The equation Cm presubscript 96 presuperscript 250 rightwards arrow ? plus Sr presubscript 38 presuperscript 98 plus 4 subscript 0 superscript 1 n is correctly completed as Cm presubscript 96 presuperscript 250 rightwards arrow Ce presubscript 59 presuperscript 148 plus Sr presubscript 38 presuperscript 98 plus 4 subscript 0 superscript 1 n. True or False

The equation Cm presubscript 96 presuperscript 250 rightwards arrow ? plus Sr presubscript 38 presuperscript 98 plus 4 subscript 0 superscript 1 n is correctly completed as Cm presubscript 96 presuperscript 250 rightwards arrow Ce presubscript 59 presuperscript 148 plus Sr presubscript 38 presuperscript 98 plus 4 subscript 0 superscript 1 n. True or False

Appl Of Ms Excel In Analytical Chemistry

2nd Edition

ISBN:9781285686691

Author:Crouch

Publisher:Crouch

Chapter2: Basic Statistical Analysis With Excel

Section: Chapter Questions

Problem 12P

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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
using exactly 5.00 mL of 0.0400 M stock CuSO4 solution. Add 100 mL of water. Data for Part IMass of empty dish: 32.470 g empty dishVolume of 0.0400 M solution: 5.00 mL CuSO4 solution.Mass of dish and 0.0400 M solution: 37.497 g dish and solutionMass of dish and CuSO4 solid: 32.503 g dish and CuSO4 solidCalculations for Part I1. Calculate the mass of solution2. Calculate the mass of solid CuSO4 dissolved in the solution.3. Calculate the number of moles of solid CuSO4 dissolved in the solution.4. Calculate the mass of water evaporated from the solution.5. Calculate the density of solution, (g solution/mL solution).6. Calculate the % by mass, CuSO4 in solution (100 x g CuSO4/g solution).7. Calculate the molality of solution (moles CuSO4/kg solvent).8. Calculate the molarity of solution (moles CuSO4/L solution).9. Given that the true molarity is 0.0400 M, calculate the percent error of your result.
using exactly 5.00 mL of 0.0400 M stock CuSO4 solution. Add 100 mL of water. Data for Part IMass of empty dish: 32.470 g empty dishVolume of 0.0400 M solution: 5.00 mL CuSO4 solution.Mass of dish and 0.0400 M solution: 37.497 g dish and solutionMass of dish and CuSO4 solid: 32.503 g dish and CuSO4 solidCalculations for Part I1. Calculate the mass of solution? 5.027 g 2. Calculate the mass of solid CuSO4 dissolved in the solution? 0.033 g 3. Calculate the number of moles of solid CuSO4 dissolved in the solution? 2.07 x 10-4 mol4. Calculate the mass of water evaporated from the solution? 0.0414 M5. Calculate the density of solution, (g solution/mL solution)?6. Calculate the % by mass, CuSO4 in solution (100 x g CuSO4/g solution)?7. Calculate the molality of solution (moles CuSO4/kg solvent)?8. Calculate the molarity of solution (moles CuSO4/L solution)?9. Given that the true molarity is 0.0400 M, calculate the percent error…
the linear trendline equation y= 4.7033x + 0.0045
Calcium Chloride Ammonium Chloride Trial 1 Trial 2 Trial 3 Trial 1 Trial 2 Trial 3 Mass of water 100g 100g 100g 100g 100g 100g Mass of salt 5.36g 10.28g 15.10g 5.29 10.43 15.31 Moles of salt Initial T 21.0 21.0 21.0 20.0 20.0 20.0 Final T 29.6 36.2 43.5 17.1 13.2 10.1 ΔT 8.6 15.2 22.5 -2.9 -6.8 -9.9 ΔH(Joules) ΔH(kj/mol)
Help with questions 2-3 on work sheet using data please
Please show a clear picture of the spreadsheet /excel file of the data including the equations used in excel.
When answering, please not only provide the calculation work, but the reasoning and basic concepts behind the answer. This will significantly help me out. Thanks!
A volumetric calcium analysis on triplicate samples of the blood serum of a patient believed to be suffering from a hyperparathyroid condition produced the following data: mmol Ca/L = 3.55, 3.65, 3.14. What is the 95% confidence interval for the mean of the data, assuming(a) No prior information about the precision of the analysis?(b) s → σ = 0.056 mmol Ca/L?
1. Calculate the experimental density of a salt solution and the percent error (same as relative error percent) using some or all the data given below. solubility of NaCl salt in water: 0.357 g/mLmass of empty graduated cylinder: 25.19g mass of graduated cylinder + salt solution: 30.47g total volume of salt solution: 4.98 mLtrue density of salt solution: 1.07 g/mL
Chemistry (i) For T = 298K ΔG° = -8.314 x 298 x ln (2.83x10-3) = 14.524 KJ/mol (ii) For T = 308.15K ΔG° = -8.314 x 308.15 x ln (9.619x10-3) = 11.891 KJ/mol (iii) For T = 318.15K ΔG° = -8.314 x 318.15 x ln (3.14x10-2) = 9.147 KJ/mol (iv) For T = 328.15K ΔG° = -8.314 x 328.15 x ln (8.82x10-2) = 6.621 KJ/mol (v) For T = 338.15K ΔG° = -8.314 x 338.15 x ln (2.42x10-1) = 3.990 KJ/mol Based on the measured ΔG° values, is this equilibrium spontaneous at room temperature? Which factor, entropy, or enthalpy, has the greater impact on spontaneity in this case? Explain your answers.
Chemistry Fraction no. Elution Vol. (mL) Abs (λ = 410) Fraction no. Elution Vol. (mL) Abs (λ = 410) 1 (PB pH 6.0) 1 0.0050 11 11 0.0041 2 2 0.0032 12 12 0.0060 3 3 0.0040 13 13 0.0064 4 4 0.0064 14 14 0.0063 5 5 0.0067 15 15 0.1109 6 6 0.0067 16 16 2.8902 7 7 0.0050 17 17 0.8500 8 (PB pH 8.0) 8 0.0049 18 18 0.2019 9 9 0.0052 19 19 0.0378 10 10 0.0050 20 20 0.0073 Use the data above to plot an elution profile, in which the absorbance is on Y-axis, and elution volume (volume = fraction number) is on X-axis. The elution profile must indicate at (1) which point the pH change occurred, (2) the fraction(s) which contain the highest myoglobin content, and (3) the maximum absorbance detected.