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Data Entry - No Scoring Enter the precise mass in GRAMS of the potassium iodate used

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5) Data Entry - No Scoring;Enter the precise mass in GRAMS of the potassium iodate used to prepare your primary standard solution.;Your mass precision should be reported to a thousandth of a gram, i.e. 0.302. (Use 3 significant figures.);Your Answer: 0.5340;No Points Possible;6) Scoring Scheme: 3-3-2-1;If one had weighed out precisely 0.500 g of KIO 3 for the primary standard solution and dissolved it in;enough deionized water to make a 250 mL solution, the molarity of that solution would be 0.00935 M.;Enter your calculated molarity of the primary standard KIO 3 solution. Please use 3 significant figures.;Your Answer: 0.009980;You Scored 3 points out of 3 Possible;7) Data Entry - No Scoring;Standardization of the sodium thiosulfate solution using the potassium iodate primary standard solution.;We must examine each of the three acceptable trials.;First, let's consider the analyte volume. You were instructed to pipet two 10.00 mL aliquots of potassium;iodate into an Erlenmeyer flask for each analyte sample to be titrated. Therefore, your analyte volumes of;the potassium iodate should be 20.00 mL.;For each trial, enter the precise volume in mL of potassium iodate solution used in the standardization of;sodium thiosulfate (e.g. 20.00 mL or 10.00 mL);Your Answer: 20;No Points Possible;Your Answer: 20;No Points Possible;Your Answer: 20;No Points Possible;8) Data Entry - No Scoring;Standardization of the sodium thiosulfate solution using the potassium iodate primary standard solution.;Now we will examine the volumes of titrant, sodium thiosulfate, used in each trial.;Your titration volumes of the sodium thiosulfate solution should be approximately 20 - 30 mL.;For each trial, enter the precise volume in mL of sodium thiosulfate solution used in the titration of your;potassium iodate solution (e.g. 20.34 mL);Your Answer: 26;No Points Possible;Your Answer: 23.70;No Points Possible;Your Answer: 26.10;No Points Possible;9) Scoring Scheme: 3-3-2-1;Using the volumes of sodium thiosulfate solution you just entered and the molarity of the potassium iodate;primary standard solution, calculate the molarity of the sodium thiosulfate secondary standard solution for;each trial.;Enter your calculated molarity of the sodium thiosulfate solution for each of the trials. Be sure to enter your;calculated molarities in the corresponding order that you entered your sodium thiosulfate volumes;previously.;The sodium thiosulfate volume you entered for entry #1 above should correspond to the sodium thiosulfate;molarity that you enter for entry #1 here.;Your Answer: 0.0461;You Scored 3 points out of 3 Possible;Your Answer: 0.0506;No Points Possible;Your Answer: 0.0459;No Points Possible;10) Scoring Scheme: 3-3-2-1;The molarity of the sodium thiosulfate solution is taken as the average of the three trials.;Please enter the average molarity.;Average =;Your Answer: 0.0475;You Scored 3 points out of 3 Possible;11) Scoring Scheme: 3-3-2-1;Please enter the standard deviation of the sodium thiosulfate molarity.;Standard Deviation =;Your Answer: 0.0030;You Scored 3 points out of 3 Possible;13) Scoring Scheme: 3-2-1-1;For the reaction of hypochlorite anion with iodide anion, the iodide anion acts as the reducing agent;according to the oxidation half-reaction, 2I-(aq) -> I2(aq) + 2e-.;Which of the following reduction half-reactions is correct to give the overall reaction of hypochlorite anion;with iodide anion?;Your Answer: c. 2H+(aq) + ClO-(aq) + 2e- -> Cl-(aq) + H2O(l);You Scored 3 points out of 3 Possible;14) Scoring Scheme: 3-2-1-1;For the reaction of thiosulfate anion with iodine, the thiosulfate anion acts as the reducing agent according;to the oxidation half-reaction, 2S2O32-(aq) -> S4O62-(aq) + 2e-.;Which of the following reducing half-reactions is correct to give the overall reaction of thiosulfate anion with;iodine?;Your Answer: a. I2(aq) + 2e- -> 2I-(aq);You Scored 3 points out of 3 Possible;15) Scoring Scheme: 3-2-1-1;Notice that the effective analytical connection for analysis of hypochlorite anion by the thiosulfate anion