Part III: Examining Changes in Terms of Energy in VS .NET

Drawing UPC - 13 in VS .NET Part III: Examining Changes in Terms of Energy

Part III: Examining Changes in Terms of Energy
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839 kJ are released Solve this problem with a chain of conversion factors Convert from grams of CO to moles, and then from moles to kJ Be sure to adjust the sign of the enthalpy and incorporate the stoichiometry of the given reaction equation: 967g CO mol CO 486 kJ = 839 kJ 1 280g CO 2 mol CO Because the sign is negative, 839 kJ are released (not absorbed) 260 105J (or 260 102 kJ) The total heat required is the sum of several individual heats: heat to warm the ice to the melting point (273K), heat to convert the ice to liquid water, heat to warm the liquid water to the boiling point, and heat to convert the liquid water to steam Be careful to match your units (J versus kJ): q (warm ice to melting point) = (477 mol)(381J mol 1 K 1)(273K 268K) = 909J q (convert ice to liquid water) = (477 mol)(601 103J mol 1) = 287 104J q (warm water to boiling point) = (477 mol)(753J mol 1 K 1)(373K 273K) = 359 104J q (convert liquid water to steam) = (477 mol)(4068 104J mol 1) = 194 105J The sum of all heats is 260 105J (or 260 102 kJ)
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146 kJ Reverse Reaction 1 to get Reaction 1' ( 1032kJ) Reverse Reaction 2 to get Reaction 2' (886kJ) Add Reactions 1' and 2', yielding H = 146 kJ 197 kJ Reverse Reaction 2 and divide it by 2 to get Reaction 2' (483kJ) Reverse Reaction 3 and divide it by 2 to get Reaction 3' (319kJ) Add Reactions 1, 2', and 3', yielding H = 197 kJ You need to reverse Reactions 2 and 3 to put the right compounds on the reactant and product sides of the equation Reaction 1 already has the right orientation Reactions 2 and 3 must be divided by 2 so the stoichiometry of the final equation matches the stoichiometry of the target equation (namely, 1 mole each of H2O, CO, and H2) Even though this division temporarily creates noninteger coefficients for O2 in these equations, the noninteger coefficients cancel out when you add reactions 1, 2', and 3'
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Part IV
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Swapping Charges
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In this part
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toms are built with three kinds of particles, two of which are charged This means that charge is important in chemistry Swapping or transferring charge between reactants alters their properties Two critical classes of reactions revolve around such movements of charge: acid-base reactions and oxidation-reduction reactions In this part, we give you the tools to deal with these charge-centric reactions In addition, we describe nuclear chemistry, which deals with transformations in atomic nuclei, the positively charged hearts of atoms
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16
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Giving Acids and Bases the Litmus Test
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In This
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Checking out the many definitions of acid and base Calculating acidity and basicity Determining the strength of acids and bases with the help of dissociation
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f you ve read any comic books, watched any superhero flicks, or even tuned in to one of those fictional solve-the-crime-in-50-minutes shows on TV, you ve likely come across a reference to acids being dangerous substances Acids are generally thought of as something which evil villains intend to spray in the face of a hero or heroine, but somehow usually manage to spill on themselves However, you encounter and even ingest a wide variety of fairly innocuous acids in everyday life Citric acid, present in citrus fruits such as lemons and oranges, is very ingestible, as is acetic acid, also known as vinegar Strong acids can indeed burn the skin and must be handled with care in the laboratory However, strong bases can burn skin as well Chemists must have a more sophisticated understanding of the differences between an acid and a base and their relative strengths than simply their propensity to burn This chapter focuses on how you can identify acids and bases, as well as several ways to determine their strengths
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