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Study Guide Chem 121 Exam 1 (Van Vo)

by: Lauren Eastham

Study Guide Chem 121 Exam 1 (Van Vo) CHEM 121A

Marketplace > University of Nevada - Las Vegas > Chemistry > CHEM 121A > Study Guide Chem 121 Exam 1 Van Vo
Lauren Eastham

GPA 3.5

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These notes will cover chapters 1-3 and embody the terms that Dr. Van Vo has specifically indicated for our first exam.
General Chemistry 1 (Dr. Van Vo)
Dr. Van Vo
Study Guide
Chemistry, percent yield, ChemicalFormulas, stoichiometry, Limiting Reactants, atoms, thompson, Dalton's Atomic Theory, empirical formula
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This 16 page Study Guide was uploaded by Lauren Eastham on Saturday September 24, 2016. The Study Guide belongs to CHEM 121A at University of Nevada - Las Vegas taught by Dr. Van Vo in Fall 2016. Since its upload, it has received 129 views. For similar materials see General Chemistry 1 (Dr. Van Vo) in Chemistry at University of Nevada - Las Vegas.


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Date Created: 09/24/16
Wednesday, September 21, 2016 CHEMISTRY 121 STUDY GUIDE 1 Chapter One: Matter and Measurement - Matter — anything that has mass and takes up space (atoms and compounds are composed of matter) • Physical properties — can be observed or measured without changing the composition of the matter - examples: appearance, color, texture, odor, melting point, freezing point, solubility, polarity • Physical changes — changes that do not alter the chemical nature of matter - examples: tearing paper into small pieces, sorting a pile of plastic and metals, ice melting into water, mixing sand and water, dissolving sugar in water, sublimating dry ice, melting solid sulfur into liquid sulfur • Chemical properties — changes that alter the chemical nature of matter - examples: reactivity with water, heat of combustion, pH, electromotive force, flammability, toxicity, oxidation states, chemical stability • Chemical changes —change that is characterized by the formation of a new chemical substance - mixtures —substance created by mixing other substances together defined by the uniformity of their composition 1 Wednesday, September 21, 2016 • heterogenous — composed of different substances that is made of different substances that remain physically separate from one another, mixtures of various proportions of molecules and atoms • homogenous — compounds that make up the mixture are uniform throughout, only 1 of phase of matter observed, atoms/ions are presented in fixed proportions throughout - examples: air, sugar water, rain water, vodka, vinegar, dishwashing detergent, steel - Math in Chemistry — • measurement — Terra, Giga, Mega, Kilo, Hecto, Deca, Deci, Centi, Milli, Micro (symbol = u), Nano, Pico - Pneumonic Device: The Great Midieval Killer Hates Dragons, Dragons Cry More Musically Near Pregnancy - we are only responsible for knowing Mega-Pico*** • significant figures —IT’S ALLABOUT THE ZEROS!! - 1. ALL non-zero numbers in a measurement are significant - 2. ALL middle zeros between non-zero numbers are significant - 3. Zeros after a decimal point ONLY are significant or a final zero AFTER the decimal point - 4. bar notations indicate a significant 0 2 Wednesday, September 21, 2016 • dimensional analysis —used to convert between units of measurement - Quick tip: To ensure that your final answer is in the units requested write the units on the far right side of the paper where your answer should be. For this problem I would have written km/hr to indicate the units that my final answer should be in. • density — is equal to mass per unit volume or Density = Mass/Volume - Quick tip: I simply remember the density as DMV… D=M/V. - Volume = Length * Width * Height (for measured shapes where volume isn’t provided but needs to be calculating) 3 Wednesday, September 21, 2016 Chapter Two: Atoms, Molecules and Ions - Laws that led to the Atomic view of Matter • Law of Conservation of Mass — total mass of the reactants will equal the mass of the products in every chemical reaction • Law of Definite (or Constant) Composition — the number of atoms and relative numbers are constant in a given compound • Law of Multiple Proportions — if 2 elements combine together to form 1 compound, A+B will be equivalent to small WHOLE numbers - History of Atoms - Dalton’s Atomic Theory — John Dalton introduced the idea that all matter is made up of atoms • Dalton expressed that atoms of 1 element have the same mass and property make- up than atoms of a different element • Dalton believed atoms are the fundamental building blocks of nature that could not be split apart into smaller particles (we now know that atoms can be split into smaller particles such as electrons, neutrons, protons, cathode rays and radioactivity) in chemical