Use the information provided to answer questions 1-4. All calculated answers must include units and correet significant figures.In an ideal calorimeter, 40.0 mL of cold (19.8 \deg C) water was mixed with 60.0 mL hol (85.0 \deg C) water, resulting in a final temperature of 36.2 \deg C (Table 1). Because the calorimeter is ideal, there is no heat exchange with the system.This means you do not need to calculate the heat capacity of the calorimeter (Ceal). Table 1. Vol H20 cold (ml.)40.0Vol H20 hot (mL )60.0H2O cold Ti (\deg C)19.8H20 hot ti (\deg C85.0H2O Tf(\deg C)36.21 NEED HELP WITH QUESTION 3,6,9 PLEASEQuestion 1: Calculate the heal exchange (a) of the cold water. Include the appropriate sign (-) if needed, and appropriate significant figures. Answer to question 1 is : 2,740 Question 2: Did the initially cold water gain or lose heat? Answer: gained heatQuestion 3: Calculate the heal exchange (q) of the hot waler. Include units, sign (-) if needed, and appropriate significant figures.Answer: unsure please help !Question 4: Did the initially hot water gain or lose heal? Answer: Lost heatQuestion 5: Complete the table according to the lab procedureTable 2, AHneutralizationVol 3.00 M NH3(mL.)60.0Vol 3.00 M HNO3(mL.)60.0 Total Solutionmass (g)Ti (\deg C23.5Tf(\deg C)32.6AT (\deg C)Answer: 120.0 and 9.1Question 6: Calculate q for this neutralization in joules. Include the correct sign (-) if needed. and appropriate significant figures. Answer: unsure please help !! Question 7: Is the neutralization reaction of NH, with HNO, an exothermic process. endothermie process, or neither?* exothermic endothermic* neitherAnswer: exothermic Question 8: Complete the table and the according to the lab procedureTable 3, AHdissolutionMass of NH4 NO3 (9)9.15 Vol. H20 (mL)75.0 Total solutionmass ( g)Ti (\deg C)21.5Tf (\deg C)4.50A (\deg C)answer: 84.15 and -17.0Question 9: Calculate q for this dissolution of ammonium nitrate in joules. Include sign (-) if needed, and appropriate significant figures. Answer: Unsure please help !!Question 10: Is the dissolving (dissolution) of ammonium nitrate (NH, NO,) an exothermic process, endothermie process, or neither?* exothermic endothermie neitherAnswer: endothermic Use the information provided to answer questions 1-4. All calculated answers must include units and correct significant figures. In an ideal calorimeter. 40.0 ml. of cold (19.8 °C) water was mixed with 60.0 mL hot (85.0 °C) water, resulting in a final temperature of 36.2 °C (Table 1). Because the calorimeter is ideal, there is no heat exchange with the system. This means you do not need to calculate the heat capacity of the calorimeter (Ccal). Table 1. Vol H₂O cold (ml) 40.0 Vol H₂O hot (ml) 60.0 H₂O cold T (°C) 19.8 W H₂O hot T (°C) 85.0 H₂OTT (°C) 36.2

Use the information provided to answer questions 1-4. All calculated answers must include units and correet significant figures.In an ideal calorimeter, 40.0 mL of cold (19.8 \deg C) water was mixed with 60.0 mL hol (85.0 \deg C) water, resulting in a final temperature of 36.2 \deg C (Table 1). Because the calorimeter is ideal, there is no heat exchange with the system.This means you do not need to calculate the heat capacity of the calorimeter (Ceal). Table 1. Vol H20 cold (ml.)40.0Vol H20 hot (mL )60.0H2O cold Ti (\deg C)19.8H20 hot ti (\deg C85.0H2O Tf(\deg C)36.21 NEED HELP WITH QUESTION 3,6,9 PLEASEQuestion 1: Calculate the heal exchange (a) of the cold water. Include the appropriate sign (-) if needed, and appropriate significant figures. Answer to question 1 is : 2,740 Question 2: Did the initially cold water gain or lose heat? Answer: gained heatQuestion 3: Calculate the heal exchange (q) of the hot waler. Include units, sign (-) if needed, and appropriate significant figures.Answer: unsure please help !Question 4: Did the initially hot water gain or lose heal? Answer: Lost heatQuestion 5: Complete the table according to the lab procedureTable 2, AHneutralizationVol 3.00 M NH3(mL.)60.0Vol 3.00 M HNO3(mL.)60.0 Total Solutionmass (g)Ti (\deg C23.5Tf(\deg C)32.6AT (\deg C)Answer: 120.0 and 9.1Question 6: Calculate q for this neutralization in joules. Include the correct sign (-) if needed. and appropriate significant figures. Answer: unsure please help !! Question 7: Is the neutralization reaction of NH, with HNO, an exothermic process. endothermie process, or neither?* exothermic endothermic* neitherAnswer: exothermic Question 8: Complete the table and the according to the lab procedureTable 3, AHdissolutionMass of NH4 NO3 (9)9.15 Vol. H20 (mL)75.0 Total solutionmass ( g)Ti (\deg C)21.5Tf (\deg C)4.50A (\deg C)answer: 84.15 and -17.0Question 9: Calculate q for this dissolution of ammonium nitrate in joules. Include sign (-) if needed, and appropriate significant figures. Answer: Unsure please help !!Question 10: Is the dissolving (dissolution) of ammonium nitrate (NH, NO,) an exothermic process, endothermie process, or neither?* exothermic endothermie neitherAnswer: endothermic Use the information provided to answer questions 1-4. All calculated answers must include units and correct significant figures. In an ideal calorimeter. 40.0 ml. of cold (19.8 °C) water was mixed with 60.0 mL hot (85.0 °C) water, resulting in a final temperature of 36.2 °C (Table 1). Because the calorimeter is ideal, there is no heat exchange with the system. This means you do not need to calculate the heat capacity of the calorimeter (Ccal). Table 1. Vol H₂O cold (ml) 40.0 Vol H₂O hot (ml) 60.0 H₂O cold T (°C) 19.8 W H₂O hot T (°C) 85.0 H₂OTT (°C) 36.2

General Chemistry - Standalone book (MindTap Course List)

11th Edition

ISBN:9781305580343

Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell

Publisher:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell

Chapter6: Thermochemisty

Section: Chapter Questions

Problem 6.27QP: Hypothetical elements A2 and B2 react according to the following equation, forming the compound AB....

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