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- qsystem = -1834.3J, with a mZn = 34.8g (cZn = 0.390J/goC), What is the ∆T given the heat loss?The standard heat of combustion of liquid cyclopentane, C5H10(l),C5H10(l), was measured to be −3291.6 kJ/mol.−3291.6 kJ/mol. What is Δ^H∘f C5H10(l),ΔH^f C5H10(l)∘, the standard heat of formation of liquid cyclopentane?1. Use the following reactions to determine the Delta Hrxn for P4O10(s) + 6H2O(l) 4H3PO4(l) 4P(s) + 5O2(g)---> P4O10(s). Delta H = -2984kJ 2H2(g) + O2(g)---> 2H2O(l) Delta H = -570kJ 2P(s) + 4O2(g) + 3H2(g)---> 2H3PO4(l). Delta H = -2534kJ B. Is the reaction exothermic or endothermic? C. Given Sof P4O10(s) = 228.9 J mol-1 K-1; Sof H2O(l) = 69.9 J mol-1 K-1; Sof H3PO4(l) = 241.98 J mol-1 K-1 calculate the DGrxn at 28.0oC. D. Is this reaction spontaneous or nonspontaneous?
- Calculate the enthalpy change for the reaction C(s) + 2H2 (g)--->CH4(g) from the following data. C(s) + O2 (g) ---> CO2(g). ΔH = – 393.5 kJ/mol H2(g) + 0.5 O2 (g) ---> H2O(l). ΔH = – 285.8 kJ/mol CH4(g) + 2 O2 (g)---> CO2(g) + 2 H2O(l). ΔH = – 890.2 kJ/molwhat should i put for the numinator of the Kb equation. Because the value of Kw is not in the options for me tu put in the numinator ?A sample of biphenyl (C12H10) weighing 0.526 g was ignited in a bomb calorimeter initially at 25°C, producing a temperature rise of 1.91 K. In a separate calibration experiment, a sample of benzoic acid (C6H5COOH) weighing 0.825 g was ignited under identical conditions and produced a temperature rise of 1.94 K. For benzoic acid, the heat of combustion at constant pressure is known to be 3226 kJ mol–1. Use this information to determine the standard enthalpy of combustion of biphenyl.
- Calculate ΔEΔE, if q=q= 0.768 kJkJ and w=w= -880 JJA sample of 39.44 mg of liquid phenyl acetylene, C6H5CCH (mol. wt. = 102.14 g/mol) is burned in pure oxygenin a bomb calorimeter at 25°C. Thermal energy is released, equivalent to 1656 J of electrical energy. Per mole,what is qV, w, ΔU? Assuming the products are CO2(g) and H2O(l), calculate ΔH.Given the following data: 4C(s) + 4H2(g) + O2(g) → CH3CH2OCOCH3(l) ΔH°=-480.0 kJ CH3CH2OH(l) + O2(g) → CH3COOH(l) + H2O(l) ΔH°=-492.0 kJ 2C(s) + 3H2(g) + 1/2O2(g) → CH3CH2OH(l) ΔH°=-278.0 kJ H2(g) + 1/2O2(g) → H2O(l) ΔH°=-286.0 kJ calculate ΔH° for the reaction:CH3COOH(l) + CH3CH2OH(l) → CH3CH2OCOCH3(l) + H2O(l)
- An English chemistry professor needs a cup of hot tea before grading 50 homeworkassignments but his electric water heater is broken . In his lab, however, he doeshave some ethanol (CH3CH2OH, MW=46.07gmol-1) and 1 gram of benzoic acid(C6H5COOH, MW=122.12 gmol-1). He knows that the benzoic acid has an enthalpyof combustion, H, of -3226.7 kJmol-1. In a constant volume (bomb) calorimeterinitially at 293.15 K, he finds that the combustion of 1.000 g of benzoic acidincreased the temperature to 297.67 K. In the exact same calorimeter, 1.000 g ofethanol raised the temperature from 293.15K to 298.38K.(a) What is the enthalpy of combustion for 1 mole of ethanol? I am confused about this question and I need help.Using Kopp's formula, estimate the heat capacity of 2‑methylheptane at 20.0C How much heat is required to heat 8.75 kg of 2‑methylheptane from 18.0 °C to 23.0 °C?C4H10O+ O2−→ CO2+ H2O Rewrite the Equation and show work