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question:A chemistry student needs to calculate the heat of hydration for the reaction between anhydrous magnesium chloride (MgCl2) and water (H2O) using the following data: the enthalpy change for the dissolution of anhydrous magnesium chloride in water is -63.3 kJ/mol, and the enthalpy change for the hydration of magnesium ions (Mg2+) and chloride ions (Cl-) is -192 kJ/mol and -131 kJ/mol, respectively. Calculate the overall heat of hydration (in kJ/mol) for the reaction: MgCl2(s) + 6H2O(l) → MgCl2.6H2O(s)

answer:To calculate the overall heat of hydration for the reaction, we need to consider the enthalpy changes for the dissolution of anhydrous magnesium chloride and the hydration of magnesium and chloride ions. The reaction can be broken down into three steps: 1. Dissolution of anhydrous magnesium chloride in water: MgCl2(s) → Mg2+(aq) + 2Cl-(aq); ΔH1 = -63.3 kJ/mol 2. Hydration of magnesium ions: Mg2+(aq) + 6H2O(l) → Mg(H2O)6^2+(aq); ΔH2 = -192 kJ/mol 3. Hydration of chloride ions (2 moles of Cl- ions per mole of MgCl2): 2[Cl-(aq) + H2O(l) → Cl-(H2O)(aq)]; ΔH3 = 2 * -131 kJ/mol = -262 kJ/mol Now, we can calculate the overall heat of hydration by adding the enthalpy changes for each step: ΔH_total = ΔH1 + ΔH2 + ΔH3 ΔH_total = -63.3 kJ/mol + (-192 kJ/mol) + (-262 kJ/mol) ΔH_total = -517.3 kJ/mol So, the overall heat of hydration for the reaction is -517.3 kJ/mol.

question:What is the enthalpy change when 50 grams of anhydrous copper (II) sulfate reacts with water to form 100 mL of a hydrated copper (II) sulfate solution with a concentration of 2 M? Given the molar enthalpy of hydration of anhydrous copper (II) sulfate is -90.5 kj/mol and the density of the solution is 1.2 g/mL.

answer:First, we need to determine the number of moles of anhydrous copper (II) sulfate in 50 grams. The molar mass of anhydrous copper (II) sulfate (CuSO4) is: Cu: 63.5 g/mol S: 32.1 g/mol O: 16.0 g/mol * 4 = 64.0 g/mol Total: 63.5 + 32.1 + 64.0 = 159.6 g/mol Now, we can find the number of moles: moles = mass / molar mass moles = 50 g / 159.6 g/mol = 0.313 mol Next, we can calculate the enthalpy change for the hydration of 0.313 mol of anhydrous copper (II) sulfate: ΔH = moles * molar enthalpy of hydration ΔH = 0.313 mol * (-90.5 kJ/mol) = -28.3 kJ Therefore, the enthalpy change when 50 grams of anhydrous copper (II) sulfate reacts with water to form 100 mL of a hydrated copper (II) sulfate solution with a concentration of 2 M is -28.3 kJ.

question:Calculate the heat of hydration of copper (II) sulfate pentahydrate given that 5.00 g of the hydrate was dissolved in 100.0 mL of water at 25°C, and the resulting solution temperature increased from 25°C to 30°C. The molar mass of copper (II) sulfate pentahydrate is 249.685 g/mol and the specific heat capacity of water is 4.184 J/g°C.

answer:First, we need to calculate the moles of copper (II) sulfate pentahydrate that were dissolved: moles = mass / molar mass moles = 5.00 g / 249.685 g/mol moles = 0.02003 mol Next, we need to calculate the heat absorbed by the water: q = mass x specific heat capacity x change in temperature q = 100.0 g x 4.184 J/g°C x (30°C - 25°C) q = 100.0 g x 4.184 J/g°C x 5°C q = 2092 J Now, we can calculate the heat of hydration per mole of copper (II) sulfate pentahydrate: heat of hydration = q / moles heat of hydration = 2092 J / 0.02003 mol heat of hydration = 104,450 J/mol The heat of hydration of copper (II) sulfate pentahydrate is approximately 104,450 J/mol.

question:What is the heat of hydration for the reaction between 50 grams of anhydrous copper(II) sulfate and excess water, given that the final solution mass is 250 grams and the temperature change observed during the reaction is -30 °C?

answer:To calculate the heat of hydration, we need to use the formula: q = mcΔT where q is the heat absorbed or released, m is the mass of the solution, c is the specific heat capacity of the solution, and ΔT is the temperature change. Given the mass of the solution (m) is 250 grams, the temperature change (ΔT) is -30 °C, and assuming the specific heat capacity (c) of the solution is similar to water (4.18 J/g·°C): q = (250 g)(4.18 J/g·°C)(-30 °C) q = -31,350 J Since the temperature change is negative, the reaction is exothermic and releases heat. Therefore, the heat of hydration for the reaction between 50 grams of anhydrous copper(II) sulfate and excess water is 31,350 J (released).