Purpose: The purpose of the experiment is to calculate the enthalpy change occurring in the first of reaction of sodium hydroxide with hydrochloric acid and sodium hydroxide with ammonium chloride. Using Hess’ Law and the results for the enthalpy change of the first two reactions should give the enthalpy change of the third reaction of ammonia with hydrochloric acid. Procedure: Part 1: Construct a calorimeter of two nested stereophony cups where the cover has a hole to fir a thermometer. Measure
Premium Ammonia Sodium chloride Chlorine
ISSN 0254-4725 Food energy – methods of analysis and conversion factors FAO FOOD AND NUTRITION PAPER 77 Ingested energy (IE) = gross energy (GE) Faecal energy (FE) Combustible gas (GaE) (from microbial fermentation) Digestible energy (DE) Urinary energy (UE) Surface energy (SE) Metabolizable energy (ME) Heat of microbial fermentation Obligatory thermogenesis‚ i.e. excess heat relative to glucose during ATP synthesis Net (metabolizable) energy (NME) Non-obligatory dietary thermogenesis
Premium Amino acid Nutrition Metabolism
III In a certain process‚ 8.00 kcal of heat is furnished to the system while the system does 6.00 kJ of work. By how much does the internal energy of the system change during the process? We have tlQ = (8000 eal )(4.184 1leal) Therefore‚ from the First Law tlQ tlU 20.2 III = 33 .5 k1 tlW and = 6.00 kJ = tlU + tl w. = tlQ - tlW = 33.5 kJ - 6.00 k1 = 27.5 k1 The specific heat of water is 4184 J /kg· K. By how many joules does the internal
Premium Thermodynamics Heat Energy
ice at -10 ºC to liquid water at 10 ºC? (The molar heat capacity of liquid water is 75.4 J/mol ºC; the molar heat capacity of ice is 40.2 J/mol ºC. The molar heat of vaporization of water is 40.7 kJ/mol. The molar heat of fusion of water is 6.02 kJ/mol.) (A) 71.8 kJ (B) 419 kJ (C) 64.2 kJ (D) 64.6 kJ (E) 11‚620 J 2. Which physical property of a liquid is NOT the result of strong intermolecular forces? (A) high vapor pressure (B) high boiling point (C) high heat of fusion (D) high
Premium Chemical reaction Chemical kinetics Reaction rate
constant volume and that cv for water may be taken as 4.18 kJ/kg.K. Problem 4: A 0.0283 m3 container is filled with air at 1.365 bar and 37.77oC. Calculate the final pressure in the container if 10544.82 J of heat are added. Assume ideal gas behaviour‚ with constant specific heats. Problem 5: Nitrogen is to be heated at constant pressure from 310.77 K to 1921.88 K. Calculate the heat transfer per mole. C p 39.65 8071 1.5 106 kJ for N 2 T T 2 kg.K Problem 6: Air is contained
Premium Thermodynamics Heat Energy
Temperature Changes in Reactions 1-3 Condition Reaction 1 Reaction 2 Reaction 3 Tmax (°C) 32.5 21.3 33.6 Tinitial (°C) 21.9 20.9 21.0 ΔT (°C) 10.6 0.4 12.6 Heat Energy Absorbed by Surroundings in Reactions 1-3 Reaction q(kJ) 1 4.5680 kJ 2 0.1724 kJ 3 5.4300 kJ The products and reactants in reactions 1 and 2 were manipulated. In reaction 2‚ the equation was flipped so NH4+ was a product and NH3 was a reactant used later in reaction 3. Since the second reaction was flipped‚ the sign must
Premium Thermodynamics Energy Enthalpy
A. WORK-OUT PROBLEMS: Write formulas for the following: calcium nitrate phosphorous pentafluoride aluminum carbonate strontium hydroxide methane potassium oxide lithium chloride barium sulfate phosphate ion dinitrogen tetroxide Give the complete electron configurations of: S‚ O2-‚ and Mn. For the following molecules/ions‚ give the Lewis structure‚ molecular geometry‚ and electron pair geometry: NO2- SF4 Write Lewis structures to represent all resonance forms of CO32-
Premium Oxygen Sodium chloride Chlorine
As well‚ the re-sealable features prevent insects from entering the bag. 4. CP of apple cider = 3.651 Amount of energy in 1 kg of 100°C water = 419.04 KJ Amount of energy in 1 kg of 100°C water = 2676.1 KJ Energy difference = 2676.1 KJ – 419.04 KJ = 2257.06 KJ Enthalpy to convert 1kg of 100°C water to dry steam = 2257.06 KJ
Premium Temperature Fahrenheit Boiling point
the combustion of cyclododecatriene. The heat of combustion for cyclododecatriene was determined using bomb calorimetry and used to solve for the stabilization energy of benzene. The bond stabilization energy of benzene is found to be 167.6 ± 388.3 kJ/mol. This evidence that benzene has a resonance stabilization energy is possibly necessary for determining the structure. Introduction The structure of benzene was a focus for many years. Although it was agreed that benzene contained three double
Free Enthalpy Thermodynamics Aromaticity
Final Biochemistry Exam Select the best answer for the following questions. 1. Which group of single-celled microorganisms has many members found growing in extreme environments? A) Bacteria B) Archaea C) Eukaryotes D) Heterotrophs E) None of the above 2. The three-dimensional structure of macromolecules is formed and maintained primarily through noncovalent interactions. Which one of the following is not considered a noncovalent interaction? A)
Premium Amino acid