Chat with us, powered by LiveChat Multiple Choice Questions A sample of 3.00 moles of liquid water is heated from 10°C to 80°C | acewriters
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A sample of 3.00 moles of liquid water is heated from 10°C to 80°C. Determine q for this sample of water during this process. (The specific heat of water is 4.18 J/g°C.) –878 J – wrong answer 878 J 18.1 kJ – wrong answer –15.8 kJ 15.8 kJ______________________________________________________________________________________________________________________________8.00 grams of CH4 gas are completely combusted in the presence of excess oxygen gas in a bomb calorimeter. The heat capacity of the calorimeter and its contents is 18.2 kJ/°C, and the calorimeter rises in temperature from 25.0°C to 49.5°C. Determine the enthalpy of combustion of CH4 gas in units of kJ/mol.–56 kJ/mol–446 kJ/mol – wrong answer 446 kJ/mol – wrong answer –892 kJ/mol 892 kJ/mol____________________________________________________________________________________________________________________________Gas grills typically use the combustion of propane, C3H8 (g), to heat up the grill. The combustion of propane is shown in the following balanced chemical reaction:C3H8 (g) + 5 O2 (g) –> 3 CO2 (g) + 4 H2O (l)Using the information given below, determine the ∆H° for this reaction.Useful Information:∆H°f (C3H8 (g)) = –104 kJ/mol∆H°f (H2O (l) = –286 kJ/mol∆H°f (CO2 (g) = –393 kJ/mol –575 kJ/mol –2219 kJ/mol – wrong answer 2219 kJ/mol –2427 kJ/mol____________________________________________________________________________________________________________________________Consider the following chemical reaction and associated change in enthalpy:H2 (g) + F2 (g) ––> 2 HF (g) ∆H = –542 kJ/molDetermine ∆H for the following reaction:HF (g) ––> 12 H2 (g) + 12 F2 (g) ∆H = ???–271 kJ/mol 271 kJ/mol –542 kJ/mol – wrong answer 542 kJ/mol –1084 kJ/mol – wrong answer 1084 kJ/mol__________________________________________________________________________________________________________________________ Consider the following data:NO (g) + 1/2 O2 (g) –> NO2 (g) ∆H = –56 kJ/molN2O4 (g) –> 2 NO2 (g) ∆H = 58 kJ/molDetermine the ∆H for the following reaction:2 NO (g) + O2 (g) –> N2O4 (g) ∆H = –170 kJ/mol ∆H = –54 kJ/mol – wrong answer ∆H = 2 kJ/mol ∆H = 54 kJ/mol ∆H = 170 kJ/mol – wrong answer

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