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Substances A and B have retention times of 16.4 and 17.63 min, respectively, on a 30 cm column. An unretained species passes through the column in 1.3 min. The peak width at base for A and B are 1.11 and 1.21 min, respectively. Calculate:
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- From the data in Problem 26-14, calculate for species C and D) (a) the resolution. (b) the length of column necessary to separate the two species with a resolution of 2.5.Substances A, B, C and D were found to have retention times of 8.3 min (A), 10.6 min (B), 12.4 min (C) and 13.5 min (D), respectively on an 15.0 cm long column. There was also an unretained species which appeared at 1.5 min. The widths of the peak bases were 0.90 min (A), 1.1 min (B), 1.3 min (C) and 0.80 min (D). Given this information listed above, please choose the most appropriate answer for the six questions listed below for the list of possible answers. Part A Calculate the capacity factor (k') for peak B and peak D. Part B. Calculate the resolution (R) between peaks B and D. Part C. What do the capacity factor and resolution tell you about the separation of peaks B and D on this column? Part D What are the theoretical plate number (N) for peaks Band D? Part E. What is the average plate height (H), based on the information in Part D and the column length listed above? Part F What would be the required length of a column to achieve a Rs of 2.07Let 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 cm
- (b) For column 1(i) What are the unadjusted retention times for components A and B?Let the retention factor of species A be 0.2 while the retention factor of species B be 0.4. If species A moves 0.4 cm how far does species B moveConsider the separation of components A and B on a GC column. The retention time for component A is 1.4 min and for component B is 2.1 min. The width at the base for the peak due toA is 0.38 min and for the peak due to B is 0.53 min. Calculate the resolution, R, and tell whetherthe two peaks are considered to be resolved and why.
- A chromatogram of a mixture of species A, B, and C provided the following data: a) Define "Retention time". b) Calculate the resolution between species A and B. c) Calculate the retention time of species B necessary to separate it from species A with a resolution of 1.5.The retention times of two compounds A and B are 16.40 and 17.63 minutes. A species that is not retained passes through the column in 1.30 minutes. The peak widths (at the base) for A and B are 1.11 and 1.21 minutes, respectively. Calculate:a) The resolution of the columnb) The resolution between compounds A and BA solute with a retention time of 407 seconds has a width at the base of 13.0 seconds on a column 12.2 meters long. Find the number of plates and plate height. The half-width of the same peak is 7.6 seconds. Find the plate height.
- A solute with a retention time of 400.0 s has a width at half-height of 8.0 s on a column 12.2 m long. Find the number of plates and the plate height.Two substances A and B have retention times of 16.40 and 17.63 min respectively on 30.0 cm column. An unsuccessful species passed over the column in 1.30 min. The widths (at the base) of the peaks ofA and B are respectively 1.11 min and 1.21 min: Calculate: a) The Resolution Rs of the column. b) The average number (Nav of theoretical plates in the column c) The equivalent height (H) d) length of column required to achieve a resolution of 1.5, e) time required to elute substance B on the longer column.With reference to the Van Deemter equation, explain thoroughly whether separation may be expected to improve in an LSC column with a 5 μm particle size or the one with a 20 um particle size