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Calculate the standard enthalpy change for the formation of carbon dioxide (CO2) from its elements carbon (C) and oxygen (O2) given the following information:- The standard enthalpy change for the combustion of carbon to form carbon dioxide is -394 kJ/mol.- The standard enthalpy change for the combustion of hydrogen gas (H2) to form water (H2O) is -286 kJ/mol.- The standard enthalpy change for the reaction of oxygen gas (O2) with hydrogen gas (H2) to form water (H2O) is -572 kJ/mol.Note: All values are at standard temperature and pressure (298 K and 1 atm).

asked Feb 3 in Chemical thermodynamics by CandyDemaio (2.3k points)

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Calculate the standard enthalpy change for the formation of ammonium chloride (NH4Cl) from its constituent elements, nitrogen (N2), hydrogen (H2), and chlorine (Cl2), given the following information:Reaction equation: N2(g) + 4H2(g) + 2Cl2(g) → 2NH4Cl(s)Standard enthalpy of formation for NH4Cl: -314.4 kJ/molStandard enthalpy of formation for N2(g): 0 kJ/molStandard enthalpy of formation for H2(g): 0 kJ/molStandard enthalpy of formation for Cl2(g): 0 kJ/mol Express your answer in kilojoules per mole (kJ/mol).

asked Feb 3 in Chemical thermodynamics by LaunaBeyers (1.8k points)

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Calculate the standard enthalpy change for the formation of ammonia gas from nitrogen gas and hydrogen gas using Hess's Law, given the following equations and their corresponding enthalpy changes:N2(g) + 3H2(g) → 2NH3(g) ∆H = -92.4 kJ/molN2(g) → 2N(g) ∆H = 941 kJ/molH2(g) → 2H(g) ∆H = 436 kJ/mol

asked Feb 3 in ThermoChemistry by CarmellaBibl (2.0k points)

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Calculate the standard enthalpy change for the formation of 2 moles of water vapor (H2O(g)) if the combustion of 1 mole of methane gas (CH4(g)) results in the formation of 2 moles of water vapor and 1 mole of carbon dioxide gas (CO2(g)). Given that the standard enthalpy change of combustion of methane is -890.4 kJ/mol.

asked Feb 3 in Chemical thermodynamics by OrenDarwin6 (1.9k points)

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Calculate the standard enthalpy change for the following reaction:2Fe(s) + 3/2O2(g) → Fe2O3(s)Given that the standard enthalpy of formation of Fe2O3(s) is -824 kJ/mol and the standard enthalpy of formation of Fe(s) and O2(g) are 0 kJ/mol.

asked Feb 3 in Chemical thermodynamics by StantonPearl (1.8k points)

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Calculate the standard enthalpy change for the following reaction: [CoCl(NH3)5]2+ (aq) + 4H2O (l) → [Co(H2O)6]2+ (aq) + 2Cl- (aq) + 5NH3 (g)Given that the standard enthalpy of formation of [Co(H2O)6]2+ (aq) is -1518.5 kJ/mol, [CoCl(NH3)5]2+ (aq) is -426.9 kJ/mol, and the enthalpy of vaporization of water is 44.0 kJ/mol.

asked Feb 3 in Chemical thermodynamics by HeleneHoward (2.0k points)

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Calculate the standard enthalpy change for the following reaction: Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g) given the following standard enthalpy of formation values: ΔHf°[Fe2O3(s)] = -824.2 kJ/mol ΔHf°[CO(g)] = -110.5 kJ/mol ΔHf°[Fe(s)] = 0 kJ/mol ΔHf°[CO2(g)] = -393.5 kJ/mol

asked Feb 3 in Chemical thermodynamics by AdelaideYwt4 (1.6k points)

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Calculate the standard enthalpy change for the following reaction: 2NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + 2H2O(l) Given the following information: - The standard enthalpy of formation of NaOH(aq) is -469.20 kJ/mol - The standard enthalpy of formation of H2SO4(aq) is -814.50 kJ/mol - The standard enthalpy of formation of Na2SO4(aq) is -1388.10 kJ/mol Note: Make sure to balance the equation and use Hess's Law if necessary.

asked Feb 3 in Chemical thermodynamics by VenusU65079 (1.8k points)

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Calculate the standard enthalpy change for the following reaction: 2H2(g) + O2(g) → 2H2O(l)Given the standard enthalpies of formation for H2O (l), H2 (g) and O2 (g) are -285.8 kJ/mol, 0 kJ/mol and 0 kJ/mol, respectively.

asked Feb 3 in Chemical thermodynamics by Latonya89F54 (2.2k points)

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Calculate the standard enthalpy change for the following reaction: [Fe(H2O)6]2+(aq) + SO4 2-(aq) → [Fe(H2O)5 SO4]-(aq) + H2O(l) given that the standard enthalpies of formation of [Fe(H2O)6]2+, [Fe(H2O)5SO4]-, and H2O are -360 kJ/mol, -950 kJ/mol, and -286 kJ/mol, respectively.

