7 B. Measurement of 25 mL of DI water30-mL beaker#312.99 129931.30 377024.31 24.7124.8512.99385325.5415] Mass of empty Beaker16] Mass of Beaker and DI waterMass of DI waterAverage mass of DI waterShow calculation of average:Deviation of massesAverage deviation:Show calculation of deviations and average of deviation:21.4°C171 Temperature of DI water:18) Density of DI water from Table 2:Volume of DI water in beaker:Show calculation of volume:

Density is defined as mass of substance contained in per unit volume of that substance. Its unit is…

Answered: #2 12.99 12.99 12.99 31.30 3770 3853…

Principles of Instrumental Analysis

7th Edition

ISBN:9781305577213

Author:Douglas A. Skoog, F. James Holler, Stanley R. Crouch

Publisher:Douglas A. Skoog, F. James Holler, Stanley R. Crouch

ChapterA1: Evaluation Of Analytical Data

Section: Chapter Questions

Problem A1.12QAP

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An experiment with a pycnometer resulted in the following data: Mass of Pycnomter = 25.428 g Mass of Pycnometer filled with water = 44.021 grams Mass of Pycnometer filled with liquid X = 41.421 grams a) What is the specific gravity of liquid X? b) If methanol (Specific Gravity = 0.790) was used rather than water in measuring the density of irregular objects, how would that affect the result? Why?
Chemistry find absolute and relative standard deviation of this data.
Weight of coin 1 (g) 11.4286 Weight of coin 2 (g) 11.4633 Weight of coin 3 (g) 11.4632 Weight of coin 4 (g) 11.5468 Weight of coin 5 (g) 11.5691 Mean (g) 11.4942 SD 0.0604068290841 %RSD 0.53% Check for suspicious outliers by using the Grubb’s test.
1. 500.0 mL 0.2 M KI2. 500.0 mL 0.2 M KCl3. 500.0 mL 0.1 M K2S2O84. 500.0 mL 0.1 M K2SO45. 500.0 mL 4.0 mM Na2S2O3 (from Na2S2O3∙5H2O)pls compelete the 2nd table given the data
confused in this part why is the answer 0.0005666? Therefore, average error/ deviation = d1 + d2 + d33= (0.0006+0.0008+0.0003)3= 0.0005666 ≈ 0.00057 error/ deviation in parts per thousand(ppt) = average error or deviationaverage concentration×1000 = 0.000570.9548×1000 = 0.596 So, the final answer is = 0.9548 M ± 0.596 ppt
Wallie weighed 3 different cylinders that all looked the same. He noticed that the masses were all different and wondered how much they deviated from the average. Determine the average deviation, in g, of the cylinder masses below. Cylinder Mass (g) 1 31.8 2 31.39 3 31.96
STATISTICAL DATA IN CHEMICAL ANALYSIS: Calculate the mean and standard deviation of the data below. Write the proper reported data. 0.624, 0.613, 0.596, 0.607, 0.582
Standard deviation of data: 206.3, 205.5, 205.8 Mean-205.9
A group of five students attempted to estimate 100 g of a substance by balancingthe amount of the substance with a 100-g standard mass and obtained the followingmasses after each estimated amount was weighed: 95.8634 g, 80.8125 g, 106.5078g, 98.2865 g, and 86.4453 g. Create a data table as shown below and use the datato determine the answers to the following questions:Mass (g) Deviation (g) % Deviation Error (g) % Errora) What is the mean? And what is the true value?b) What are the deviation and % deviation from the mean for each value?c) What are the error and % error for each value?
In the UV assay of paracetamol you are required to weigh tablet powder “…containing 0.150 g of Paracetamol…”. If the stated content of paracetamol per tablet is 500 mg and the weight of 20 tablets is 11.223 g then the amount of paracetamol tablet powder to be weighed is
1. σ=[(Σd2)/(n-1)]½ n= number of trials d= measured value - mean σ=[(8.50mL-8.74mL)2+(9.72mL-8.74mL)2+(8.00mL-8.74mL)2/(3-1)]½ 2. σ=[(Σd2)/(n-1)]½ n= number of trials d= measured value - mean σ=[(10.69-11.52)2+(11.80-11.52)2+(12.07-11.52)2/(3-1)]½
The following densities of water were obtained by Student A: 1.09g/mL, 1.14g/mL, and 1.15g/mL.Determine the average and standard deviation of the density.

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