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IDIO-W2 . ppo 0 en 14 Part I 1 Unknown metal # B. Appearance of Metal (shape X color) red ball 2 52.957 1009 -46.459 33.

Posted: Wed May 18, 2022 12:52 pm
by answerhappygod
Idio W2 Ppo 0 En 14 Part I 1 Unknown Metal B Appearance Of Metal Shape X Color Red Ball 2 52 957 1009 46 459 33 1
Idio W2 Ppo 0 En 14 Part I 1 Unknown Metal B Appearance Of Metal Shape X Color Red Ball 2 52 957 1009 46 459 33 1 (42.19 KiB) Viewed 58 times
Idio W2 Ppo 0 En 14 Part I 1 Unknown Metal B Appearance Of Metal Shape X Color Red Ball 2 52 957 1009 46 459 33 2
Idio W2 Ppo 0 En 14 Part I 1 Unknown Metal B Appearance Of Metal Shape X Color Red Ball 2 52 957 1009 46 459 33 2 (42.82 KiB) Viewed 58 times
Idio W2 Ppo 0 En 14 Part I 1 Unknown Metal B Appearance Of Metal Shape X Color Red Ball 2 52 957 1009 46 459 33 3
Idio W2 Ppo 0 En 14 Part I 1 Unknown Metal B Appearance Of Metal Shape X Color Red Ball 2 52 957 1009 46 459 33 3 (39.62 KiB) Viewed 58 times
Idio W2 Ppo 0 En 14 Part I 1 Unknown Metal B Appearance Of Metal Shape X Color Red Ball 2 52 957 1009 46 459 33 4
Idio W2 Ppo 0 En 14 Part I 1 Unknown Metal B Appearance Of Metal Shape X Color Red Ball 2 52 957 1009 46 459 33 4 (59.51 KiB) Viewed 58 times
IDIO-W2 . ppo 0 en 14 Part I 1 Unknown metal # B. Appearance of Metal (shape X color) red ball 2 52.957 1009 -46.459 33.691 52.969 82.651 33.688. 48.963 22.3 22.5 110.5. 109.5 in of metal ) m of calonmeter + water (9) moj empty Olomote (9) m of water (g) Temp. initial of water (O Temp initial of hot metal (6) Final temp of water & vetal (c) Que Part I Trial To of HOICO 22.2 Ti of Naotic) 22.8 Aug Ti (0) 22.5 28-1 Q=mcat CE mat 2 22.3 22.7 22.5 1000L Voor = ho +VNOOH (m) 50450 100ml If (max T)CC) 28.8 AT (TA-D)) 6.3 29.0 6.5
Experiment 11: CALORIMETRY Show your calculations on a separate sheropper and enter the results here. Provide the units with your values. Trial 2 Trial Part II: Molar Heat of Neutralization Total Volume of Solution After Mixing Mass of Solution After Mixing of Reactants of Soln) - Volume of HCI used in neutralization Molarity of HCI used in neutralization #mol HCI- Molar Heat of Neutralization Average Molar Heat of Neutralization 148
Experiment 11: CALORIMETRY Data and Results Date Name CHEM 131 See: Shower calculations on a separate sheet of paner and enter the results here. Part 1: Specific Heat of an Unknown Metal Trial 2 Trial 1 Unknown Metal Mass of metal (g) - Mass of water (8) Initial Temp of Water (°C) - Initial Temp of Hot Metal (°C) - Final Temp of Water & of Metal ("C)- st of water ("C)- 47 of metal ("C) - quatur (J) - (J) Few (og'ch- Average Sms (gc)- Conclusion with justifications: What is the identity of the unknown metal? 147
equate them? 6. In Part I of the experiment, we utilize the equation meat = 3 m AT. If we change the mass of the metal, what would you expect to change (g2 s? m? A7?). Explain your answer. ffftt Post-lab Questions 1. Examine the initial and final temperatures in Part I. Explain how the temperatures tell you what type of reaction was involved (endothermic or exothermic). Are the signs of your qwater and metal consistent with this? Explain. 2. In Part I, we see that copper and zinc have the same specific heat (See table in the Introduction. If you obtained an experimental value of 0.39 3.g .cl how might you determine which metal you have as an unknown? Explain. 3. We assumed that no heat is lost to the surroundings beyond the nested coffee cups. In Part 1, obviously there would have been some loss in heat as the hot metal is transferred to the calorimeter. How does that unavoidable heat loss affect your calculated specific heat of the metal? Would your calculated specific heat be too high or too low due this error? Explain fully. 4. In Part II, we assume that the density and specific heat of the solution is the same as that of water. What justifications do we have to make that assumption? Explain. 5. Using the molar heat of neutralization obtained in your experiment (assuming it is correct), calculate how much heat you would expect to be produced if you mixed 50.0 mL of 0.250 M HCl with by 50.0 mL of 0.250 M NaOH. Show your calculations. (Hint: How many moles of HCl are involved?)