CH101 Chapters 1 and 2
CH101 Chapters 1 and 2 CH 101
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This 5 page Class Notes was uploaded by Lauren Dutch on Wednesday September 7, 2016. The Class Notes belongs to CH 101 at University of Alabama - Tuscaloosa taught by Dave Nikles in Fall 2016. Since its upload, it has received 216 views. For similar materials see General Chemistry in Chemistry at University of Alabama - Tuscaloosa.
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Date Created: 09/07/16
CH101 Sections 1.5-1.9 (scroll to next page) I. Modern Atomic Theory and Its Laws A. Law of Conservation of Mass- In a chemical reaction, matter is neither created nor destroyed. B. Law of Definite Proportions- All samples of a given compound, regardless of how they were made, have the same proportions of their constituent elements. C. Law of Multiple Proportions- When two elements A and B form two different compounds, the masses of element B that combine with 1g of element A can be expressed as a ratio of a small whole numbers. II. Discovery of Electron A. J.J. Thomson’s Cathode Rays 1. Cathode rays travel through a glass cathode ray tube away from the negative cathode towards the positive anode. 2. Discovery of a negatively charged, low mass particle present in all atoms called the electron B. Millikan’s Oil Drop Experiment 1. Calculated charge on oil droplets falling in and electric field 2. Discovered that the charge on an electron is -1.60x10^-19 C. Ernest Rutherford’s Gold Foil Experiment 1. Bombarded thin gold foil with positive particles a) Most passed through but some were deflected b) Denounced the plum pudding model that an atom is a positive sphere with negative electrons in it 2. Discovered the structure of the atom and explained it with the nuclear theory a) Most of the atom’s mass and all of its positive charge is contained in the nucleus b) Most of the volume is empty space, where the tiny negative electrons are found c) There are the same number of protons in nucleus and electrons outside the nucleus so the atom is neutrally charged III. Nuclear Model A. Almost all mass resides in the nucleus 1. Number of protons determines the atomic number 2. Protons and neutrons determine atomic mass (amu) 3. Isotopes have same atomic number (number of protons) but different atomic masses (different number of neutrons) a) Carbon-14 is an isotope of carbon. There are still 6 protons but there are two extra neutrons. 4. An element’s atomic mass is calculated by adding together all the different isotopes masses times their abundance a) Chlorine has two naturally occurring isotopes. Chlorine-35 has a mass of 34.97 amu and an abundance of 75.77%. Chlorine-37 has a mass of 36.97 amu and an abundance of 24.23%. (1) Atomic mass= 0.7577(34.97amu) + 0.2423(36.97amu) = 35.45amu B. Mass spectrometry calculates the mass of isotopes and their abundances by separating particles according to mass 1. Creates a mass spectrum a) X axis indicates mass b) Y axis indicates relative abundance c) Mass spectrum of chlorine CH101 Chapter 2 I. Scientific Measurements A. Accuracy- how close you are to a true or accepted value B. Precision- reproducibility of a measurement C. Resolution- distinguishing parts of an object D. Significant figures 1. Multiplication and division- answer with the lowest number of sig figs from the problem a) Example: 2.5 * 3.42=8.6 2. Addition and subtraction- answer with lowest number of decimal places from the problem a) 34.56-5.4=29.2 E. Must have a value AND a unit 1. SI unit of length is the meter (m) 2. Prefixes a) Macroscale- human eye can see unaided b) Microscale- need an optical microscope to see c) Nanoscale- need an electron microscope to see II. Converting between moles and atoms A. Avogadro’s number: 1 mol=6.022x10^23 atoms B. 1 mole of an element equals the amu of that element in grams C. Example: Calcuate the number of carbon atoms in a 0.035 gram pencil lead. 0.035 g C (1 mol C / 12.01 g C) (6.022x10^23 atoms C / 1 mol C) = 1.75 atoms C Grams and moles cancel out, leaving atoms
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