Laboratory Questions (Page 298 - Calorimetry) Part A.1 The 200mm test tube also contained some water (besides the metal) that… Answer: The temperature change will be lower Water has higher specific heat It means that it can absorb higher amounts of heat and rise very litte in temperature Hence, the heat which was initially present is being absorbed by water which doesn’t give required rise in temperature owing to it’s higher specific heat Part A.4 Part B Answer: Yes, in the case of all strong acid-strong base condition, all will give same amount of enthalpy of neutralization Where as in case of any the acid/base or weak acid – weak base case, some amount of energy is used up for the dissoaciation of weak acid/weak base then automatically enthalpy of neutralization is less compared to that of strong acid – strong base condition Enthalpy of neutralization is the heat change when one equivalent of an acid reacts with equivalent of base One equivalent of an acid produces one mole of H+ and one equivalent of base produces one mole of OH- ion Thus in another words heat of neutralization is the heat change when one mole H+ and one mole of OH- ion respectively Hence heat of neutralization for all strong acid-base reactions is same Part B Answer: q = mc∆ T Where q = heat loss (in J) ; m = mass of substance ; c = specific capacity of substance q = 2.35 x 1.34 J/g.℃ x 6.22℃ q = 19.6 J Heat loss to the inner 2.35g, Styrofoam cup is 19.6 J Part B.3 Answer: 40 ml will neutralize 44 ml of NaOH using N1V1 = N2V2 or say it will neutralize 0.88 ml NaOH, unreacted ml or 0.12M NaOH Part B.3 Answer: This factory error cause the reported energy of neutralization, (delta) Hn will be unaffected because since the volume is kept constant and when the concentration of the acid increases, the acid particles also increase So the concentration will be directly proportional to the temperature change Because neutralization is an exothermic reaction, during the reaction, the temperature will increase until the point of neutralization (at which point the temperature is at its highest), and after this point, the temperature will start to decrease, so effect of temperature nullified Part C.3 Answer: If some salt remains in the paper, this means that you will not include that specific mass of salt Therefore, the heat measured will not be the heat of the total mass but the heat of a smaller mass The enthalpy will be lower than that calculated or expected Part C Answer: Since it is not perfect, enthalpy would be lower since same of the heat will be lost from calorimeter NH4NO3 + Heat - NH4+ + NO3It is endothermic reaction In endothermic reaction heat energy is absorbs So more amount of heat energy absorb the calorimeter, higher ∆ H would be higher than one Laboratory Questions (Page 174 – Molar Mass of a Volatile Liquid) Part A.1 a It would be too high because the waters weight would be mistaken for the mass of the vapor b If the mass of the water vapor is reported as too high, then the reported molar mass will be too high Part A.1 Answer: Too high because the oil from the fingers will increase the amount of mass overall on the flask, and during calculations it'll seem like theres a greater mass of vapor than there actually is Part B.2 Answer: The reported molar mass of the liquid will be unaffected because the vapor will remain inside the flask just in a different state The weight will not be negatively affected by this and the molar mass will be unaffected as well Part B.2 Answer: Too high because when calculating moles of vapor using PV=nRT if the temperature is too high then the calculated moles will be too low Then when you divide the mass by these too low moles then the molar mass will be too high Part C.1 Answer: The calculated molar mass would be too low Because of this assumption, there would be a decrease in the volume So, as the volume decreases, the moles calculated would also be too low due to the direct relationship between volume and moles Part C.2 Answer: Too low because when calculating moles of vapor using PV=nRT if the pressure if too high then the calculated moles will be too high Then when you divide the mass by these too high moles then the molar mass will be too low