Assignment 3.4 Part F – The Reaction of Iron Nails with a Copper Solution Introduction: The experiment in this activity involves the reaction between a copper (II) chloride solution with iron nails and the mole ratios involved in the reaction. Measurements are taken to determine the moles of each reactant involved in the reaction and thus the number of atoms or molecules involved. Apparatus and Materials: Refer to the reaction of iron nails with a copper solution assignment in Module 3‚ Section
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(%) magnesium by mass: Balance the equation get the moles of KMnO4 get moles of MnO4- get moles of Fe2+ get grams of Fe2+ get molar mass of Fe(NH4)2(SO4)2● 6H2O find grams per mole ratio of Fe2+ and Fe(NH4)2(SO4)2● 6H2O multiply by 100% % of Fe2+ Experimental errors: 1. Volume of KMno4 lost because it splashed and stayed at the side of the beaker: Increased volume of KMno4 used increased moles of KMno4 increased moles of Fe2+ increased grams of Fe2+ increased percentage
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One can find an empirical formula by taking a sample of a compound and dividing the number of moles of one element in the compound by the number of moles of another element in the compound to form a small wholenumber formula. For example‚ in a sample of a made up compound of oxygen and lead‚ one mole of lead has a molar mass of 207.2 g/mole‚ and oxygen has a molar mass of 16 g/mole. If this compound forms in a one to one atom ratio‚ then the ratio of moles will be 13 moles of oxygen to 1 mole of lead. In the experiment‚ the students
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Title: Acids‚ Bases & Buffers Objectives: 1. To understand the acid-base chemistry. 2. To prepare and evaluate a buffer system 3. To measure the buffering capacity of two types of isotonic drinks. Introduction: There are acid-forming‚ basic forming and neutral food‚ however the acid or alkaline properties of a food is unable to judge by the actual acidity of the food itself. For example‚ citrus fruits such as lemon are acidic‚ but they are alkaline-forming when we consume and digest it. Therefore
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INTRODUCTION Differences between acids and bases An acid-base reaction is based on the reaction involving the ionization of water H2O -> H+ + OH- This means that water can break apart into a hydrogen ion and a hydroxide ion. These two ions can also join together to form a water molecule. When a strong acid is placed in water‚ it will ionize completely‚ and break down into its constituent ions in which one of it a hydrogen ion. When a strong base is placed in water‚ it will ionize
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Exercise 1: Alcohol Dehydrogenase (ADH) Reaction Bich 413 section 508 Abstract The Alcohol Dehydrogenase Reaction in Oxidized Nicotinamide Adenine Dinucleotide. The oxidation-reduction of Nicotinamide adenine Dinucleotides (NADH) are very Important reaction in the biological system. When the Oxidized Nicotinamide Adenine Dinucleotide (NAD+) is react with a alcohol‚ catalyze by Alcohol Dehydrogenase (ADH)‚ the result products are NADH‚ H+‚ and aldehyde.In the biological system‚ the
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Heat of Combustion of Magnesium Background: The students were given full instructions on how to experimentally determine the enthalpy of reaction (ΔHrxn) for the combustion of magnesium ribbon‚ using Hess’s Law. Data Collection: | |Reaction 1 |Reaction 2 | | |(MgO) |(Mg)
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delivery of 23 ml. of titrant Part A) Reaction is: HNO2 + OH- NO2- + H2O Ka for Nitrous acid = 7.2E-4 Kw = Ka*Lb Kb = 1.0E-14/7.2E-4 Kb = 1.38E-11 Moles of base: (0.15)*(0.022) = 3.3E-3 moles 1:1 Ratio‚ So‚ Moles of acid = 3.3E-3 moles NaOH: (0.005)*(0.15) = 7.5E-4 moles HNO2: 3.3E-3 – 7.5E-4 = 2.55 E-3 moles Using Henderson- Hassel Balch equation: Ph = 4.752 + log(7.5E-4/2.55E-3) Ph =4.21 Part B) Ph = 4.752
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water = (Mass of hydrate) – (mass of dehydrate) Mass of water = (answer to #1) – (answer to #3) Mass of water = (5.000g) – (4.500g) Mass of water = 0.500g 5. Convert the mass of water to moles of water. To do this‚ we need the molar mass (from 2.04 and 3.09-molar mass is the mass‚ in grams‚ of one mole of your
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the mass of water and dividing it by the total mass of the hydrate and then multiplying that answer by 100%. The number of moles of water in a hydrate was determined by taking the mass of the water released and dividing it by the molar mass of water. The number of moles of water and the number of moles of the hydrate was used to calculate the ratio of moles of water to moles of the sample. This ratio was then used to write the new and balanced equation of the dehydration process. The sample was then
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