精華區beta NTU-Exam 關於我們 聯絡資訊
課程名稱︰分析化學丙 課程性質︰化工系必修 課程教師︰林萬寅 開課系所︰化工系 考試日期(年月日)︰2010/11/10 考試時限(分鐘):130 mins 是否需發放獎勵金:是 (如未明確表示,則不予發放) 1.Complete the follwing calculations. 8% (a) (16.84+3.756)/2.00 (b) log[(4.00*10^(-8)) * (1.625*10^(-6))] 2.A spherical ball has a diameter of 20.0 ±0.1 cm and a mass of 8.00 ±0.02 kg Calculate the density (including uncertainty) of the ball in g/mL. 8% 3.Two students analyzed the nitrogen content (in wt%) in a milk sample known to contain 16.10% N and yielded the following results. Answer the follwing questions. Student # of measurements Confidence limit (95%) __________________________________________________________________________ A 4 15.10~15.9% B 6 15.68~16.32% (a)Do the results of the two students differ significantly from the theoretical value at 95% confidence level? 4% (b)Calculate the mean and standard deviation for the result of each student. 6% (c)Which student is more accurate? Why? 2% (d)Which student is more precise? Why? 2% (e)Calculate the pooled standard deviation from the results of the two students. 4% (f)Do the results of the two students differ from each other at 95% confidence level? 5% (t=3.18, 2.57, 2.31 for 3, 5, and 8 degrees of freedom, respectively) 4.For the five data 4.08, 4.17, 4.19, 4.22, and 4.24, answer the following question. (a)Find the mean and standard deviation for the entire set of data. 4% (b)Use Grubbs test to decide whether 4.08 should be discarded or not? 4% (G=1.672 for 5 measurements at 95% confidence level) 5.A standard addition method was used to determine the calcium content in a water sample and yielded the following results. The best-fit line for the analysis is y = 54.6 + 3.142x , where y = signal and x = added Ca in ppb. Added Ca (ppb) 0 5 10 15 20 X Signal (μA) 56.0 68.3 85.6 103.0 117.2 95.0 (a)What is the concentration of Ca (in ppb) for the original sample? 4% (b)Find the amount of Ca added to the sample for "X" in the above Table. 4% 6.The density of pentane is 0.626 g/mL at 25℃. Find the true mass of pentane if it weighs 96.85 g in air at 25℃. The density of air and weight are 0.0012 g/mL and 8.0 g/mL, respectively. 5% 7.A 1.00 g molybdate sample (containing K2MoO4) was dissolved in 50.00 mL solution and passed through a column containing Zn(s) to convert molybdate to Mo^3+. The resulting Mo^3+ solution required 24.60 mL of 0.0120 M KMnO4 to reach the end point according to the following reaction: 3MnO4^- + 5Mo^3+ + 4H+ → 3Mn^2+ + 5MoO2^2+ + 2H2O A blank titration required 0.10 mL of KMnO4 solution. (a)Find the molarity of molybdate in the solution. 5% (b)Calculate the percentage of K2MnO4 (Fw = 238.0) in the sample. 5% 8 Answer the follwing questions. 30% (a)Describe the meaning of systematic error and random error (b)Describe how to convert molality to (i) molarity and (ii) ppm. (c)Distinguish range, linear range, and dynamic range (d)Define the limit of detection and limit of quantification (e)Describe the mechanism of color change for an adsorption indictor used in Fajans method. (f) Describe the general steps in a chemical analysis -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 114.36.60.15 ※ 編輯: vvjames 來自: 114.36.60.15 (11/14 15:43)