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Practice Exercise 2
Use Table 5.3 to calculate the enthalpy change for the combustion of 1 mol of ethanol:
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Chapter 5 Solutions
Chemistry: The Central Science, Books a la Carte Edition & Solutions to Red Exercises for Chemistry & Mastering Chemistry with Pearson eText -- Access Card Package
- Practice Exercise 2 Use the average bond enthalpies in Table 5.4 to estimate AH for the combustion of ethanol. H H H-C-C-0-H H Harrow_forwardHomework 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° = ?arrow_forwardPractice Exercise 1The coinage metals (Group 1B) copper, silver, and gold havespecific heats of 0.385, 0.233, and 0.129 J/g-K, respectively.Among this group, the specific heat capacity andthe molar heat capacity as the atomic weightincreases. (a) increases, increases (b) increases, decreases(c) decreases, increases (d) decreases, decreasesarrow_forward
- A 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_forwardPractice 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_forwardExercise 9.42 Enhanced with Feedback and Hints MISSED THIS? Watch KCV 9.4, IWE 9.2; Read Section 9.4. You can click on the Review link to access the section in your e Text. Part A How much heat is required to warm 1.30 kg of sand from 30.0 °C to 100.0 °C? Express the heat in joules to three significant figures. » View Available Hint(s) ΑΣφ Jarrow_forward
- Course 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 543.0 kJ : 2BRF3(g)- →Br2(g) + 3 F2(g) Based on this value and the standard enthalpies of formation for the other substances, the standard enthalpy of formation of Br2(g) is kJ/mol. Submit Answer Try Another Version 1 item attempt remainingarrow_forwardQuestion 11 Burning 1.29 g of a fuel causes the water in a calorimeter to increase by 13.0°C. If the calorimeter has a heat capacity of 3.09 kJ/°C, what is the energy density of the fuel (in kJ/g)? Type answer: CHECKarrow_forward
- 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
- General Chemistry - Standalone book (MindTap Cour...ChemistryISBN:9781305580343Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; DarrellPublisher:Cengage Learning
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