i need the answer quickly Need both table solution if u don't know both table please skip wrong answer gives u downvotelMaterialSample 150.00mlInitial volume of water (ml)60.00mlFinal volume of water (ml.)Volume of solid (ml)25.95gMass, m (0)Density, D (experimental) (g/ml)Density, D (accepted) (g/mL)% ErrorData Table D: Density of a LiquldUnknown1Unknown 210.00ml10.00mlVolume of liquid (ml)50.00g57.75g50.00gMass of empty beaker (0)60.10gMass of beaker with liquid (g)Mass of liquid (a)Density, D (experimental) (giml)MaterialOmitOmitDensity D(accepted) (giml)OmitOmitErrorSpecific Gravity (calculated)0.9900.780Specific Gravity (hydrometer)

Answered: Image /qna-images/answer/17fb8d0d-51f8-43e3-bf76-4388705ddc7c.jpg

Answered: Material Sample 1 50.00ml Initial…

Principles of Instrumental Analysis

7th Edition

ISBN:9781305577213

Author:Douglas A. Skoog, F. James Holler, Stanley R. Crouch

Publisher:Douglas A. Skoog, F. James Holler, Stanley R. Crouch

ChapterA1: Evaluation Of Analytical Data

Section: Chapter Questions

Problem A1.22QAP

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What is the molar mass of a solution if 2.00g of the solute is dissolved onto 9.00g of solvent? The Kf is 4.5° C and the temp drops from 70.00° C ( pure solvent) to 65 00° C (solution). You Must show all work and include unit and you Must get answer to correct sig figs.
Volume of an unknown used was 30 mL, Initial Buret volume was 0 and the Final Buret volume was 8.5 mL. What is the molarity of the unknown solution if the Net volume of NaOH being used was 8.5 mL and Millimoles (mmoles) NaOH reacted was 0.791? Then, what is the Mass (g) of Acetic Acid in unknown sample and thr average percent (%) Acetic Acid? (assume density = 1g/mL)
0.02 M NaOH reagent was standardized using 0.1000 g Potassium hydrogen phthalate (204.22 g/mol) and it required 25.8 mL to reach end point. Answer must be in 4 decimal places.
Data collected Measurement Value Initial Volume of NaCl solution in cylinder, (mL) 7.54 Volume of NaCl solution remaining in cylinder, (mL) 1.09 Volume of NaCl solution transferred to dish, (mL) Mass of dish, (g) 73.421 Mass of dish and NaCl solution, (g) 80.412 Mass of NaCl solution, (g) Mass of dish and dried NaCl , (g) 75.312 Mass of dried NaCl, (g) Density of solution, (g/mL) Percent NaCl (m/m), (%) Percent NaCl (m/v), (%) Molarity of NaCl, (moles/L)
A 25.0 mL solution containing acidified Na2C2O4 requires 15.0 mL of 0.0500 M KMnO4 solution to reach endpoint. 5 C2O42- + 2 MnO4- + 16 H+ → 10 CO2 + 2 Mn2+ + 8 H2O Given the balanced chemical reaction above, what is the concentration (in molarity) of the Na2C2O4 solution? Answer: _____ M
What's the differences between two questions? Q1) How much calcium would you ingest by drinking one 8 oz glass of your tap water? Show all calculations. -->Tap water 8Oz = 8 x 0.0296L = 0.2368L Hardness = 66.73ppm = 66.73mg/L CaCO3 1L has 66.73mg CaCO3 0.2368L has 66.73mg x 0.2368 = 15.8017mg MW of CaCO3 = 100g/mol MW of Ca = 40g/mol 100g CaCO3 has 40gf of Ca 15.8017mg CaCO3 has 40/100 x 15.8017mg Ca We would ingest 6.321mg of Ca. Q2) What percentage of the recommended daily dose of calcium (1,150 mg/day) does 1.0 L of your water provide? Show all calculations. --> 66.73mg/1150mg x 100 = 5.80% My Question) Why this calculation is wrong? I think this calculation is same with question 1. Isn't it? CaCO3 = 100g/mol, Ca = 40g/mol 100g CaCO3 has 40g Ca. 66.73 CaCO3 has 40/100 x 66.73mg Ca Ca = 26.70mg 26.70mg/1150mg x 100 = 2.32%
For results/conclusions Using the images attached Write 2 paragraphs 1st paragraph: Starts with Objective. Results of DeltaT and i stated 2nd paragraph: sources of errors discussed The objective is The colligative properties are the properties that undergo a change when a solute is introduced to a solvent. Freezing point depression is also a colligative property. The addition of a solute to a solvent reduces the freezing point of the resulting solution. The freezing point of the solution is always lower than the freezing point of the solvent. This difference in the freezing point between solvent and solution is called freezing point depression. The freezing point depression is directly proportional to the molality of the solution or the number of moles of solute present in the solution.
If a pharmacist adds 3g of hydrocortisone (HC) to 60 g of a 5% (w/w) hydrocortisone cream, what is the final percentage strength of hydrocortisone in the product. Hint: identify first the amount of HC present in 60 grams then add the 3 grams to both the amount of solute and solution then compute for the new percentage. Answer has 2 decimal places.
Submit a clean, dry, and properly labeled 50-mL reagent bottle for your unknown solution. Pipet 20.00 mL of the sample in 250-mL Erlenmeyer flask. Add 5 mL of buffer and 5 drops of indicator. Titrate the solution until it turns light blue. If the titration consumes more than 50 mL of the titrant, dilute the sample accordingly. Compute for ppm of CaCO3 using the following table. Mean Molarity = 2.487 x 10^-3 MBlank correction = 0.015 Formula = V (mL) titrant x Mean Molarity of Titrant x Molecular weight of CaCO3 x 1/V (L) sample
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Tabulated Data: Trial I Trial II Final reading HCl 26. 40 mL 27.10 mL Initial reading HCl 1.2 mL 1.8 mL Volume of HCl used ? Final reading NaOH 33.7 mL 32.8 mL Initial reading NaOH 0.8 mL 0.5 mL Volume of NaOH used ? ? Calculations and solutions: Normality of NaOH : 0.25 N Solve for the average Normality of HCl : ? Calculations and solutions: Trial I. NHCl = Trial II. NHCl = Average NHCL =

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