B. Weight of Na2SO4 Sample 0.3512 g C. Total Vol. of Precipitant (BaCl2) used WITH 1М ВаCI2 PLEASE FIND THE calculated volume of BaCl2 precipitant will be calculated using this formula: NVBaCl2 = (weight of sample) (1000)(EW) V(ml)BaCl2 = (weight sample) %3D (1000)N(EW)
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- Hexanoic acid was added to an immiscible biphasic solvent system, water and CCl4 at 20.0OC and the equilibrium concentrations of hexanoic acid were determined to be 3.66 g/L in H2O and 67.0 g/L in CCl4. Caluclate the distrubution coeffiecent (D2) of hexanoic acid in water with respect to CCl4.An aqueous waste stream that has a maximum concentrationof 0.50 M H₂SO₄ (d=1.030 g/mL at 25C) will be neutralized by controlled addition of 40% caustic soda (NaOH; d=1.430 g/L) before it goes to the process sewer and then to the chemical plant waste treatment facility. However, a safety re-view finds that the waste stream could meet a small stream of animmiscible organic compound, which could form a flammable vapor in air at 40.°C. The maximum temperature of the causticsoda and the waste stream is 31°C. Could the temperature in-crease due to the heat of neutralization cause the vapor to ex-plode? Assume the specific heat capacity of each solution is 4.184 J/gK.A 100 g soil containing 20% smectitie mineral matter was mixed with 150 mL solution containing 0.10 M Pb. CEC of pure smectitie is 110 emol(+)/kg. CEC of the soil (cmol(+)/kg) is:
- A chemist receive different mixtures for analysis with the statement that it contain NaOH, NaHCO3 , Na2CO3 or compatible mixtures of these substances together with the inert material. From the data given, identify the respective materials and calculate the percentage of each component. 1.000g samples and 0.2500 N HCl were used in all cases. Sample 1 With phenolphthalein as the indicator, 24.32 ml of HCl was used. A duplicate sample required 48.64 ml HCl using methyl orange as the indicator. Sample 2. With phenolphthalein as the indicator it uses 28.2 ml of HCl to make it colorless and added with methyl orange indicator and uses 11.3 ml of HCl to reach the end point.By the use of Henderson Hasselbalch equation; pH = pKa + log{[acetate ion]/[acetic acid]} 4.5 = 4.75 + log{[0.10 M]/[acetic acid]} -0.25 = log{[0.10 M]/[acetic acid]} [Acetic acid] = 0.10 M/ 10-0.25 [Acetic acid] = 0.10 M/0.56 [Acetic acid] = 0.1786 M Moles of sodium acetate dissolved in 250 mL buffer solution = 0.10 M× (250mL/1000mL) × 1L = 0.025 mol Weight (w) of sodium acetate (purity 100%) dissolved to prepare 250 mL of solution with buffer concentration of 0.10 M is calculate as follow; w100% = 0.025 mol × 82.0343 g/mol = 2.051 g Weight (w) of sodium acetate (purity 99%) is calculate as follow; w99% = 2.051 g× (100/99) = 2.072 g What was the volume of 6.12 M acetic acid HC2H3O2 needed to prepare the 250 mL acetic acid/acetate ion buffer solution required in this part? Show your calculations.One litre of a saturated aqueous solution of Ag2SO4 (MW = 311.79 g mol- 1) at 25 °C is evaporated to dryness. 4.844 g of Ag2SO4 residue was produced. What is the solubility product (Ksp)?
- It is known that acid content has a major effect on theflavor of vinegars, but most cheaper vinegars are diluted similarly to 5% acidity Wt./vol. % is equivalent to gsolute per 100mL solution (so 5% is equivalent to 5 g acid/100 mL solution). a.) First, calculate the approximate molar concentration of acetic acid in the 5% wt./vol vinegar. b.) Next, calculate the expected molarity of acetic acid in the solution upon dilution by a factor of 5. Thank you!The student then determined ΔH neutralization for the reaction of sodium hydroxide and acetic acid, using the procedure described in this module. The student added 100.0 mL of 0.8500M NaOH to 100.0 mL of 0.8404M acetic acid. Prior to and following the mixing of the acid and base solutions, the following temperature-time data were collected. g) Identify the limiting reagent and briefly explain why it is limiting. h) Find ΔH neutralization for the reaction.Study the tabulated data during an gravimetric analysis of BaSO4 experiment Trial T1 T2 T3 Mass of sample (g) 1.0040 1.0100 1.0050 Constant weight of empty crucible (g) 10.2453 10.2454 10.2454 Constant weight after ignition (g) 10.3253 10.3252 10.3253 1. Calculate the mean mass of precipitate as BaSO4 a. 0.0799 g b. 0.0800 c. 0.08 g 2. Calculate the mean weight of sulphate (MM BaSO4 = 233.39 g/mol, So42- = 176 g/mol)? a. 0.0603 g b. 0.06 g c. 0.06025 g 3. Calculate the mean % SO42- in the sample a. 6.00% b. 5.991% c. 5.9908%
- If 500mL of 0.10M Ca2+ is mixed with 500mL of 0.10M SO42-, what mass of calcium sulfate will precipitate? Ksp for CaSO4 is 2.40 • 10-5. express answer to 3 significant figures. And include units.A powdered crude sample of Na2CO3 containing only inert impurities is to be determined by reacting 225.0 mg of the crude sample to 10.0 mL of 3.00 M HCl solution, and bubbling the resulting CO2(g) product in water that is at exactly 29 °C. After the reaction has completed, the level of the liquid inside the eudiometer rests 4.30 cm above the water level in the beaker. The graduation on the eudiometer indicates that the trapped gas is 44.37 mL. The experiment was done under a barometric pressure of 755.2 torr.What is the %purity of the sample to the nearest whole number?The thiourea in a 1.455 g sample of organic material was extracted into a dilute sulfuric acid solution and titrated with 37.31 mL of 0.009372 M Hg2+ via reaction: 4(NH2)2CS + Hg2+ →[(NH2)2CS]4 Hg2+ P.S. Answer only the last two letters of the following questions. (Only C and D) a. Is this an example of total analysis technique or concentration technique? Explain. b. Calculate the percent (NH2)2CS ( 76.12 g/mol) in the sample. c. What is classification of the analysis based on the amount of sample and amount of analytes present? Explain. d. If the true value is 10.00%, calculate the absolute and relative error.