in;this experiment is the sum of the two reactions;1) hypochlorite anion with iodide anion to form iodine, and;2) thiosulfate anion with the iodine formed in reaction one.;In your laboratory notebook sum these two reactions to find the stoichiometric factor that relates moles of;thiosulfate anion needed to react with each mole of hypochlorite anion in the bleach sample.;Then select the correct response.;Your Answer: b. Two moles of thiosulfate anion needed to react You Scored 3 points out of 3 Possible;with one mole of hypochlorite anion.;16) Data Entry - No Scoring;For each of 3 acceptable trials, enter the mass of your bleach sample in GRAMS. Your mass precision;should be reported to a thousandth of a gram.;Your Answer: 1;No Points Possible;Your Answer: 1;No Points Possible;Your Answer: 1;No Points Possible;17) Data Entry - No Scoring;For each of the 3 acceptable trials, enter the precise volume in milliliters of sodium thiosulfate solution;used in the titration of your bleach samples (e.g. 20.34 mL).;Be sure to enter your trial volumes in the corresponding order that you entered your masses of your bleach;samples previously.;The bleach sample mass you entered for entry #1 above should correspond to the sodium thiosulfate;volume you enter for entry #1 here.;Your Answer: 27;No Points Possible;Your Answer: 24.25;No Points Possible;Your Answer: 24.65;No Points Possible;18) Scoring Scheme: 3-3-2-1;Using the volumes of sodium thiosulfate solution you just entered, the mass of bleach sample, and the;average molarity of the sodium thiosulfate solution entered earlier, calculate the mass percent of NaClO for;each bleach sample.;Enter the calculated mass percent of NaClO in each of the 3 acceptable trials. Be sure to enter your mass;percent values in the corresponding order that you entered your masses of your bleach samples and;volumes of your sodium thiosulfate previously.;The bleach sample mass you entered for entry #1 and the volume of sodium thiosulfate you entered above;should correspond to the mass percent of NaClO you enter for entry #1 here.;Your Answer: 4.77;You Scored 3 points out of 3 Possible;Your Answer: 4.29;No Points Possible;Your Answer: 4.36;No Points Possible;19) Scoring Scheme: 3-3-2-1;Using your 3 trial values, calculate the average mass percent of NaClO.;Enter the average mass percent of NaClO in the bleach sample. (Use 3 significant figures.);Average =;Your Answer: 4.47;You Scored 3 points out of 3 Possible;20) Scoring Scheme: 3-3-2-1;Please enter the standard deviation of the average mass percent of NaClO in your bleach samples.;Standard Deviation =;Your Answer: 0.26;You Scored 3 points out of 3 Possible;21) Scoring Scheme: 3-3-2-1;Please enter the calculated 90% confidence limit for the average mass percent of the NaClO in your bleach;samples.;90% confidence limit =;Your Answer: 0.44;You Scored 3 points out of 3 Possible;22) Press continue to see the percent error in your analysis of the percent mass of NaClO in your bleach;sample.;Your Percent Error calculated by program: 23 %;Your Score: 1 out of 5 points;23) Concluding Remarks: Briefly discuss interpretations of your observations and results. Include in your;discussion, any conclusions drawn from the results and any sources of error in the experiment.;Your Answer: In this experiment, my lab partner and I used a Text Answers to be Scored by your TA;pair of oxidation reduction reactions determine the;concentration of a bleach unknown. We did this by;standardizing a solution of sodium thiosulfate. We completed;three trials and our average molarity for the thiosulfate was;0.0475 with a standard deviation of 0.003. For the next part of;our experiment, we used our sodium thiosulfate solution to;analyze a bleach unknown. The bleach we used was sodium;hypochlorite (NaClO). Using our three trials, our average mass;percent of NaClO was 4.47 with a standard deviation of 0.26;and a 90% confidence limit of 0.44. By titrating our solutions;we received a percent error of 23. Sources of error in this;experiment could have been that we may have mistakenly;overshot the equivalence point and added more titrant then we;really needed

 

Paper#15924 | Written in 18-Jul-2015

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