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The following table summarizes the five steps and their results. Each step number is a link back to the explanation of the calculation. Converting to kJ gives us this: 1.4832 kJ 24.08 kJ 30.1248 kJ 162.8 kJ 2.9088 kJ Step q 72.0 g of H2O 1 1483.2 J Δt = 10 (solid) 2 24.08 kJ melting 3 30124.8 J Δt = 100 (liquid) 4 162.8 kJ boiling 5 2908.8 J Δt = 20 (gas) Summing up gives 221.3968 kJ and proper significant digits gives us 221.4 kJ for the answer. Notice how all units were converted to kJ before continuing on. Joules is a perfectly fine unit; it's just that 221,396.8 J is an awkward number to work with. Usually Joules is used for values under 1000, otherwise kJ is used. By the way, on other sites you may see kj used for kilojoules. I've also seen Kj used. Both of these are wrong symbols. kJ is the only correct symbol. Enthalpy When a process occurs at constant pressure, the heat evolved (either released or absorbed) is equal to the change in enthalpy. Enthalpy (H) is the sum of the internal energy (U) and the product of pressure and volume (PV) given by the equation: H=U+PV When a process occurs at constant pressure, the heat evolved (either released or absorbed) is equal to the change in enthalpy. Enthalpy is a state function which depends entirely on the state functions T, P and U. Enthalpy is usually expressed as the change in enthalpy (ΔH) for a process between initial and final states: ΔH=ΔU+ΔPVΔ If temperature and pressure remain constant through the process and the work is limited to pressurevolume work, then the enthalpy change is given by the equation: ΔH=ΔU+PΔV Also at constant pressure the heat flow (q) for the process is equal to the change in enthalpy defined by the equation: ΔH=q By looking at whether q is exothermic or endothermic we can determine a relationship between ΔH and q. If the reaction absorbs heat it is endothermic meaning the reaction consumes heat from the surroundings so q>0 (positive). Therefore, at constant temperature and pressure, by the equation above, if q is positive then ΔH is also positive. And the same goes for if the reaction releases heat, Page 140
then it is exothermic, meaning the system gives off heat to its surroundings, so q (the 4..18 came from the stage it where it was) -19340 = 2832.789 (T1-179173.7145) --> 159833.7145 = 2832.786T --> divide to find t so 56.42279879. Use the Problem least amount 3 of significant figures, so 4 which is 56.42 degrees Celsius is your final temperature. Calculate the enthalpy (ΔH) for the process in which 45.0 grams of water is converted from liquid at 10° C to vapor at 25° C. To calculate entha Solution 1 amount of heat= (mass of substance) x (specific heat capacity) x (change of temperature) amount of heat=(1gram) (4.18 joules)/ (grams °C) (10°C) amount of heat=41.8 joules Solution 2 Q=mcΔT -19340 J = (677.7g) (4.18J) (T f – 63.25°C) -19340 J = (2832.789) (T f – 63.25°C) -19340 J = 2832.789 T f – 179173.7145 159833.7145 = 2832.786T f 56.42279879 = T f T f = 56.42 °C Solution 3 Part 1: Heating water from 10.0 to 25.0 °C ΔkJ = 45.0g H 20 x (4.184J/gH 2O °C) x (25.0 - 10.0) °C x 1kJ/1000J = 2.82 kJ Part 2: Vaporizing water at 25.0 °C ΔkJ = 45.0 g H 2O x 1 mol H 2O/18.02 g H 2O x 44.0 kJ/1 mol H 2O = 110 kJ Part 3: Total Enthalpy Change ΔH = 2.82 kJ + 110kJ ΔH = 112.82 kJ Page 141
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Honors Chemistry Class Policies and
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Unit 1 (22 days) Chapter 1 Introduc
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Lorenzo Walker Technical High Schoo
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Chapter 1 Unit 1 Introduction to Ch
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K H D B D C M KING HENRY DIED DRINK
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2. One cereal bar has a mass of 37
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1. How many meters are in one kilom
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The Learning Goal for this assignme
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RULE #3: To add/subtract in scienti
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Rule 3: A final zero or trailing ze
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For addition and subtraction, look
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Dimensional Analysis This is a way
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Chapter 4 Unit 2 Atomic Structure T
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Looking at Ions We haven’t talked
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Neutron Madness We have already lea
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In the space below, write the unabb
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Create groups for these Scientist a
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The Learning Goal for this Assignme
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Atomic Size Define Atomic Size: The
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Define Electronegativity: Electrone
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Unit 3 Chapter 25 Nuclear Chemistry
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Name ____________________ Juan Roqu
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Learning Goal for this section: The
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Beta Particle Decay Beta decay is a
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Ion Charge? What is the charge on f
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C 2 H 6 O Ethanol CH 3 CH 2 O Step
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Linear Molecular Geometry Orbital E
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Trigonal planar Molecular Geometry
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Trigonal Pyramidal Molecular Geomet
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Trigonal Bi pyramidal Molecular Geo
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Octahedral Molecular Geometry Orbit
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Orbitals Equation Lone Pairs Angle
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Page 96 and 97: 6) ___ 1 Mn(NO2)2 + ___ 1 BeCl2 __
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Page 102 and 103: Step 1: 200g C3H8 is equal to 4.54
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Page 106 and 107: %yield = 6.1 tons × 100 = 64% 9.6t
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Page 136 and 137: Step Two: solid ice melts Now, we c
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