endpoint of the titration is reached. In an acid-base titration the experimenter is trying to determine the equivalence point of the reaction‚ which is the point when the amount of base added was exactly the correct amount to have the moles of base completely react with the moles of acid. If the experimenter chooses the correct indicator‚ the endpoint of the reaction will be as close as possible to the actual equivalence point. The indicator most commonly used for acid base titrations is phenolphthalein
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Outlet stream S3 contains 21.6 lbmol/hr of component F The overall conversion of Reactions 1 and 2 based on A is 80% Column specifications: Light key = A Heavy key = B Mole-recovery of B in bottom = 98% Process Feed: Molar flow rate of A in feed = 100 lbmol/hr Molar flow rate of B in feed = 150 lbmol/hr Other information: Mole fraction of A in column overhead = 24.06% Total molar flow ratio of stream S5 to S4 = 9.26 3 Molar ratio of component
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sodium carbonate and 1.00 g of lead II nitrate was given to start the lab out with. To calculate the percent yield of lead II carbonate‚ the limiting reactant must be identified first. 1.00 g Na2CO3 x 1 mole106 g = 0.00943 mole Na2CO3 1.00 g Pb(NO3)2 x 1 mole331 g = .00302 mole Pb(NO3)2 Limiting reactant: lead II nitrate Next‚ the theoretical yield of lead II
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10-5 mole/ L.s | |2 |0.10M |0.050M |2.623 x 10-5 mole/ L.s | |3 |0.05M |0.10M |2.326 x 10-5 mole/ L.s | |4 |0.10M |0.025M |9.661 x 10-6 mole/ L.s | |1T2 |0.050M |0.050M |3.158 x 10-5 mole/ L.s | | The rate of each reaction was determined through this table. The slope for each experiment was determined. An example for such experiment A follows: Slope = ∆S2O82-/∆T = [(10.0-2.0) x 10-4 mole]/(22.5-4.8)sec = [8.0 x 10-4 mole]/17.7 sec = 4.5 x 10-5 mole/sec The rate would be : [4.5 x 10-5 mole/sec]/0
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| Calculations were completed as follows: Vol. of NaOH added (mL) * 1L/1000mL * 0.981mol NaOH/L = moles of NaOH added For monoprotic acid (R-COOH): Moles of Acid = Moles of NaOH Mass of acid (g)/moles of Acid = Acid’s molar mass (g/mol) For Diprotic acid (R(COOH)2): Moles of acid = ½ * Moles of NaOH Mass of acid/moles of acid = Acid’s molar mass (g/mol) To further enforce the conclusions drawn regarding the acids identity from the titration (crotonic acid
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Abstract. This report is about how to standardize a Sodium Hydroxide (NaOH) solution by titrating it with pure sample of Potassium acid Phthalate (KHC8H4O4). This experiment has two sections. The first section is to standardize the Sodium Hydroxide by titration. Three sample of 0.7 – 0.9 g of solid KHP are place into each of the three numbered Erlenmeyer flasks. 50 ml of distilled water are added to each three of it from graduated cylinder and constantly shake it until the KHP solution are completely
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Though a person has many risk factors‚ does not mean that they will end up getting cancer‚ and vise-versa‚ if a person has little or no risk factors‚ that does not mean that they will never be diagnosed. Risk factors for melanoma include; sun exposure‚ moles‚ fair skin‚ freckles‚ light hair color‚ a family history of melanoma‚ a personal history with melanoma‚ immune suppression‚ age‚ gender‚ and Xeroderma Pigmentosum
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of moles and grams of urea and the amount of urea used originally used in the experiment‚ so the ammonium chloride will be primarily compared to the corrected amount of urea to make sure that the amount of solvent is equal. Graph 1 and 5 and Data Table 2 show that the mass of solutes were overall very similar: 8.735 grams of ammonium chloride and 8.99 grams of urea. Overall‚ this experiment proved that the freezing point depression is a colligative property that depends on the number of moles of
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The limiting reagent it CuCl2 because it is the one that will run out first compared to the .009 moles of Na2Cl3‚ which it the excess reagent because it is a higher amount of moles compared to the .007 moles of the CuCl2. The amount of excess reagent in grams that should remain in solution if the theoretical yield of CuCO3 is produced is: 1 mole of CuCl2 (63.55) +(2*35.45) =134.45 1mole of Na2Cl3 (22.99 *2) +(12.01) +(3*16) =105.99 (this is the excess
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charge on an ion formed when an atom gains two electrons? -2 The Mole 1) Find the molar mass of zinc 65g/mol 2)Find the molar mass of sulfuric acid H2 SO4 98g/mol 3) How many moles of sodium chloride are present in 117g of Na2 CI 2 moles 4) I have 54g of water (H2O) and 84 g of iron (Fe) do I have more moles of water or of iron I have more miles of water (54 g = 3 moles H20 ‚ 84g = 1.5 moles Fe) Arrangement of Electrons 1) Draw diagrams to show the electron
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