I have 1 gram of sample, and add 0.1 mL of a 50 ug/mL solution of atrazine. I extract with 10 mL of methanol. I take a1 mL aliquot of the methanol extract, dilute with 10 mL water, and add it to a C-18 SPE column. I wash with 10 mL10% methanol in water, and elute with 2 mL methanol. I take a 1.0 mL aliquot and add 100 ul of simazine internalstandard (5 µg/mL).Since I am lazy, I decide to do a single point calibration. I take 1.0 mL of a 1 µg/mL atrazine standard and add 100 µμlsimazine (5 ug/mL).I inject 10 µL of the standard, spike and check on the LC/MS using positive mode ESI, and get these areas:AtrazineSimazineStandard1000100Spike20050Check100200What was the atrazine spike recovery?

Ans. Spike Recovery or spike and recovery is a method used to determine that whether an analyte…

I have 1 gram of sample, and add 0.1 mL of a 50 ug/mL solution of atrazine. I extract with 10 mL of methanol. I take a 1 mL aliquot of the methanol extract, dilute with 10 mL water, and add it to a C-18 SPE column. I wash with 10 mL 10% methanol in water, and elute with 2 mL methanol. I take a 1.0 mL aliquot and add 100 ul of simazine internal standard (5 ug/mL). Since I am lazy, I decide to do a single point calibration. I take 1.0 mL of a 1 ug/mL atrazine standard and add 100 μl simazine (5 ug/mL). I inject 10 µL of the standard, spike and check on the LC/MS using positive mode ESI, and get these areas: Atrazine Simazine Standard 1000 100 Spike 200 50 Check 100 200 What was the atrazine spike recovery?

I have 1 gram of sample, and add 0.1 mL of a 50 ug/mL solution of atrazine. I extract with 10 mL of methanol. I take a 1 mL aliquot of the methanol extract, dilute with 10 mL water, and add it to a C-18 SPE column. I wash with 10 mL 10% methanol in water, and elute with 2 mL methanol. I take a 1.0 mL aliquot and add 100 ul of simazine internal standard (5 ug/mL). Since I am lazy, I decide to do a single point calibration. I take 1.0 mL of a 1 ug/mL atrazine standard and add 100 μl simazine (5 ug/mL). I inject 10 µL of the standard, spike and check on the LC/MS using positive mode ESI, and get these areas: Atrazine Simazine Standard 1000 100 Spike 200 50 Check 100 200 What was the atrazine spike recovery?

