What is the answer for the last question? (Question "C") mol=mass (g)/ molar mass (g/mol)-> 1.46 (g)/143 (g/mol)= 0.01 mol.84/10000 Word Li(b) Assuming that all of the Ag* (aq) ions precipitated, calculate the original molarity of Ag+ (aq) in the solution before HC1(aq) was added.B IUx? X, 5 C nequation to use-> molarity= mol / volume (L)--->if all the Ag precipitated we can conclude; mol AgCl=mol Ag0.01/0.2 (L) = 0.05 M24/10000 Word Lim(c) The calculated concentration of Ag+(ag) from the student's data is higher than the actual concentration of the original solution. The student claims that the calculated concentration is too high becausthe original solution contained some Na+ (aq) impurities that resulted in some NaCl(s) precipitating along with the AgC1(s). Do you agree or disagree? Justify your answer.В I Ux x, 5 ĆX2ΩG,0/10000 Word LimitAfter the experiment, the student learns that the precipitate, AgCl(s), is slightly soluble in water. To determine the solubility, the student mixes a sample of the precipitate with warm, distilled water to make asaturated solution. The student finds the following equation for the dissolution of AgCl(s) in water and a Ksp value at the temperature of the water.AgCl(s) 2 Ag (aq) + Cl¯ (aq)Ksp = 4.0 × 10-10(d) Calculate [Ag+lin the student's solutionIIII

I disagree that NaCl is precipitate out.NaCl is ionic salt which is completely dissociate in water…

(c) The calculated concentration of Ag+ (ag) from the student's data is higher than the actual concentration of the original solution. The student claims that the calculated concentration is too high bec the original solution contained some Na+ (ag) impurities that resulted in some NaClI(s) precipitating along with the AgCl(s). Do you agree or disagree? Justify your answer. B x? x, 5 C 1 E E N 0/ 10000 Word L After the experiment, the student learns that the precipitate, AgCl(s), is slightly soluble in water. To determine the solubility, the student mixes a sample of the precipitate with warm, distilled water to ma saturated solution. The student finds the following equation for the dissolution of AgCl(s) in water and a Kn value at the temperature of the water. AgCI(s) 2 Ag* (aq) + Cl (aq) Kp = 4.0 x 10-10 (d) Calculate LAg+lin tho ctudont'o nolu

(c) The calculated concentration of Ag+ (ag) from the student's data is higher than the actual concentration of the original solution. The student claims that the calculated concentration is too high bec the original solution contained some Na+ (ag) impurities that resulted in some NaClI(s) precipitating along with the AgCl(s). Do you agree or disagree? Justify your answer. B x? x, 5 C 1 E E N 0/ 10000 Word L After the experiment, the student learns that the precipitate, AgCl(s), is slightly soluble in water. To determine the solubility, the student mixes a sample of the precipitate with warm, distilled water to ma saturated solution. The student finds the following equation for the dissolution of AgCl(s) in water and a Kn value at the temperature of the water. AgCI(s) 2 Ag* (aq) + Cl (aq) Kp = 4.0 x 10-10 (d) Calculate LAg+lin tho ctudont'o nolu

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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