asked Feb 3 in Chemical thermodynamics by LillyHoskins (1.9k points)

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Calculate the standard enthalpy change for the following reaction, given the standard enthalpies of formation:2Fe (s) + 3Cl2 (g) → 2FeCl3 (s)ΔH°f(FeCl3) = -399.4 kJ/molΔH°f(Fe) = 0 kJ/molΔH°f(Cl2) = 0 kJ/mol

asked Feb 3 in Chemical thermodynamics by EfrainAlarco (2.2k points)

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1 answer 37 views

Calculate the standard enthalpy change for the following reaction involving solids at 298 K:Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g) Given the following standard enthalpies of formation in kJ/mol: Fe2O3(s) = -825.5 Fe(s) = 0CO2(g) = -393.5 CO(g) = -110.5

asked Feb 3 in Chemical thermodynamics by ArnoldJanous (2.1k points)

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Calculate the standard enthalpy change for the following reaction involving liquids at 298 K: C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l)Given the standard enthalpies of formation of C2H5OH(l), CO2(g) and H2O(l) are −277.6 kJ/mol, −393.5 kJ/mol and −285.8 kJ/mol respectively.

asked Feb 3 in Chemical thermodynamics by ArleneLeichh (2.1k points)

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Calculate the standard enthalpy change for the following reaction at 298 K:Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)

asked Feb 3 in Quantum Chemistry by Cathern2110 (2.0k points)

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Calculate the standard enthalpy change for the following reaction at 298 K: 2NaHCO3(s) + MgCl2(aq) → MgCO3(s) + 2NaCl(aq) + H2O(l)Given the following standard enthalpies of formation: ΔHf°(NaHCO3) = -950.7 kJ/molΔHf°(MgCl2) = -641.8 kJ/molΔHf°(MgCO3) = -1128.2 kJ/molΔHf°(NaCl) = -411.2 kJ/molΔHf°(H2O) = -285.8 kJ/mol

asked Feb 3 in Chemical thermodynamics by DeboraMccomb (1.9k points)

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1 answer 53 views

Calculate the standard enthalpy change for the following reaction at 298 K using the standard enthalpies of formation:2SO2(g) + O2(g) → 2SO3(g)

asked Feb 3 in Chemical thermodynamics by GarnetFjo447 (1.8k points)

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1 answer 61 views

Calculate the standard enthalpy change for the following reaction at 25°C given the enthalpies of formation (in kJ/mol) of the compounds involved:CaCO3(s) + 2HCl(aq) → CaCl2(aq) + CO2(g) + H2O(l)Hf(CaCO3) = -1206.9 kJ/molHf(CaCl2) = -795.8 kJ/molHf(CO2) = -393.5 kJ/molHf(H2O) = -285.8 kJ/molHf(HCl) = -92.31 kJ/mol

asked Feb 3 in Chemical thermodynamics by CherylDooley (1.7k points)

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Calculate the standard enthalpy change for the following neutralization reaction where hydrochloric acid (HCl) reacts with sodium hydroxide (NaOH) to form sodium chloride (NaCl) and water (H2O).HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)Given the following tabulated standard enthalpies of formation values: ΔHf°(HCl) = -167 kJ/mol, ΔHf°(NaOH) = -469 kJ/mol, ΔHf°(NaCl) = -411 kJ/mol, ΔHf°(H2O) = -286 kJ/mol.

asked Feb 3 in Chemical thermodynamics by GavinMott60 (1.8k points)

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Calculate the standard enthalpy change for the following neutralization reaction between hydrochloric acid (HCl) and sodium hydroxide (NaOH):HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)Given: - The standard enthalpy of formation (ΔHf°) of NaCl(aq) is -407.3 kJ/mol.- The standard enthalpy of formation (ΔHf°) of H2O(l) is -285.8 kJ/mol.- The specific heat capacity (c) of the solution is 4.18 J/g°C.- The temperature change (ΔT) during the reaction is 15°C. - The mass (m) of the solution is 50.0 g. What is the standard enthalpy change (ΔH°) for the neutralization reaction between HCl and NaOH?

asked Feb 3 in Chemical thermodynamics by AlineBacon1 (1.5k points)

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Calculate the standard enthalpy change for the following chemical reaction involving solutions:2HNO3(aq) + Ba(OH)2(aq) → Ba(NO3)2(aq) + 2H2O(l)Given the following information:ΔHf° [HNO3(aq)] = -207.5 kJ/molΔHf° [Ba(OH)2(aq)] = -994.0 kJ/molΔHf° [Ba(NO3)2(aq)] = -537.5 kJ/mol

asked Feb 3 in Chemical thermodynamics by AguedaD83620 (2.6k points)

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