Chemistry: The Central Science (13th Edition)
13th Edition
ISBN: 9780321910417
Author: Theodore E. Brown, H. Eugene LeMay, Bruce E. Bursten, Catherine Murphy, Patrick Woodward, Matthew E. Stoltzfus
Publisher: PEARSON
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Textbook Question
Chapter 5.7, Problem 5.12.2PE
Practice Exercise 2
Use Table 5.3 to calculate the enthalpy change for the combustion of 1 mol of ethanol:
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Practice Exercise 1If the heat of formation of H2O1l2 is -286 kJ>mol, whichof the following thermochemical equations is correct?(a) 2 H1g2 + O1g2¡H2O1l2 ΔH = -286 kJ(b) 2 H21g2 + O21g2¡2 H2O1l2 ΔH = -286 kJ(c) H21g2 + 12 O21g2¡H2O1l2 ΔH = -286 kJ(d) H21g2 + O1g2¡H2O1g2 ΔH = -286 kJ(e) H2O1l2¡H21g2 + 12O21g2 ΔH = -286 kJ
Practice
TEXT ANSWER
A calorimeter holds 75 g water at 24.0°C. A sample of hot
silver is added to the water. The final temperature of the
water and silver is 28.0°C. What is the change in enthalpy
associated with the change in the water's temperature?
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Note: The specific heat of water is 4.18 gº
Use the formula ΔΗ = = - cm ^T
Show your work.
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X₂ X²
2 Ix
Enter your answer here
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Homework 3
• Use standard enthalpies of formation from
Table 7.2 to determine the enthalpy change
at 25 °C for the following reaction.
2 Cl2(g) + 2H20(1)
4 HCI( g) + O2(g)
AH° = ?
Chapter 5 Solutions
Chemistry: The Central Science (13th Edition)
Ch. 5.2 - Prob. 5.1.1PECh. 5.2 - Prob. 5.1.2PECh. 5.3 - Prob. 5.2.1PECh. 5.3 - Prob. 5.2.2PECh. 5.3 - Prob. 5.3.1PECh. 5.3 - Prob. 5.3.2PECh. 5.4 - Prob. 5.4.1PECh. 5.4 - Prob. 5.4.2PECh. 5.5 - Prob. 5.5.1PECh. 5.5 - Prob. 5.5.2PE
Ch. 5.5 - Practice Exercise 1 Suppose you have equal masses...Ch. 5.5 - Prob. 5.6.2PECh. 5.5 - Prob. 5.7.1PECh. 5.5 - Prob. 5.7.2PECh. 5.6 - Prob. 5.8.1PECh. 5.6 - Prob. 5.8.2PECh. 5.6 - Prob. 5.9.1PECh. 5.6 - Prob. 5.9.2PECh. 5.7 - Prob. 5.10.1PECh. 5.7 - Prob. 5.10.2PECh. 5.7 - Prob. 5.11.1PECh. 5.7 - Prob. 5.11.2PECh. 5.7 - Prob. 5.12.1PECh. 5.7 - Practice Exercise 2 Use Table 5.3 to calculate the...Ch. 5.8 - Prob. 5.13.1PECh. 5.8 - Practice Exercise 2 Given the following standard...Ch. 5.8 - Prob. 5.14.1PECh. 5.8 - Prob. 5.14.2PECh. 5 - Prob. 1DECh. 5 - Prob. 1ECh. 5 - Prob. 2ECh. 5 - Prob. 3ECh. 5 - Practice Exercise 2
Using Table 20.1, rank...Ch. 5 - Prob. 5ECh. 5 - Prob. 6ECh. 5 - Prob. 7ECh. 5 - Prob. 8ECh. 5 - Prob. 9ECh. 5 - Prob. 10ECh. 5 - Prob. 11ECh. 5 - Prob. 12ECh. 5 - Prob. 13ECh. 5 - Prob. 14ECh. 5 - Prob. 15ECh. 5 - Prob. 16ECh. 5 - Prob. 17ECh. 5 - Prob. 18ECh. 5 - Prob. 19ECh. 5 - Prob. 20ECh. 5 - Prob. 21ECh. 5 - Prob. 22ECh. 5 - Prob. 23ECh. 5 - Prob. 24ECh. 5 - Prob. 25ECh. 5 - Prob. 26ECh. 5 - Prob. 27ECh. 5 - In chemical kinetics, the entropy of activation is...Ch. 5 - Prob. 29ECh. 5 - Prob. 30ECh. 5 - Prob. 31ECh. 5 - The following data compare the standard enthalpies...Ch. 5 - Prob. 33ECh. 5 - Prob. 34ECh. 5 - Prob. 35ECh. 5 - What is the reducing agent in the following...Ch. 5 - Prob. 37ECh. 5 - Prob. 38ECh. 5 - Prob. 39ECh. 5 - Prob. 40ECh. 5 - Prob. 41ECh. 5 - Prob. 42ECh. 5 - Prob. 43ECh. 5 - The standard cell potential is 1.46 V for a...Ch. 5 - Prob. 45ECh. 5 - Prob. 46ECh. 5 - Prob. 47ECh. 5 - Prob. 48ECh. 5 - Prob. 49ECh. 5 - Practice Exercise 1
Which of the following...Ch. 5 - Prob. 51ECh. 5 - Prob. 52ECh. 5 - Prob. 53ECh. 5 - Prob. 54ECh. 5 - Prob. 55ECh. 5 - Prob. 56ECh. 5 - Prob. 57ECh. 5 - Prob. 58ECh. 5 - Prob. 59ECh. 5 - Prob. 60ECh. 5 - What is the connection between Hess’s law and the...Ch. 5 - Prob. 62ECh. 5 - 20.2 You may have heard that “antioxidants” are...Ch. 5 - Prob. 64ECh. 5 - Prob. 65ECh. 5 - Prob. 66ECh. 5 - Prob. 67ECh. 5 - Prob. 68ECh. 5 - Prob. 69ECh. 5 - Prob. 70ECh. 5 - Prob. 71ECh. 5 - Prob. 72ECh. 5 - 20.13
What is meant by the term oxidation?
