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- From a 10-mL sample, a 1-mL aliquot was taken and diluted to 100mL. From this, a 5-mL aliquot was taken and diluted to 20mL.The final 20mL was found to have a concentration of 0.004M analyte X.What is the concentration of analyte X in the 10-mL sample?Standardization of Sodium Thiosulfate Solution Primary Standard used: Potassium Dichromate Formula mass of 1o standard: __________________ % Purity of 1o standard: 99.80% Trials 1 2 3 Weight of K2Cr2O7 (g)Weight of K2Cr2O7 (g) 0.0315 0.0331 0.0380 Final Volume Reading Na2S2O3 (ml) 27.50 27.50 34.90 Initial Volume Reading Na2S2O3 (ml) 1.10 0.00 1.00 Net Volume Na2S2O3 used (ml) Molarity of Na2S2O3How analytes are recovered from SFC and SFE
- what is a chemical reason that a sample could have less than 100% purity? A) Impurity was an acid B) Impurity were not acids C) Overshot end pointshow the complete solution and write answers in 3 significant figures. DYE CALCULATED Rf Tartrazine (yellow #5) ? Sunset yellow (yellow #6) ? Allura red (red #40) ? Brilliant blue (blue #1) ? M&M's color COMPOSITION Orange ? Green ? Blue ? Brown ? CHROMATOGRAM BEFORE: (image 1) CHROMATOGRAM AFTER: (image 2)5.00 mL of stock solution is diluted to 25.00 mL, producing solution ALPHA. 10.00 mL of solution ALPHA is diluted to 25.00 mL, resulting in solution BETA. 10.00 mL of solution BETA is then diluted to 25.00 mL, producing solution GAMMA. dilution factor for ALPHA from stock solution = 0.167 dilution factor for BETA from ALPHA solution = 0.0476 part c and d?
- Convert the amount of spiperone in each sample to the concentration in the 1 ml total volume of assay tube. Express this concemtration in molar in standard form.Data Sheet Extraction ODD EVEN ODD EVEN GROUP Single Extraction 1 Single Extraction 2 Multiple Extraction 1 Multiple Extraction 2 Weight of tea leaves 10.002 (g) 10.123 (g) 10.013 (g) 10.007 (g) Weight of evaporating dish + caffeine 123.689 (g) 124.334 (g) 121.815 (g) 126.523 (g) Weight of empty evaporating dish 123.513 (g) 124.147 (g) 121.614 (g) 126.246 (g) Weight of caffeine Identify the two layers in the procedure. In which part does caffeine can be extracted?A student was given a stock aluminum(III) solution with a concentration of 5.000 parts per million (ppm). (ppm are defined as mg/L for dilute aqueous solutions.) The student prepared five 100 mL standard solutions and an unknown as described in the procedure section of the experiment. The absorbance of each solution was read at 565 nm, using the blank to set zero absorbance. The results are tabulated below. Solution Volume of Al(III) stock solution used (mL) Absorbance 1 8.00 0.701 2 6.00 0.548 3 4.00 0.378 4 2.00 0.208 5 1.00 0.123 Unknown --- 0.534 3. What is your best estimate of the Al(III) concentration of the unknown sample in the cuvet based upon where its absorbance falls on the standard curve?
- 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.A 500.00 mg vitamin C (MW176.12g/mol) tablet was ground, acidified, and dissolved in H2O to make a 250.0 mL solution. A 50.00 mL aliquot containing vitamin C, KI and starch was analyzed and titrated with 12.31 mL of 0.01042 M KIO3. What is the % (w/w) vitamin C in the tablet? KIO3 + 5 KI + 6 H+ = 3 I2 + 6 K+ + 3 H2O C6H8O6 + I2 = C6H6O6 + 2I- + 2H+ A 67.77 % B 33.89 % C 22.59 % D 3.39 %Normality of a solution was determined by five different titrations. The results were 9.5 mL, 8.5 mL, 9.1 mL, 9.3 mL, and 9.1 mL. Calculate the mean, median and standard deviation of this data
