This preview shows pages 1–2. Sign up to view the full content.

1 ANSWER KEY: EXPERIMENT #4c - Iron Analysis PRE-LAB QUESTIONS: 1. (1 point) Let's determine the %T from an Absorbance value of 2.0: %T = (1/10 A ) × 100% = (1/10 2.0 ) × 100% = 1% Transmittance!!! Clearly, a % Transmittance value of 1% is extremely low , meaning that virtually no light escapes the sample (i.e. almost all light is absorbed by the sample). This, in turn, means that, because so little light gets through the sample, errors are very large , and it is difficult to extract meaningful data from the sample. 2. (1 point) Calculate the Absorbance values from the %T: A = log(100%/%T) = log(100%/12.5%) or log(1.0/0.125) = 0.903 or 0.90 A = log(100%/%T) = log(100%/80%) or log(1.0/0.80) = 0.0969 or 0.097 3. (1 point) Beer's Law: A = ε bc ε = A/bc ε = (1.020) / [(2.00 cm)(15.0 × 10 -5 M)] ε = 3400 M -1 cm -1 (or 3.40 × 10 3 M -1 cm -1 ) 4. (3 points) Do the Plot: Beer's Law Plot y = 6686.8x + 0.0196 0.000 0.200 0.400 0.600 0.800 1.000 1.200 1.400 0.0E+00 5.0E-05 1.0E-04 1.5E-04

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 2

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online