.
Explanation:
10. What is the concern of the Dorchester,Ltd case
Answer:
It was concerned that the date for its response of 28 January 2008 was only 18 working days after 5 January and was simply not long enough to respond to the detailed claim submitted given the new evidence and revised figures. The Dorchester therefore sought declarations from the Court through Part 8 proceedings that:
"there is a serious risk of a breach of natural justice in the conduct of the adjudication … and, unless the parties agree a realistic timetable …, any decision issued by the Adjudicator against the existing timetable … would be unenforceable by reason of breach of natural justice".
Explanation:
hope it helps..keep safe po
11. my name is kj Baconguis 12 year old tektok account kj Baconguis
Answer:
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12. Arnold vs willets and patterson ltd case digest
Answer:
hi wht is itd?
Explanation:
brainliest pls ty
13. What is PTE int he laboratory T.L.E
Answer:
A. The standard PTE Academic exam fee is Rs 14,700. In case the candidate is booking his PTE exam within 48 hours of the PTE test date, the candidate would be required to pay the PTE Late Booking Fee.14. almost ready ltd cash flow
Answer:
hello
Step-by-step explanation:
15. Given the following data: Substance ΔH°f , kJ/mol SO2Cl2 (g) ‒ 364 H2SO4 (l) ‒ 814 H2O (l) ‒ 286 If the ΔH°rxn for the following reaction is ‒ 62.0 kJ, SO2Cl2 (g) + 2 H2O (l) → H2SO4 (l) + 2 HCl (g) the ΔH°f in kJ/mol for the HCl (g) is _______. a. − 184 kJ/mol b. − 62.0 kJ/mol c. − 92.0 kJ/mol d. − 368 kJ/mol
Answer:
Use the ΔH°f information provided to calculate ΔH°rxn for the following:ΔH°f (kJ/mol) SO2Cl2(g) + 2 H2O(l) → 2 HCl(g) + H2SO4(l) ΔH°rxn = ?SO2Cl2(g) -364 H2O(l) -286 HCl(g) -92 H2SO4(l) -814a. +161 kJb. -422 kJc. -256 kJd. +800 kJe. -62kJ16. The machine does work of 49 kJ as it realeses 107 kJ of heat. How much work in KJ must be done on the machine to maintain the internal heat at 30 kJ?
Answer:
polygons nonagon hexagon heptagon yun thx
Answer:
The first law of thermodynamics states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system:
AU=Q-W
where AU is the change in internal energy, Q is the heat added, and W is the work done.
In this problem, the machine releases 107 kJ of heat, so Q = -107 kJ (negative because heat is released). The work done by the machine is 49 kJ, so W = 49 kJ. We want to find the work that must be done on the machine to maintain the internal heat at 30 kJ, so AU = 30 kJ - (-107 kJ) = 137 kJ.
Therefore, we can rearrange the equation
to solve for the work done:
W=Q-AU
W = -107 kJ - 137 kJ
W = -244 kJ
The negative sign indicates that work must be done on the machine to maintain the internal heat at 30 kJ, and the magnitude of the work is 244 kJ. Therefore, 244 kJ of work must be done on the machine to maintain the internal heat at 30 kJ.
17. What is the kinetic energy of a 500-kg car moving at 3.0 m/s? a.) 1.50 kJ b.) 1.75 kJ c.) 2.00 kJ d.) 2.25 kJ
Answer:
d.) 2.25 kJ
Explanation:
hope it help thanks Love you
18. 1. A 2 kW electric resistance heater in a laboratory dryer turned on and kept on for 30 min the amount of energy transferred is? a) 1 kJ b) 60 kJ c) 1800 kJ d) 3600 kJ e) 7200 kJ
Answer:
Letter D or 3600 KJ
Explanation:
19. carpenters and construction workers make as a hammer in driving emails in what way can they protect their fingers from being hammered
Answer:
b. use leather gloves
Explanation:
para baga
20. EPE LTD NUILOLON Yaus mo
Answer:
ano pu yon?
Explanation:
i dont understand.
Answer:
Not sure what do you mean?
21. Calculate the internal energy of the system that absorbs 50 kJ of heat and 65 kJ work done by the system. A. + 15 J B. -15 kJ C. +20 kJ D. -20 kJ
SOLUTION:
Step 1: Determine the signs for heat and work.
Since heat is absorbed, then
q = +50 kJ
Since the work is done by the system, then
w = -65 kJ
Step 2: Calculate the change in internal energy.
