2. Boil 1 Lof distilled water for 5 min to expel CO2. Pour the water into a polyethylene bottle, which should be tightly capped whenever possible. Calculate the volume of . NAOH needed to prepare 1 L of 0.1 M NAOH. (The o 1.22 NaOH is per milliliter of solution.) Use a graduated cylinder to transfer this density of &-
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- 23A 0.20 g sample of primary standard Na2C204 (134 g/mol) needed 37.22 mL of of KMNO4 solution to reach the end point. What is the molarity of KMN04 reagent Please explain the solution in detail O a. 0.016 M O b. 0.055 M Oc 0.008 M O d. 0.100 MForeattt part of iTIS qtestfoni; tonsider FOHOWIHg Btetat: You need to make a stock solution of Ca( C2H3O2 )2 with a concentration of 6.50 M. You have a 850.0 mL volumetric flask available to make this stock solution in. You are then tasked with using this stock solution to prepare a sample for use in an experiment. The experimental sample should have a final volume of 15.0 mL and a concentration of 0.350 M Ca(C2H3O2)2. a) What is the molar mass of Ca(C2H3O2)2? b) How many moles of the solute would be needed to create the stock solution using the volumetric flask available? c) How many grams of the solute should be added to the volumetric flask to create the stock solution? d) What volume, in mL, of the stock solution should be used to prepare the experimental solution through dilution? US 目 MThe ascorbic acid, C,H;Os (176.124 g/mmol) content of a bottle of peach juice containing 2.50 L was determined by titration with a standard solution of potassium iodate, KIO3. The KIO; solution was prepared by dissolving 100.4 mg primary standard grade KIO; (214.001 mg/mmol) to a final volume of 250.0 mL. To a 200.00-mL aliquot of the sample, 10 mL of KI, 10 mL of HCl and 5 mL of starch solution were added and subsequently titrated, requiring 18.19 mL of the KIO; titrant. The relevant reactions are given below: KIO; + 5KI + 6H* → 3l2 + 6K* + 3H;0 CH;O, + l2→ CH,O, + 21- + 2H* a. How many millimoles of ascorbic acid is in the titrated sample? b. What is the mass of ascorbic acid per bottle of the peach juice?
- ne concentrated solution. 3- Describe the preparation of 400 mL of 6 F H;PO, from the commercial solution which is 85% (w/w) H;PO4 and has density of 1.69 g/mL.A 0.8250 g of soda ash sample was dissolved in distilled H₂O and diluted to 250-mL using a volumetric flask. This solution was labeled as dilute soda ash solution (DSAS). a. A 50.00 mL aliquot portion of the dilute soda ash solution needed 5.20 mL of 0.1200 M HCI to reach the phenolphthalein end point. About 2-3 drops of methyl orange to the same mixture and it needed 10.00 mL of 0.1200 M HCI to reach the methyl orange end point. 1. Calculate the number of millimoles of HCI needed to reach the phenolphthalein end point. 2. Calculate the number of millimoles of HCI needed to reach the methyl orange end point. 3. What is the %Na₂CO, and %NaHCO, present in the soda ash sample? b. From the dilute soda ash solution (DSAS) two 50-mL portions were obtained. The first portion needed 5.20 mL of 0.1200 M HCI to reach the phenolphthalein end point. The second 50-ml portion needed 15.20 mL to reach the methyl orange end point. 4. Calculate the number of millimoles of HCI needed to reach the…nohesup och towana no bas moldog sniwallot om besl odi sisluola) (1) obrolde (II)noni y 12 Tamisanos noituloa zucoups Jm-0.28 nA noituloz od to vinelom fatuloz or lo 22ant orb 21 JW 2 Solution 1 2 3 4 Part 1 Data related to aqueous solutions of sodium chloride (NaCl) and aqueous solutions of ethanol (C₂H5OH) are provided in the table below. Use the table to answer the following questions. Circle the letter of the choice that best answers the question. NaCl 3.0 3.0 3.0 3.0 Mass (g) H₂O 100.0 200.0 300.0 400.0 Solution 5 6 7 8 Snoitulor odi to smudov odi 21 IndW 2. Which of the following solutions is the most dilute? a. Solution 1 b. Solution 2 Volume (mL) C₂H5OH 2.0 5.0 9.0 15.0 c. Solution 3 H₂O 100.0 100.0 100.0 100.0 d. Solution 4
- 5. A solution containing 15 mmole of CaSO, is diluted to 250 mL. The number of milligrams of CasO4 2H;0 per ml of the final solution is (Atomic weights: Ca 40, S =32, 0 = 16, H = 1):Give detailed Solution (no need Handwritten1. How do you prepare 500 mL of a ~0.10 M NaOH solution? What mass of NaOH(s) should you use? 2. If you wish to use 20 mL of 0.10 M NaOH to react fully with potassium hydrogen phthalate (KHP), what mass of KHP should you use in each titration trial?, g (formula for KHP is: C8H5KO4.)
- Determinati an 25 ml by dissolving 110n (111) per 1 This (as ml 750ml experiment Solution The to of was The a and process Q on added The portion layer taken 1 volumetric diluted Litre Colution Two These of in diluted ml soo 1% ETA of Question A. reaction an Solution 0₁ 446 9 of A. R. hydrated sulphate, (NH4) Fe (504) ₂ · 12H₂0 2 of distilled then of wertel) 0,0446 g/L of lion Absorbance was 250 ml in to ·layers frask was Is 2. 3. Was Chlore form and were 5 ml, 10 ml, 15 ml por tions 4. Iron • iron (111) four oxine it mixture the done. (111) seperatory conducted Volume SMT as 10 ml 15 ml 20 ml Plo + absorbance formed readings a (A.R) Jayer measured the por tions Solution was Water diluted out 20 ml were then Each of these mark against funnel as was after with for 8- ity droxy quinolate. follows. to iron (111) solution was prepared of then Calibration was ) 10 obtain collected. ammonium F. W-481,979 then these Absorbance 0,076 0,169 9185 0₁234. Shook. the shaking added portions…3. Analysis of a mixture consisting of NaOH + Na2CO3 + inert matter gives the following data: 10.00 g. Its aqueous solution is diluted to 250.0 mL and two separate 25.00 mL sample portions are titrated. With one portion, an end point with phenolphthalein is obtained in cold solution, with 44.52 mL of 0.5000 N HCI. The other portion requires 46.53 mL of the acid for an end point with methyl orange. Calculate the percentage composition of the original sample.6. The phosphorus content in a 0.3004-g sample was precipitated as (NH4);PO4 .12M0O3 a slightly soluble precipitate. The precipitate was filtered, washed, and further redissolved in acid. The resulting solution was treated with an excess of Pb²* resulting in the formation of 0.3017 g of PbMoO4. Calculate the % w/w of P2Os.||