4) The following reaction in solution has been studied: Reagents + OH- à products reactivos + OH → productos t (min) V (ml) 120 240 420 600 900 1440 11,45 9,63 8,11 6,22 4,79 2,97 1,44 Aliquots (10 mL), whose reaction was interrupted, were extracted at different times and titrated with 0.01 M HCl. Check that the reaction order with respect to OH- is 1 and determine the rate constant.
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- 6-20. A 4.476-g sample of a petroleum product was burned in a tube furnace, and the SO, produced was col- lected in 3% H,O2. Reaction: SO-(g) + H;O, → H,SO4 A 25.00-mL portion of 0.00923 M NaOH was intro- duced into the solution of H,SO,, following which the excess base was back-titrated with 13.33 mL of 0.01007 M HCI. Calculate the parts per million of sulfur in the sample.The Merck Index indicates that 10 mg of guanidine,CH5N3, may be administered for each kilogram of bodymass in the treatment of myasthenia gravis. The nitrogenin a 4-tablet sample that had a total mass of 7.50 g wasconverted to ammonia by a Kjeldahl digestion, followedby distillation into 100.0 mL of 0.1750 M HCl. Theanalysis was completed by titrating the excess acid with11.37 mL of 0.1080 M NaOH. How many of thesetablets represent a proper dose for a patient who weighs(a) 100 lb, (b) 150 lb, and (c) 275 lb?In this double titration experiment determining composition of soda ash, you are given the following data: M HCl titrant: 0.0501 M +/- 0.0507 Mass of soda ash sample, g: 0.2505 ± 2.0e-4 (Analytical Balance) Total volume of Sample Stock, mL: 75 ± 0.05 (100mL graduated cylinder) Volume of Sample Stock Aliquot, mL: 25 ± 0.3 Find the missing information. Show solutions
- The ethyl acetate concentration in an alcoholic solution was determined by diluting a 10.00-mL sample to 100.00 mL. A 20.00-mL portion of the diluted solution was refluxed with 40.00 mL of 0.04672 M KOH:CH3COOC2H5 + OH- → CH3COO- + C2H5OHAfter cooling, the excess OH2 was back-titrated with 3.41 mL of 0.05042 M H2SO4. Calculate theamount of ethyl acetate (88.11 g/mol) in the original sample in grams7-11. Limestone consists mainly of the mineral calcite, CaCO3. The carbonate content of 0.541 3 g of powdered limestone was measured by suspending the powder in water, adding 10.00 mL of 1.396 M HCI, and heating to dissolve the solid and expel CO2: CACO3 (s) +2H* → Ca²* +CO,T +H2O Calcium carbonate FM 100.086 The excess acid required 39.96 mL of 0.100 4 M NaOH for complete titration to a phenolphthalein end point. Find the weight percent of calcite in the limestone.M 23 (H) Inorganic Analytical Chemistry pard My courses CHEM 23 (H) FS-AY:20-21 Week 4-5: Quantitative Analysis by Gravimetry and Titration Quiz 4 Qu 10 The 500.0-mg sample of impure Na2 CO3 (FM: 105.96) required 22.00 mL of the HCI standard solution (the obtained concentration of HCI from the previous question, problem 5: Titrimetry short answer type) for complete neutralization. ut of 1. Give the balanced chemical equation between the sample, Na2CO3 and the standard, HCI. 2. What is the mass (g) of Na, CO3 in the sample? 3. Calculate the % (w/w) purity of a 500.0-mg sample of impure Na2 CO3 Note: Type your solution here or upload its pic in Jpeg or pdf format. Another option is to send it to me via email (subject: Fam name-Quiz 4) or messenger, privately. Do not forget to box the final answer and write your name. В I 1.
- UEid- base tit ration the Consentrat ion of ammonia Solution excess HCL was titrated ith o.07om Na, cO3 Solutian. The volume of Na, Coz Solution reguired was 15.50 ml using the following eguations: 2HCLem> + Na, COgcaps CO2 Cays + HzO1) t NaCL Calculate the concentration (molarity and ppm) of ammonia in the Original Solution.(MM of NHzis 17.031.9/4) 10.0 was by of o.10OMI HCE to Then10-27. A 10.231-g sample of window cleaner containing ammonia was diluted with 39.466 g of water. Then 4.373 g of solution were titrated with 14.22 mL of 0.106 3 M HCl to reach a bromocresol green end point. (a) What fraction of the 10.231-g sample of window cleaner is contained in the 4.373 g that were analyzed? (b) How many grams of NH3 (FM 17.031) were in the 4.373-g sample? (c) Find the weight percent of NH3 in the cleaner.1. Ph ОН Н Ill OH Me 0.1 equiv TSOH CHCl3, rt, 30 min 0.1 equiv TsOH CHCl3, reflux, 30 min
- ALL INFORMATION NEEDED IS PROVIDED, I NEED FIRST QUESTIONS ANSWERED FIRST BEFORE OTHERS CAN BE ANSWEREDRate = -(Delta[RCl])/Delta t andln[RCl]t=-k*t+ln[RCl]0 < [R-Cl]-time equation Test tube A = 0.30 of 0.10M NaOH(aq), 6.70ml DI H2O & 1 drop of bromophenol bue Test Tube B = 3.00M of 0.10M R-CL in acetone. Mix together in 3 water baths Bath A Trial 1= 20C, 53.6s Trial 2= 20C, 49s for color to changeBath B Trial 1= 32C, 18s Trail 2= 32C, 20s Bath C Trial 1= 10C, 204s Trial B= 10C, 200s average out both trials for each bathcolor change happens after 10% of R-Cl has reacted so In terms of R-Cl this means that 10% of the R-Cl has reacted and ____ of the R-Cl is left Q1. at the time of the color change [R-Cl]t=____[R-Cl]o (Fill in the blank) Q4. Substitute the value of [R-Cl]t into appropriate equation listed above and solve for k in terms of [R-Cl] and t. Q5. Use average time for each temp and determine the value of k (with units) at each temperature Q6. Rearrange the Arrhenius…5. A 300.0 mg sample containing Na,CO3, NaHCO3 and NaOH and inert material either alone or in some combination was dissolved and titrated with 0.1000 M HCI the titration required 24.41 mL to reach the phenolphthalein endpoint. And an additional 8.67 mL to reach the methyl red endpoint. Determine the composition of the sample and calculate the percent of each titrated component.100 ml g 0:4 Nof H sq mixed aiith Final Normaliliy og Result toletian is 100m) op 02 N NEOH, Then A) 0.1 B) 0-2 c) 0 3 D) 0.4