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{"ops":[{"insert":"Substances A and B have retention times 8.20 and 8.815 min, respectively, on a 30.0 cm\u00a0\ncolumn. An unretained species passes through the column in 1.30 min. The peak withs (at\u00a0\nbase) foe A ana B are 0.555 and 0.605 min, respectively. Calculate:\na. The column resolution \nb. The average number of theoretical plates in the column. \nc. The plate height \nd. The length of the column required to achieve a resolution of 1.5. Note that\u00a0\nRs = \u221a\ud835\udc75\/\ud835\udfd2 x (\ud835\udf36 \u2212\ud835\udfcf\/\ud835\udf36) x ( \ud835\udc72\/\ud835\udfcf+ \ud835\udc72) and the quantities K and \u03b1 do not change greatly with\u00a0\nincreasing N and L. Substitute N1 and N2 in this equation and divide. \ne. The time required to elute substance B on the column that Rs value of 1.5. Note that\u00a0\n\ud835\udc61\ud835\udc45 =(\ud835\udfcf\ud835\udfd4 \ud835\udc79 x \ud835\udc7a\/\ud835\udfd0 x \ud835\udc6f)\/\ud835\udc96  x  (\ud835\udf36\/(\ud835\udf36 \u2212 \ud835\udfcf ))squared\n x  ((\ud835\udfcf+\ud835\udc72)cubed\/\ud835\udc72 squared) and the quantities K and \u03b1 do not change greatly\u00a0\nwith increasing N and L. Substitute Rs1 and Rs2 in this equation and divide.\u00a0\n"}]}

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