On...Ch. 5 - Prob. 74ECh. 5 - Prob. 75ECh. 5 - Prob. 76ECh. 5 - Prob. 77ECh. 5 - Prob. 78ECh. 5 - Prob. 79ECh. 5 - Prob. 80ECh. 5 - Prob. 81ECh. 5 - Prob. 82ECh. 5 - Prob. 83ECh. 5 - Prob. 84ECh. 5 - Prob. 85ECh. 5 - The heat of combustion of ethanol, C2H5OH(l) is...Ch. 5 - Prob. 87ECh. 5 - Prob. 88ECh. 5 - Prob. 89ECh. 5 - The automobile fuel called E85 consists of 85%...Ch. 5 - Prob. 91AECh. 5 - Prob. 92AECh. 5 - Prob. 93AECh. 5 - Prob. 94AECh. 5 - 5.95 Consider a system consisting of the following...Ch. 5 - A sample of gas is contained in a...Ch. 5 - Prob. 97AECh. 5 - Prob. 98AECh. 5 - A house is designed to have passive solar energy...Ch. 5 - Prob. 100AECh. 5 - Prob. 101AECh. 5 - Prob. 102AECh. 5 - Burning methane in oxygen can produce three...Ch. 5 - Prob. 104AECh. 5 - Prob. 105AECh. 5 - The hydrocarbons acetylene (C2H2) and benzene...Ch. 5 - Prob. 107AECh. 5 - Three common hydrocarbons that contain four...Ch. 5 - Prob. 109AECh. 5 - The Sun supplies about 1.0 kilowatt of energy for...Ch. 5 - It is estimated that the net amount of carbon...Ch. 5 - Prob. 112IECh. 5 - Prob. 113IECh. 5 - Prob. 114IECh. 5 - Prob. 115IECh. 5 - Prob. 116IECh. 5 - Prob. 117IECh. 5 - The methane molecule, CH4, has the geometry shown...Ch. 5 - Prob. 119IE
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- Practice Exercise 2 Carbon occurs in two forms, graphite and diamond. The en- thalpy of the combustion of graphite is -393.5 kJ /mol, and that of diamond is -395.4 kJ/mol: C(graphite) + O2(g) → CO2(8) AH = -393.5 kJ - C(diamond) + O2(8) CO2(8) AH = -395.4 kJ - %3D Calculate AH for the conversion of graphite to diamond: C(graphite) C(diamond) AH = ? - %3Darrow_forwardCourse dashboard A bomb calorimetry experiment is performed with xylose, C5H1005 (s), as the combustible substance. The data obtained are mass of xylose burned: 2.059g, heat capacity of calorimeter 4.728 kJ/ C, initial calorimeter temperature: 23.29°C; final calorimeter temperature 27.19°C. What is the heat of combustion of xylose, in kilojoules di per mole? aretle Lütfen birini seçin: O a. -2.38x103 kJ/mol b. -2.17x103 kJ/mol Uy O c. -5.64x10 kJ/mol Ka d. -1.82x103 kJ/mol O e. -1.34x103 kJ/molarrow_forwardA scientist measures the standard enthalpy change for the following reaction to be -773.2 kJ : 2C0(g) + 2 NO(g)- →2CO2(g) + N2(g) Based on this value and the standard enthalpies of formation for the other substances, the standard enthalpy of formation of CO2(g) is kJ/mol. Submit Answer Try Another Version 2 item attempts remainingarrow_forward
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- Exercise 3. Solve this extra challenging problem. A large paraffin candle has a mass of 96.83 gram. A metal cup with 100.0 mL of water at 16.2°C absorbs the heat from the burning candle and increases its temperature to 35.7°C. Once the burning is ceased, the temperature of the water was 35.7°C and the paraffin had a mass of 96.14 gram. Determine the heat of combustion of paraffin in kJ/gram. GIVEN: density of water = 1.0 g/mL. Consider the problem above. Identify at least 1 source of error as you can. Indicate as well the direction of error that would have resulted. That is, identify whether the error would have caused the experimentally derived value to be less than or more than the accepted value.arrow_forward- Practice Exercise 1 If the heat of formation of H,O(1) is –286 kJ/mol, which of the following thermochemical equations is correct? (a) 2 H(g) + O(g) → H2O(1) (b) 2 H2(8) + O2(8) (c) H2(8) + O2(8) (d) H2(g) + O(g) → H2O(g) (e) H2O(1) AH = -286 kJ AH = -286 kJ → 2 H,O(1) H20(1) %3D AH = -286 kJ AH = -286 kJ - %3D H2(8) + O2(8) AH = -286 kJ - %3Darrow_forwardWe rite is co - Practice Exercise 1 Calculate the enthalpy change for the reaction deies of 2 H2O2(1) → 2 H20(I) + O2(8) elsg rffwepb - using enthalpies of formation: elua eloubon AHf[H2O2(1)] = -187.8 kJ/mol AHF[H2O(1)] = -285.8 kJ/molarrow_forward
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