answer:To find the standard enthalpy change for the formation of carbon dioxide (CO2) from its elements, we need to find the reaction that represents this formation and then use the given data to calculate the enthalpy change. The formation of one mole of CO2 from its elements can be represented as follows: C(s) + O2(g) -> CO2(g) (1) However, we don't have the enthalpy change directly for this reaction. Instead, we can use the combustion reactions of carbon and hydrogen to find the required enthalpy change. First, let's consider the combustion of carbon: C(s) + O2(g) -> CO2(g) ΔH° = -394 kJ/mol (2) This is exactly the same as reaction (1), so we know that the enthalpy change for reaction (1) is -394 kJ/mol. Now, let's consider the combustion of hydrogen gas: 2H2(g) + O2(g) -> 2H2O(l) ΔH° = -572 kJ/mol (3) We can rearrange this equation to give the formation of water from its elements: H2(g) + 1/2 O2(g) -> H2O(l) ΔH°f = -286 kJ/mol (4) This is not the same as the formation of CO2, but we can use it to find the required enthalpy change. First, we need to balance the atomic ratio between hydrogen and carbon in reactions (1) and (4): C(s) + O2(g) -> CO2(g) (1) H2(g) + 1/2 O2(g) -> H2O(l) (4) To make the number of oxygen atoms equal in both reactions, we multiply reaction (4) by 2: C(s) + O2(g) -> CO2(g) (1) 2H2(g) + O2(g) -> 2H2O(l) (4 x 2) Now, we can subtract reaction (4 x 2) from reaction (1) to find the enthalpy change for the formation of CO2: C(s) + O2(g) -> CO2(g) (1) -2H2(g) - 2O2(g) -> -2H2O(l) (-4 x 2) --------------------------------------------- C(s) + 2H2(g) -> CO2(g) + 2H2O(l) The enthalpy change for this overall reaction is the sum of the enthalpy changes for reactions (1) and (-4 x 2): ΔH°f = ΔH°(1) + ΔH°(-4 x 2) = -394 kJ/mol + (-2 x -286 kJ/mol) = -394 kJ/mol + 572 kJ/mol = 178 kJ/mol So, the standard enthalpy change for the formation of carbon dioxide (CO2) from its elements is 178 kJ/mol.,To calculate the standard enthalpy change for the formation of carbon dioxide (CO2) from its elements carbon (C) and oxygen (O2), we can use the given information and Hess's Law. Hess's Law states that the total enthalpy change for a reaction is the same, regardless of the pathway taken. We are given the following reactions: 1. C (s) + O2 (g) → CO2 (g) ΔH1 = -394 kJ/mol 2. H2 (g) + 1/2 O2 (g) → H2O (l) ΔH2 = -286 kJ/mol 3. H2 (g) + 1/2 O2 (g) → H2O (l) ΔH3 = -572 kJ/mol We are interested in the enthalpy change for the formation of CO2, which is given by reaction 1. Since reaction 1 is already in the desired form, we don't need to manipulate the other reactions. Therefore, the standard enthalpy change for the formation of carbon dioxide (CO2) from its elements carbon (C) and oxygen (O2) is: ΔH = -394 kJ/mol,To calculate the standard enthalpy change for the formation of CO2 from C and O2, you only need the standard enthalpy change for the combustion of carbon (which is given). The other values provided are for other reactions involving hydrogen and water, which are not relevant to the formation of CO2. The standard enthalpy change for the formation of CO2 from its elements carbon (C) and oxygen (O2) can be calculated as follows: 1. The combustion of carbon to form carbon dioxide can be represented by the following balanced equation: C (s) + O2 (g) → CO2 (g) 2. The standard enthalpy change for this reaction is given as -394 kJ/mol. This is the energy released when one mole of carbon reacts with one mole of oxygen to form one mole of carbon dioxide. Therefore, the standard enthalpy change for the formation of carbon dioxide from its elements carbon and oxygen is -394 kJ/mol.

