Table 4.2 Determination of acetic acid content in vinegar sample. mL of Volume of NaOH, mL % (w/v) Replicate Sample Initial Final Volume added (mean±SD) 1 10 mL 0.00 7.8 250 mL 10 mL 7.8 15.3 25 mL 10 mL 15.3 22.4 3 50 mL
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- Which of the following methods of preparing 0.0010M HCl solution from 1.0M stock HCl Solution willyield the smaller overlall uncertainty? a. A one step Dilution that uses 1mL(±0.006) pipet and 1000mL(±0.30) Volumetric Flask, or b. A two-step dilution that uses 20mL (±0.03) pipet and a 1000mL(±0.30) volumetric flask and a25 mL (±0.03) pipet and a 500mL(±0.20) volumetric flask for second dilution.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 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.
- The gravimetric factor used to express CoCBr6·H20 in a sample that is finally weighed as PbClBr is choose below: FW PbClBr / 6 x FW CoCBr6·H20 FW CoCBr6·H20 / FW PbClBr FW CoCBr6·H20 / 6 x FW PbClBr 6 x FW PbClBr / FW CoCBr6·H20Study 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%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)
- Q3 / The solubility of sodium chloride NaCl in water at 290 Kis 35.8 kg / 100 kg of water. Express the solubility as the following: 1. Mass fraction and the mass percent of Naci 2. Mole fraction and mole percent of NaCl 3. kmol of NaCl per 1000 kg of water Note: molecular weight of NaCl = 135.8, and water = 18.016Gypsum (CaSO4) is a common percipitant in water desalination. CaSO4 <=> Ca2+ + SO42- where Ksp=10-4.6. Assuming that: [Ca2+]=2x10-2 M; [SO42-]=2x10-3M: a) Find Qsp or IAP for the given water b) Find the solubility index (SI) and determine whether CaSO4 is under, super, or at saturation in water. Feel free to make any assumptions you wish, as long as they're stated.Bristol Community College Fall River, Massachusetts Experiment 6: Molar Mass of a Molecular Solid from Freezing Point-Depression Measurement Name: __________________________________ Date: _______________ Approved: ___________ DATA SHEET Mass of lauric acid (in Part II) Mass of benzoic acid (in Part II) Freezing temperature of pure lauric acid (from Part I) data from Video 2 of Part I Freezing temperature of solution (from Part II) data from Video referenced in Part II Freezing point depression, Tf ( = Tf, lauric acid – Tf, solution) Molality (m) of solution ( Eq. 1) Moles of benzoic acid ( Eq. 2) Experimental molar mass of benzoic acid (Eq. 3) Calculate the molar mass of benzoic acid, C6H5COOH. Percent error Summary Questions A student determines…
- 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%Given the following data and the λmax determined in the previous questions, determine:1) the molar absorptivity of copper at λmax of cobalt 2) the molar absorptivity of copper at λmax of copper 3) the molar absorptivity of cobalt at λmax of cobalt 4) the molar absorptivity of copper at λmax of cobalt 5) the concentration of cobalt in mM 6) the concentration of copper in mM Copper concentration, mM λmax cobalt λmax copper 20.00 0.177 0.238 40.00 0.382 0.487 60.00 0.518 0.731 80.00 0.765 1.134 100.00 0.902 1.391 Cobalt concentration, mM λmax cobalt λmax copper 20.00 0.186 0.163 40.00 0.412 0.354 60.00 0.637 0.477 80.00 0.754 0.694 100.00 0.961 0.831 unknown, Abs at λmax cobalt 0.893 unknown, Abs at λmax copper 0.997EX 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.
