3, 4, 5, 6, 9, 10, 11, 12, 13, 15,

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. 1 / 3

Analytical Chemistry Hage/Carr

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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

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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.

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