When 23.6 of calcium chloirde CaCl2 was dissolved in water in a calorimeter, the temperature rose from 25.0 degrees C to 21.56 degrees C. If the heat capacity of the solution and the calorimeter is 1071 J/ degrees C, what is the enthalpy change when 1 mol of sodium nitrate dissolves in water? The solution process is
NaNO3 + Na + NO3
Delta H = ?
sorry it is supposed to say
When 23.6 of calcium chloirde CaCl2 was dissolved in water in a calorimeter, the temperature rose from 25.0 degrees C to 38.7 degrees C. If the heat capacity of the solution and the calorimeter is 1258 J/ degrees C, what is the enthalpy change when 1 mol of sodium nitrate dissolves in water? The solution process is
CaCl2 ----> Ca2+ + 2Cl
Delta H=?
Sorry I read the wrong question my fault I appoligize
I can GUESS that the 23.6 is GRAMS and not mols. I can also GUESS that the next to last sentence should read "when 1 mol calcium chloride dissolves in water?" If my guesses are correct, then
q = heat evolved on solution = Cp x delta T.
q = 1258 J/oC x delta T for 23.6 g CaCl2.
Therefore, you know q (delta H) for 23.6 g CaCl2. You want to know the q for 1 mol CaCl2 or 111 g CaCl2. A proportion. You need to obtain the exact molar mass of CaCl2 since I used round numbers. I just estimated it.
To determine the enthalpy change when 1 mole of sodium nitrate (NaNO3) dissolves in water, you can use the concept of Hess's Law and the enthalpy of formation values for the given compounds involved.
Hess's Law states that the overall enthalpy change of a reaction is independent of the pathway taken, as long as the initial and final conditions are the same. In this case, we can break down the process of dissolving sodium nitrate into individual steps involving the formation of the corresponding ions.
The given equation for the solution process is:
NaNO3 + Na + NO3
To calculate the enthalpy change, you need the following information:
1. The enthalpy of formation values for NaNO3, Na, and NO3.
2. The enthalpy change for the dissolution of Na+ and NO3- ions.
First, let's calculate the heat change for dissolving calcium chloride (CaCl2) in water using the information given:
Mass of CaCl2 = 23.6 g
Heat capacity of the solution and calorimeter = 1071 J/°C
Temperature change (ΔT) = 25.0°C - 21.56°C = 3.44°C
Using the formula:
q = m * c * ΔT
where q is the heat change, m is the mass, c is the specific heat capacity, and ΔT is the temperature change, we can calculate the heat change for the dissolution of CaCl2.
q = 23.6 g * 1071 J/°C * 3.44°C = 856.71 J
Next, we will use this value to calculate the enthalpy change for dissolving 1 mole of NaNO3.
Since CaCl2 contains two chloride ions (Cl-), and the enthalpy change is for 23.6 g (which is approximately the molar mass of CaCl2), we can calculate the enthalpy change per mole of chloride ions:
ΔH/(2 moles of Cl-) = 856.71 J
To calculate the enthalpy change when 1 mole of NaNO3 dissolves, we need to determine the enthalpy change for 1 mole of chloride ions.
ΔH/(1 mole of Cl-) = (ΔH/(2 moles of Cl-)) * (1 mole of Cl-) / (2 moles of Cl-)
ΔH/(1 mole of Cl-) = 428.35 J
Now that we have the enthalpy change per mole of chloride ions, we can determine the enthalpy change per mole of NaNO3. Since NaNO3 contains one nitrate ion (NO3-), the enthalpy change for the dissolution of NaNO3 is the same as the enthalpy change for the dissolution of one chloride ion.
ΔH/(1 mole of NO3-) = ΔH/(1 mole of Cl-) = 428.35 J
Therefore, the enthalpy change when 1 mole of sodium nitrate dissolves in water is 428.35 J.