reactions, atoms would rearrange themselves and combine with other • atoms in new ways 4 Wednesday, September 21, 2016 • J.J.Thomson’s plum pudding model (“Raisin Muffin” Model) — a positive sphere (pudding) of matter with negative electrons (plums) spread in layers throughout the sphere • Thompson measured ratio of cathode ray particles mass to its charge using magnetic and electric fields - Ernest Rutherford’s a- scattering experiment gained him the discovery of the atomic nucleus - 5 Wednesday, September 21, 2016 - Atomic Symbols and Atomic Mass — Found using the periodic table of elements - The Atomic Mass often translated into the measurement grams represents the average mass of the atomic element (7 g Li) - The Atomic Number represents the number of (3) protons and (3) electrons in an elements atom (Li) - Neutrons are calculated by taking the atomic number and subtracted it by the atomic mass (Lithium: 7 - 3 = 4 neutrons) - Quick Tip: It is highly recommended that the most used elements’ names and symbols are memorized. Even better if you can name them all! This is important to finding the correct mass number on a periodic table that does not have the full name of the element on it. The one provided for our exam will not have element’s names printed on it - Periodic Table — Discovered by Mendeleev the father of the periodic table of elements • Periodic Law of the Elements — elements of similar properties occur at periodic intervals when arranged in a particular order 6 Wednesday, September 21, 2016 - Groups — vertical columns • 1 A = Alkali Metals • 2 A = Alkaline Earth Metals • 6 A = Chalcogens • 7 A = Halogens • 8 A = Noble gases - Periods — horizontal rows - metalloids —Boron, Silicon, Geranium, As, Sb and Te - nonmetals —Hydrogen, Group 8A, Group 7 A, Oxygen, Sulfur, Carbon, Nitrogen, and Phosphorus - metals — transitional metals are located predominantly in the middle Group 3 B through 2B with Group 1 A (excluding Hydrogen), Group 2 A, Group 3 A (not including Boron), Sn, Pb, Bi and Po making up the main-group metals - Ionic Compounds — bonds formed through charge transfer (unequal electron sharing) - forms an ion primarily between a metal and a nonmetal • predicting formulas —to predict a compounds formula it is imperative that one knows the charge of each of the most popular elements used in compounds 7 Wednesday, September 21, 2016 • neutral species of compounds have electrons lost and gained at equal values (negative and positive ions will cancel each other out creating neutrality) • Cation (+) is named first followed by the anion (-) • charges are not shown in the compound name • subscripts in the formula represent the combining ratio for the ions - Cations are positively charged ions • Common cations to know: H+, Li+, Na+, K+, Cs+, Ag+, NH4 +, Cu +, Co 2+, Cu 2+, Fe 2 +, Mn 2+, Hg2 2+, Hg 2+, Ni 2+, Pb 2+, Sn 2+, Mg 2+, Ca 2+,Sr 2+, Ba 2+, Zn 2+, Al 3+, Cr 3 +, Fe 3+ - Quick tip: The bolded cations are the most used throughout this course - Anions are negatively charged ions • Common Anions to know: H-, F-, Cl-, Br-, I-, CN-, OH-, CH3COO-, ClO3-, ClO4 -, NO3-, MnO4-, O 2-, O2 2-, S2-, CO3 2-, CrO4 2-, Cr2O7 2-, SO4 2-, N 3-, PO4 3- - Quick Tip — the bolded terms represent the most used anions throughout this course Trends on the periodic table can be used to determine ionic charges • - Group 1A has a 1+ or just + charge, 2 A has a 2+ charge, N has a 3-, most of Group 6 A has a 2- charge, most of Group 7 A has a - charge, Al has a 3+ charge 8 Wednesday, September 21, 2016 - Group 8 A consists of noble gases which are stable on their own and therefor have a neutral charge • nomenclature — for two binary elements (metals with fixed charge) - First word — full name of metal - Second word — base name or stem word in element followed by “-ide” • Ex: (Mg(OH)2) = Magnesium Hydroxide nomenclature — metals with unfixed charge • - Frist word — full name of metal followed by the roman numeral of the charge on the metal - Second word — stem word of anion followed by “-ide” • Ex: iron (III) sulfide (Fe2S3) • polyatomic ions — groups of atoms held together by covalent bonds MUST MEMORIZE PER DR. VO!!! • 9 Wednesday, September 21, 2016 - NH4 + Ammonium, H3O + Hydronium, CH3COO- Acetate, CN - Cyanide, OH- Hydroxide, ClO3- Chlorate, ClO4- Perchlorate, NO2 - NItrite, NO3 - Nitrate, MnO4- Permanganate, CO3 2- Carbonate, Cr2O7 2- dichromate, SO4 2- Sulfate, SO3 2- Sulfite, PO4 3- Phosphate - Quick tip: I recommend making flash cards for the polyatomic ions to ensure that the name, formula and charge is memorized because they will be used later in stoichiometry problems, balancing equations, etc…. - Covalent compounds — bonds formed through sharing electrons • primarily involving nonmetals • nomenclature - First word — Greek prefix + name of 1st element • prefix “mono” is not included • when the element name begins