[tex]\begin{aligned} \Delta E & = q + w \\ & = \text{+50 kJ} + (-\text{65 kJ}) \\ & = \boxed{-\text{15 kJ}} \end{aligned}[/tex]
Hence, the answer is B. -15 kJ.
[tex]\\[/tex]
#CarryOnLearning
22. pumili ng isang yugto o panahon ng pte historiko at iguhit ito.
Answer:
awashmewepwep washmenene
23. Color the spaces below that contain the answers to find a letter.2173180061137214729214002125430942044386237011102114313214254147348116141250423961341506312415633140The letter ise Education International (S) Pte Ltd ISBN 978-981-47-6941-925
Answer:
Ano po ibig sabihin nyan paturo nalang po sa comments
24. For the reaction 2NaHCO3 → Na2CO3 + CO2 + H2O ∆H = 129.37 kJ/mole Find the standard heat of formation of NaHCO3 given the following ∆H of formation: Na2CO3 = -1,130.94 kJ/mol; CO2 = -393.5 kJ/mol; H2O = -241.8 kJ/mol.
Answer:
Chemistry
5 points
5.0
2
Answer
shineri1234 • Beginner
Answer:
∆H°_{rxn} = -5314 \: \frac{kJ}{mol}
Explanation:
Given:
∆H°_{f}(\text{C₄H₁₀}, \: g) = -126 \: \frac{kJ}{mol}
Required:
∆H°_{rxn}
Solution:
Step 1: Write the balanced chemical equation
2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O
Explanation:
pa mark brainliest po ako please
25. When gas absorbs 28 KJ of heat and has 13 KJ of work done on it
Answer:DU= +13 KJ
Explanation:
Pa brainliest po plsss :<
26. the PTE presents vast amount of information regarding the atoms of the elements, TRUE OR FALSE?
Answer:
True po.
Explanation: Base po sa pinagaralan namin
27. The machine does work of 49 kJ as it realeses 107 kJ of heat. How much work in KJ must be done on the machine to maintain the internal heat at 30 kJ?
The first law of thermodynamics states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system. Mathematically,
ΔU = Q - W
where ΔU is the change in internal energy, Q is the heat added to the system, and W is the work done by the system.
In this problem, the machine releases 107 kJ of heat, so Q = -107 kJ (negative because heat is leaving the system). The internal energy needs to be maintained at 30 kJ, so ΔU = 30 kJ. We can rearrange the equation to solve for the work done on the machine:
W = Q - ΔU
W = -107 kJ - 30 kJ
W = -137 kJ
Therefore, 137 kJ of work must be done on the machine to maintain the internal heat at 30 kJ.
Answer:
The first law of thermodynamics states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system:
AU=Q-W
where AU is the change in internal energy, Q is the heat added, and W is the work done.
In this problem, the machine releases 107 kJ of heat, so Q = -107 kJ (negative because heat is released). The work done by the machine is 49 kJ, so W = 49 kJ. We want to find the work that must be done on the machine to maintain the internal heat at 30 kJ, so AU = 30 kJ - (-107 kJ) = 137 kJ.
Therefore, we can rearrange the equation
to solve for the work done:
W=Q-AU
W = -107 kJ - 137 kJ
W = -244 kJ
The negative sign indicates that work must be done on the machine to maintain the internal heat at 30 kJ, and the magnitude of the work is 244 kJ. Therefore, 244 kJ of work must be done on the machine to maintain the internal heat at 30 kJ.
28. Calculate ΔG0 is {free energy} kJ at T is 338.8 K for the reaction for which ΔH0 is 224.1 kJ and ΔS0 is -0.71 kJ/K.
Answer:
tom decides to keep and the toys he,ll dpnate
29. What theISmaincausethese problem of ABC LTD?
Explanation:
thanks sa points thanks uli❤️
30. 7. A heat engine has an efficiency of 31.4% and receives 8.72 kJ of heat per cycle. How much work does it perform in each cycle? a) 2.7 Kj, 6 kJ 2.7 kJ, 10 kJ b) 5 kJ, 3.0 k 5.98 kJ, 2.7 kJ c) d) and receives 8.72 kJ of
The efficiency of a heat engine is given by the equation:
efficiency = work output / heat input
Multiplying both sides by heat input, we get:
work output = efficiency x heat input
Substituting the given values, we have:
efficiency = 31.4% = 0.314 (in decimal form)
heat input = 8.72 kJ
work output = 0.314 x 8.72 kJ
work output = 2.74 kJ
Therefore, the work performed by the heat engine in each cycle is 2.74 kJ.
Answer: a) 2.7 Kj, 6 kJ 2.7 kJ, 10 kJ (rounded to two significant figures)
