A student needs 25.0 mL of 9.60x10 4 M FD&C Yellow Dye No. 6 solution. If the student has 2.00 mM FD&C YellowDye No. 6 stock solution to work with, how much of the 2.00 mM solution should be added to the 25.0 mL volumetricflask? Hint, 2.00 mM is 0.00200 M.8.330 mL12.00 mL2.083 mL0.08330 mLO 10.50 mLClick Save and Submit to save and submit. Click Save All Answers to save all answers.Save All Answe

tudent needs 25.0 mL of 9.60x10 e No. 6 stock solution to work with, how much of the 2.00 mM solution should be added to the 25.0 mL volumetric sk? Hint, 2.00 mM is 0.00200 M. ) 8.330 mL -4 M FD&C Yellow Dye No. 6 solution. If the student has 2.00 mM FD&C Yellow 12.00 mL O 2.083 mL 0.08330 mL 10.50 mL ick Save and Submit to save and submit. Click Save All Answers to save all answers. Save All Answe

tudent needs 25.0 mL of 9.60x10 e No. 6 stock solution to work with, how much of the 2.00 mM solution should be added to the 25.0 mL volumetric sk? Hint, 2.00 mM is 0.00200 M. ) 8.330 mL -4 M FD&C Yellow Dye No. 6 solution. If the student has 2.00 mM FD&C Yellow 12.00 mL O 2.083 mL 0.08330 mL 10.50 mL ick Save and Submit to save and submit. Click Save All Answers to save all answers. Save All Answe

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...

See similar textbooks

Similar questions

What is the mass percent of Fe for each trial ? Mass of (NH4)2Ce(NO3)6 (in 100.00 mL) = 4.5038 g Mass of unknown (in 250.00 mL) = 10.000 g Trial # Volume of unknown solution (mL) Titration Volume (mL of Ce4+) Mass Percent(% Fe) 1 25.00 mL 7.20 mL 2 25.00 mL 7.40 mL 3 25.00 mL 7.60 mL 4 25.00 mL 7.50 mL
help please answer in text form with proper workings and explanation for each and every part and steps with concept and introduction no AI no copy paste remember answer must be in proper format with all working
I need help calculating solubility for this question. thank you C. Calcium Fluoride, CaF2 (aka, Fluorite), in a solution containing 19 mg/L of LiF (aka, Griceite), with all of the LiF being dissolved into solution. Please assume that LiF dissolves such that [Li] = [F] = 7.33 x 10-4 m. After adding CaF2 to the solution and allowing it to dissolve, what are final values of [Ca], [Li] and [F]? Note: you are given the concentrations of [Li] and [F] that result from dissolving LiF. No solubility calculation is required for dissolving LiF; just use the concentration given above. Steps: - First, write the stoichiometric equation for dissolution of CaF2, with the solid on the left and the two ions in solution on the right, separated by a "two-way" arrow. Hint: when CaF2 dissolves, you get 2 moles of F for every 1 mole of Ca. - Second, set up a law of mass action equation for dissolution of CaF2. There is a common ion effect for F, so if [Ca] = x, then [F]= 2x + [FLiF], where [FLiF] is from…
e * OWLV2 | Onlin x * MindTap - Cen x C The SolubilityX b Answered: Ma G is Ags insolubl genow.com/ilrn/takeAssignment/takeCovalentActivity.do?locator=assignment-take [References] The solubility of magnesium fluoride, MgF2, in water is 1.5 x 10-2 g/L. What is the solubility (in grams per liter) of magnesium fluoride in 0.39 M of sodium fluoride, NaF? Solubility = g/L Submit Answer Try Another Version 3 item attempts remaining C OWLV2 | Online tea.. Homo Insert
Students performed three KHP titrations in Part 1 to standardize their NaOH solution. They reported the following measurements: Determination #1 Determination #2 Determination #3 Mass of KHP + Vial 14.0862 g 14.5621 g 14.3998 g Mass of 'Empty' Vial 13.2492 g 13.7695 g 13.6887 g Final Burette Reading 45.10 mL 38.53 mL 37.17 mL Initial Burette Reading 5.08 mL 0.52 mL 3.19 mL How many moles of NaOH did they use in their first titration? Report your answer to the correct number of significant figures
A 0.1824-g sample that contains both NaCl (MW= 58.442) and NaBr (MW= 102.89) is titrated with 0.1052 M AgNO, requiring 26.48 mL. Calculate the percentages of NaCl and NaBr in the sample. (Answer: 75.13% NaCI, 24.87% NaBr)
Complete the following solubility table. (Simply enter S for soluble or I for insoluble. B) Develop a scheme to separate CO3 2-, NO3 -, and Br- anions in solution and draw a flow chart that depicts your scheme.
Example / What volume of 0.40 M Ba(OH)2 must be added to 50 mL of 0.30 M NaOH to give a solution 0.50 M in OH. Solution: Let x= mL Ba (OH)2 and the final volume is (50+x) mL mmole OH- = mmole NaOH +2 x mmol Ba(OH)2 0.5 M x (50 + x)mL 0.30 M x 50 mL + 2 x (0.40 M x (x) mL) x = 33 mL Ba(0H)2 Example / Prepare 500 mL of 0.010 M solution of Na from solid Na,CO3. /-
Volume vinegar used (mL)= 5.00 mL % from labe= l5 % Molarity of NaOH = 0.0976 M Titration volume measurements Trial 1 Trial 2 Trial 3 Inital volume NaOH (mL) 0.69 0.86 0.16 Final volume NaOH (mL) 43.47 43.36 43.03 Volume NaOH used (mL) Average volume NaOH used (mL) Average volume NaOH used (L) Moles NaOH used (from avg volume) Moles CH3COOH Molarity CH3COOH (mol/L) Mass CH3COOH (g) % (m/v) CH3COOH in vinegar
A 2.00 g sample of KCl is dissolved in 38.0 mL of water. The resulting solution is then added to 53.0 mL of a 0.320 M CaCl2 (ag) solution. Assuming that the volumes are additive, calculate the concentrations of each ion present in the final solution. Concentration of K+ ions = M Concentration of Ca2+ jons = M Concentration of Cl ions = M Submit Answer Try Another Version 4 item attempts remaining
How many grams of Pb(10₂)2 (Kap-3.7x10-13) will dissolve in 500. mL of a 0.0042 M PbCl₂ solution? (Ksp -1.7x10-5)
in text form with proper workings and explanation for each and every part and steps with concept and introduction no AI no copy paste remember answer must be in proper format with all working!!!!!!!