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Calculate how much by volume of each concentrated stock solution you will need. Your starting materials are 1M HEPES 1 M midsole and 5 M salt.
![Elution Buffer (50 ml)
20 mM HEPES 7.4
150 mM NaCl
250 mM imidazole](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2e1cb03f-91d2-4991-91c8-37c8965f1bd7%2F86d52bb4-9945-46a6-aca3-42c75a6c4455%2Ftcyhxq_processed.png&w=3840&q=75)
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- An unknown sample containing mixed alkali (NaOH, NaHCO3, or NazCO3) was analyzed using the double flask method. A 250 mg sample was dissolved in 250 mL CO2 free water. A 20.0 mL aliquot of this sample required 11.3 mL of 0.009125 M HCI solution to reach the phenolphthalein end point. Another 29.0 mL aliquot of the sample was titrated to the bromocresol green endpoint using 31.1 mL of the standard acid. How many millimoles of the components are there in the original solid sample? O 0.28 mmol Na2CO3 O 1.29 mmol NAOH, 0.97 mmol Na2CO3 O 01 mmol NaOH, 0.078 mmol NazCO3 O 1.29 mmol NażCO3, 0.97 mmol NaHCO3 O Cannot be determinedhow to find the concentration in mg/100ml btw the concentration of NBS is 0.0806 MA 0.300-g feed sample is analyzed for its protein content by the modified Kjeldahl method. If 25.0 mL of 0.100 M HCl is required for titration, what is the percent protein content of the sample?
- An unknown sample containing mixed alkali (naoh, nahco3, or na2co3) was analyzed using the double flask method. a 250 mg sample was dissolved in 250 ml co2 free water. a 20.0 ml aliquot of this sample required 11.3 ml of 0.009125 m hcl solution to reach the phenolphthalein end point. another 29.0 ml aliquot of the sample was titrated to the bromocresol green endpoint using 31.1 ml of the standard acid. how many millimoles of the components are there in the original solid sample14 12 10 8 + Equivalence point 6. 4 2 20.0 40.0 60.0 80.0 mL of 0.100 M NaOH added наStandardization of 1N H2SO4 Titration 1 Titration 2 Titration 3 Titration 4 Mass of flask and Sodium Carbonate 1.5067 g 1.5098 g 1.5076 g 1.5077 Final buret reading (mL) 22 mL 22mL 21.5 mL 22.5 mL Volume of Sulphuric acid used (mL) 30 mL 30 mL 30 mL 30 mL What is the result of my titration?
- Determine the total hardness of water (ppm CaCO3; MM: 100.09 g/mole) if 100.0mL sample of drinking water required 38.41mL of 4.65 x 10-3 M EDTA for titration. 179 ppm CaCO3 357 ppm CaCO3 452 ppm CaCO3 733 ppm CaCO3We want to make 400 ml of 1 d running buffer from the 20x stock , you will add ——-ml Of 20c stock to ——- ml waterQuantity of Acetic Acid in Salad Dressing Titration Results: 1000 μL aliquots of salad dressing against a sodium hydroxide solution Please only fill in 3 concordant results! Titration number: 1 2 3 Error Final burette reading (mL) 7.9 15 22.2 ±0.05 mL Initial burette reading (mL) 0.9 7.9 17 ±0.05 mL Titre (mL) 7 7.1 7.2 ±0.10 mL Average (mean) volume of concordant titres: 7.1 Concentration of Acetic Acid (M) (to 3 sig figs): 6.88 mL IM Concentration of Acetic Acid (as %w/v) (to 3 sig figs): % Note: %w/v = (g/mL) x 100%
- A 1.43 x 10-4 L solution of 45.7 mg/mL protein was analyzed using the Kjeldahl procedure. After digestion of the protein, the liberated NH3 was collected in 9.80 mL of 0235 M HCl. The unreacted acid required 10.75 mL of a standard solution of 0.0149 M NaOH for complete titration. Calculate the weight percent of nitrogen in the protein. (Molar mass N = 14.00674 g/mol)A local company sent you their green alternative for window cleaner to be tested for percent (w/v) acetic acid content. For your experiment, you first standardized your NaOH titrant with 0.8053 g of (99.80 % purity) KHP. You used 40.60 mL of NaOH for your standardization. After that you then analyzed a 50.00 mL sample and found that you needed 33.20 ml NaOH to reach the end point. Summary of results: Standardization Sample analysis KHP Weight (g) Purity NaOH (mL) used Determine the following: 0.8053 g Volume of sample 50.00 mL 99.80% NaOH (mL) used 33.20 mL 40.60 mL a. Molarity of NaOH b. % (w/v) acetic acidppm Pb^2+ in 10^-6 % Pb (NO3)2