**Hint:** This is the diluted : heat capacity = 3.184 J/g °C and D = 1.00 g/mole**Heat of reaction** = - (heat of solution) - heat of calorimetry**c. Calculate the heat of reaction for the following chemical reaction:**\[ \text{NaOH (s)} + \text{HCl} \text{(aq)} \rightarrow \text{NaCl} \text{(aq)} + \text{H}_2\text{O(l)} \]**Using the following reactions:**\[\text{NaOH(s)} \rightarrow \text{NaOH(aq)} \quad \Delta H = -102 \text{ kJ/mole}\]\[\text{NaOH(aq)} + \text{HCl(aq)} \rightarrow \text{NaCl(aq)} + \text{H}_2\text{O(l)}\]Use the heat of reaction from your calculations. ### Determination of Heat Capacity of a Calorimeter**Procedure:**1. **Experiment Setup:** - 50.0 mL of H₂O at an initial temperature of 23.5°C is added to a calorimeter (coffee cup). - Stirred for 5 minutes. - Then, 50.0 mL of hot water at 98.0°C is added to the calorimeter containing the initial 50.0 mL of water. - The temperature of the mixture is recorded every 30 seconds. The data is as follows:| Time (seconds) | Temperature (°C) ||----------------|------------------|| 30 | 55.0 || 60 | 50.0 || 90 | 45.5 || 120 | 42.0 || 150 | 38.0 || 180 | 37.5 || 210 | 37.5 || 240 | 37.0 || 270 | 37.0 || 300 | 37.0 || 330 | 37.0 |2. **Constants:** - Density (D) of H₂O = 1.00 g/mL. - Heat capacity of H₂O (l) = 4.184 J/g°C. **Tasks:**a. **Plot Temperature vs. Time:** - Graph temperature over time to determine the final equilibrium temperature.b. **Calculate Heat Capacity of the Calorimeter:** - Use the formula: heat lost by hot water = heat gained by cold water + heat gained by calorimeter.---### Determination of Heat of Reaction**Procedure:**1. **Experiment Setup:** - 50.0 mL of 0.100 M NaOH(aq) at 22.0°C is added to a calorimeter. - 50.0 mL of HCl(aq) at 21.0°C is then added to the calorimeter. - Temperature is recorded every 30 seconds. The data is as follows:| Time (seconds) | Temperature (°C) ||----------------|------------------|| 30 | 25.0 || 60 |

“Since you have asked multiple question, we will solve the first question for you. If you want any specific question to be solved then please specify the question number or post only that question.” Calorimeter measures amount of involved in a system. This can be obtained from mass of solution, specific heat of solution and temperature changes. If the reaction is exothermic then there will be increase in temperature in calorimeter as the heat is absorbed by the solution. If the reaction is endothermic then there will be decrease in temperature in calorimeter as the heat is absorbed by the solution thermal energy.Final temperature is calculated from the graph which is drawn between time and temperature data. The calorimeter value is obtained by heat balance. First we will plot the data and obtain equation from graph The final temperature will be at time is zero seconds Atx=0y=-0.0539x+50.936y=-0.0539(0)+50.936y=50.936°C The final temperature is obtained 50.936°C Heat Balance Calculation heat lost by hot water will be equal to heat gained by both water and calorimeter. -qhot water=qwater+qcalorimeterq=mC∆T-qhot water=-(mhot water×Cwater×(Tfinal-Thot))=-50g×4.184Jg°C×(50.936-98)°C=9845.79 Jqwater=m water×Cwater×(Tfinal-Twater)=50g×4.184Jg°C×(50.936-23.5)°C=5739.61 Jqcalorimeter=Ccalorimeter×(Tfinal-Tinital)=Ccalorimeter(50.936-23.5)°C=Ccalorimeter(27.436°C)9845.79J=5739.61J+Ccalorimeter(27.436°C)9845.79J-5739.61J=Ccalorimeter(27.436°C)4106.18J=Ccalorimeter(27.436°C)4106.18J27.436°C=Ccalorimeter149.66 J/°C=Ccalorimeter The final answer a)The final temperature from graph 50.936°C b)Heat of calorimeter 149.66J/°C