Beakers #: 2 Contents in beakers 10 mL of 0.5 M HC:H;O 8 mL of 0.5 M HCSH302 6 mL of 0.5 M HC5H302 + 2 mL of 0.5 M NaC2H3O2 4 mL of 0.5 M NaC2H3O2 4 mL of 0.5 M HC5H302 + 6 mL of 0.5 M NaC2H302 5 2 mL of 0.5 M HC5H302 + 8 mL of 0.5 M NaC2H302 6 10 mL of 0.5 M NaC2H302 part 1 pH (Buffer 2.66 3.80 4.50 4.85 5.35 7.71 preparations) Part 2pH 1.56 3.23 4.00 4.00 4.83 (Strong Acid (5 drops) (5 drops) (12 drops) (15 drops) (7 drops) 5.90 (5 drops) added) Part 3 pH 3.83 4.38 4.99 5.40 5.98 (Strong Base (5 drops) (5 drops) (10 drops) (10 drops) (5 drops) 12.73 (5 drops) added)

Beakers #: 2 Contents in beakers 10 mL of 0.5 M HC:H;O 8 mL of 0.5 M HCSH302 6 mL of 0.5 M HC5H302 + 2 mL of 0.5 M NaC2H3O2 4 mL of 0.5 M NaC2H3O2 4 mL of 0.5 M HC5H302 + 6 mL of 0.5 M NaC2H302 5 2 mL of 0.5 M HC5H302 + 8 mL of 0.5 M NaC2H302 6 10 mL of 0.5 M NaC2H302 part 1 pH (Buffer 2.66 3.80 4.50 4.85 5.35 7.71 preparations) Part 2pH 1.56 3.23 4.00 4.00 4.83 (Strong Acid (5 drops) (5 drops) (12 drops) (15 drops) (7 drops) 5.90 (5 drops) added) Part 3 pH 3.83 4.38 4.99 5.40 5.98 (Strong Base (5 drops) (5 drops) (10 drops) (10 drops) (5 drops) 12.73 (5 drops) added)

Fundamentals Of Analytical Chemistry

9th Edition

ISBN:9781285640686

Author:Skoog

Publisher:Skoog

Chapter7: Statistical Data Treatment And Evaluation

Section: Chapter Questions

Problem 7.21QAP

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Beaker 0.00200 M Fe(NO3)3, mL 0.00200 M NaSCN, mL total volume, mL 1 3.000 2.000 10.00 2 3.000 3.000 10.00 3 3.000 4.000 10.00 4 3.000 5.000 10.00 5 (blank) 3.000 0.000 10.00 In solutions 1-4 you are adding 3.00ml of 0.00200M Fe3+ to some SCN- solution and dilutiing to a final volume of 10.00ml. What is the new concentration of the Fe3+ due to dilution? Your answer should have 3 sig figs.
Seoyeon needs to prepare a 250 mL solution of 0.1000 M Na2CO3. Which glassware should she use to prepare the solution? 250-mL volumetric flask 250-mL graduated cylinder 250-mL Erlenmeyer flask 250-mL beaker For the given data set: [6.7, 7.3, 7.5, 7.2, 7.9, and 7.1] Determine the outlier using Grubbs test. Gcrit = 1.887 6.7 7.9 7.3 7.1 There is no outlier. The results of an analysis for a certain sample are given (in ppm): 20.1, 24.1, 23.5, 22.2, and 21.6. Calculate the 95% confidence interval. 20.3 ppm ≤ x̄ ≤ 24.3 ppm 20.5 ppm ≤ x̄ ≤ 24.1 ppm 20.8 ppm ≤ x̄ ≤ 24.8 ppm 21.0 ppm ≤ x̄ ≤ 24.6 ppm
a. If 55mg of CaC2O4 is placed in a 250mL pack of H2O nanopure at 25 celcius, how much (in mg) of CaC2O4 remains undesolved? b. If 100mg of CaC2O4 is placed in a 500mL pack of HCl at 25 celcius, how much (in mg) of CaC2O4 remains undesolved? FM: 128g/mol Pksp= 7.90
2 CoCl2(aq) + 8 NH3(aq) + H2O2(aq) + 2 NH4Cl(aq)  2 [Co(NH3)5Cl]Cl2(s) + 2 H2O(l) In a 50-mL Erlenmeyer flask, cobalt(II) chloride hexahydrate (1.075 g), an aqueous solution of ammonia (1.2 mL, 15 M), ammonium chloride (0.505 g), and hydrogen peroxide (1.0 mL, 30.0% (w/w), density = 1.10 g/mL) were combined. After completion of the reaction, [Co(NH3)5Cl]Cl2 (0.450 g) was isolated. What was the percent yield of the product?
he following five beakers, each containing a solution of sodium chloride (NaClNaCl, also known as table salt), were found on a lab shelf: Beaker Contents 1 200. mLmL of 1.50 MM NaClNaCl solution 2 100. mLmL of 3.00 MM NaClNaCl solution 3 150. mLmL of solution containing 25.5 gg of NaClNaCl 4 100. mLmL of solution containing 25.5 gg of NaClNaCl 5 300. mLmL of solution containing 0.450 molmol NaClNaCl Arrange the solutions in order of decreasing concentration. Rank from most concentrated to least concentrated. To rank items as equivalent, overlap them.
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19 g of unknown organic sample was dissolve in 640 mL of Dicloromethane (DCM). The boiling point of benzene was increased by 3.78oC. Determine the molecular weight of the unknown sample? Kb of DCM = 2.42oC/m Bb of benzene = 39.6 oC density of benzene = 1.33 g/mL at 25 °C Round your answer to the nearest whole number, no units required.
Beaker 0.00200 M Fe(NO3)3, mL 0.00200 M NaSCN, mL total volume, mL 1 3.000 2.000 10.00 2 3.000 3.000 10.00 3 3.000 4.000 10.00 4 3.000 5.000 10.00 5 (blank) 3.000 0.000 10.00 In Solutions 1-4 you are adding successively larger volumes of 0.00200M SCN- to the Fe3+ solution and diluting to 10.00 ml. Calculate the final diluted molarity of SCN- in solution #1 Your answer should have 3 sig figs =
Table of caffeine standards concentration . Sample Conc, ppm Std1 16 Std2 32 Std3 48 Std4 64 Std5 80 If the volume used to make 100 mL of std 1 is 2 uL what is the concentration in M used to make std calibration curve ? The standards are going to be used to build calibration curve to analyze caffeine in an energy drink. if 500 mL of the energy drink has target of 400 mg , how will you prepare the sample if you need 10 mL for the analysis ? Caffeine MM=194.19 g/mol.
During lab, a student used a Mohr pipet to add the following solutions into a 25 mL volumetric flask. They calculated the final volumes added, which are recorded below. The student then followed the directions in the manual to make the stock solution. Volumes Used to Create Stock Solution Volume 0.200 M Fe(NO3)3 (mL) Volume 0.00200 M KSCN (mL) Stock Solution 8.99 mL 2.38 mL Using the stock solution above, the student made additional dilutions, with the final volumes below. Calculate the [FeSCN2+] in Standard 2. Report your answer in mM. Volumes Used to Create Standard Solutions Solution Volume Stock solution (mL) Volume Water (mL) Standard 1 7.04 3.47 Standard 2 4.72 5.13 Standard 3 2.71 7.43 Standard 4 1.41 9.14 Note: Reporting your answer in mM is for grading purposes only. It is not necessary during lab. Report your answer to three places after the decimal.
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