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© Laurier Department of Chemistry and Biochemistry. No permissions for wider dissemination . WILFRID LAURIER UNIVERSITY CH110 ASSIGNMENT 1 – To Excel in Chemistry NAME: Fauzia Ali Hassan STUDENT NUMBER: 169-068-519 DATE SUBMITTED: 24 01,2024 Click here to enter a date. LAB SECTION: L6 Instructions : Fill in the blanks and submit all sheets and required graphs/tables in ONE document only to be uploaded (no .pages or .heic files) to your online CH110 Lab MyLS Excel Assignment dropbox. No other lab report criteria are required for this assignment. Part 1: Simple Linear Plot • Which set of data is plotted on the y -axis? Volume of one mole of helium (L) the x -axis? Temperature (K) • Record the following information: The equation of the fitted trendline y=0.1026x -6.8464 The value of the slope of this line 0.1026 The value of the y -intercept of this line 6.08464 • Is the fit of the trendline to your data a good fit? (Yes or No) Yes Briefly explain your response: As my R^2 is close to 1 then it will be considered to a good fit for my data. Another reason is my data has a liner line and no a sctherefore it indicts that it a good fit. • Determine the temperature, in Kelvin, of the gas in the cold room when it has a measured volume of 10.5 L using: a) Extrapolating by graphical means 169 K b) Extrapolating by using the equation of the trendline 6.36K Show your calculations for b) below: y=0.1026x-6.8464 10.5 L= 0.1026x-6.8464 10.5 L = -6.7438x Answer= -1.56 • Include (copy and paste) your graph in this report. Be sure that your axes are properly labeled,

© Laurier Department of Chemistry and Biochemistry. No permissions for wider dissemination . with units, the equation of the line and the R 2 value are visible, and that your graph has an approprite title. Click here to enter text. Part 2: Using Functions • For the Time of Diffusion versus Temperature plot, why is time plotted on the y -axis? Since all the variables are held constant, diffusion time is shown on the y axis and since the temperature is a dependent variable as it can change at anytime. • For the Time of Diffusion versus Temperature plot, which set of data is placed on the left in the Excel worksheet? Time of diffusion is placed on the left in the excel worksheet. • What are the R 2 values for the four graphs? Fill-in the table below. Time vs. Temperature Time vs. ln(Temp) Time vs. (Temp) 0.5 Time vs. (Temp) − 0.5 R 2 0.913 0.995 0.965 0.9977 • Which plot of the four graphs produces the most linear relationship? How do you know? I believe the 4st graph has the most liner because it close to 1 more than the other graphs record. • Record the following information: The equation of the trendline for the most linear plot y=107.94x + 1.008 The value of the slope of this line 107.94 y = 0.1026x - 6.8464 R² = 0.987 y = 0.1026x - 6.8464 R² = 0.987 0 5 10 15 20 25 30 35 150 200 250 300 350 400 Volume Of One Mole Of Helium (L) Tempeture (K) Tempeture (K) Verses Volume Of One Of Helium (L)

© Laurier Department of Chemistry and Biochemistry. No permissions for wider dissemination . • Include your graphs in this report. Be sure that your axes are properly labeled, with units, the equation of the line and the R 2 value are visible, and that your graph has an appropriate title. y = 107.94x + 1.008 R² = 0.9977 0.0000 2.0000 4.0000 6.0000 8.0000 10.0000 12.0000 0.0000 0.0100 0.0200 0.0300 0.0400 0.0500 0.0600 0.0700 0.0800 0.0900 Time Of Diffusion (ms) Temperature Dependent Inverse Square Root Temperature - Dependent Inverse Square Root Diffusion Trend Chart y = -0.0076x + 10.045 R² = 0.9134 0 2 4 6 8 10 12 Time Of Diffusion (ms) Temperature (K) Thermal Diffusion: Temperature (K) Verses Time Of Diffusion ( ms) trend chart.

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© Laurier Department of Chemistry and Biochemistry. No permissions for wider dissemination . y = -0.3132x + 13.076 R² = 0.965 0.0 2.0 4.0 6.0 8.0 10.0 12.0 0.0 5.0 10.0 15.0 20.0 25.0 30.0 Time of diffusion (ms) Square Root of the Temperature (K)^2 The Square Root: Temperature verses Time of Diffusion In Millisecond. y = -3.0139x + 24.662 R² = 0.995 0.00 2.00 4.00 6.00 8.00 10.00 12.00 0.00 1.00 2.00 3.00 4.00 5.00 6.00 7.00 Time of diffusion (ms) Nature Logarithm Of The Temperature Nature Logarithm: Temperature Verses Time of diffusion (ms)

