chapt02_lec

chapt02_lec - 2-1Dr Wolf’s CHM 101Chapter 2The Components...

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Unformatted text preview: 2-1Dr. Wolf’s CHM 101Chapter 2The Components of The Components of MatterMatter2-2Dr. Wolf’s CHM 101Chapter 2: The Components of Matter2.1Elements, Compounds, and Mixtures: An Atomic Overview2.2The Observations That Led to an Atomic View of Matter2.3Dalton’s Atomic Theory2.4The Observations That Led to the Nuclear Atom Model2.5The Atomic Theory Today2.6Elements: A First Look at the Periodic Table2.7Compounds: Introduction to Bonding2.8Compounds: Formulas, Names, and Masses2.9Mixtures: Classification and Separation2-3Dr. Wolf’s CHM 101Definitions for Components of MatterElement- the simplest type of substance with unique physical and chemical properties. An element consists of only one type of atom. It cannot be broken down into any simpler substances by physical or chemical means.Molecule- a structure that consists of two or more atoms which are chemically bound together and thus behaves as an independent unit.2-4Dr. Wolf’s CHM 101Compound- a substance composed of two or more elements which are chemically combined.Mixture- a group of two or more elements and/or compounds that are physically intermingled.Definitions for Components of Matter2-5Dr. Wolf’s CHM 101Law of Conservation of Mass:The total mass of substances does not change during a chemical reaction.reactant 1 + reactant 2producttotal masstotal mass=calcium oxide + carbon dioxidecalcium carbonateCaO + CO2CaCO356.08g + 44.00g100.08gThe Mass Laws2-6Dr. Wolf’s CHM 101Law of Definite (or Constant) Composition: No matter what its source, a particular chemical compound is composed of the same elementsin the same parts (fractions) by mass.CaCO340.08 amu12.00 amu3 x 16.00 amu1 atom of Ca1 atom of C3 atoms of O100.08 amu40.08 amu100.08 amu= 0.401 parts Ca12.00 amu100.08 amu= 0.120 parts C48.00 amu100.08 amu= 0.480 parts O2-7Dr. Wolf’s CHM 101mass(kg) of pitchblendeSample Problem 2.1Calculating the Mass of an Element in a CompoundPROBLEM:Pitchblende is the most commercially important compound of uranium Analysis shows that 84.2g of pitchblende contains 71.4g of uranium, with oxygen as the only other element. How many grams of uranium can be obtained from 102kg of pitchblende?PLAN:The mass ratio of uranium/pitchblende is the same no matter the source. We can the ratio to find the answer.SOLUTION:mass(kg) of uraniummass(g) of uranium= 86.5 kg uranium= 102kg pitchblende xmass(kg) pitchblende xmass(kg) uranium in pitchblendemass(kg) pitchblende71.4kg uranium84.2kg pitchblendemass (kg) of uranium =86.5 kg uranium x1000gkg= 8.65 x 104g uranium2-8Dr. Wolf’s CHM 101Calculating the Mass of an Element in a Compound Ammonium Nitrateammonium nitrate = NH4NO3How much nitrogen(N) is in 455kg of ammonium nitrate?...
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chapt02_lec - 2-1Dr Wolf’s CHM 101Chapter 2The Components...

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