Another student in the class repeated this experiment. The student's calculated values for ΔH1 and ΔH2 are shown below. ΔH1 PART B1 (heat of neutralisation of NaOH) = -42.54 kJ mol-1ΔH2 PART B2 (heat of solution + neutralisation of NaOH) = -120.42 kJ mol-1Use these values to determine this students experimentally determined answer for the enthalpy of solution for NaOH (ΔH3) in kJ mol-1.Provide your answer to 1 decimal place. INCLUDE A SIGN. Do not include units.

Input your answer for the enthalpy of solution of NaOH (ΔH3) here. Your answer should be in kJ mol-1 to 2 decimal places. INCLUDE A SIGN. Do not include units.

(ii) Show that the heat energy change, Q, in the student’s experiment is about 1100 J.[for the mixture, c = 4.2 J/g/°C][massof1.0 cm 3of mixture = 1.0 g] (3)(iii) The student uses 0.020 mol of potassium hydroxide in his experiment.Calculate the enthalpy change (ΔH) in kJ/mol, for 1.0 mol of potassium hydroxide.Include a sign in your answer.

If you performed a similar experiment in the lab and used the following data to calculate the enthalpy of solution + neutralisation for sodium hydroxide. How many significant figures should you present your answer to? (Pay special attention to your value for temperature change, this may have fewer significant figures than you are expecting.) [1 mark]Volume HCl(aq): 40.0 mLVolume H2O: 60.0 mLMass of NaOH: 1.511 gInitial temperature: 20.5 oCFinal temperature: 29.9 oCQuestion 26Answera.3 significant figuresb.5 significant figuresc.4 significant figuresd.1 significant figuree.2 significant figures

The correct sign on the heat of solution + neutralisation of NaOH is

ΔH for the following reaction is -180 kJ. How much heat is released when 100 g of Na are allowed to react? The molar mass of sodium is 23.0 g/mol. 2 Na(s) + 2 H2O(l) –> 2 NaOH(aq) + H2(g) Group of answer choices491 kJ691 kJ591 kJ391 kJ