68OpenVellumHMAC=defce5f03204e95efa2fcf11b1ecff98#10001I Review | Constants | PeriPart AE 18.250.0 mL of 0.15 M HCHO2 with 80.0 mL of 0.12 M NaCHO2mExpress your answer using two decimal places.• View Available Hint(s)να ΑΣφpH =SubmitPart B120.0 mL of 0.11 M NH, with 240.0 mL of 0.11 M NH, CIExpress your answer using two decimal places.> View Available Hint(s)Πνα ΑΣφpH =P Pearsonright 2022 Pearson Education Inc. All rights reserved. Terms of Use | Privacy Policy I Permissions | Contact Us |f9f124+DDI&

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I Review | Constants | Part A 50.0 mL of 0.15 M HCHO2 with 80.0 mL of 0.12 M NaCHO2 Express your answer using two decimal places. > View Available Hint(s) Hνα ΑΣφ ? pH = Submit Part B 120.0 mL of 0.11 M NH3 with 240.0 mL of 0.11 M NH,CI Express your answer using two decimal places. > View Available Hint(s) ? pH = P Pearson

I Review | Constants | Part A 50.0 mL of 0.15 M HCHO2 with 80.0 mL of 0.12 M NaCHO2 Express your answer using two decimal places. > View Available Hint(s) Hνα ΑΣφ ? pH = Submit Part B 120.0 mL of 0.11 M NH3 with 240.0 mL of 0.11 M NH,CI Express your answer using two decimal places. > View Available Hint(s) ? pH = P Pearson

Fundamentals Of Analytical Chemistry

9th Edition

ISBN:9781285640686

Author:Skoog

Publisher:Skoog

Chapter31: Introduction To Analytical Separations

Section: Chapter Questions

Problem 31.19QAP

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Patients undergoing an upper gastrointestinal tract laboratory test are typically given an X-ray contrast agent that aids with the radiologic imaging of the anatomy. One such contrast agent is sodium diatrizoate, a nonvolatile water-soluble compound. A 0.378-m solution is prepared by dissolving 38.4 g sodium diatrizoate (NaDTZ) in.l.60 102 mL water at 3 1.2C (the density of water at 31.2C is 0.995 g/cm3). What is the molar mass of sodium diatrizoate? What is the vapor pressure of this solution if the vapor pressure of pure water at 31.2C is 34.1 torr?
A 18 g of unknown organic sample was dissolve in 756 mL of benzene. The boiling point of benzene was increased by 3.36oC. As the first step of analysis, determine the moecular weight of the unknow sample? Kb of benzene= 2.64oC/m Bb of benzene = 80.09 oC density of benzene = 0.874 g/mL at 25 °C Answer in whole number, no units required.
A 0.1 M solution of acid was used to titrate 10 ml of 0.1 M solution of alkali and thefollowing volumes of acid were recorded: 9.88 10.18 10.23 10.39 10.21 Calculate the 95% confidence limits of the mean and use them to decide whetherthere is any evidence of systematic error.
Prepare 1500mL of 0.5N H2SO4 solution with specific gravity of 1.84g/mL and assay of 97%
1) How many milliliters of 70% ethanol and 20% ethanol must be mixed to prepare 500ml of 30% ethanol? 2) A gaseous blend of nitric oxide and nitrogen (NOmax) contains 0.09% v/v nitric oxide (NO) and 99.92% v/v nitrogen. Express the strength of nitric oxide in PPM (round to whole number) 3) The ratio strength of a 10mL vial of neostigmine methylsulfate injection USP is 1:1500.Calculate the amount of neostigmine methylsulfate in milligrams contained in the 10mL vial. (Round to the whole number).
Q1. Dissolved 0.273 grams of pure sodium oxalate (Na,C,O.) in distilled water and added sulfuric acid and titration the solution at 70 ° C by using 42.68 ml of KMNO, solution and has exceeded end point limits by using 1.46 ml of standard oxalic acid (H; C;O.) with 0.1024 N. Calculate the normlity of KMN0.. Note that the molecular weight of sodium oxalate (Na,C,O.) = 134 and its equivalent weight = 67
A solution of Ca(OH)2 (FM = 74.09) was used in determining the equivalent weight of an unknown acid. To do so, the said solution was standardized against 0.1235 g KHP (FM = 204.22), requiring 7.45 mL to reach the phenolphthalein endpoint. On the other hand, a 0.0543 g unknown acid requires 12.85 mL standard Ca(OH)2 to reach the desired endpoint.a. Calculate the standard normality of Ca(OH)2.b. Calculate the equivalent weight of the unknown acid.c. From the information in the table, what is the most probable identity of the unknown acid?
Determine the percentage Fe in a sample of limonite from the following data:Sample : 0.5000g ; KMnO4 used = 50.00 ml ; 1.000ml of KMnO4 is equivalent 0.005317 g Fe,FeSO4 used = 6.00 ml; 1.000ml FeSO4 is equivalent 0.009200 g FeO ( ans 44.59 %)
A 50.00 (±0.03) mL portion of an HCl solution required 29.71(±0.03) mL of 0.01963(±0.0030) M Ba(OH)2 to reach an end point with bromocresol green indicator. The molar concentration of the HCl is calculated using the equation below (attached image): a.) Calculate the uncertainty of the result (absolute error). M=0.02333(±?????) M b.) Calculate the coefficient of variation for the result. CV= (Sy/y) x 100%
EX 2.Given-> Volume of Na+ = 500 ml Molarity of Na+= 0.0100M Molar mass of Na2CO3 = 105.99 gm/mole Millimole of Na+ = molarity × volume Number of millimole = 0.0100 × 500 = 5 millimole Na2CO3 ---> 2Na+ + CO32- Millimole of Na2CO3 = millimole of Na+/2 Millimole of Na2CO3 =5/2 = 2.5 millimole Mole of Na2CO3 = 2.5 × 10-3mole (1 mole = 10^3 millimole) Weight of Na2CO3 required = mole × molar mass = 2.5 × 10-3 × 105.99 =0.26 gm Hence, 0.26 gm Na2CO3 must dissolve in 500 ml of water.
5.00 mL of stock solution is diluted to 25.00 mL, producing solution ALPHA. 10.00 mL of solution ALPHA is diluted to 25.00 mL, resulting in solution BETA. 10.00 mL of solution BETA is then diluted to 25.00 mL, producing solution GAMMA. dilution factor for ALPHA from stock solution = 0.167 dilution factor for BETA from ALPHA solution = 0.0476 part c and d?
An unknown sample of Cu2+gave an absorbance of 0.262. Then 1.00 mL of solution containing 100 ppm (ug/mL)Cu2+was mixed with 95.0 mL of the unknown and the mixture was diluted to 100 mL in a volumetric flask with deionized water. The absorbanceof the new solution was 0.500. a.Denoting the initial unknown concentrationas [Cu2+]i,write an expression for the final concentrationafter dilution [Cu2+] b.Find [Cu2+]in the unknown.