answer:To calculate the standard enthalpy change for the dissolution of 10 grams of NaCl, we first need to determine the number of moles of NaCl in 10 grams. The molar mass of NaCl is approximately 58.44 g/mol (22.99 g/mol for Na and 35.45 g/mol for Cl). Number of moles = mass / molar mass Number of moles = 10 g / 58.44 g/mol ≈ 0.171 mol Now, we can use the molar enthalpy of dissolution to calculate the standard enthalpy change. Standard enthalpy change = number of moles × molar enthalpy of dissolution Standard enthalpy change = 0.171 mol × (-3.9 kJ/mol) ≈ -0.667 kJ The standard enthalpy change for the dissolution of 10 grams of NaCl in water at 25°C is approximately -0.667 kJ.,To calculate the standard enthalpy change for the dissolution of 10 grams of NaCl in water at 25°C, we first need to determine the number of moles of NaCl present. The molar mass of NaCl is 58.44 g/mol (22.99 g/mol for Na + 35.45 g/mol for Cl). Now, we can find the number of moles of NaCl using the following formula: moles = mass / molar mass moles = 10 g / 58.44 g/mol = 0.171 mol Now that we know the number of moles, we can calculate the standard enthalpy change for the dissolution process using the given molar enthalpy of dissolution of NaCl, which is -3.9 kJ/mol. Standard enthalpy change = moles × molar enthalpy of dissolution Standard enthalpy change = 0.171 mol × -3.9 kJ/mol = -0.6669 kJ So, the standard enthalpy change for the dissolution of 10 grams of NaCl in water at 25°C is -0.6669 kJ., First, we need to find the number of moles of NaCl present in 10 grams. The molar mass of NaCl is 58.44 g/mol, so: Number of moles = mass / molar mass = 10 g / 58.44 g/mol ≈ 0.171 mol Now, we can calculate the standard enthalpy change for the dissolution of 0.171 moles of NaCl using the given molar enthalpy of dissolution: Standard enthalpy change = molar enthalpy of dissolution * number of moles = -3.9 kJ/mol * 0.171 mol ≈ -0.67 kJ So, the standard enthalpy change for the dissolution of 10 grams of NaCl in water at 25°C is approximately -0.67 kJ.

answer:To calculate the standard enthalpy change, we can use the following equation: ΔH = q / n where ΔH is the standard enthalpy change, q is the heat absorbed, and n is the number of moles of NaOH dissolved. First, we need to find the number of moles of NaOH dissolved: moles of NaOH = mass / molar mass moles of NaOH = 10.0 g / 40.00 g/mol = 0.25 mol Now, we can calculate the standard enthalpy change: ΔH = q / n = 9.14 kJ / 0.25 mol = 36.56 kJ/mol The standard enthalpy change for the dissolution reaction of NaOH in water at 25°C is 36.56 kJ/mol.,To calculate the standard enthalpy change for the dissolution reaction of NaOH, we first need to determine the number of moles of NaOH and then use the heat absorbed by the solution to find the enthalpy change per mole of NaOH. 1. Calculate the number of moles of NaOH: moles of NaOH = mass of NaOH / molar mass of NaOH moles of NaOH = 10.0 g / 40.00 g/mol = 0.25 mol 2. Convert the heat absorbed by the solution from kJ to J: 9.14 kJ = 9.14 * 1000 J = 9140 J 3. Calculate the standard enthalpy change per mole of NaOH: ΔH = heat absorbed by the solution / moles of NaOH ΔH = 9140 J / 0.25 mol = 36560 J/mol 4. Convert the enthalpy change from J/mol to kJ/mol: ΔH = 36560 J/mol * (1 kJ / 1000 J) = 36.56 kJ/mol The standard enthalpy change for the dissolution reaction of NaOH is 36.56 kJ/mol., First, we need to find the moles of NaOH dissolved in the solution. We can do this using its molar mass: Moles of NaOH = mass / molar mass Moles of NaOH = 10.0 g / 40.00 g/mol = 0.250 mol Next, we need to determine the change in temperature (ΔT) of the water due to the dissolution reaction. Since heat is absorbed by the solution, the temperature should increase. We can calculate ΔT using the formula: q = mcΔT where q is the heat absorbed, m is the mass of the substance, c is its specific heat capacity, and ΔT is the change in temperature. Rearranging for ΔT, we get: ΔT = q / (mc) For the water, m = 100.0 g c = 4.18 J/g·°C q = 9.14 kJ = 9140 J (since 1 kJ = 1000 J) So, ΔT = 9140 J / (100.0 g * 4.18 J/g·°C) = 21.85°C Now, let's assume that all the heat absorbed by the water comes from the dissolution of NaOH. This means that the enthalpy change for the dissolution reaction (ΔH_diss) is equal to the negative value of q (the heat absorbed) because the reaction is endothermic. ΔH_diss = -q = -9140 J = -9.14 kJ The standard enthalpy change for the dissolution reaction is -9.14 kJ. Note that this value is negative because the dissolution process is endothermic, meaning it absorbs heat from the surroundings.

