Aim: To Find and test the Iron content in different food sources INTRODUCTION: A Redox titration was used in order to perform this experiment. Reduction/oxidation (redox) process occurs when electrons are transferred from a donor species (the reducing agent) to another acceptor species (the oxidizing agent). It happens between an analyte and a titrant. A redox titration is done just as a normal titration is done‚ however instead of titrating an acid against a base‚ an oxidizing agent is titrated
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KHC8H4O4 (aq) + NaOH (aq) KHC8H4O3 - Na+ (aq) + H2O (l) Average volume of NaOH titrate used = ½ (5.00 + 5.00) = 5.00 cm3 No. of moles of KHC8H4O4 reacted = 0.01 × 0.05044 = 5.044 x 10-4 mol From the equation‚ mole ratio of NaOH ≡ KHC8H4O4 is 1:1‚ No. of moles of NaOH reacted = 5.044 x 10-4 mol
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EXPERIMENT 3 Name: Flame Tests & Electron Configuration Pre-Laboratory Questions and Exercises Due before lab begins. Answer in the space provided. 1. Write electron configuration for the alkali metals Li‚ Na‚ K‚ and Rb. Li ____He 2s1_____________________________________________ Na ____Ne 3s1______________________________________________ K _______Ar 4s1___________________________________________ Rb _______Kr 5s1___________________________________________ 2. Write the electron configuration
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MA‚ PAULINE Date Performed: 10 SEPTEMBER 2014 NACIONGAYO‚ DANIELLE Date Submitted: 17 SEPTEMBER 2014 TEDERA‚ YVES HEAT EFFECTS AND CALORIMETRY Experiment No. 2 I. RESULTS A. Determination of Heat Capacity In this experiment‚ an improvised calorimeter was used to determine the heat capacity. The calorimeter weighed 4.47 grams prior to the addition of water. Tap water‚ 40 mL to be exact‚ was added to the calorimeter which increased the weight to 43.87 grams. The water was measured using
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mL | 46.6-34.8=11.8 mL 46.6-34.8=0.0118 L | 1) Number of moles of NaOH used=0.0118 L of NaOH0.1 M of NaOH 1) Number of moles of NaOH used=0.00118 mol 2) The number of moles of NaOH used is equal to the number of moles of H+ions present in the solution‚ which is equal to the number of moles of the unknown salt. Moles of NaOH=Moles of H+=Moles of Unknown Salt=0.00118 mol 3) Molecular weight is equal to grams per mole. The mass of the salt was measured before the titration was carried
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Indigo Synthesis Author: Carolina Morales Group #1 Lab Partners: Jennifer Capeloto‚ Samantha Hoffman and Adriana Garibay Instructor: Saehan Park Chem. 152‚ Section 43 Date Work Performed: February 10‚ 2010 Date Work Submitted: February 24‚ 2010 Abstract: The percent yield of the color Indigo that is the color of blue jeans for the three trials was determined to be 13.50%‚ 15.30% and 14.80%. A series of experiments were performed to a cloth dye comparison. The absorbance
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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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Introduction: The objective of this experiment is to utilize the techniques of weighing and titrating to determine the total amount of MgSO4 present in an unknown sample. Standardized EDTA will be used to titrate the unknown solution. This type of reaction is a complexation reaction‚ which usually involves Lewis acids and bases. EDTA in complexation reactions serves as a chelating ligand. The base‚ EDTA‚ will bind to the metal ions‚ which serve as the Lewis acid‚ thus playing a role as a ligand
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WORKSHEET MOLE By Sir Shah Murtaza Q1. Potassium reacts with water as shown in the equation. a. Describe what you would see when potassium reacts with water. b. A sample of 0.195 g of potassium was added to 500 cm3 of cold water. When the reaction was finished‚ 100 cm3 of 0.100mol/dm3 hydrochloric acid was added to form solution X. (i) Calculate the number of moles of hydroxide ions formed when the potassium was added to water. (ii) Calculate the
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large quantity of solute in a given amount of solvent. Dilute solutions contain relatively little solute in a given amount of solvent. There are two specifics term to express concentration‚ which molarity and percent by mass. Molarity is the number of moles of solute per liter of solution. Percent by mass is the mass in grams of solute per 100 grams of solution. Vinegar is a dilute solution of acetic acid. The molecular formula for acetic acid is CH3COOH. Both molarity and percent by mass of acetic
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