A ceric sulfate solution that was standardized using 200 mg of arsenic trioxide (MW197.84g/n) consuming 38.5ml of the solution. The solution was the used to assay 2,000 mg sample of Ferous Carbonate (FeCO3 = 115 85
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- The salt content was extracted from a 12.0000g junk food sample. The extract was diluted to 100.00mL. From this solution, 15.00mL was taken and required 23.75mL of a 0.08943M AgNO solution to reach the endpoint. What is the percentage by mass of salt as NaCI (58.45 g/mol) in the junk food sample? (Answer: 6.897% NaCl)Barium chloride is reacted with sodium sulfate to produce barium sulfate and sodiumchloride. How many moles of the precipitate is produced from 0.8 moles of BaC12?A. 0.4 moleB. 0.8 moleC. 1.6 molesD. None of the above A 315mL of water was added to 2100 mL of 19M NaCI solution. What is the newconcentration of the solution?(Use the given information: MW: Na = 23 g/mol, C1 = 35g/mol)A. 2.85 MB. 2.48 MC. 18.60 MD 19.0 M 500g of Al2S3 is dissolved in H20 to make 15.0 L of solution. What is its normality? (MW:Al=27g/mol, S=32g/mol, H=1g/mol, 0=16g/molA. 0.11 NB. 0.07 NC. 0.22 ND. 0.04 NVolume 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)
- In one experiment, 0.0995M HCl was used to analyze soda ash and the following data were obtained: 0.2980 g Mass of Na2CO3 Initial reading of HCI Final reading of HCI 0.20mL 20.00mL What is the net volume in liters of HCl used? How many moles of Na2CO3 was used in the analysis? What is the corresponding Normality of the HCl used? What is the percent Na2CO3 in the soda ash sample?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 =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
- If 50mL of NaOH (mw=40g/mol) solution required 30mL of sulfuric acid (mw=98g/mol) solution in a titration and 30mL of sulfuric acid solution were required in the titration of 0.5038g of pure sodium carbonate (mw=106g/mol). What was the normality of the NaOH solution? (use 4 significant figures)What's the average concentration of NH3 in the commercial cleaning solution using the data Volume of Cleaning solution 10.00 for all 3 trials, molarity of HCl 0.0530 for all 3 trials, Volume of HCl first trial is 4.8 second trial is 4.4 and third trial is 4.4, Molarity of NH3 in cleaning solution first trial is 0.0254 second trial is 0.0233 and third trial is 0.0233. What's the diluted solution in all 3 trials?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%
- A solution containing 25.0 mL of H2C2O4 required 13.78 mL of 0.04162 N KMnO4 for titration forming Mn2+ and CO2 as products. Calculate the normality and molarity of the H2C2O4..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) sampleHow many milliliters of an injection containing 1 mg of drug per milliliterof injection should be administered to a 6-month-old child weighing 16 lb toachieve a dose of 0.01 mg/kg? answer: 0.073 mL. what is the full solution?