Caffeine eine benze, and aspartame content of mountain dew soda was determined using reverse phase HPLC. The concentration of the standard used are as follows, caffeine: 0.70 mg/mL, benzoate: 1.6 mg/mL, and aspartame 5.0 mg/mL. The following volume of the standards were taken to a 50.0 mL volumetric flask and diluted to the calibration curve:
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- Question 16 Caffeine, benzoate, and aspartame content of mountain dew soda was determined using reverse phase HPLC. The concentration of the standard used are as follows, caffeine: 0.70 mg/mL, benzoate: 1.6 mg/mL, and aspartame 5.0 mg/mL. The following volume of the standards were taken to a 50.0 mL volumetric flask and diluted to the mark to construct a calibration curve: Caffeine Benzoate Aspartame 1 mL 1 mL 1 mL 2 mL 2 mL 2 mL 3 mL 3 mL 3 mL 4 mL 4 mL 4 mL 5 mL 5 mL 5 mL The 12-oz can mountain dew sample were treated as follows: The 12- oz can were has been left out overnight to get rid the carbonation then, 2 mL were drawn into a plastic syringe, filtered, and placed into a vial. An equal amount of deionized water was added. A 100 μL sample were injected into a sample loop using same parameters with the standards. Assume 12 oz 354.9 mL The following data were obtained: Peak Area Benzoate Caffeine Aspartame Std 1 2.30 3.54 3.33 Std 2 4.10 5.59 5.83 Std 3 7.37 9.98 9.63 Std 4 9.56…Question 15 Caffeine, benzoate, and aspartame content of mountain dew soda was determined using reverse phase HPLC. The concentration of the standard used are as follows, caffeine: 0.70 mg/mL, benzoate: 1.6 mg/mL, and aspartame 5.0 mg/mL. The following volume of the standards were taken to a 50.0 mL volumetric flask and diluted to the mark to construct a calibration curve: Caffeine Benzoate Aspartame 1 mL 1 mL 1 mL 2 mL 2 mL 2 mL 3 mL 3 mL 3 mL 4 mL 4 mL 4 mL 5 mL 5 mL 5 mL The 12-oz can mountain dew sample were treated as follows: The 12- oz can were has been left out overnight to get rid the carbonation then, 2 mL were drawn into a plastic syringe, filtered, and placed into a vial. An equal amount of deionized water was added. A 100 µL sample were injected into a sample loop using same parameters with the standards. Assume 12 oz = 354.9 mL The following data were obtained: Peak Area Benzoate Caffeine Aspartame Std 1 2.30 3.54 3.33 Std 2 4.10 5.59 5.83 Std 3 7.37 9.98 9.63 Std 4 9.56…Question 17 Caffeine, benzoate, and aspartame content of mountain dew soda was determined using reverse phase HPLC. The concentration of the standard used are as follows, caffeine: 0.70 mg/mL, benzoate: 1.6 mg/mL, and aspartame 5.0 mg/mL. The following volume of the standards were taken to a 50.0 mL volumetric flask and diluted to the mark to construct a calibration curve: Caffeine Benzoate Aspartame 1 mL 1 mL 1 mL 2 mL 2 mL 2 mL 3 mL 3 mL 3 mL 4 mL 4 mL 4 mL 5 mL 5 mL 5 mL The 12-oz can mountain dew sample were treated as follows: The 12- oz can were has been left out overnight to get rid the carbonation then, 2 mL were drawn into a plastic syringe, filtered, and placed into a vial. An equal amount of deionized water was added. A 100 μL sample were injected into sample loop using same parameters with the standards. Assume 12 oz = 354.9 mL The following data were obtained: Caffeine Peak Area Benzoate 3.54 Aspartame Std 1 2.30 3.33 Std 2 4.10 5.59 5.83 Std 3 7.37 9.98 9.63 Std 4 9.56…
- Question 12 Caffeine, benzoate, and aspartame content of mountain dew soda was determined using reverse phase HPLC. The concentration of the standard used are as follows, caffeine: 0.70 mg/mL, benzoate: 1.6 mg/mL, and aspartame 5.0 mg/mL. The following volume of the standards were taken to a 50.0 mL volumetric flask and diluted to the mark to construct a calibration curve: Caffeine Benzoate Aspartame 1 mL 1 mL 1 mL 2 mL 2 mL 2 mL 3 mL 3 mL 3 mL 4 mL 4 mL 4 mL 5 mL 5 mL 5 mL The 12-oz can mountain dew sample were treated as follows: The 12- oz can were has been left out overnight to get rid the carbonation then, 2 mL were drawn into a plastic syringe, filtered, and placed into a vial. An equal amount of deionized water was added. A 100 μL sample were injected into a sample loop using same parameters with the standards. Assume 12 oz = 354.9 