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    2424.4 J Now to find the enthalpy change per mole or (kJ/Mol) for the reaction we divide the enthalpy change value obtained with the number of moles of a solvent which is not in excess using (Enthalpy change )/(number of mole ) Now‚ Enthalpy change(∆H) = - 2424.4 J Number of moles = 0.0125 mol Using the equation provided above‚ we can plug in the values necessary to find the enthalpy change value per mole of the reaction ( - 2424.4 KJ)/(0.0125 mol ) - 193952 J/Mol This value is obtained

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    Enthalpy Lab Background

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    Enthalpy‚ represented by the sign ∆H in kJ/mol‚ is the heat change in a reaction. It shows whether how much heat is released or absorbed during the reaction. If the reaction is endothermic‚ the enthalpy would be positive and if the reaction is exothermic‚ the enthalpy would be negative. During a chemical reaction‚ which consists of breaking and creating bonds‚ heat is either absorbed or released. In this lab‚ the reaction uses the disassociation of an ionic compound ammonium nitrate shown in the

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    Monopoly Market Structure

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    A monopoly is a market structure where there is merely one manufacturer/supplier for a product. The lone business is the industry. Entrance into such a market is controlled based on elevated costs or additional obstacles‚ which may be‚ political social or economic. In an oligopoly‚ there are simply a limited number of firms that create an industry. This top quality assemblage of firms has control over the price in addition to a‚ monopoly; an oligopoly also has extraordinary obstacles to admittance

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    Chemestry

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    is relevant to this question. What could be the proton number of an element that has three unpaired electrons in each of its atoms? A 5 B 13 C 15 D 21 4 Given the following enthalpy changes‚ I2(g) + 3Cl2(g) → 2ICl3(s) I2(s) → I2(g) ∆Ho = –214 kJ

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    Practice Final

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    Last Name ____________________First Name _______________________ Ignore: 5‚ 7‚ 19‚ 30 1. | How many grams of CaCl2 (molar mass = 111.0 g/mol) are needed to prepare 4.44 L of 0.500 M CaCl2 solution? | | A) 369 g B) 271 g C) 258 g D) 296 g E) 246 g | 2. | An aqueous solution of ammonium sulfate is allowed to react with an aqueous solution of lead(II) nitrate.The complete ionic equation contains which of the following species (when balanced in standard form)? | | A)

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    Carbon Dioxide Flash-Freezing Applied to Ice Cream Production by Teresa Susan Baker B.S. Mechanical Engineering B.S. Earth‚ Atmospheric‚ and Planetary Sciences Massachusetts Institute of Technology‚ 2003 SUBMITTED TO THE DEPARTMENT OF MECHANICAL ENGINEERING IN PARTIAL FULLFILLMENT OF THE REQUIREMENTS FOR THE DEGREE OF MASTER OF SCIENCE IN MECHANICAL ENGINEERING AT THE MASSACHUSETTS INSTITUTE OF TECHNOLOGY FEBRUARY 2006 © 2006 Massachusetts Institute of Technology. All rights reserved

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    Consider the following parallel reactions k 1 A −→ B k 2 A −→ C The activation energies are 45.3 kJ/mole for k1 and 69.8 kJ/mole for k2 . If the rate constants are equal at 320 K‚ at what temperature will k1 /k2 = 2.00? 6. [15 points] The pre-exponential factor and activation energy for the unimolecular reaction CH3 NC(g) −→ CH3 CN(g) are 4.0 × 1013 s−1 and 272 kJ/mole‚ respectively. Calculate the values of ∆S ◦‡ ‚ ∆H ◦‡ ‚ and ∆G◦‡ at 300 K. 7. [20 points] An enzyme-catalyzed

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    Chayote

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    average nutritional composition of fruits per 100 g edible portion is: water 94 g‚ energy 80 kJ (19 kcal)‚ protein 0.8 g‚ fat 0.1 g‚ carbohydrate 4.5 g‚ dietary fibre 1.7 g‚ Ca 17 mg‚ Mg 12 mg‚ P 18 mg‚ Fe 0.3 mg‚ Zn 0.7 mg‚ vitamin A 56 IU‚ thiamin 0.03 mg‚ riboflavin 0.03 mg‚ niacin 0.47 mg‚ folate 93 μg‚ ascorbic acid 7.7 mg (USDA‚ 2002). The young leaves and shoots contain per 100 g: water 90 g‚ energy 251 kJ (60 kcal)‚ protein 4.0 g‚ fat 0.4 g‚ carbohydrate 4.7 g‚ fibre 1.2 g‚ Ca 58 mg‚ P 108 mg

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    Script of Dev. D

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    ijkBs D;k ?kkl ds lkFk [kkmaxk---- Paro [kk ldrs gks rks [kk yks----Dev csodwQ---Paro rqe---Dev mYyw dh iB~Bh---Paro rw mYyw dk iB~Bk---He looks at her……… Paro [kkuk gS rks [kkvks ;k Qsd nks ukys esa----a He takes it and eats it. Paro vkt ?kj is rq>s fNRrj iM+us okays gS----- yky & yky VqYys ysdj ?kwesxk dy-----He stops eating. Paro LkRrw wB cksy jgs Fks rwus flxjsV ih-----Dev igys vki ;g r; dj yks vki eq>s ekj fdlfy;s jgs gks] flxjsV ihus ds fy‚ ;k lRrw cqykus ds fy‚----Dev‟s older brother

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    Heat Pump

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    Tout Heat Exchanger water flow rate‚ m & bar bar deg C deg C litres/min 1 2 3 4 5 6 Calculate (a) The heat transfer rate from the condenser q1 = m & C (T o – Ti) kJ/s where m & is the water mass flow rate (kg/s) and C = 4.2 kJ/kg K (b) The measured coefficient of performance (c) The ideal coefficient of performance using T a and T b Plot the measured and ideal COPs against the water mass flow rate (litres/min). 5. Discussion

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