Density of object analysis Density (g/cm³) Diameter Volume Height (cm) Mass (g) (cm) (cm³) Object A 1.14 1.71 3.175 Object B 1.14 1.88 3.468 Object C 1.19 1.94 3.932 Object D 1.12 1.51 2.665 Table view List view Density statistics Density (g/cm³) Average Standard deviation Coefficient of variation (CV) True density 1.810 Absolute error % error

Density of object analysis Density (g/cm³) Diameter Volume Height (cm) Mass (g) (cm) (cm³) Object A 1.14 1.71 3.175 Object B 1.14 1.88 3.468 Object C 1.19 1.94 3.932 Object D 1.12 1.51 2.665 Table view List view Density statistics Density (g/cm³) Average Standard deviation Coefficient of variation (CV) True density 1.810 Absolute error % error

Fundamentals Of Analytical Chemistry

9th Edition

ISBN:9781285640686

Author:Skoog

Publisher:Skoog

Chapter6: Random Errors In Chemical Analysis

Section: Chapter Questions

Problem 6.1QAP

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What are the steps in the Chapman cycle? Explain the importance.
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Systematic error causes the mean of a data set to differ from the accepted value. True False
Time(s) [X3] (mole/L) 0 0.600 500 0.481 1000 0.392 1500 0.336 2000 0.273 2500 0.230 3000 0.181 3500 0.156 4000 0.120 4500 0.106 5000 0.072 plot ln[X3] vs time plot 1/[X3] vs. time
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
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
Conversion of units using the DimensionalAnalysis(Factor Label Method) I – Give the equivalent units of measurement. 1. 1m = ____ cm = _____ mm =_____ dm =______km = _____in =____ ft =___yd=______mi2. 1kg = ____ g =______mg = _____cg = _____ dg = ____lb 3. 1h = _____s = _____ ms = ______ min = ____ yr =______ day 4. 1A = ____ MA = ____V = ______ W = ______ kW = ____mA 5. 1mol = ____ mmol = _____kmol = ______Mmol 6. 1K = ____ oC= ______ oF=_______oR 7. 1g/mL = ______g/cm3 = _____lb/in3 = _____ lb/cm3 =______kg/m3 8. 1Pa=________kPa =_______mPa = ____ GPa = _____ atm 9. 1J = ________kJ = ______ nJ = _____cal = ____k cal =_____dcal 10.1Byte = _____ GB =______ kB = _____TB=_____MB =_____
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.
Pls answer the question linked below: https://pasteboard.co/0afMgF1vi1jX.png
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Identify and provide an explanation of the differencesbetween homogeneous and heterogeneous sampling in the context of samplingmethods.
An N-type sample of silicon has uniform density (Nd = 1019cm-3) of arsenic. and a P-type