Pen-Problem Solu�ons for Analy�cal Chemistry and Quan�ta�ve Analysis 1e David Hage James Carr (Selected Chapters, 100% Original Verified, A+ Grade) Chapters Included in SM: 3, 4, 5, 6, 9, 10, 11, 12, 13, 15, 16, 17, 18, 20, 21 and 23.
Analytical Chemistry Hage/Carr
Solution Pen-Problem Chapter three (1) Change 1.04 g/mL to grams per Liter: 1.04 g/mL x 1000 mL/1L = 1004 grams/L. Then determine the grams of CH2O in the liter of solution: 1004 x 0.0552 = 55.42 g Next, change the gram/L to moles/L: 55.42 x 1 mol/30.03g = 1.846 M = 1.85 M (2) Determine the grams of water in the liter of the solution: 1004 – 55.42 = 948.6 g. Then convert grams of water to kg: 948.6 g x 1Kg/1000g = 0.9486 Kg. Divide moles in a liter by the kg of water in the liter: 1.85 moles/0.9486 Kg = 1.950 m
2
Pen Problem Chapter 4
SOLUTIONS TO PEN PROBLEM CHAPTER 4: (A) The value for five data points found for 99% confidence level in Table 4.9 is 0.959. The experimental value for r (0.99) exceeds the table value therefore there is a 99% confidence that the calculated line of best-fit does represent the data. (B) Y1 = 0.0067 (410) – 2.4 = 0.347 Y2 = 0.0067 (420) – 2.4 = 0.414 Y3 = 0.0067 (440) – 2.4 = 0.548 Y4 = 0.0067 (450) – 2.4 = 0.615 Y5 = 0.0067 (500) – 2.4 = 0.950 (C) Residual values (Measured absorbance – calculated absorbance) 0.40 - .347 = +0.05 0.45 - .414 = +0.04 0.55 - .550 = 0 0.60 – 0.615 = - 0.02 1.0 – 0.95 = + 0.05 (D) The graph of this data (residual versus mg quercetin) reveals a non-random pattern. This suggested pattern indicates a problem, for at least this range of the data, in assuming that the linear line of best-fit should be used to represent this data. To be considered supportive of the linear fit model, the residual graph should show a random pattern.
Pen Problem Ch 21
SOLUTIONS FOR PEN PROBLEM Chapter 21 (A) First, Determine the ratio of caffeine peak heights to deuterated peak heights for each standard: 0 ppm 5/155 = 0.0323 50 ppm 42/159 = 0.264 100 ppm 86/154 = 0.558 150 ppm 134/157 = 0.854 200 ppm 185/162 = 1.14 250 ppm 222/158 = 1.41 Next, apply a linear regression treatment with concentration as the x – axis and the peak ratio as the y – axis: y = ax + b; y = 0.00561x – 0.00882; correlation = 0.9995 (B) The caffeine peak height to deuterated caffeine peak height is (175/158) = 1.11 (C) 1.11 = 0.00561x – 0.00882 ; solving for “x” ; (1.11 –0.00882)/0.00561 = 196 ppm
Pen Problem Ch 23
SOLUTIONS FOR PEN PROBLEM Chapter 23 (A) Electrophoretic mobility () =
v (Ld /t m ) (44.0 cm/10.5 min) = 6.98 cm2 /kV min E (V/L) (30.0 kV/50.0 cm)
(B) velocity = distance/time; 22.0 cm/10.5 min = 2.09 cm/min
(C) N =
EL d 2DL
(6.98 cm 2 /kVmin)(30 .0 kV)(44.0 cm) 2(1.8 x10 10 m 2 /sec)(60 s/min)(10 4 cm 2 /1m 2 )(50.0 cm) 8.53 x 105 theoretical plates