answer:To calculate the standard enthalpy change for the neutralization reaction between hydrochloric acid and sodium hydroxide, we can use the formula: ΔH°rxn = ΣΔH°f(products) - ΣΔH°f(reactants) where ΔH°rxn is the standard enthalpy change for the reaction, ΣΔH°f(products) is the sum of the standard enthalpies of formation of the products, and ΣΔH°f(reactants) is the sum of the standard enthalpies of formation of the reactants. For this reaction, the reactants are HCl(aq) and NaOH(aq), and the products are NaCl(aq) and H2O(l). Therefore, we have: ΣΔH°f(products) = -407.3 kJ/mol (for NaCl(aq)) - 285.8 kJ/mol (for H2O(l)) ΣΔH°f(products) = -693.1 kJ/mol ΣΔH°f(reactants) = -167.2 kJ/mol (for HCl(aq)) - 470.1 kJ/mol (for NaOH(aq)) ΣΔH°f(reactants) = -637.3 kJ/mol Now we can plug these values into the formula to find the standard enthalpy change for the reaction: ΔH°rxn = ΣΔH°f(products) - ΣΔH°f(reactants) ΔH°rxn = -693.1 kJ/mol - (-637.3 kJ/mol) ΔH°rxn = -55.8 kJ/mol Therefore, the standard enthalpy change for the neutralization reaction between hydrochloric acid and sodium hydroxide is -55.8 kJ/mol. This means that the reaction releases 55.8 kJ of heat energy for every mole of HCl(aq) and NaOH(aq) that react.,To find the standard enthalpy change for the neutralization reaction, we can use the following equation: ΔH_reaction = Σ ΔH_f(products) - Σ ΔH_f(reactants) where ΔH_reaction is the standard enthalpy change for the reaction, and ΔH_f is the standard enthalpy of formation for each compound. For the given reaction: ΔH_reaction = (-407.3 kJ/mol) + (-285.8 kJ/mol) - (-167.2 kJ/mol) + (-470.1 kJ/mol) ΔH_reaction = (-693.1 kJ/mol) - (-637.3 kJ/mol) ΔH_reaction = -55.8 kJ/mol The standard enthalpy change for the neutralization reaction between hydrochloric acid and sodium hydroxide is -55.8 kJ/mol.,To find the standard enthalpy change for the neutralization reaction between hydrochloric acid and sodium hydroxide, we can use the following equation: ΔH_reaction = Σ ΔH_f(products) - Σ ΔH_f(reactants) In this case, the products are NaCl(aq) and H2O(l), and the reactants are HCl(aq) and NaOH(aq). The standard enthalpy of formation for each compound is given in the question. Plugging in the values, we get: ΔH_reaction = -407.3 kJ/mol (NaCl) - 285.8 kJ/mol (H2O) - -167.2 kJ/mol (HCl) - 470.1 kJ/mol (NaOH) ΔH_reaction = (-407.3 + 285.8) + (-167.2 + 470.1) ΔH_reaction = -121.5 kJ/mol + 298.9 kJ/mol ΔH_reaction = 177.4 kJ/mol So, the standard enthalpy change for the neutralization reaction between hydrochloric acid and sodium hydroxide is 177.4 kJ/mol.