Q: 5. Draw a structural example of a primary, secondary and tertiary alcohol.
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Q: 9. For the reaction: 2 CH3OH(g) + H2(g) CaH«(g) + 2 H;O(g) a) Calculate AG° for this reaction using…
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Q: D. Enthalpy is negligible B. Enthalpy change is negative 7. Which represents an exothermic reaction?…
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Q: An analytical chemist is titrating 206.5 mL of a 0.9600M solution of cyanic acid (HCNO) with a…
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Q: e formula of ionic compounds
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A: Polar molecule: Polar molecules will form when two atoms don't share electrons equally in a covalent…
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Q: -tBuo O-Bu H,N. HO, + PBUOH + CO, NEt, CH,CI, Ph
A: Note : Boc protection of amines.
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A: Major organic product of the given reaction = ?
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Q: Predict the product of each reaction below and indicate if the mechanism is likely to be SN1, Sx2,…
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Q: the voltage of the cell is 0.09223 V at 25.0°C when [Cd 2+( aq)] = 0.1036 M and the pressure of H 2…
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Q: 1. PBr, 2. NaNg, THF HO 3. Pd/C, H2 (Excess) 4. Mel (Excess) 5. Ag,0, H20 `NH2 NH2 A B D
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- In order to improve the peptide separation by using a HPLC system, trifluoroacetic acid acts as mobile-phase modifier was added during the preparation of mobile phase. The preparation was performed by a postgraduate student as following:“2.851 g trifluoroacetic acid (MW: 114.02 g/mol) was made up to 500 cm3 in a graduated flask. To this solution, 50 cm3 of ethanol was added, and after mixing the mobile phase was placed in the solvent reservoir and pumping was commenced at 1.5 cm3 min-1.”Based on the given preparation procedure, identify THREE mistakes that were made.A 0.0200 gram blood sample was decomposed by a microwave digestion technique followed by dilution to 100.0 mL in a volumetric flask. Aliquots of the sample solution were treated with a lead complexing reagent and water as follows: Solution 1: 10.0 ml blood sample + 20.0 mL complexing agent + 30.0 mL H20. Solution 2: 10.0 ml blood sample + 20.0 mL complexing agent + 26.0 mL H20 + 4.00 mL of 78 ppb Pb2+ standard. The resulting solutions were analyzed by UV/Vis at 375 nm. Absorbance for solution 1 = 0.155 and for solution 2 = 0.216. Calculate the concentration of lead (ppb) in the original sample.analyte concentration(C)(mg/ml) injection volume (ul) elution time (time) peak DAD signal(mAU) caffeine 1 1 4.67 302.85 aspartame 5 1 7.53 15.83 benzoic acid 1 1 8.14 89.98 saccharin 1 1 1.91 84.86 mixture(add everything above with 1:1:1:1 ratio) 1 4.47 69.58 How to get the concentration of the mixture in this case?
- The following is an HPLC separation of deuterated benzenes on a C18 column that 440 cm long. The mobile phase was 30% acetonitrile:70% water at 30oC. What is the mobile phase velocity? Find the retention factor k' for C6D6. Find the plate number and plate height for C6D6.. Assuming that the peak widths for C6H5D and C6H6 are the same as that of C6D6., find the resolution for C6H5D and C6H6. Retention times for C6H5D and C6H6 are 193.3 and 194.3 respectively, Find the relative retention time and unadjusted relative retention time between C6H5D and C6H6.Column Plate Number: 16Retention Factor: 3.1Sample Peak Retention Time: 1.2minutesSample Peak Width: 1.53First Peak Retention Factor: 2.10Second Peak Retention Factor: 1.37Efficiency: 0.43Pressure: 375barrColumn: 28cm x 5mmViscosity: 2.3poiseFlow Rate: 1.2Column Permeability: 2.8 FIND: a & ba. Resolution b. Pressure if flow rate is adjusted to 2.3 c. Particle Size d. Selectivity Factor e. The retention time of the unretained peakYou obtained the following raw data when setting up a Bradford standard curve: BSA (mg/ml) Absorbancy 595nm 0 0.225 1 0.310 2 0.420 3 0.510 4 0.610 5 0.720 6 0.810 7 0.915 8 0.950 9 0.980 10 0.990 After blanking against a bradford-dH2O sample, the protein concentration of an unknown sample was determined using the same method and an absorbancy of 0.523 was obtained. Set up a standard curve, excluding outliers (experimental and statistical) and determine the protein concentration in the unknown sample in mg / ml (up to 3 significant figures).
- PART A: Substance A was found to have a retention times of 3.15 minutes on a 70.00-cm column. An unretained species passed through the column in 0.50 minutes. The peak width (at base) for A was 0.60 minutes. Using this information, determine the following: 1. What is the retention factor for analyte A? 2. What is the plate count for this column? 3. What is the height equivalent of a theoretical plate? PART B: Match each statement with one of these terms. A Partition coefficient B Mobile phase C Elution D Longitudinal diffusion E Stationary phase F Eddy diffusion G Retention factor H Plate number I Dead time J Retention time K Selectivity factor L Column resolutionWhat is the retention factor of a sample when after injectiion the void time (tM) took 4.18 sec? The retention time (TR) is 21.19 secFT-IR technique can be utilized for the analysis of unknown analytes by matching it with__________. Both choices are correct Commercial library/database Reference standard
- Let the retention factor of species A be 0.5, while the retention factor of species B be 0.3. If species A moves 1.0 cm how far does species B move? 0.60 cm 0.80 cm 1.0 cm 1.2 cm 1.6 cmTwo species A and B are known to have water/hexane partition coefficient of 5.99 and 6.16.They are separated by elution on silica gel with hexane as eluent. The ratio for the packingVS/VM =0.425a. Calculate the retention factor for each soluteb. Calculate the selectivity factorc. How many plates are needed to provide a resolution of 1.5?d. What column length should be used to provide a resolution of 1.5 if the plate height ofthe packing is 1.5 ×10-3cm?e. If the flow rate is 6.75 cm min-1, how long will it take to elute the two species? (the question wanted to be answered)A 5.00-mL sample of blood was treated with trichloroacetic acid to precipitate proteins. After centrifugation, the resulting solution was brought to pH 3 and extracted with two 5-mL portions of methyl isobutyl ketone containing the lead-complexing agent APCD. The extract was aspirated directly into an air/acetylene flame and yielded an absorbance of 0.527 at 283.3 nm. Five-milliliter aliquots of standard solutions containing 0.400 and 0.600 ppm of lead were treated in the same way and yielded absorbances of 0.396 and 0.599. Find the concentration of lead in the sample in ppm assuming that Beer’s law is followed.