Data Part A Indicate the absorbencies at the time points you used for the varyin
ID: 98064 • Letter: D
Question
Data Part A Indicate the absorbencies at the time points you used for the varying enzyme volumes Enzyme Volume (mL) S1545"5 9005 120 0.0943) 01781 0,2590 03414 04175 0 5972 0,581 0656 0.1 0.2 0.3 0.09430.1781 0.2590 0.3414 0.4175 0.50720.581 0.6567 0.7294 0.1967 0.3385 0.4891 0.6272 0.7632 0.8903 0.9775 0.9775 0.9698 0.2064 0.4179 0.6293 0.82980.9759 0.9775 0.9744 0.9759 0.9728 0.3965 0.6883 0.9489 0.9667 0.9667 0.9667 0.9592 0.9607 0.9533 0.4722 0.8422 1.0081 1.0098 1.0081 1.0115 0.9949 0.9933 0.9917 385 0,4891 06372 0703) 0 8908 05975 097 2064 04179 06393 08298 0979 0,9775 0974 09759 03972 0.4 472 08462 1091 0m 101 1015 09p9p Note: If start value is not zero, subtract 'start value' from each value and get the corrected data.Explanation / Answer
1. ii)iii)iv)v)
Question-2--- i)A/t for 0.1ml =0.00529.,(ii)A/t for 0.2ml,=0.009248.,(iii)A/t for 0.3ml=0.0128.,(iv)A/t for 0.4ml =0.01841.,(v)A/t for 0.5ml =0.01786.
3) By lambert-beers law= A= e x c x l,(i)e=3600M-1cm-1,= change in Absorbance /e xtime=0.012245.(ii) 0.034251 for 0.2 ml (iii)0.059258 for 0.3 ml (iv)0.13712 for 0.4 ml (v)0.01786/3600 x30 x 10-6=0.16537 for 0.5 ml
4) 0.1 ml =0.2 x 0.1 mg enzyme=0.02 mg enzyme,0.04 mg, 0.6mg, 0.8mg, 0.10mg
5)
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