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    Ideal Gas Law Lab

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    Introduction: The goal of this experiment was to measure the amount of gas produced in various reactions. The Ideal Gas Law was needed in order to calculate the mass of reactants and moles of gas produced: ‚ where is the pressure in atm‚ is the volume in Liters‚ is the number of moles‚ is the ideal gas constant [0.082 (Latm)/(Kmol)]‚ and is the temperature in Kelvins. Considering the units on R‚ it was important to convert pressure‚ volume‚ and temperature to atm‚ L‚ and K‚ respectively. In

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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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    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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    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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    the truth

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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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    CHE Problems

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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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    Organic Compound Analysis

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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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    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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    Rate Law Lab

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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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    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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