a solution that was 0.002M in I2 showed an absorbance of 0.84 units on an arbitr

Question

a solution that was 0.002M in I2 showed an absorbance of 0.84 units on an arbitrary scale. the aq solution by itself was used at the control bank and set to an absorbance of 0.00 on the same scale. Draw up a graph of I2 concatenation against the absorbance assuming the relationship is linear using the grid below.

A solution that was 0.002 M showed absorbance of0.84 units on an arbitrary scale. The aqueous olution by itself was used a the control blank and set to an absorbance of0,00 on the same scale. Draw up a graph of 12 concentration (the independent variable) against absorbance (the dependent variable) assuming the relationship is linear using the grid below

Explanation / Answer

change in concentration from 0 to 0.002 = 0.002M, change in absorbance = 0.84-0 =0.84

so 0.002x= 0.84

x =0.84/0.002=420 this is the slope of absorbance vs concentration

choose value between 0 and 0.002M and from slope calculate the absorbance

concentration (M) Absorbance(A) 0 0 0.0001 0.042 0.0002 0.084 0.0003 0.126 0.0004 0.168 0.0005 0.21 0.0006 0.252 0.0007 0.294 0.0008 0.336 0.0009 0.378 0.001 0.42 0.0011 0.462 0.0012 0.504 0.0013 0.546 0.0014 0.588 0.0015 0.63 0.0016 0.672 0.0017 0.714 0.0018 0.756 0.0019 0.798 0.002 0.84

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