purse Homen/course.html?courseld%3D15769386&HepID3D169f9f4bdc9f2df8dc5b7368af32865e#10001e Hometroduction to MasteringChemistry2 of 15roduction to Numeric AnswersPart BWhen entering large numbers in the answer box, do not use commas. For example, enter 1276400 for the number 1,276,400. Do not enter 1, 276, 400. If you accidentallyenter commas, you will receive feedback as a reminder. Answer the following question by typing the numeric answer into the answer box. What is the sum of 9260 and3240?Express your answer numerically to at least three significant figures.• View Available Hint(s)Eν ΑΣφsum =SubmitScientific notationYou may want to enter 367.000 000 in scientific notatinn as 3 67 x 10° There are twn ways to do thisP PearsonDormicdonc Contact l.Copuriaht 0207 0 Doarcen Education Inc Alrightr rorornd Armr12:58hp

Significant figures are the digits having a meaning in a given number.The number of decimal places…

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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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Chemistry pls help, I dont understand how this software wants me to enter the answer. its asking lfr the starting amount but its not taking it in the way Im putting it
Using a 25mL pipette, you prepare a standard solution by pipetting 25.00mL of the standard iron solution[0.25g/L] into a 500mL volumetric flask. Then, using a 10mL pipette you transfer a 4mL aliquot of this solution into a 50mL volumetric flask and dilute up to the mark. Calculate the % uncertainty.
Using a 25mL pipette, you prepare a standard solution by pipetting 25.00mL of the standard iron solution[0.25g/L] into a 500mL volumetric flask. Then, using a 10mL pipette you transfer a 4mL aliquot of this solution into a 50mL volumetric flask and dilute up to the mark. Calculate the % uncertainty. Question 1 options: 0.65 0.75 0.0075 0.45
Sample IdentificationCodeConcentration of M%TA=2-log(%T)Q50004.00 x 10-417.90.75Q50013.20 x 10-425.00.6Q50022.40 x 10-435.70.46Q50031.60 x 10-450.20.3Q50048.000 x 10-570.80.15SampleIdentificationCode%TA=2-log(%T)AMQ0210150143.70.359518560.360.000192Q0210150244.10.355561410.360.00018Q0210150343.80.358525890.360.00017Q0210150444.10.355561410.360.00018Q0210150543.80.358525890.360.00017What was their percent error?43%Does Batch 021015 meet legal requirements?No, because it is not between 2.85 * 10(4) and 3.15 * 10(4)Well #DropsBluedye1234567891012345678910Drops 9Distilled water876543210Concne 0.26tration0.52Test Tube #0.781.041.3Solutions3Concentration (M)2.082.32.6Concentration (ppm)1:1 dilution11.82Starting Dilution21.562:1 dilution0A.Zero standard0Was your calibration curve as linear as you expected?B.Did you experience any â€œdriftâ€ of the resistance readings?C.What is the equation of your best-fit line?D.What commercial drink did you analyze?E.Assuming…
4. Ksp=4.1x10-36 for Pb3(AsO4)2Calculate Eofor Pb3(AsO4)2(s) + 6e-  3Pb(s) + 2AsO43-Pb2+ +2e-  Pb(s) Eo= -0.126 V
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.
( plz do all with explanation
If exactly 200.0 mL of an aqueous solution that is 1.037E3 ppb Pb(NO3)2 (331.2 g/mol) is mixed with 100.0 mL of an aqueous solution that is 3.090E9 fM KI (166.00 g/mol), how many micrograms of solid PbI2 (461.01 g/mol, 6.16 g/mL) are formed? Assume complete reaction.
5.00 mL of stock solution is diluted to 25.00 mL, producing solution ALPHA. 10.00 mL of solution ALPHA is diluted to 25.00 mL, resulting in solution BETA. 10.00 mL of solution BETA is then diluted to 25.00 mL, producing solution GAMMA. dilution factor for ALPHA from stock solution = 0.167 dilution factor for BETA from ALPHA solution = 0.0476 part c and d?
answe all 3 parts as the policy allows us to asl 3 questions each time please and thank you:))
W00d chips are t0 be dried in a r0tary drier fr0m 35% m0isture to 12% m0isture. If 10 kg of wet w00d is charged/min, h0w much pr0duct can be 0btained 0ver an 8hr peri0d? Find the rati0 of kg water evap0rated/kg wet w00d charged.
Please answer super fast and answer all questions and show calculations For the image attached For 1. a Mass of metal: Trial 1 is 35.0228 g Trial 2 is 35.0915 g Trial 3 is 34.0821 g Mass of water: Trial 1 is 20.0177 g Trial 2 is 20.0250 g Trial 3 is 20.0168 g For delta t of water: Trial 1 is 15.5 C Trial 2 is 15.7 C Trial 3 is 15.1 C For delta t of metal Trial 1 is 80.1 C Trial 2 is 80.2 C Trial 3 is 79.5 C For B my calculated Specific heat is: Trial 1 is 0.462 Trial 2 is 0.467 Trial 3 is 0.466

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