Sample Blank $2 $3 $4 $5 $6 $7 $8 $9 $10 Stock of Fe Stock of SC Volume Fe3+, mL 2.000 2.000 2.000 2.000 2.000 2.000 2.000 2.000 2.000 2.000 [Fe3+] M 0.200 M 2.00E-03 M 0.200 0.198 0.195 0.193 0.190 0.188 0.186 0.184 0.182 0.180 Volume SCN-, mL [SCN-] M 0.0000 0.0250 0.0500 0.0750 0.1000 0.1250 0.1500 0.1750 0.2000 0.2250 0.00E+00 2.47E-05 4.88E-05 7.23E-05 9.52E-05 1.18E-04 1.40E-04 1.61E-04 1.82E-04 2.02E-04 Absorbamce 0 0.097 0.214 0.313 0.408 0.509 0.599 0.688 0.771 0.858

Sample Blank $2 $3 $4 $5 $6 $7 $8 $9 $10 Stock of Fe Stock of SC Volume Fe3+, mL 2.000 2.000 2.000 2.000 2.000 2.000 2.000 2.000 2.000 2.000 [Fe3+] M 0.200 M 2.00E-03 M 0.200 0.198 0.195 0.193 0.190 0.188 0.186 0.184 0.182 0.180 Volume SCN-, mL [SCN-] M 0.0000 0.0250 0.0500 0.0750 0.1000 0.1250 0.1500 0.1750 0.2000 0.2250 0.00E+00 2.47E-05 4.88E-05 7.23E-05 9.52E-05 1.18E-04 1.40E-04 1.61E-04 1.82E-04 2.02E-04 Absorbamce 0 0.097 0.214 0.313 0.408 0.509 0.599 0.688 0.771 0.858

Fundamentals Of Analytical Chemistry

9th Edition

ISBN:9781285640686

Author:Skoog

Publisher:Skoog

Chapter5: Errors In Chemical Analyses

Section: Chapter Questions

Problem 5.9QAP

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A. B. and C. already solved (https://www.bartleby.com/questions-and-answers/chemistry-question/a89691d6-2162-4677-b0cc-dc84fcceda08). Please answer E, F and GCircle in pencil in the image on th eirght cide of the diagram is a "2"( Pb(NO3)2 ).
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
Pheromones are compounds secreted by females of manyinsect species to attract mates. Typically, 1.0 × 10−8gof a pheromone is sufficient to reach all targeted maleswithin a radius of 0.50 mi. Calculate the density of thepheromone (in grams per liter) in a cylindrical air spacehaving a radius of 0.50 mi and a height of 40 ft (volumeof a cylinder of radius r and height h is πr2h).
pls complete the 2nd table given the data1. 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)
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.
ᴀ ꜱᴛᴇᴇʟ ᴛʀᴀɪɴ ʀᴀɪʟ ɪꜱ 400 ᴍᴇᴛᴇʀꜱ ʟᴏɴɢ, ɪɴ ᴍᴀʀᴄʜ ɪᴛ ɪꜱ -30 ᴅᴇɢʀᴇᴇ ᴄ ᴀɴᴅ 40 ᴅᴇɢʀᴇᴇ ᴄ ɪɴ ᴊᴜʟʏ. ᴡʜᴀᴛ ɪꜱ ᴛʜᴇ ᴄʜᴀɴɢᴇ ɪɴ ʟᴇɴɢᴛʜ ɪɴ ᴍᴍ? ᴋ ꜰᴏʀ ꜱᴛᴇᴇʟ = 11.7 x 10^-6 ᴍ/ᴍ-ᴄ
2 CoCl2(aq) + 8 NH3(aq) + H2O2(aq) + 2 NH4Cl(aq)  2 [Co(NH3)5Cl]Cl2(s) + 2 H2O(l) In a 50-mL Erlenmeyer flask, cobalt(II) chloride hexahydrate (1.075 g), an aqueous solution of ammonia (1.2 mL, 15 M), ammonium chloride (0.505 g), and hydrogen peroxide (1.0 mL, 30.0% (w/w), density = 1.10 g/mL) were combined. After completion of the reaction, [Co(NH3)5Cl]Cl2 (0.450 g) was isolated. What was the percent yield of the product?
Sample Physical Appearance 0.1 M Na2CO3 0.1 M CaCl2 Na2CO3 + CaCl2 mixture Part II filtrate + CaCl2
Two aqueous hydrogen bromide solutions containing 1.00 wt% HBr (SG=1.0041) and 60.0 wt% HBr (SG=1.6731) are mixed to form a 3.54 M HBr solution (SG=1.1935). What feed rate of the 60.0 wt% HBr solution would be required to produce 1350 kg/hr of product?
How can I convert 728.75468cm/s2 to N/kg ????
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…
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.