What's Cs NaCl = 3.91 J/g mean?
Is dT delta T.
Naoh (aq) + HCl (aq) --> NaCl (aq) + H2O (l)
Using the dT determined in part 1, calculate the heat capacity of the calorimeter. Enthalpy of reaction= -58.3 kj/mol, Cs (NaCl = 3.91j/g C)density of 1 M NaCl = 1.037g/mL. dT= 12.88 degrees Celcius.50 mL of 2.00 M HCl and 50 ml of 2.05 M NaOH were used
8 answers
yes dt is delta T and i think it is the specific heat capacity of NaCl but im not sure
You have 100 mL solution.
Use density to calculate mass from mass = volume x density.
mols NaCl formed = M x L = 0.2M x 0.050 = 0.1 mol. delta H rxn = 58.3 kJ/mol or 5.83 kJ for this reaction which is q in the following..
q = [mass NaCl soln x specific heat of NaCl soln x dT] + [Ccal x dT] = 0
Substitute and solve for Ccal.
Use density to calculate mass from mass = volume x density.
mols NaCl formed = M x L = 0.2M x 0.050 = 0.1 mol. delta H rxn = 58.3 kJ/mol or 5.83 kJ for this reaction which is q in the following..
q = [mass NaCl soln x specific heat of NaCl soln x dT] + [Ccal x dT] = 0
Substitute and solve for Ccal.
Originally the enthalpy is negative then how does it become postive and I have followed the steps and I came up 47.36 J/g degrees C. Also I don't understand why enthalpy is q if you could please explain that, and thankyou so much.
I had to let that equation equal something and I didn't want to type that line again so I let it equal q. The usual expression for specific heat and dT calculations is
mass H2O x specific heat H2O x (Tfinal-Tinitial) = 0. We plug in the numbers and solve for Tinitial or Tfinal or specific heat or mass H2O. In this case, however, we have all of those values and the equation equals q or any other letter I choose. I changed the sign because we are adding heat to the solution, not taking heat out; + signs mean we add heat. Another way of looking at it is that the - sign means the neutralization is an exothermic one and that it is being used to release heat and that ADDS heat to the solution.
I went through the calculations in a hurry and came up with 47.17 J/g for Ccal I would round that to 47.2.
mass H2O x specific heat H2O x (Tfinal-Tinitial) = 0. We plug in the numbers and solve for Tinitial or Tfinal or specific heat or mass H2O. In this case, however, we have all of those values and the equation equals q or any other letter I choose. I changed the sign because we are adding heat to the solution, not taking heat out; + signs mean we add heat. Another way of looking at it is that the - sign means the neutralization is an exothermic one and that it is being used to release heat and that ADDS heat to the solution.
I went through the calculations in a hurry and came up with 47.17 J/g for Ccal I would round that to 47.2.
Thankyou so much for all the help and the explanation and your time.
I have one more question, why is the volume of NaCl 0.05 L when we are calculating the mols and 0.1 L when we are calculating mass?
The volume of NaCl is never 0.100.
The mass of the SOLUTION (it is a solution of NaCl) comes from 50 mL + 50 mL = 100 mL, then
mass of solution = volume x density = 100 mL x 1.037 g/mL = about 104 or so grams for the salt solution.
The volume of NaCl is never 0.05 L.
The reaction is HCl + NaOH ==> NaCl + H2O
You have mols HCl = M x L = 2.00M x 0.050 L = 0.100 mol NaCl formed from this and the HCl is the limiting reagent.
For the NaOH we have 2.05M x 0.050 = 0.102 mols. Of course only 0.100 mol NaCl is formed.
The mass of the SOLUTION (it is a solution of NaCl) comes from 50 mL + 50 mL = 100 mL, then
mass of solution = volume x density = 100 mL x 1.037 g/mL = about 104 or so grams for the salt solution.
The volume of NaCl is never 0.05 L.
The reaction is HCl + NaOH ==> NaCl + H2O
You have mols HCl = M x L = 2.00M x 0.050 L = 0.100 mol NaCl formed from this and the HCl is the limiting reagent.
For the NaOH we have 2.05M x 0.050 = 0.102 mols. Of course only 0.100 mol NaCl is formed.
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