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- 6 Concanavalin (ConA) is a 25.5KDa protein with pI of 4.7 and optical absorbance (A 0.1% 289) of 1.14. calculate the concentration of an aqueous solution of pure BSA protein with an absorbance of 0.66 at 280nm. give your answer in units of mg/ml ?How many grams of glucose (C6H2O6 molecular mass =180daltons) would be present in one liter of a 1M (molar) solution of glucose?The concentration of cholesterol (C27H46O) in blood is approximately 5.0 mM. How many grams of cholesterol are in 250 mL of blood?
- A solution with a density of 0.876 g>mL contains 5.0 g of toluene 1C7H82 and 225 g of benzene. Calculate the molarity of the solution.Glucose-1-phosphate has a ΔG°′ value of −20.9 kJ/mol, whereas that for glucose-6-phosphate is −12.5 kJ/mol. After reviewing the molecular structures of these compounds, explain why there is such a difference in these values.The much-abused drug cocaine is an alkaloid. Alkaloids are noted for their bitter taste, an indication of their basic properties. Cocaine, C17H21O4N, is soluble in water to the extent of 0.17g/100mL solution, and a saturated solution has a pH = 10.08. What is the value of Kb for cocaine?
- Propanamide and methyl acetate have about the same molar mass, both are quite soluble in water, and yet the boiling point of propanamide is 486 K, whereas that of methyl acetate is 330 K. Explain.The-carotene molecule has λmax 450 nm and ɛ = 15.000 m2 / mol. Calculate the expected absorbance value for a solution in which 0.1 mg is dissolved in 10 mL of water. ( β-carotene, C40H56, 536 g/mol path length 1 cm)Given the titration curve of the hypothetical polyprotic acid X at 0.100 M concentration (pKa1=4.0, pKa2=8.0, pKa3=12.0) titrated with 0.600 M NaOH, identify the pH at point C, H, E, and M.
- Beyond the differences in molecular mass, what is different about determining the molarity of a solution containing a salt (e.g. NaCl, MgCl2) versus a covalently linked molecule (e.g. glucose)?Explain how you would prepare 100ml of 10mMProline solution given that the molecular mass of proline= 115.13 g/molpH 3.5 4.5 5.5 6.5 7.5 8.5 Absorbance 0.098 0.027 0.068 0.028 0.032 0.054 Concentration in diluted supernatant (mg/ml) 0.196 0.054 0.136 0.0056 0.0064 0.0108 Concentration in undiluted supernatant (mg/ml) 0.98 0.27 0.68 0.28 0.32 0.54 Formula for the amount (g) of soluble proteins in the soy flour extract: In 15 ml of soy flour extract (with 1/50 dilution), Soluble protein (g) = (C x V x F) / 1000 C = concentration (mg/ml); F = dilution factor; V = volume of solution (ml) Calculate the % solubility of protein in the soy flour (Assume the soy flour contains 35% protein (w/w))