mL The following data were obtained: Peak Area Benzoate Caffeine Aspartame Std 1 2.30 3.54 3.33 Std 2 4.10 5.59 5.83 Std 3 7.37 9.98 9.63 Std 4 9.56…Question 14 Caffeine, benzoate, and aspartame content of mountain dew soda was determined using reverse phase HPLC. The concentration of the standard used are as follows, caffeine: 0.70 mg/mL, benzoate: 1.6 mg/mL, and aspartame 5.0 mg/mL. The following volume of the standards were taken to a 50.0 mL volumetric flask and diluted the mark to construct a calibration curve: Caffeine Benzoate Aspartame 1 mL 1 mL 1 mL 2 mL 2 mL 2 mL 3 mL 3 mL 3 mL 4 mL 4 mL 4 mL 5 mL 5 mL 5 mL The 12-oz can mountain dew sample were treated as follows: The 12- oz can were has been left out overnight to get rid the carbonation then, 2 mL were drawn into a plastic syringe, filtered, and placed into a vial. An equal amount of deionized water was added. A 100 μL sample were injected into a sample loop using same parameters with the standards. Assume 12 oz = 354.9 mL The following data were obtained: Caffeine Peak Area Benzoate 3.54 Aspartame Std 1 2.30 3.33 Std 2 4.10 5.59 5.83 Std 3 7.37 9.98 9.63 Std 4 9.56…3:36 1 ull LTE O Send a chat What is the number of moles of glucose (C,H,206) in 0.500 of a 0.40 M solution? STARTING AMOUNT ADD FACTOR ANSWER RESET *( ) 36.04 0.020 1 0.40 0.20 1.3 180.18 0.500 6.022 x 1023 * Reflect in ePortfolio
- Sheet 5 From the following data determine the following: Define outlier and then identify the outlier in the data. Determine if the outlier should be kept of discarded from the data. Trial 1 2 3 4 HCl (mL) 22.3 28.4 29.8 29.3 NaOH (mL) 35.2 35.1 34.9 35.022:12 Question 23 of 30 Submit How many grams of a solution that is 5.2 % sucrose by mass are needed to obtain 18 g of sucrose X STARTING AMOUNT ADD FACTOR ANSWER RESET *( ) 18 100 g solution 3.5 6.022 x 1023 %m sucrose 0.052 5.2 g sucrose 94 350 0.0094 0.94 1 Tap here or pull up for additional resourcesTable 1: Volumes in of iodine in mL used to titrate mixtures of 10 mL of reaction mix and 10 mL of urine. student trial 1 trial 2 trial 3 total volume (mL) a 12.6 11.8 12.5 300 290 b. 10.6 10.7 10.6 11.1 118 12. 10.9 405 13.0 12.5 349 mg vit. C mL iodine TV (mL/hour) 10 mL urine aliquot TVC (mg/hour) =7 (mL) · CF !! What is the hourly vitamin C excretion rate (mg/h) of student b if the CF is 1.06? (include no more than two decimals in your answer, e.g.: 123.45, and do not type the units) 10.6 need ASAP
- Should be 3 answers in total. Will thumbs up! Next the student conducts two different runs: In RUN 1, 10.0 mL of 5.00 M NAOH is combined with 10.0 mL of 0.010 M CO and 80.0 mL of wate In RUN 2, 20.0 mL of 5.00 M NAOH is combined with 10.0 mL of 0.010 M CO and 70.0 mL of wate The following plots were obtained for each of the runs: [crystal orange] vs time (RUN 1) In[crystal orange] vs time (RUN 1) 0.0012 0.001 -2 y--0.4463x- 7.3334 y =-0.0002x + 0.0007 R- 0.6565 R = 0.893 0.0008 4 0.0006 -6 0.0004 -8 0.0002 0.0002 -12 time (hr) time (hr) 1/[crystal orange] vs time (RUN 1) 1/[crystal orange] vs time (RUN 2) 12000 25000 10000 20000 8000 15000 6000 y = 2000x + 1000 R =1 10000 y 4000x + 1000 R =1 4000 2000 5000 time (hr) time (hr) Figure 4. Kinetic traces for the reaction between crystal orange and sodium hydroxide. What is the order with respect to CO? | Select ] What is the order with respect to NaOH? | Select ] What is the value of the averaged rate constant for the reaction above? Give…✓ Question Completion Status: Shown below is a calibration curve for the analysis of theobromine and caffeine in chocholate. if an unknown sample has a caffiene peak height of 3.50 cm what is the concentration of caffeine in the sample in ppm? 20 Peak height (centimeters) 15 10 5 0 Theobromine y = 0.1977x-0.2104 25 Unknown Figure 0-7 Gumantative Chemical Analysis, Seventh Edition 2007 Freeman and Company Caffeine y = 0.088 4x-0.030 3 50 Analyte concentration (parts per million) 100 M ED Feb 2 8:3The sample containing 4.2 % NaCl will be determined gravimetrically as AgCl. What weight (in milligrams) of the sample must be taken to obtain a precipitate that will weigh about 0.39 mg? FM: NaCl = 58.44 AgCl = 143.3 %3D Answer in two decimal places. Do not include units in the answer. Answer: