Experiment II. Separation Of a Sample Mixture By Liquid-Liquid Extraction Reading assignment: Techniques in Organic Chemistry 2nd ed pages 75-99. 3rd ed pages 113-140. Topics and Techniques i) identification of solvent layers of two immiscible solvents ii) partioning of a compound between two immiscible solvents and determination of KD iii) liquid-liquid extraction with aqueous acids and bases with organic solvents. iv) use of drying agents Introduction Liquid-liquid extraction is a method
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many different factors. This gives rise to the concept of percent yield. Sodium bicarbonate and hydrochloric acid was used in the experiment. The yields of sodium chloride reaction were determined and then compare it to the theoretical yield. HCl(aq) + NaHCO3(s) -> NaCl(aq) + CO2(g) + H2O(I) Materials and Methods In order to determine the percent yield of a reaction they will have to perform a series of instructions leading the students to the completion of the experiment. Weighted a clean dry
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Module 4: 10. Strong Bases: NaOH‚ KOH‚ Na2CO3‚ NaHCO3‚ Na3PO4‚ NaHPO4‚ Strong Acids: Fe(NO3)3‚ HCL‚ HNO3‚ Al(NO3)3‚ NiCl2‚ H2SO4 11. Acidic Neutral Basic NaCl KNO3 NaOH HC2H3O2 Na2CO3 NaC2H3O2 NaHCO3 Fe(NO3)3 NaNO3 Na3PO4 HCl MgSO4 KOH HNO3 Na2SO4 NaHPO4 CuSO4 NaNO2 CoCl2 Al(NO3)3 NiCl2 H2SO4 KCl NH4Cl 12. CuSO4 CuCO3.Cu(OH)2(s) + 2 H2SO4(aq) ----> 2 CuSO4(aq) + CO2(g) + 3 H2O(l) Na3PO4 Na3PO4 (aq) + 3H2O (l) --> H3PO4 (aq) + 3NaOH
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provided pre-diluted chemicals. Utilizing the 96-well plate‚ 2 pipet drops of each chemical were added to the wells in the following combinations: a) NaHCO3+HCl b) HCl+Bromthymol blue c) NH3+1 drop Bromthymol blue d) HCl+blue dye e) Blue dye+NaOCl followed by HCl f) NaOCl+KI followed by various test foods g) KI+Pb(NO3)2 h) NaOH+phenolphthalein i) HCl + phenolphthalein j) NaOH+AgNO3 k) AgNO3+NH3 l) NH3+CuSO4. Along the way‚ observations were made pertaining to the reactions witnessed. This experiment
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+ 2 HCl (aq) → 2NaCl(aq) + H2O(l) + CO2 (g)KA3 + KA1 : NaHCO3 (s) + HCl (aq) → NaCl (aq) + H2O (l) + CO2 (g)(e) Energy Na2CO3 (s) + 2 HCl (aq) + H2O (l) + CO2 (g)HKA2 / kJ H 2 x HKA3 2 NaCl (aq) + 2 H2O (l) + 2 CO2 (g) 2 NaHCO3 (s) + 2 HCl (aq)(f) Na2CO3(s) + 2 HCl (aq) → 2NaCl(aq) + H2O(l) + CO2 (g) [1] HKA2 = - P kJ NaHCO3 (s) + HCl (aq) → NaCl (aq) + H2O (l) + CO2 (g) [2] HKA3 = + Q kJ Reverse equation [1]2NaCl(aq) + H2O(l) + CO2 (g) → Na2CO3(s) + 2 HCl (aq)
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by gravity and set it aside to dry. The filter paper weighed about 1.320g and sucrose weighed out at 1.028g. Sucrose also ended up on the outside rim of the filter paper. We then proceed to separate the aspirin. We measured 30mL of NaHCO3 and mixed it with 7mL of 6M HCl. We drained the organic layer into a pre-weighed Erlenmeyer flask and save it for recovery of acetanilide. We took the combined aqueous extracts in an
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chemical reactions. Balancing Chemical Equations Balance the equations below: 1) ____ N2 + ____ H2 ____ NH3 2) ____ KClO3 ____ KCl + ____ O2 3) ____ NaCl + ____ F2 ____ NaF + ____ Cl2 4) ____ H2 + ____ O2 ____ H2O 5) ____ Pb(OH)2 + ____ HCl ____ H2O + ____ PbCl2 6) ____ AlBr3 + ____ K2SO4 ____ KBr + ____ Al2(SO4)3 7) ____ CH4 + ____ O2 ____ CO2 + ____ H2O 8) ____ C3H8 + ____ O2 ____ CO2 + ____ H2O 9) ____ C8H18 + ____ O2 ____ CO2 + ____ H2O 10) ____ FeCl3 + ____ NaOH ____
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Classification for Functional Groups Alcohols Lucas test (Differentiates primary‚ secondary‚ and tertiary alcohols) Reagent: ZnCl2 in conc. HCl Observation: Rate of reaction (tertiary alcohol> secondary alcohol> primary alcohol) Procedure: 2mL Lucas Reagent in test tube+ 3-4 drops of alcohol‚ stopper‚ shake vigorously‚ NOTE time required (less than 10m mins only) to form an emulsion or separate layers. Oxidation (Confirms if alcohol is oxidizable: presence of H in C-OH bond) Reagent:
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sodium carbonate‚ soda ash‚ a titration is done using a standardized solution of HCl. Aqueous HCl is a strong acid and therefore almost completely disassociates into H+ and CL-. Therefore‚ when HCl is used in a titration‚ the H+ is the titrant. Carbonate in aqueous solution is able to accept a proton‚ i.e. it acts as a base. When carbonate accepts the H+ a bicarbonate ion is formed. Na2CO3(aq) + HCl(aq) NaHCO3(aq) + NaCl(aq) This is not the complete reaction for the titration because
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Section: 1411-76426| Data Tables: Part 1: |Chemicals|Well No.|Observations of the Reaction| A.|NaHCO3 + HCl|1 |clear bubbles formed| B.|HCl + BTB|2 |turned orange after BTB was added‚ transparent‚ acidic| C.|NH3 + BTB|3 |turned dark blue after BTB was added‚ opaque‚ basic| D.|HCl + blue dye| |blue dye was missing from kit| E.|Blue dye + NaOCl| |blue dye was missing from kit| | with the 1 drop of HCl| || F.|NaOCl + KI|4 |small white precipitate‚ color changed to black after starch was added
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