204.22 g/mol No. of moles of KHP = Mass of KHP used / Molar mass = 0.42 g / 204.22 g/mol = 0.0021 moles Concentration of NaOH = No. of moles / Volume = [0.0021 mol / {(22.50 + 25) / 1000} L] * 100 = 4.4 M Trial 2 Mass of KHP transferred = 0.4139 g Volume of Distilled water = 25 mL Volume of NaOH used = 22.80 mL Molar mass of KHP = 204.22 g/mol No. of moles of KHP = Mass of KHP used / Molar mass = 0.4139 g / 204.22 g/mol = 0.0020267 moles Concentration of NaOH = No. of moles / Volume = [0.0020267
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Stoichiometry of a Precipitation Reaction March 20‚2013 Amber McCollum Introduction Stoichiometry is a branch of chemistry that deals with the quantitative relationships that exist among the reactants and products in chemical reactions To predict the amount of product produced in a precipitation reaction using stoichiometry‚ accurately measure the reactants and products of the reaction‚ determine the actual yield vs. the theoretical yield and to calculate the percent yield. The equation
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and good-natured Mole loses patience with spring cleaning. He flees his underground home‚ heading up to take in the air. He ends up at the river‚ which he has never seen before. Here he meets Ratty (a water rat)‚ who at this time of year spends all his days in‚ on and close by the river. Rat takes Mole for a ride in his rowing boat. They get along well and spend many more days boating‚ with Rat teaching Mole the ways of the river. One summer day shortly thereafter‚ Rat and Mole find themselves near
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of moles of toluene = Mass/Mr = 43.35g/(92.15g/mol) = 0.470428648 moles ≈ 0.470 moles Volume of Sulphuric acid = 10mL (10cm3) Density of Sulphuric acid = 1.84g/cm3 Mass of Sulphuric acid = Density x Volume = 1.84g/cm3 x 10cm3 = 18.4g Mr of Sulphuric acid = 98.09g/mol No. of moles of Sulphuric acid = Mass/Mr = 18.4g/(98.09g/mol) = 0.187582832 moles ≈ 0.188 moles Limiting reagent: Sulphuric acid No. of moles of p-Toluenesulphonic acid = 0.188 moles Mr of p-Toluenesulphonic
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solvent kg of solvent = 10 mL x 1 g/mL x 1kg/1000g = 0.01 kg molality = moles of solute / kg solvent moles of solute = molality x kg solvent Isopropyl alcohol trial 1 moles of solute = 5.91 m x 0.01 kg = 0.0591 moles Isopropyl alcohol trial 2 moles of solute = 5.38 m x 0.01 kg = 0.0538 moles Ethyl alcohol trial 1 moles of solute = 4.30 m x 0.01 kg = 0.0430 moles Ethyl alcohol trial 2 moles of solute = 3.76 m x 0.01 kg = 0.0376
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38 grams | Mass of hydrated MgSO4 | 3.00 grams | Mass of crucible‚ lid‚ and anhydrous MgS04 | 12.82 grams | Mass of anhydrous MgSO4 | 1.45 grams | Mass of water in hydrated MgSO4 | 1.55 grams | Moles of anhydrous MgSO4 | 0.0120 moles | Moles of water in hydrated MgSO4 | 0.0860 moles | Observation of anhydrous MgSO4 | Different shade of white‚ more thicker and solid; powder-like | 3. To obtain the mass of water‚ first measure the mass of the crucible with the lid and anhydrous MgSO4
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AP Chemistry Analysis of Alum * Purpose: * In this lab we performed several tests to determine if our crystals were actually aluminum potassium sulfate. * Procedures: * Materials: * Chemicals: * Aluminum potassium sulfate‚ 2.5 g * Equipment – Part 1: * 150 mL beaker * Bunsen burner * 2 capillary tubes * Mortar and pestle * Notched stopper to hold thermometer *
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Marie Alessandra T. Reyes Group 4 18L Quantitative Determination of the Acidity of Soft Drinks I. Introduction Soft drinks are well known beverages among the young that are consumed mostly for pleasure. These beverages normally contain flavoring‚ sweeteners coloring‚ carbonic acid and acids. Carbonic acid and acids play an important part in the formulation of soft drinks. They enhance the flavor and give a pleasant refreshing ’lift ’ to the drink. The type of acid used can even affect the palatability
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Quantities The Mole General‚ Organic‚ and Biological Chemistry Copyright © 2010 Pearson Education‚ Inc. 1 Collection Terms A collection term states a specific number of items. 1 dozen donuts = 12 donuts 1 ream of paper = 500 sheets 1 case = 24 cans General‚ Organic‚ and Biological Chemistry 2 A Mole of Atoms A mole is a collection that contains the same number of particles as there are carbon atoms in 12.0 g of carbon 12C 6.02 x 1023 atoms of an element (Avogadro’s number) 1 mole of Element
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BSc Pharmaceutical Technology Practical - No 5 Preparation of a Primary Standard Solution Aim: to prepare a standard solution of pure silver nitrate and use it to determine the concentration of chloride ions in a sample of tap water and another sample of bottled water. Chemicals: * High grade purity silver nitrate * Potassium chromate indicator * Tap water * Bottled water * Distilled water Apparatus: * Laboratory oven * Dessicator * Conical flask
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