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- Calculate AH° and AE° for the following reactions at 25 °C. Use standard enthalpies of formation from the table below. Round your answers to four significant digits. Substance AH; (kJ mol-1) Substance AH: (kJ mol-1) Co,(g) -393.5 226.75 Ni(g) C,H,(g) (CH3),CO(/) H,O(g) HC(g) NH,(g) -46.19 -248.1 NH.CIG) N,O(g) O(g) -315.4 -241.8 81.57 -92.30 (a) 2C2H2(g) + 502(g)4CO2(g) + 2H20(g) AH° = i kJ AF° = kJ eTextbook and Media (b) C2H2(g) + 5N20(g) → 2CO2(g) + H2O(g) + 5N2(3) AH° = i kJ AE° = i kJ6 S+0₂ 250₂ 25 + - 30₂ 0₂-2503 S0₂ - 2503 AH--196 KJ AH-790kJc) Calculate the standard enthalpy change for the reaction 3C(s) + 4H2 (g) C;H8 (g) Given that C(s) + O2 (g) CO2 (g) AH° = -394 kJmol·' H2 (g) + ½O2 (g) H2O (I) AH° = -286 kJmol C3H8 (g) + 502 (g) 3CO2 (g) + 4H2O (I) AH° = -2220 kJmol·'
- 2 ने \S h CoefRdlen+ 02 A०० e what is balanding the coefficient Agel tHhe follawing equuation? के AgNO,Ca AgCl[Relerences] Acetaldehyde, CH; CHO, is a colorless, flammable liquid used in the manufacture of acetic acid, perfumes, and flavors. What is the heat of vaporization of CH3CHO? CH3 CHO(1) → CH3 CHO(g); AH° =? AH;(CH3CHO(1)) = – 191.8 kJ/mol AH;(CH3 CHO(g)) = - 166.1 kJ/mol Heat of vaporization = kJ Submit Answer Try Another Version 6 item attempts remaining Previc e to searchRank the elements or compounds in the table below in decreasing order of their boiling points. That is, choose 1 next to the substance with the highest boiling point, choose 2 next to the substance with the next highest boiling point, and so on. substance A B с D H :0: H | || | H с C -C-H 174 H H H chemical symbol, chemical formula or Lewis structure H H H | | C I 1 I H H H (II | Ag - Ar - :O: -O-H boiling point (Choose one) (Choose one) ✓ (Choose one) (Choose one) ✓6Hello tutor please I need your help with this exercice . Thank youОН Me 220 °C ОН Н I6. From the following heats of combustion, AH° = -3119.6 kJ/mol kJ/mol CH3OH(1) + 1/2O2(g) → CO2(g) + 2H2O(I) C(graphite) + 0,(g) → co,(g) H2 (g) + 1/202(g) → H2O1) AH:° = -393.5 kJ/mol AH.º= -285.8 kJ/mol Calculate the enthalpy of formation of methanol from its elements: C(graphite) + 2H,(g) + 1/2O2(g) → CH;OH (g)2 NOCI(g) → 2 NO(g) +Cl2(g) AH° =+75.56 kJ 2 NO(g) + 02(g) → 2 NO2(g) AH° =-113.05 kJ 2 NO2(g) → N2O4(g)AH° =-58.03 kJ Compute AH° of N2O4(g) +Cl2(g) → 2 NOCI(g) + O2(g) in kJ. +246.65 -95.52 -246.65 +95.52 +47.76Constants: • c = 2.9979 x 108 m/s h = 6.626 x 10-34 J s per one photon • R = 8.314 J/(K mol) = 0.08206 L atm/(K mol) NA = 6.022 x 1023 particles/mol RH = 1.097 x 107 m1 = 2.178 x 10-18 J . . You have an aqueous Glucose solution that is 12.0% Glucose by mass. What is the molality of Glucose in the solution? Molar Mass of Water = 18.015 g/mol Molar Mass of Glucose 180.16 g/mol =Given the following data: Pa(s) + 6 Clz9) → 4PCI3(g) AH = -1225.6 kJ Pa(s) + 5 O2(g) → P4010(5) AH = -2967.3 kJ PCI3(g) + Clz(g) → PCI5(a) AH = -84.2 kJ PCI3(9) + O2lg) → ClaPO(9) AH = -285.7 kJ Calculate AH for the reaction P,O1015) + 6 PCI5(9) → 10 Cl3PO(g)SEE MORE QUESTIONS