with a vowel, the A or O at the end of Greek prefix is dropped for pronunciation reasons - Second Word — Greek prefix + base/stem name of 2nd element ending in “ide” • Ex: CO2 — Carbon dioxide When hydrogen is first in a compound, it takes the place as a metal and does • not need the greek prefix added to its name - Ex: Hydrogen phosphate (H3PO4) 10 Wednesday, September 21, 2016 - Acids to memorize • HCl (aq) hydrochloric acid, HNO3(aq) nitric acid, H2SO4 (aq) sulfuric acid, H3PO4 (aq) phosphoric acid - (properties above are all acids made up of compounds dissolved in water) Chapter 3: Stoichiometry - The Mole —1 mole of anything is equal to Avogadro’s Number = 6.022 * 10^23 of anything • this number makes working with large numbers easier • Ex: 12 g C is equal to 1 mol which is equal to 6.022 * 10^23 particles of C - Empirical Formulas —formula giving the simplest positive whole integer ratio of atoms in a compound 1. — take each of the given masses and convert to moles • • 2. — Divide each element’s mol by the lowest mol value of the elements • 3. — If each number is a whole number or very close then these numbers represent the ratio formula for each element within the compound • 4. — If one or more numbers aren’t whole numbers, multiply each by a number that will bring the decimals to whole numbers (as in the example below) 11 Wednesday, September 21, 2016 • 5. — Write out the formula C14H18N2O5 - Molecular Formulas (combustion) — - molecular formulas- the formula for a compound • ex: A compound is 75.46% carbon, 20.10% ocygen, and 4.43% hydrogen by mass. It's molecular weight is 318.31 g/mol. - 1. Find empirical formula • Find mass of each element by fixing each overall mass sample with 100 g (you can use any mass but 100 works best with percentages) • (.7546) (100 g) = 75.46 g C • (.0443) (100 g) = 4.43 g H (.2010) (100 g) = 20.10 g O • • Convey mass of each element to moles 12 Wednesday, September 21, 2016 • (75.46 g C) (1 mol/ 12.00 g C) = 6.289 mol C • (4.43 g H) (1 mol/ 1.008 g H) = 4.39 mol H • (20.10 g O) (1 mol/ 16.00 g O) = 1.256 mol O • Ratio of moles of each element • (1.256 mol O)/ (1.256) = 1 mol O • (6.289 mol C)/ (1.256) = 5.007 mol C • (4.39 mol H)/ (1.256) = 3.50 mol H • Multiply each value by 2 to achieve whole numbers for each value • The empirical formula is C10H7O2 - 2. Calculate the mass of the empirical formula • 10(12.00) + 7(1.008) + 2(16.00) = 159.06 g/mol - 3. Determine how many empirical units are in a molecular unit by dividing the molecular mass by the empirical one • (318.31 g/mol)/(159.06 g/mol) = 2.001 empirical units per molecular - 4. New formula for molecular unit • C20H14O4 - Chemical Equations — • writing from scratch —writing chemical formula from scratch - when writing chemical equations it is important to identify the charge of each element within a compound as well as which compound or element is the cation (+) and which is the anion (-) 13 Wednesday, September 21, 2016 • Ex: sodium hydroxide and iron (II) chloride react with each other to form sodium chloride and iron (II) hydroxide - 1. Write out each compound and add subscripts to balance the charges NaOH (aq) + FeCl2 (aq) ----> NaCl(aq) + Fe(OH)2 (s) • he products were formed simply by taking the cations, rewriting them in the same order and then switching the anions. A cation will always be with an anion within a compound - 2. Now simply balance the equation so that each side has an equal amount of elements on both sides 2NaOH(aq) + FeCl2 ----> 2NaCl (aq) + Fe(OH)2 (s) - balancing — As seen above, placing numbers in front of entire compound so that each element has the same amount in both the products and the reactants • Quick tip — Subscripts within a chemical equation should NEVER be changed • Stoichiometry — - 1. Write the correct balanced chemical equations with the correct formulas - 2. Identify the given and unknown quantities for each • Units given must be converted to moles - 3. Use coefficients within the chemical equation to set up mole ratios - 4. After solving, convert quantities into the required units (typically grams) 14 Wednesday, September 21, 2016 - 5. Make sure that the answer is reasonable and the units are correct - basic — - limiting reactant — Perform multiple stoichiometry problems with the different amounts of grams given to determine which elements limits the overall product yield. 15 Wednesday, September 21, 2016 • Quick tip: In this limiting reactant problem, we were given only 2 values to work with (FeO3 and CO). Some problems may involve 3 or more values. The process would still be the same with the smallest value representing the limiting reactant - percent yield —Usually the total yield is calculated in grams using a stoichiometry problem method and then divided by a theoretical yield that will be given within a problem 16


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