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- 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. 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.To determine the organic material in a dried lake bed, the percent carbon by mass is measured at two different locations. To compare the means of the two different locations, it must first be determined whether the standard deviations of the two locations are different. Location 1 standard deviation: 0.050 Location 2 standard deviation: 0.34 What is the calculated ? value for comparing the standard deviations?calculate mean,SD and 95%confidence limit. Use t=4.3.Compare the two means using the student's t test and precision of the two methods using F test at 95% confidence level . For two sets of three replicates use t=2.78 and fcrit=19
- Mean, standard deviation, Coefficient variable, and 95% confidence limits of aspirin in mg *0.3090g *0.3030g *0.3023gA powder was prepared containing 3.00% NaCN and 97.00% NaCl. A sample obtained from that mixture containing 7.374×10^5 particles weighs 10.0 g. Determine the number and percent relative standard deviation of NaCN particles from a sample of the mixture weighing 6.30 g. I need nNaCN particles and %RSDThe results obtained by two analysts for the lead content of samples obtained at different certified points in Rustenburg are as follows: Analyst 1 – 79.22, 79.41,79.66, 79.45 Analyst 2 – 79.09, 79.08, 79.25, 79.13, 79.10, 79.19 Using the equation, calculate the following 1.the standard deviations 2 .the mean values of the two sets of data and 3. s-pooled
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