Equilibrium Solutions Volume (ml) Trial # 2.00 x 10-3 M Fe(NO3)3 (ml) in 5.00 M HNO3 2.00 x 10-3 M HSCN(ml) in 5.00 M HNO3 0.5000M HNO3(ml) Absorbance [H+] 1 5.00 ml 1.00 ml 4.00 ml 0.085 0.500 2 5.00 ml 2.00 ml 3.00 ml 0.195 0.500 3 5.00 ml 3.00 ml 2.00 ml 0.345 0.500 4 5.00 ml 4.00 ml 1.00 ml 0.432 0.500 5 5.00 ml 5.00 ml 0.00 ml 0.506 0.500 What is the [Fe3+]initial (moles /L x 10 -4) for trial 1? What is the [HSCN]initial (moles /L x 10 -4) for trial 1?

Equilibrium Solutions Volume (ml) Trial # 2.00 x 10-3 M Fe(NO3)3 (ml) in 5.00 M HNO3 2.00 x 10-3 M HSCN(ml) in 5.00 M HNO3 0.5000M HNO3(ml) Absorbance [H+] 1 5.00 ml 1.00 ml 4.00 ml 0.085 0.500 2 5.00 ml 2.00 ml 3.00 ml 0.195 0.500 3 5.00 ml 3.00 ml 2.00 ml 0.345 0.500 4 5.00 ml 4.00 ml 1.00 ml 0.432 0.500 5 5.00 ml 5.00 ml 0.00 ml 0.506 0.500 What is the [Fe3+]initial (moles /L x 10 -4) for trial 1? What is the [HSCN]initial (moles /L x 10 -4) for trial 1?

