Volume of the original unknown AGNO3 solution =mLMass of AgCl (s) =,Mole of AgCl =(mole = mass /molar mass)Calculate mole of AgNO3 in the 10 mL unknown solution based on the following reactionequation:NaCl(aq)+ AGNO3(aq) → NaNO(aq) + AgCls)Mole of AGNO3 in the 10 mL unknown solution =use the coefficient ratio)_(hint,Mass of AGNO, in the 10 mL unknown solution =(mass = mole x molar mass)Calculate mass of AGNO, in the original unknown solution as follows:Mass of AGNO, in the original unknown solution= Mass of AGNO3 in the 10 mL unknown solution xvolume of the unknown AGNO3 solution in mL10 ml i.Watch the Introductory Video and Support Information on the upper left to learn how touse this virtual lab.Place a bottle of 0.1 M NaCl solution, the unknown AGNO, solution, a 250 mL Erlenmeyerflask, a 10 mL pipette, and a burette (in Glassware→Other) on the workbench. If thebench is said to be too full, set the items along the right side the bench to save space.Record the volume of the original unknown AGNO3 solution in the Data sheet.ii.Workbench 1Unknown AgNO,52.500 ml. @ 25.0°C0.1M NaCI100.00 ml. 25.0C250 ml Flask0.0000 ml @25.0°C10 mL Pipette0.0000 ml@ 25.0°C50 mL Burette0.0000 ml 25.0°Cii.Transfer 50 mL of 0.1 M NaCl solution into the burette.iv.Withdraw 10 mL of the unknown AgNO, solution into the pipette and transfer the solutioninto the empty Erlenmeyer flask.v.Deliver 1 mL of the 0.1 M NaCI solution from the burette into the 10 mL unknown AGNO,solution. Pay attention to mass of AgCl displayed in the information windowInformationName: 250 mL. FlaskVolume: 11.007 mlSpecies (s)ABCIgrams0.01433210 ml PipetteTemperature: 25.00°C0.0000 mle 25.ordnknown AgNO,500 ml@ 25.0°C250PrediseSig FigRealisticpHt: 7.00Pour 0From 50 ml. Burette to 250 ml. Flask1.00 mL transferred.250 ml. Flask0.1M NaCI11.007 ml@25.0'C50.000 mle 25.0cvi.Repeat addition of 1 mL of the 0.1 M NaCl solution from the burette into the unknownsolution until the mass of AgCl doesn't change any more. Record the mass of theprecipitate AgCI on the Data Sheet.

Volume of the original unknown AgNO3 solution = mL %3D Mass of AgCI (s) = g Mole of AgCI = . (mole = mass /molar mass) Calculate mole of AgNO, in the 10 mL unknown solution based on the following reaction equation: NaCljag) + AGNO3(aq) → NANO3(aq) + AgCl) Mole of AGNO, in the 10 mL unknown solution = use the coefficient ratio) _(hint, Mass of AGNO, in the 10 mL unknown solution = (mass = mole x molar mass) Calculate mass of AGNO, in the original unknown solution as follows: Mass of AGNO, in the original unknown solution volume of the unknown AgNO3 solution in ml = Mass of AgNO, in the 10 mL unknown solution x 10 ml

Volume of the original unknown AgNO3 solution = mL %3D Mass of AgCI (s) = g Mole of AgCI = . (mole = mass /molar mass) Calculate mole of AgNO, in the 10 mL unknown solution based on the following reaction equation: NaCljag) + AGNO3(aq) → NANO3(aq) + AgCl) Mole of AGNO, in the 10 mL unknown solution = use the coefficient ratio) _(hint, Mass of AGNO, in the 10 mL unknown solution = (mass = mole x molar mass) Calculate mass of AGNO, in the original unknown solution as follows: Mass of AGNO, in the original unknown solution volume of the unknown AgNO3 solution in ml = Mass of AgNO, in the 10 mL unknown solution x 10 ml

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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https://app.101edu.co ... Question 12 of 37 How many liters of a 0.209M KI solution is needed to completely react with 2.43 g of Cu(NO3)2 according to the balanced chemical reaction: 2 Cu(NO3)2(aq) + 4 KI(aq) → 2 Cul(aq) + I2(s) + 4 KNO3(aq) STARTING AMOUNT ADD FACTOR ANSWER RESET *( ) %D 0.124 1 0.0310 0.0620 0.00542 2.43 166 2 187.57 6.022 x 1023 0.209 mol Cu(NO3)2L Cu(NO3)2 L KI mL KI g KI M KI g Cu(NO3)2 mol KI mL Cu(NO3)2 DEC 21 1
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[References] a. An aqueous solution contains 3.45 g of iron(III) sulfate, Fe2 (SO4)3, per liter. What is the molarity of Fe2 (SO4)3 ? Molarity = M b. When the compound dissolves in water, the Fet ions and the SO4 ions in the crystal go into the solution. What is the molar concentration of each ion in the solution? Molar concentration = M Fe+ Molar concentration = M SO Submit Answer Try Another Version 6 item attempts remaining Visited Email l
34.In an experiment 2 mol of Pb(NO3)2 and 3 mol of NaI are mixed and the reaction represented by the balanced equation occurs:Pb(NO3)2 + 2 NaI → PbI2 + 2 NaNO3.The solid produced is separated from the solution, while more water is added to NaNO3 until a volume of 1000 mL is completed. The limiting reagent in the reaction is the:Immersion reader. Pb(NO3)2+2NaI------PbI2 + 2NaNO3 Pb(NO3)2 , because stoichiometrically fewer moles are required for the reaction NaI, because the reacting mass is lower NaI, because according to stoichiometry, it is completely consumed Pb(NO3)2 because it is in smaller amount than NaI
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