answer:To determine the standard enthalpy of formation for ethanol (C2H5OH), we can use Hess's Law, which states that the enthalpy change of a reaction is the same whether it occurs in one step or several steps. In this case, we can use the given combustion reaction and the standard enthalpies of formation for CO2 and H2O to find the standard enthalpy of formation for ethanol. The general equation for the standard enthalpy of formation is: ΔH°f(reaction) = Σ [ΔH°f(products)] - Σ [ΔH°f(reactants)] For the given combustion reaction: C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l) We have the standard enthalpies of formation for CO2 and H2O: ΔH°f(CO2) = -393.5 kJ/mol ΔH°f(H2O) = -285.8 kJ/mol We can plug these values into the equation: ΔH°f(reaction) = [2 * ΔH°f(CO2) + 3 * ΔH°f(H2O)] - [ΔH°f(C2H5OH) + 3 * ΔH°f(O2)] Since the standard enthalpy of formation for an element in its standard state is zero, ΔH°f(O2) = 0. Therefore, the equation becomes: ΔH°f(reaction) = [2 * (-393.5) + 3 * (-285.8)] - ΔH°f(C2H5OH) Now, we can solve for ΔH°f(reaction): ΔH°f(reaction) = (-787) + (-857.4) - ΔH°f(C2H5OH) ΔH°f(reaction) = -1644.4 kJ/mol - ΔH°f(C2H5OH) To find the standard enthalpy of formation for ethanol, we need to rearrange the equation: ΔH°f(C2H5OH) = -1644.4 kJ/mol - ΔH°f(reaction) However, we are not given the value for ΔH°f(reaction). We cannot determine the standard enthalpy of formation for ethanol without this value.

answer:To calculate the standard enthalpy of formation (ΔH°f) for butane (C4H10), we can use the following equation: ΔH°f (C4H10) = Σ ΔH°f (products) - Σ ΔH°f (reactants) The complete combustion of butane can be represented by the following balanced chemical equation: C4H10(g) + 13/2 O2(g) → 4 CO2(g) + 5 H2O(l) Now, we can calculate the standard enthalpy of formation for butane using the given data: ΔH°f (C4H10) = [4 × ΔH°f (CO2) + 5 × ΔH°f (H2O)] - [ΔH°f (C4H10) + 13/2 × ΔH°f (O2)] Since the standard enthalpy of formation for elements in their standard state is zero, ΔH°f (C(s)) = ΔH°f (H2(g)) = ΔH°f (O2(g)) = 0. We are given the standard enthalpies of combustion for C4H10, CO2, and H2O as -2877.2, -393.5, and -285.8 kJ/mol, respectively. We can use these values to find the standard enthalpies of formation for CO2 and H2O: ΔH°f (CO2) = -393.5 kJ/mol ΔH°f (H2O) = -285.8 kJ/mol Now, we can plug these values into the equation: ΔH°f (C4H10) = [4 × (-393.5) + 5 × (-285.8)] - [ΔH°f (C4H10)] ΔH°f (C4H10) = [-1574 + (-1429)] - [ΔH°f (C4H10)] ΔH°f (C4H10) = -3003 - ΔH°f (C4H10) Now, we can use the given standard enthalpy of combustion for C4H10: -2877.2 = -3003 - ΔH°f (C4H10) ΔH°f (C4H10) = -3003 + 2877.2 ΔH°f (C4H10) = 125.8 kJ/mol So, the standard enthalpy of formation for butane (C4H10) is 125.8 kJ/mol.

answer:To find the heat capacity of ammonia gas, we can use the formula: q = mcΔT where q is the heat capacity, m is the mass of the substance, c is the specific heat capacity, and ΔT is the change in temperature. For ammonia gas (NH3), the specific heat capacity at constant pressure (c_p) is approximately 4.70 J/(g·°C). The mass (m) of ammonia gas is given as 50 grams, and the change in temperature (ΔT) is 25 °C. Now, we can plug these values into the formula: q = (50 g) × (4.70 J/(g·°C)) × (25 °C) q = 50 × 4.70 × 25 q = 5875 J The heat capacity of 50 grams of ammonia gas at a constant pressure of 1 atm with a temperature increase of 25 °C is 5875 Joules.

answer:To calculate the heat capacity of 50 grams of water, we can use the formula: Heat capacity (Q) = mass (m) × specific heat (c) × change in temperature (ΔT) In this case, we are given the mass (m) as 50 grams, the specific heat (c) as 4.184 J/g°C, and the change in temperature (ΔT) is not given, but we can still calculate the heat capacity per degree Celsius. Heat capacity (Q) = 50 g × 4.184 J/g°C Heat capacity (Q) = 209.2 J/°C So, the heat capacity of 50 grams of water at 25°C is 209.2 Joules per degree Celsius.