Fundamentals Of Analytical Chemistry

9th Edition

ISBN:9781285640686

Author:Skoog

Publisher:Skoog

Chapter21: Potentiometry

Section: Chapter Questions

Problem 21.22QAP

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ALL INFORMATION NEEDED IS PROVIDED, I NEED FIRST QUESTIONS ANSWERED FIRST BEFORE OTHERS CAN BE ANSWEREDRate = -(Delta[RCl])/Delta t andln[RCl]t=-k*t+ln[RCl]0 < [R-Cl]-time equation Test tube A = 0.30 of 0.10M NaOH(aq), 6.70ml DI H2O & 1 drop of bromophenol bue Test Tube B = 3.00M of 0.10M R-CL in acetone. Mix together in 3 water baths Bath A Trial 1= 20C, 53.6s Trial 2= 20C, 49s for color to changeBath B Trial 1= 32C, 18s Trail 2= 32C, 20s Bath C Trial 1= 10C, 204s Trial B= 10C, 200s average out both trials for each bathcolor change happens after 10% of R-Cl has reacted so In terms of R-Cl this means that 10% of the R-Cl has reacted and ____ of the R-Cl is left Q1. at the time of the color change [R-Cl]t=____[R-Cl]o (Fill in the blank) Q4. Substitute the value of [R-Cl]t into appropriate equation listed above and solve for k in terms of [R-Cl] and t. Q5. Use average time for each temp and determine the value of k (with units) at each temperature Q6. Rearrange the Arrhenius…
Xanthurenic acid stock solution Absorbance 25 0.052 50 0.128 75 0.202 100 0.310 150 0.448 200 0.621 Plot a graph, from graph, y=mx (force the trend line to pass through origin since the blank with no xanthurenic acid was set to be zero absorbance) in which, y: Absorbance, x concentration
3. Titration: Strong Acid / Weak Base. (I warned you to be ready!) Here's the cool chart! Analyte: 20.00 mL 0.100 M Hydrazine (N₂H4) Hd A: initial Vertical (Value) Axis Title 0 B: half equivalence C: equivalence 5 10 15 20 Titrant: mL of 0.100 M HCI added 25 20.00 mL of 0.100 M hydrazine (N₂H4, pKb = 5.77) is titrated with 0.100 M HCl. a) Find the initial pH (point A) and the pH at the equivalence point (point C). b) Find the pH at the half equivalence point (point B).
MISSED THIS? Read Section 18.6 (Pages 823-826); Watch IWE 18.12. Determine the minimum concentration of the precipitating agent on the right to cause precipitation of the cation from the solution on the left. 9.0x10-2 M Cala; K2SO4 Express your answer using two significant figures. [K₂SO4]= ΜΕ ΑΣΦ Submit Request Answer Part C 1.7x103 M AgNO3; RbCl Express your answer using two significant figures. [RbCl] ΜΕ ΑΣΦ Submit Request Answer M M
4
Mmole of NaOH?
Calculate the average concentration of Vitamin C in each sample
20 mL of 0.1M titrand titrated with 0.1M titrant. Data points at 1 mL increments. 14.0 12.0 10.0 HO 8.0 6.0 4.0 2.0 0.0 0.0 10.0 20.0 30.0 Volume of titrant (mL) 40.0 444444 50.0 (a) Is the species being titrated an acid or a base? (b) Is the species being titrated strong or weak? (c) What is the value of Ka (if a weak acid) or K (if a weak base) for the species being titrated? If the species is strong (100% dissociated), enter 999.
Acid-Base titration: Standardization of 0.1 M NAOH Trial 1 Mass KHP used, g 0.297 g Final reading Volume NaOH, mL 15.59 mL Initial reading Volume NaOH, mL 0.23 mL Volume NaOH consumed, mL 15.36 mL Moles KHP used Moles NaOH used Molarity of NaOH solution
Calculate the total volume of HCl titrant delivered to the equivalence point for each trial
I did an experiment: Molar concentration of Fe (NO3)3=.002 Molar concentration of NaSCN= .002 On my 1st trial: Volume of Fe(NO3)3 = 5 mL Moles of Fe3+, initial (mol) = 1 x 10-5 Volume of NaSCN (mL)= 1 mL Moles of SCN-,initial (mol)= 2 x 10-6 Absorbance was .08 Now trying to find Calculation of Kc I already found [FeNCS2+] equilibrium, from calibration curve (mol/L)= 2.25 x 10-5 Now I need to know: 1. moles FeNCS2+ at equilibrium (10mL) (mol) 2. moles Fe3+, reacted (mol) 3. moles Fe3+, equilibrium (mol) 4. [Fe3+] equilibrium, unreacted, 10 mL (mol/L) 5. moles SCN-, reacted (mol) 6. Moles SCN-, equilibrium (mol) 7. [SCN-] equilibrium (unreacted) 10 mL (mol/L) Feel free to just give me the formula to find these. Or to charge me 2 or 3 questions if I asked too much. Thank you!!!!!!
Data for titration of 15.00 mL of vinegar with approximately 1.0 M NaOH. Note you must use the exact concentration of the standard NaOH solution to calculate the moles of NaOH. Quantity Еxample Trial 1 Trial 2 Trial 3 M NaOH (exact concentration) 0.992 M 0.953 M 0.953 M 0.953 M V initial buret reading = V, 0.20 mL 1.20 mL 0.52 mL 0.15 mL V final buret reading = V, 12.90 mL 14.22 mL 13.71 mL 13.31 mL %3D Vep = VNAOH added = V; - V, 12.70 mL 13.02 mL 13.19 mL 13.16 mL Vep = VNAOH in L 0.01270 L 0.01302 L 0.01319 L 0.01316 L moles NaOH = 0.0126 mol 0.0124 mol 0.0126 mol 0.0125 mol MNAOH X VNAOH moles AA = 0.0126 mol 0.0124 mol 0.0126 mol 0.0125 mol moles NaOH V sample = V acid 0.0150 L 0.0150 L 0.0150 L 0.0150 L (15.00 mL) (15.00 mL) (15.00 mL) Actual molarity 0.840 M 0.827 M 0.840 M 0.833 M of AA Average molarity for 3 trials => of AA 0.833 M 1. Report the acetic acid (AA) concentration of vinegar in units of molarity (M). 2. Use your average AA concentration to calculate the mass…

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Trial #	2.00 x 10^-3 M Fe(NO₃)₃ (ml) in 5.00 M HNO₃	2.00 x 10^-3 M HSCN(ml) in 5.00 M HNO₃	0.5000M HNO₃(ml)	Absorbance	[H⁺]
1	5.00 ml	1.00 ml	4.00 ml	0.085	0.500
2	5.00 ml	2.00 ml	3.00 ml	0.195	0.500
3	5.00 ml	3.00 ml	2.00 ml	0.345	0.500
4	5.00 ml	4.00 ml	1.00 ml	0.432	0.500
5	5.00 ml	5.00 ml	0.00 ml	0.506	0.500