Contents: Introduction Background Importance of Experiment Aim and Hypothesis Results Discussion Discoveries and Explinations Limitations Applications Conclusion Biography Introduction (482 words) This experiment measures the rate of change and efficiency of different burning alcohols. An alcohol‚ any of a class of organic compounds characterized by one or more hydroxyl (−OH) groups attached to a carbon atom of an alkyl group (hydrocarbon chain). Alcohols are among the most common
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SOLVED PROBLEMS OF CHAPTER # 13 TITLE: RECPROCTING INTERNAL COMBUSTION ENGINES PROBLEM # 13.1: A quantity governed four-stroke‚ single-cylinder gas engine has a bore of 146mm and a stroke of 280mm.At 475 rev/min and full load the net load on the friction brake is 433 N‚ and the torque arm is 0.45 m. The indicator diagram gives a net area of 578 mm² and a length of 70 mm with a spring rating of 0.815 bar per mm. calculate the ip‚bp‚ and mechanical efficiency. GIVEN DATA: Four stroke
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value of 60 lbft3 to kgm3‚ Ans: 961.1 kg/m3 b. Energy value of 1.7 ×103Btu to kJ‚ Ans: 1.8 x 103 kJ c. Enthalpy value of 2475 BtulbtokJkg‚ Ans: 5756 kJ/kg d. Pressure value of 14.69 psig to kPa‚ Ans: 201.88 kPa e. Viscosity value of 20 cp to Pa∙s‚ Ans: 2 x 10-2 Pa∙s Table of Conversion Factor 1 lb | 0.454 kg | 1 ft | 0.305 m | 1 Btu | 1.055 kJ | 1 psia | 1 psig‚ pound-force per square inch gauge - 14.69 | 1 psia‚ pounds
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column or page – find all choices before answering. 001 1 003 6.0 points What would be pH of a 1 M solution of a very expensive weak base‚ unobtainamine‚ whose pKb is 3.4? 6.0 points 1. 12.3 correct Consider a reaction with ∆Hrxn = 997 kJ · mol−1 . Which of the following pairs of K values and temperatures is possible for this reaction? 1. K1 = 15.7‚ T1 = 138 K; K2 = 1.57 × 1038 ‚ T2 = 153 K correct 2. K1 = 11.5‚ T1 = 128 K; K2 = 3.83‚ T2 = 145 K 3. K1 = 3.1‚ T1 = 103 K; K2 =
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Friends frequently pass comment on how quickly and easily healthy food comes out of my kitchen. For this I thank my role model mother who always had a nourishing meal or snack on the table in a flash to feed any number of people who happened to be there at the time. My country upbringing also helped me develop the skills and initiative needed to produce great meals without the convenience of nearby shops and the easy options that town folk have. I now combine these skills with the nutrition knowledge
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diplococci bacteria responsible for the sexually transmitted infection gonorrhea (Ryan KJ‚ 2004). Neiseeria Gonorrhoeae infections are obtained through sexual contact through the moist mucous membranes of the urethra in males‚ endocervix and urethra in females. About 95% of infected males are symptomatic‚ with symptoms appearing in 2-5 days‚ with pus-like discharge from urethra‚ then painful urination starts to develop (Ryan KJ‚ 2004). Many women have no symptoms‚ and those that do experience common symptoms
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Introduction The purpose to of this experiment is to carry out the alkylation of sodium saccharin with iodo-ethane and analyze the product mixture to determine the structure of the major product. Sodium saccharin is made from the base catalyzed de protonation of saccharin. This nucleophilic reaction is special because the nucleophilic atom can be oxygen or nitrogen and the leaving group is iodide ion. The solvent used in this reaction is very important for determining the rate of nucleophilic
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CHAPTER I INTRODUCTION 1.1 HISTORY: Ethylene Glycol (1‚ 2 – ethanediol)‚ HOCH2CH2OH usually called glycol is the simplest Diol. Diethylene glycol and Triethylene glycol are Oligomers of Mono ethylene glycol. Ethylene glycol was first prepared by Wurtz in 1859; treatment of 1‚2 dibromoethane with silver acetate yielding ethylene glycol diacetate via saponification with potassium hydroxide and in
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184 J n = m/M = 2/23 + 16 + 1 = 0.05 mol ΔHs = -Q/n = -2043.184 / 0.05 = -40863.68 J/mol = -40.86 kJ/molΔHS NaOH = -40.86 kJ < 0 ∴ exothermic reaction | Q = mCΔT = 52 x 4.18 x (22.2 - 25) = -608.608 J n = m/M = 2/(14 + 4 x 1) + (14+16 x 3) = 0.025ΔHs = -Q/n = 608.608 / 0.025 = 24344.32 J/mol = 24.34 kJ/molΔHS NH4NO3 = 24.34 > 0 ∴ endothermic reaction | Discussion The dissolution of NaOH is an exothermic process whereas
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and saturated vapor enters the turbine. The condenser pressure is 6 kPa. Determine (a) the thermal efficiency. (b) the back work ratio. (c) the net work of the cycle per unit mass of water flowing‚ in kJ/kg. (d) the heat transfer from the working fluid passing through the condenser‚ in kJ per kg of steam flowing. (e) Compare the results of parts (a)–(d) with those of Problem 8.6‚ respectively‚ and comment. 8.8 Plot each of the quantities calculated in Problem 8.6 versus turbine inlet temperature
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