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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To quantify the amount of ethanol in a beverage, a technician prepared the first solution which contained 4.90 grams unknown, 1.00 grams acetonitrile, and 4.10 grams water. After the GC analysis which reported the peak areas of 5,742 for ethanol and 3,829 for acetonitrile, the technician added 0.20 grams ethanol to the first solution and ran the GC analysis again. This time, the peak areas of 6,382 for ethanol and 3,709 for acetonitrile for solution 2 were obtained. What was the percent by mass ethanol in the original beverage?
Prior to fat analysis, two samples: Sample 1 and Sample 2 with the initial weight for each sample=2.50g. After pre-dried, samples were grinded and reweighted; s1-2.15g and s2=2.11g. Samples were placed in dried glass wool in each thimble and placed extractor for 4-5 hours. Samples were then dried and cooled in desiccators before another weight recorded. The weight for s1=93.51g and for s2=93.78g #Weight of round bottom flask (with glass wool+ thimble) (g) = 92.00 #Weight of round bottom flask (g) =91.00 1) Calculate the percentage of dried fat in s1 2) Calculate the percentage of dried fat in s2. 3) Sample 3 have dried fats of 50%. Identify the correct sequence from the highest to the lowest % of dried fats in sample.
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 %
Six 50.0 ml volumetric flask 1 through 6 are labeled. 10.00 ml of 2.00 x 10-1 M Fe(NO3)3 solution was then pipet into each volumetric flask. A 1.00,2.00,3.00,4.00 and 5.00ml of 2.00 x 10-3M NaSCN solution was then pipetted to flask 2 through 6 respectively. A sufficient amount of 0.10 M nitric acid was then added to each flask to make each a total volume of 50.00ml. Hint M1V1=M2V2 a)what would be the concentration of Fe(NO3)3 in 0.10M HNO3? b)what would be the concentration of NaSCN in 0.10M HNO3?
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A 1 mL microbial suspension comparable to 1 mL of 1.0 McFarland standard was subjected to a series of dilutions. An aliquot of 0.2 mL was transferred into a 100-mL volumetric flask and diluted with sterile saline solution to its final volume. Then a 0.1 mL of the resulting solution was transferred into a 50-mL volumetric flask diluted to its final volume. What would be the final concentration?
25 ml of pickled brine was taken from an apple pickled with vinegar and diluted to 500 ml, 25 ml of this was taken and acidity was determined with 0.2 N (Factor: 1.05) NaOH. In the main trials, 6.3 ml - 6.5 ml - 6.9 ml, respectively, 0.1 ml base was used in the analysis without using the sample.For example, how many ppm is its acidity?
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Rinse and fill the buret with 0.100 M HCl solution. Transfer accurately 25.0 mL of saturated calcium hydroxide solution (2g of calcium hydroxide per 100 ml of water) into two separate E-flasks. Add 2 drops of phenolphthalein to each flask and titrate each solution until the pink color disappears. Voume of HCl used: Trial 1= 6.80ml; Trial 2 = 6.30mL Determination of solubility and Ksp of calcium hydroxide 2. Compute for the moles of H+ used and the moles of OH- present. moles of H+ used = (concentration of HCl) × (volume of HCl used)moles of OH- = moles of H+ used 3. Construct an ICE table for the reaction.4. Calculate the molar solubility (in mol/L) of OH- and Ca2+.5. Determine the solubility of Ca(OH)2 in g/L. (MM of Ca(OH)2 = 74.096 g/mol)6. Calculate the Ksp of Ca(OH)2.7. Compute for the percent error of the experimental value for Ksp of Ca(OH)2 with the literature value.(Ksp of Ca(OH)2 = 8.0 × 10-6 at 25°C)
this is the gc from an experiment we did for the extraction of eugenol from clove oil. i just need the chemical compound associated with each peak labeled. thank you. A flowchart of the experiment is included as well. Thank you.
ASSAY Calibration solutions Calibration solutions of naproxen in the range 5 – 25 μg/mL were prepared. Sample preparation 20 tablets weighing 12.3819 g were crushed to a fine powder. A portion of the powder (145.4 mg) was shaken with approximately 150 mL of acetic acid (0.05 M) for 5 min and then made up to volume in a 250 mL volumetric flask (stock solution). Approximately 50 mL of the stock solution was filtered and a 25 mL aliquot was diluted to 100 mL in a volumetric flask. 10 mL of the resulting solution was further diluted to 100 mL with acetic acid (0.05 M). Analysis The standards and sample solutions were analyzed by HPLC under the following conditions: Column: octadecylsilyl (ODS), 4.6 mmx150 mm, mobile phase: acetonitrile : 0.05 M acetic acid (85:15), flow rate: 1 ml/min, UV detection at 243 nm. Results: A calibration curve of concentration versus peak area was constructed for the standard solutions and gave the straight-line equation: y = 3555.6x + 85, r = 0.9999 The area…
Preparation of 0.25N NaOH Objective: Apparatus and Materials: Procedure : Calculate the exact mass of NaOH needed to prepare 1.0L of 0.25N NaOH for standardization. Show your computations. Using the triple beam balance, weigh the desired amount of NaOH. Place the weighed amount of NaOH into the 500 mL volumetric flask and add distilled water until it reaches the mark. Agitate and transfer the solution into the reagent bottle. Add again distilled water to the same volumetric flask until it reaches the mark and pour it to the same reagent bottle, thus having a total volume of 1.0 L. Label the reagent bottle, 0.25N NaOH. Data and Computations: Conclusion: