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- An analytical determination for manganese in an aqueous solution on gives thefollowing replicate measurements in ppm as 12.1, 12.3, 12.2, 12.2, 12.4, 12.3, 12.2, 12.4,12.2, and 12.5. Determine the following;a) The range of the datab) The meanc) Mediand) Standard deviatione) Variancef) Relative standard deviation RSD (ppt)The following data represented the concentration of ofloxacin present in chicken meat samples: 4.17, 4.40, 4.35, 4.35, 4.20, 4.16,4.20 and 4.17. Estimated the % relative standard deviation.3. Seven analyses for the phosphorus content of a fertilizer resulted in 16.2, 17.5, 15.4, 15.9, 16.8, 16.3, and 17.1%. Find (a) the standard deviation, s, , and (b) the 95% confidence interval for the true value, H. 4. The following five values were obtained for the wt % of an organic acid in a sample: 30.3, 31.1, 32.6, 36.7, 28.9. Determine if the value of 36.7 may be rejected at the 90% (two-tailed) confidence level. 5. The concentration of a NaOH solution is determined in a single experiment by titrating a weighed sample of pure, dried KHP (MW = 204.229) with the NaOH. The data are: Weight of KHP = (11.6723 + 0.0001) - (10.8364 ±0.0001) g Volume NaOH = (32.68 + 0.02) - (1.24 + 0.02) mL Where the standard deviations are o, values. (a) What is the molarity of the NaOH? (b) Based on the propagation-of-error formula, what is the standard deviation of the molarity,oM ?
- The results in the table below were obtained from the analyses of Mg2+ content in an orange juice sample using a flame-AAS. Based on these results answer the following questions: (i) (ii) Replicate 1 2 3 4 Mg²+(ppm) 7.08 7.21 7.12 7.09 Calculate the mean, standard deviation and coefficient of variation (or %RSD) for the results obtained. If the certified value of the Mg2+ content of the apple juice sample is 7.45 ppm. What is the percentage of recovery for this analysis?2. A method for analysis of codeine in prescription drugs yielded the following results when applied to a codeine- free blank: 0.01, -0.2, 0.3, 0.2, 0.0, -0.1 mg codeine. Calculate the detection limit (in terms of miligrams in codeine) at 99% confidence level, based upon the mean.Felodipine calcium channel blocker standard (0.251mg/ml) and felodipine sample (0.245mg/ml) solutions were prepared and injected to the HPLC. The peak area of felodipine standard is 275428 and the sample is 272982. The potency (purity) of felodipine standard is 98.9%. What is the assay percentage of felodipine? a. 102.23 b. 98.71 c. 101.07 d. 100.42
- A volume of 250 ml of a 0.05 M solution of a reagent of formula weight (relative molecular mass) 40 was made up, the weighing being done by difference. The standard deviation of each weighing was 0.0001 g: what were the standard deviation and relative standard deviation of the weight of reagent used? The standard deviation of the volume of solvent used was 0.05 ml. Express this as a relative standard deviation. Hence calculate the relative standard deviation of the molarity of the solution. Repeat the calculation for a reagent of formula weight 392.Water samples are analyzed for water hardness with the following results: 102.2, 102.8, 102.9, 103.3 and 102.3 ppm CaCO3. (a) Calculate the average (b) Calculate standard deviation (c) Calculate and relative standard deviation or (the coefficient of variation). (d) If the known value is 103.25 ppm. Does your answer agree with the known value at confidence interval of 98%? Q3. Nitrite (NO2-) was measured by two methods (Gas chromatography and spectrophotometry) in rainwater. The average result for the gas chromatography is = 0.069 mg/L with a standard deviation of s1= 0.005 and n1= 7. The average result for the spectrophotometry is = 0.063 mg/L with a standard deviation of s2= 0.008 and n2= 5. Do the two methods agree with each other at the 95% confidence level? Q5. Two methods were used to measure fluorescence lifetime of a dye. The average result for Method 1 is = 1.382 ns with a standard deviation of s1= 0.039 and n1= 7. The average result for Method 2 is = 1.346 ns with a standard…The same chloride analysis but using a new method for which the standard deviation was not known, gave the following replicate results, mean and estimated standard deviation: Chloride (ppm) = 346, 359, 350; Mean = 351.67 ppm; Standard deviation = 6.66ppm. Solve for the CI for the 3 determinations at 95% probability level.
- . 100 ml boiled cooled and filtered water sample takes 9.6 ml of M/50 EDTA in titration. The Permanent hardness of the water sample in terms of ppm of CaCO3 equivalent isA/ The following results were obtained in the replicate determination of the lead content of a blood sample (0.752, 0.756, 0.752, 0.751 and 0.760 )ppm. Calculate the mean, standard deviation, relative standard deviation%, median, variance and confidence limits at t=1.9 of this set of data.The following data was obtained when four bottles of wine, were analyzed for residual sugar content and the results obtained are the followingB Bottles % w/v Residual Sugar 1 0.97, 0.86, 1.02 2 1.25, 1.32, 1.13, 1.20, 1.12 3 0.9, 0.92, 0.73 0.72, 0.77, 0.61,0.58 a) For each set, determine the mean, spread, standard deviation s and coefficient of variation. b) Calculate the pooled standard deviation, spooled. 2.