This is a chemistry assignment. Please follow all instructions and read the files that I have attached. This assignment includes 6 graphs (three of which are given to you in the excel file) and a completed lab report (at the end of lab manual N) ***please have it done by tonight or tomorrow morning// I will add bonus

Plot the ln[CV+] vs t graph for the second and the third kinetics runs in Experiment M. The sixth graph is the plot of ln(k) vs 1/T from the Experiment N data. For the Part I of the lab report, k’ = - slope (Question 3). To calculate the initial concentration of [OH-], use C1V1=C2V2, where C1 is the stock solution concentration from the observations, V1 is 3.00 mL, and V2 is 6.00 mL. (Question 4). Use this equation to calculate the initial concentration of CV+ for the table on page N-14. To find k, use k= k’/[OH]n (Question 6). Question 9. Calculate the ratio (moles of OH- reacted)/(moles of OH- initial)x 100%. It should be very small. Let’s find moles of OH- reacted . In the equation CV+ (aq) + OH-(aq) → CVOH(aq) there is 1:1 mole ratio between CV+ and OH-. Therefore, moles of OH- reacted = moles of CV+ reacted. Moles of CV+ reacted = moles of CV+ initial – moles of CV+ at t=120s Moles of CV+ initial = C x V , moles of CV+ at t=120s = (Abs at t=120 s)/83264 x 6.00x10-3 L. Question10. If [CV+] = [OH-], rate = k[CV+][OH-] = k[CV+]? For Part II of the lab report, plot ln([CV+]) vs t for each of the four runs, find k’ from the graphs of ln ([CV+]) vs t (recall experiment M) , calculate k values, then ln(k). Plot ln(k) vs 1/T. The slope = -Ea/R. Note! Do not include Exp. M Run 1 data in the table on the page N-17. In the equation of a trendline y= mx + b, the value of m is a slope; the value of b is an y-intercept. Units of m are derived by dividing the units of the y-axis by the units of the x-axis. Units of b should be the same as those of the y-axis. When you take the log or ln of a value with units, the units are lost. For question 3 on N-17, use R = 8.314 J/K mol.