Step 1: Pb (NO3) 2 (aq) + CaCl2 (aq) PbCl2 (aq) + Ca (NO3) 2 (aq) (double displacement reaction) According to the solubility guidelines lead (II) chloride (PbCl2) is a possible precipitate. This is because even though most chlorides are soluble‚ lead chloride is considered insoluble (p.2‚ Lesson 17). Step 2: PbCl2 ↔ Pb2+ + 2Cl- Q = [Pb2+] [Cl-] 2 Step 3: Ksp = 1.2 x 10-5 (from table 17.1‚ p.5) Step 4: V2 = 20.0 mL (volume of Pb (NO3)2) + 45.0 mL (volume of CaCl2) = 65.0
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04.05 Chemical Reactions: Combustion: Lab Worksheet and Rubric Before You Begin: You may either copy and paste this document into a word processing program of your choice or print this page. Procedure: 1. Iron (III) and copper (II) sulfate solution 2. Fill a small test tube halfway with copper (II) sulfate solution. Add a 2.0 gram iron rod to the solution and observe the reaction. 3. 4. 5. Lead (II) nitrate and potassium iodide solutions 6. Pour about 2.0 mL of lead (II) nitrate into the test tube
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Lab 5: The SN2 Reaction: 1-Bromobutane From K. L. Williamson‚ Macroscale and Microscale Organic Experiments‚ 2nd Ed. 1994‚ Houghton Mifflin‚ Boston. p247; revised 2/22/02 Prelab Exercise: Write a detailed flow sheet/ flow chart for the isolation and purification of 1-bromobutane. Designate how each minor-product is removed from the major product and which layer holds the product in each experimental step performed. [pic] Introduction This experiment utilizes SN2 chemistry
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Over the past few years the WHO (World Health Organization) concerned about the public health‚ conducted the largest global systematic study to discover the causes of major health risk factors‚ the statistics about air pollution stood out the most due to the deaths caused by air pollution. Michael Graham Richard reports that 3.5 million deaths are caused by indoor air pollution and 3.3 million deaths are caused due to outdoor pollution. This essay will be discussing the use of electrostatic precipitator
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Chemical Reactions Lab Introduction Four different types of chemical reactions were observed in this lab. Synthesis reactions occur when two different elements or compounds form a more complex compound as shown by equation (1). Single replacement occurs when one element replaces another to form a new compound as shown in equation (2). Double replacement occurs when two different atoms in different compounds trade places with one another to form two new compounds as shown in equation (3).Combustion
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Mike Morrison – SCH3U Types of Chemical Reactions Lab Intro/Purpose The purpose of this lab is to determine types of reactions with only the reactants and through observation. Through observing the experiments‚ one is able to see a visual of each type of reaction taking place and it gives them a better understanding of how chemical reactions work and what products they may form. Materials & Procedure Materials: * Wood Splints - Copper Sulfate * Sodium Sulfate - Copper
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LEARNING TASK NO. 4B Composition Stoichiometry Problems Mole Relationship from Chemical Formulas: a) Determine the number of moles of chloride ions in 2.53 mol ZnCl2. b) Calculate the number of moles of each element in 1.25 mol glucose (C6H12O6). c) How many molecules of oxygen atoms are present in 5.00 mol diphosphorus pentoxide? d) Calculate the number of moles of hydrogen atoms in 11.5 mol water. e) A sample of ethanol (C2H5OH) has a mass of 45.6 g.How many carbon atoms does the sample
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acts on a molecule called substrate and it also significantly speed up a chemical reaction by lowering the activation energy. In order to learn about the enzyme and its behaviour‚ this lab practical is conducted to examine the kinetic of the enzyme alkaline phosphatase. As illustration‚ when alkaline phosphatase is added to a substrate called p-Nitrophenyl phosphate (colourless in alkaline solution)‚ a series of reaction takes place and eventually releases a product called p-nitrophenol (yellow in alkaline
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Ch. 1 Fundamentals of Chemistry Subchapter – Stoichiometry and Units 1. An element X forms an oxide with the formula X4O10. If 31 g of X combines with 40 g of oxygen‚ what is the identity of X? A. N B. P C. C D. S Correct Answer: B: P Explanation: We can solve for the atomic mass of X to identify it. The amount in g of X for any amount in g of O is given by: Rearranging the above equation and substituting the given mass of X and O‚ we get: Thus‚ the atomic mass of X
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of steel wool into the test tube and measured the temperature again. We stored the test tubes on the test tube rack to observe the next day. A chemical reaction did occur in the lab. There were a color change and an odor change that was observed on Day 2 of the lab. On Day 1‚ the liquid was clear and the smell was very strong. On Day 2 of the lab‚ most of the liquid was gone but the color was a muddy color and the smell got milder‚ it wasn’t as strong as Day 1. Not only was there a color and odor
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