© Laurier Department of Chemistry and Biochemistry. No permissions for wider dissemination . • Include the table of values that was used to create all four plots. Include units and significant digits as appropriate. Column1 Column2 Column3 Column4 Column5 Column6 Temperature (K) Time Of Diffusion ( ms) Nature logarithm Time of diffusion Square root of the tempureture Time of di 1.50E+02 9.72 5.01 9.72 12.2 9.7 2.50E+02 7.93 5.52 7.93 15.8 7.9 3.50E+02 6.87 5.86 6.87 18.7 6.9 4.50E+02 6.15 6.11 6.15 21.2 6.2 5.50E+02 5.61 6.31 5.61 23.5 5.6 6.50E+02 5.19 6.48 5.19 25.5 5.2 7.50E+02 4.86 6.62 4.86 27.4 4.9 Click here to enter text. Part 3: Two Data Sets and Overlay • Record the equations of the trendline fitted to the plot of: Nanoprism A: y=1.078x -0.0052 Nanoprism B: y=3.28x – 2.3385 • Include your graph in this report. Be sure that your axes and title are properly labeled, with units, the equation of the line, the R 2 value and the updated legend are visible. Click here to enter text. Part 4: Choosing Correct Parameters for Graphing y = 1.076x - 0.0052 R² = 0.9991 y = 3.28x - 2.3385 R² = 0.9989 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0 0.2 0.4 0.6 0.8 1 Concentration (mol/L) Absorbance (Unitless) Nanoprism A and B: Concentration Verses Aborsorbance Trends Across Varibles.

© Laurier Department of Chemistry and Biochemistry. No permissions for wider dissemination . b) Consider this equation which relates the rate constant ( k ) of a hydrolysis reaction to temperature ( T ): A T R E k a ln 1 ln +       − = If this equation is in the slope-intercept form of a line, which term in this equation corresponds to: y? In K x? 1/T m? -E a /R b? InA c-f) Use appropriate functions to manipulate the data in Table 4 and create a linear plot. Record the equation of the trendline below. Attach your table of values for the calculated data and your linear plot in your report. Equation: y=0.0074x - 1.8255 Table: Click here to enter text. Temperature (K) Rate Constant k (s^-1 2.50E+02 0.047 2.55E+02 0.0687 2.60E+02 0.0985 2.65E+02 0.141 2.70E+02 0.197 Plot:

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© Laurier Department of Chemistry and Biochemistry. No permissions for wider dissemination . g) Using the equation of the best-fit line determine i) and ii) (see below). Show your complete work in the space provided. i) the value of E a (in kJ/mol) − 0.0616 kJ/mol ii) the value of A (in s -1 ) 0.161s−1 Part 5: Statistical Analysis a) For the WLU#1 data set, record the following values with units (determined using Excel): the mean SO 4 2- concentration 34.9 ppm the median SO 4 2- concentration 34.3 ppm the standard deviation in the data set 3.03 ppm b) Are there any outliers in the WLU#1 data set (Yes or No) YES Click here to enter text. If yes, which measurement(s) are the outliers? 42.7 Show the calculations you used to identify the outlier(s). If you determined that there were no outliers, explain how you came to this conclusion. The outliner that occurred in this set of date when it lies at beyond two standard deviaviation from the mean. As a result, this is how I come to conclusion. Taking the standard deviation and

© Laurier Department of Chemistry and Biochemistry. No permissions for wider dissemination . multiplying by 2= 6.06. Mean: 34.9 ppm and taking the outliner 42.7 - 34.9= 7.8 ppm which is greater than 6.06, therefore the outlier of the date is 42.7. c) Re-calculate the following values (using Excel) excluding the outlier(s) if appropriate : the mean SO 4 -2 concentration 34.1 ppm the median SO 4 -2 concentration 33.5 ppm the standard deviation in the data set 2.03 ppm d) Did the standard deviation change between the original and re-calculated data? If so, what does this represent for your data? In the recalculated date, an insignificant change occurred between the two sets of dates, indicating the results was more accurate. After removing the outlier, the standard deviation decrease. Insert Academic Integrity statement here: I, Fauzia Ali Hassan, understand the acedemic integrity policies at WLU and this is my own work and i will not be sharing with anyone. Copyright Infringement Notice The educational materials developed for this course, including, but not limited to, lecture notes and slides, handout materials, labs, templates and assignments, and any materials posted to MyLearningSpace, are the intellectual property of the course and lab instructor. These materials have been developed for student use only and they are not intended for wider dissemination and/or communication outside of a given course. Posting or providing unauthorized audio, video, or textual material of lecture and lab content to third-party websites violates an instructor’s intellectual property rights, and the Canadian Copyright Act.  Recording lectures in any way is prohibited in this course unless specific permission has been granted by the instructor.  Failure to follow these instructions may be in contravention of the university’s Code of Student Conduct and/or Code of Academic Conduct, and will result in appropriate penalties. Participation in this course constitutes an agreement by all parties to abide by the relevant University Policies, and to respect the intellectual property of others during and after their association with Wilfrid Laurier University.