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    Mass of empty bottle (g) 21.5924 Mass of ascorbic acid used (g) 0.1002 Number of moles: (mass of ascorbic acid used)/(molecular mass of ascorbic acid) = 0.1002g/(176.10gmol-1) = 5.69 x10-4 moles M1: concentration of stock ascorbic acid V1: volume in volumetric flask with stock solution M2: concentration of diluted ascorbic acid V2: volume in volumetric flask with diluted solution (M1 x 100)/1000 = 5.69 x10-4 moles M1: 5.69 x10-3 mol/L M1V1 = M2V2 5.69 x10-3 mol/L x (10/1000) = M2 x (100/1000)

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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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    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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    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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    Studies 265747595250 Finding suspicious moles or skin cancer early is the key to treating skin cancer successfully.Examining yourself is usually the first step in detecting skin cancer. You are a family physician and you have a few patients coming in to have their moles observed. For each patient use the ABCD chart to identify whether the mole is suspicious. Write a summary in the space provided‚ answering the following: 1. Is there a possibility the mole is cancerous and why/why not? 2. What

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    generate elemental sulfur from hydrogen sulfide. 2H2S + 3O2 2SO2 + 2H2O SO2 + 2H2S 3S + 2H2O How much sulfur (in grams) is produced from 31.2 grams of O2? A) 140.7 g B) 20.8 g C) 31.3 g D) 62.5 g E) none of these 4. How many moles of hydrogen sulfide are contained in a 69.5-g sample of this gas? A) 0.49 mol B) 2.04 mol C) 103.6 mol D) 34.5 mol E) 4.08 mol 5. How many atoms of hydrogen are present in 5.9 g of water? A) 2.0e23 B) 1.8e24 C) 7.1e24 D) 3.9e23

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    of reaction is negative heat of solution. Heat of solution was found by using the formula qsolution = mCsΔT (m= mass; Cs= 4.184 J/g°C; and ΔT = change in temperature). The enthalpy change was found by dividing the heat of reaction by the number of moles of H2O formed. Procedure: Experiment 1 was performed after getting a coffee cup calorimeter. A graduated cylinder was used to obtain 50 mL of 0.862M HCl. It was added to the empty coffee cup calorimeter‚ and its initial temperature was obtained by

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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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    BACKGROUND: Esterification is a reaction that combines an alcohol with an organic acid‚ with a water molecule is being taken out‚ and an ester is formed. A concentrated acid catalyst speeds up the esterification. In this experiment set up‚ sulphamic acid was used‚ as it is a solid acid and be added dry‚ without any water‚ which is a reactant in this experiment equilibrium. Esterification is a slow and reversible reaction. The equation for the reaction between an acid RCOOH and an alcohol RẬᶦ

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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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