Solve for the dissolved oxygen content of the water sample in Day 1 if the Na2S2O3 titrant (0.1152 M) consumed were 10.28 mL, 9.84 mL and 10.02 mL. The volume of the water sample obtained is 30 mL. Assume that the procedure for BOD determination experiment done here is same to our class experiment. 306 ppm 316 ppm 312 ppm 308 ppm
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Solve for the dissolved oxygen content of the water sample in Day 1 if the Na2S2O3 titrant (0.1152 M) consumed were 10.28 mL, 9.84 mL and 10.02 mL. The volume of the water sample obtained is 30 mL. Assume that the procedure for BOD determination experiment done here is same to our class experiment.
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306 ppm
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316 ppm
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312 ppm
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308 ppm
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- MISSED THIS? Read Section 18.6 (Pages 823-826); Watch IWE 18.12. Determine the minimum concentration of the precipitating agent on the right to cause precipitation of the cation from the solution on the left. 9.0x10-2 M Cala; K2SO4 Express your answer using two significant figures. [K₂SO4]= ΜΕ ΑΣΦ Submit Request Answer Part C 1.7x103 M AgNO3; RbCl Express your answer using two significant figures. [RbCl] ΜΕ ΑΣΦ Submit Request Answer M MTable 2. Titration data Trial 1 Trial 2 Trial 3 Initial burette reading (mL) 1.19 2.26 2.39 Molarity of NaOH (M) 0.100 0.100 0.100 Volume of vinegar sample (mL) 5.00 5.00 5.00 Final burette reading (mL) 48.55 49.43 49.99 Expected color at end point Volume of NaOH used (mL) Average Volume of NaOH used in liters Average moles of NaOH used (mol) Average moles of acetic acid (mol) Average molarity of acetic acid (M) Average mass of acetic acid (g) Average mass of vinegar (g) (assume the density of vinegar is 1.00 g/mL) Average mass % of acetic acid in vinegar Known mass % of acetic acid in vinegar is 5.45% Percent Error2A student used a pipette to add 25 cm³ of sodium hydroxide of unknown concentration to a conical flask. The student carried out a titration to find out the volume of 0.1 mol/dm3 sulfuric acid needed to neutralise the sodium hydroxide. The following is the table of results obtained. Titration Volume of sulfuric acid used/cm³ 1 27.85 27.30 27.25 27.10 27.30 a. Write down a balanced equation for the reaction between sodium hydroxide and sulfuric acid. b. Describe the titration experimental procedure (details of washings are expected) c. Determine the average titre value and explain your choice of titre values in calculating the average. d. Determine the concentration of sodium hydroxide. e. If instead of sulfuric acid, hydrochloric acid of the same concentration was used what would you expect the average titre value to be and why?
- Data for titration of 15.00 mL of vinegar with approximately 1.0 M NaOH. Note you must use the exact concentration of the standard NaOH solution to calculate the moles of NaOH. Quantity Еxample Trial 1 Trial 2 Trial 3 M NaOH (exact concentration) 0.992 M 0.953 M 0.953 M 0.953 M V initial buret reading = V, 0.20 mL 1.20 mL 0.52 mL 0.15 mL V final buret reading = V, 12.90 mL 14.22 mL 13.71 mL 13.31 mL %3D Vep = VNAOH added = V; - V, 12.70 mL 13.02 mL 13.19 mL 13.16 mL Vep = VNAOH in L 0.01270 L 0.01302 L 0.01319 L 0.01316 L moles NaOH = 0.0126 mol 0.0124 mol 0.0126 mol 0.0125 mol MNAOH X VNAOH moles AA = 0.0126 mol 0.0124 mol 0.0126 mol 0.0125 mol moles NaOH V sample = V acid 0.0150 L 0.0150 L 0.0150 L 0.0150 L (15.00 mL) (15.00 mL) (15.00 mL) Actual molarity 0.840 M 0.827 M 0.840 M 0.833 M of AA Average molarity for 3 trials => of AA 0.833 M 1. Report the acetic acid (AA) concentration of vinegar in units of molarity (M). 