You have finished two trials of the sample analysis for the experiment and you noticed that the burette is almost empty (~5 mL remaining). Should you add more titrant before the next sample analysis or not? Why or why not? Select one: YES. It will not affect the calculations and not cause error NO. It will affect the calculations and cause error Neither A nor B Cannot be determined

You have finished two trials of the sample analysis for the experiment and you noticed that the burette is almost empty (~5 mL remaining). Should you add more titrant before the next sample analysis or not? Why or why not? Select one: YES. It will not affect the calculations and not cause error NO. It will affect the calculations and cause error Neither A nor B Cannot be determined

Appl Of Ms Excel In Analytical Chemistry

2nd Edition

ISBN:9781285686691

Author:Crouch

Publisher:Crouch

Chapter3: Statistical Tests With Excel

Section: Chapter Questions

Problem 5P

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The following volumes of 0.000300 M SCN- are diluted to 15.00 mL. Determine the concentration of SCN- in each sample after dilution. These values will be used during the experiment. Sample 0.000300 M SCN- (mL) [SCN-] (M) 1 1.50 2 3.50 3 7.00 4 10.00
You want to determine the concentration of NaCl in a water. In preliminary analysis, it appears to be saline water with a NaCl content of approximately 4%. Should you dilute the sample if your titrant has a concentration of 0.1 M and you want to spend 25 mL of sample? Solution: You must dilute approximately ten times, taking 10 mL of sample and making up to 100 mL. In this way, 17.20 mL of titrant will be spent.
You want to make a solution of the primary standard sodium oxalate. It has a molecular mass of 133.999 g/mol (no error). Your analytical balance has an error of 0.2 mg. You add some weighing paper to the balance and it reads a mass of 0.23626 g. You then add sodium oxalate onto the weighing paper until the mass registers as 2.8539 g. You carefully transfer this to a 250.0 ml volumetric flask with an error of 0.1 ml and dissolve the sodium oxalate in a sufficient volume of water. What is the concentration of the oxalic acid (in M) and what is the absolute error (also in M).
Based on the following SAMPLE 2 data, calculate the concentration (M) of NaOH. Report answer with FOUR decimal places. Do not include units in response. REPORT FOR EXPERIMENT 22 Neutralization-Titration I Data Table Sample 1 Sample 2 Mass of flask and KHP 94.7994 g 95.3800 g Mass of empty flask 93.6616 g 94.2144g Mass of KHP Final buret reading 18.25 mL 36.94 mL Initial buret reading 0.15 mL 18.24 mL Volume of base used
Titration of Vinegar Experimental Data Trial 1 Trial 2 Initial Buret Reading 0.05 mL 0.02 mL Final Buret Reading 25.25 mL 23.22 mL Volume of base used Molarity of base used 0.1045 M 0.1045 M Volume of vinegar used 5.00 mL 5.00 mL
In the experiment, the amount of acetic acid in vinegar was determined by titrating with 0.1 M NaOH solution. In the analysis, three trials were performed to obtain more precise result. The table shows the initial and final burette readings for three titrations: Initial Burette Reading (mL) Final Burette Reading (mL) Trial 1 2.9 10.6 Trial 2 10.6 18.1 Trial 3 18.1 25.7 According to the given data, calculate molarity of acetic acid in the commercial vinegar. Please be sure that your answer includes followings; a) moles of NaOH for each trial b) average mole of NaOH c) average mole of acetic acid in 50 mL solution d) average mole of acetic acid in 250 mL solution e) molarity of acetic acid in vinegar.
You are asked to weigh 2 g of lithium and add it to 1500 ml of distilled water +2drops of indicator, knowing that the density of water is 1 at room temperaturewhile conducting this experiment. After calculating the amount of LiOH, you need to dilute to prepare 0.25 M of LiOH in a 3000 ml solution. Show your calculation and units.Knowing that the initial volume that you used was 1500 ml. Use this equation to calculate ( CLiOH x VLiOH) before dilution = (CLiOH x VLiOH) after dilution
5. If the label on the bottle lists the citric acid content as 1 g per serving with a serving size of 240 mL determine the percent error in the titration results. Express result with 2 significant figures.
Volumetric Analysis Liza carefully pipets 25.0 mL of 0.525 M NaOH into a test tube. She places the test tube into a small beaker to keep it from spilling and then pipets 75.0 mL of 0.355 M HCl into another test tube. When Liza reaches to put this test tube of acid into the beaker along with test tube of base, she accidentally knocks the test tubes together hard enough to break them and their respective contents combine in the bottom of the beaker. Is the solution formed from the contents of the two test tubes acidic or basic? What is the pH of the resulting solution?
A student conducts the experiment as outlined in the posted procedure, and records the following data. Calculate the concentration (M) of calcium ions in the filtered solution. [HCI] (M): 0.0103 initial buret reading (mL): 3.94 final buret reading (mL): 33.94
Based on the analysis of a vinegar solution via titration with 0.1 M NaOH. Results are as follows: Trial 1 Volume of vinegar solution used (mL) 5.00 Final burette reading (mL) 44.20 Initial burette reading (mL) 0.00 Determine the percentage mass of acetic acid in vinegar for Trial 1. The density of vinegar is 1.01 g/mL and molecular mass for acetic acid is 60 g/mol.
Average the two percentages and calculate the percentage difference. given : Concentration of acetic acid in the 100 mL sample of titration 1 = 3.238 % Concentration of acetic acid in the 100 mL sample of titration 2 = 3.099 % NaOH vs CH3COOH Burette solution is NaOH and the pipette solution is 5.0 mL of Vinegar Titration Initial burette reading Final burette reading Volume of NaOH consumed Average volume of NaOH Approximate 0.0 mL 20.1 mL 20.1 mL 20.3 mL Titration 1 0.0 mL 20.9 mL 20.9 mL Titration 2 0.0 mL 20.0 mL 20.0 mL To find the average volume Average volume = 20.1 mL + 20.9 mL + 20.0 mL320.1 mL + 20.9 mL + 20.0 mL3 = 20.3 mL Concentration of Vinegar = Volume of NaOH * Concentration of NaOHVolume of vinegarVolume of NaOH * Concentration of NaOHVolume of vinegar = 20.3 mL * 0.0647 M5.0 mL20.3 mL * 0.0647 M5.0 mL = 0.2627 M Concentration of NaOH = Volume of…