| | |Experiment 12- Calorimetry and Hess’s Law | | | | | |Jenna Hyde | |Lab partners: …show more content…
| | | | |reaction, cleared back up.| | |MgO |n/a |n/a | |.501 | Calculations Part A Calculations Trial 1 Mass of cold water: [pic] = 1.00 (g/mL)X 50.0mL= 50 g Tf from graph by extrapolation: 35.5 degrees Celsius ΔTHW=Tf-Ti= 35.5C-54.0C= -18.5C ΔTCW=Tf-Ti=35.5C-22.5C=13.0C qHw=mcΔTHW=(50.0g)(4.184 J/gC)(-18.5C)= -3870 J or -3.87 kJ qCW=mcΔTCW=(50.0g)(4.184 J/gC)(13.0C)= 2720J or 2.72kJ qcal for the cup: |qHW|=|qCW|+qcal 3870=2720+qcal qcal=1150J or 1.15kJ qcal=CcupΔT 1150J=Ccup(13.0C) Ccup=88.4J/C Trial 2 Mass of cold water: [pic] = 1.00 (g/mL)X 50.0mL= 50 g Tf from graph by extrapolation: 36.1 degrees Celsius ΔTHW=Tf-Ti= 36.1C-51.0C= -14.9C ΔTCW=Tf-Ti=36.1C-23.7C=12.4C qHw=mcΔTHW=(50.0g)(4.184 J/gC)(-14.9)= -3120 J or -3.12 kJ qCW=mcΔTCW=(50.0g)(4.184 J/gC)(12.4C)= 2590J or 2.59kJ qcal for the cup: |qHW|=|qCW|+qcal 3120=2590+qcal qcal=530J or 0.530kJ qcal=CcupΔT 530J=Ccup(12.4C) Ccup=42.7J/C Average Ccup=(88.4+42.7)/2=65.55=65.6 J/C Part B Calculations Tf from graph by extrapolation: 29.7C Mass of HCl: [pic] = 1.00 (g/mL)X 100.0mL= 100 g ΔTCW=Tf-Ti= 29.7C-23.0C=6.7.C
Procedure: Using distilled water, premeasured containers and objects determine displacement of fluids and density of objects. Use ice and heat measure temperatures in Celsius, Fahrenheit and Kelvin.
In order to measure the heats of reactions, add the reactants into the calorimeter and measure the difference between the initial and final temperature. The temperature difference helps us calculate the heat released or absorbed by the reaction. The equation for calorimetry is q=mc(ΔT). ΔT is the temperature change, m is the mass, c is the specific heat capacity of the solution, and q is the heat transfer. Given that the experiment is operated under constant pressure in the lab, the temperature change is due to the enthalpy of the reaction, therefore the heat of the reaction can be calculated.
The purpose of this lab is to test substances and to determine the physical and chemical properties of substances.
You have the mass of water from calculation #9, the specific heat of water is 4.184 J/g(oC), and the temperature change of water
The freezing point constant (Kf) of water is 1.86 °C m-1. Each mass amount and Van’t Hoff factor was calculated then analyzed in a table.
Students will carefully observe acts of aggression and prosocial behavior on television, report their observations, and analyze their data to draw conclusions.
The boiling point elevation constant for water that was experimentally determined in this analysis was 0.4396 °C/m, which was derived from the slope of the trend line in Figure 2. This is slightly lower than the constant provided in lecture of 0.51 °C/m. This could be due to further evaporation of water from the solutions tested via refractive index after the boiling temperature was recorded.
A higher volume of NaOH will result in more moles of NaOH being added to the HCl, which results in more HCl reacting. This makes the calculated molarity of the HCl be smaller than the actual molarity of the HCl.
The luminous yellow flame is smoky because no air is entering the burner and hydrocarbon is converted into carbon dioxide
Purpose: To find the relationship between the mass and the volume of the four samples.
C. An unknown, rectangular substance measures 3.6 cm high, 4.21 cm long, and 1.17 cm wide.
The purpose of Step 8 is to make sure we get the exact mass of the solution. There could be trapped molecules of water in the solution. Heating the solution would evaporate the water molecules.
have five C-H bonds, one C-C bond, one C-O bond and one O-H bond. To
Introduction: The theory behind this experiment is the heat of a reaction (∆E) plus the work (W) done by a reaction is equal to
3. Calculate the total heat released in each reaction, assuming that the specific heat of the solution is the same as for pure water (4.18J/gK). Use q=mcΔT. Show work here and record your answer in Data Table 2.