2. Use your average AA concentration to calculate the mass…Solve for the dissolved oxygen content of the water sample in Day 1 if the Na₂S2O3 titrant (0.1152 M) consumed were 10.28 mL, 9.84 mL and 10.02 mL. The volume of the water sample obtained is 30 mL. Assume that the procedure for BOD determination experiment done here is same to our class experiment. A) 316 ppm B) 312 ppm с 308 ppm (D) 306 ppmIdentify which of the graphs represent the given titration experiments best. Graph A Graph B 12 12 11- 11- 10- 10- 9- 9- 8- 8- 7- 7- 6- 6- 5- 5- 4- 3- 3- 2- 2 1 10 15 20 25 30 35 40 45 50 10 15 20 25 30 35 40 45 50 Titrant.volume Cml). Titrant.volume (mL). Answer Bank This graph does not represent a titration. a basic solution titrated with an acid an acidic solution titrated with a base Hd Hd
- Saved CHEMY 101 SEC SEM T2 2020-21 The amount of water which should be added to 0.700 M NAOH solution to make 100 mL 0.250 M solution is 19 Multiple Choice 01:25:58 79.6 mL 35.7 mL 55.2 mL 64.3 mL raw K Prev 19 of 20 Next >Solve for the dissolved oxygen content of the water sample in Day 1 if the Na₂S₂O3 titrant (0.1152 M) consumed were 10.28 mL, 9.84 mL and 10.02 mL. The volume of the water sample obtained is 30 mL. Assume that the procedure for BOD determination experiment done here is same to our class experiment. A) 312 ppm B) 316 ppm 308 ppm D) 306 ppmVolume vinegar used (mL)= 5.00 mL % from labe= l5 % Molarity of NaOH = 0.0976 M Titration volume measurements Trial 1 Trial 2 Trial 3 Inital volume NaOH (mL) 0.69 0.86 0.16 Final volume NaOH (mL) 43.47 43.36 43.03 Volume NaOH used (mL) Average volume NaOH used (mL) Average volume NaOH used (L) Moles NaOH used (from avg volume) Moles CH3COOH Molarity CH3COOH (mol/L) Mass CH3COOH (g) % (m/v) CH3COOH in vinegar
- Trial 1 Trial 2 Trial 3 Initial burette reading (mL) 2.29 1.41 1.95 Molarity of NaOH (M) 0.100 0.100 0.100 Volume of vinegar sample (mL) 5.00 5.00 5.00 Final burette reading (mL) 50.37 49.39 49.84 Table 2. Titration data Trial 1 Trial 2 Trial 3 Initial burette reading (mL) 2.29 1.41 1.95 Molarity of NaOH (M) 0.100 0.100 0.100 Volume of vinegar sample (mL) 5.00 5.00 5.00 Final burette reading (mL) 50.37 49.39 49.84 Expected color at end point Volume of NaOH used (mL) 48.08 47.98 47.89 Compute for the ff: a. Average moles of acetic acid (mol)? b. Average molarity of acetic acid (M)? c. Average molarity of acetic acid (M)?Preparation and Standardization of KMnO4 solution Experimental data Complete the table below. Trial 1 0.2001 g Trial 2 0.2065 g Trial 3 Weight of sodium oxalate (Na2C2O4, MM= 134 g/mol) Titration data 0.2050 g Final reading Initial reading 29.86 mL 0.00 mL 30.66 mL 30.52 mL 0.00 mL 0.00 mL Total vol. of KMNO4 used Computed Molarity of KMNO4 solution Mean Molarity Computed Normality of KMNO4 Mean Normality of KMNO4 solution Reaction Involved: Calculations:Part A: Standardization of a Sodium Hydroxide Solution Titration 1 Titration 2 Titration 3 Mass of 125 mL flask 45.849g 46.715g 44.953g Mass of flask and KHP 46.849g 47.745g 46.003g Initial buret reading (mL) 0.5 ml 0.5 ml 0.5 ml Final buret reading (mL) 27.8 ml 26.5 ml 26.7 ml Volume of NaOH used (mL) 45.11 ml 45.06 ml 45.14 ml Calculations Titration 1 Titration 2 Titration 3 Moles of KHP Moles of NaOH Molarity of NaOH Average Molarity of NaOH: _______________