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Units and Measurement (Section)Use appropriate metric

Chemistry: The Central Science | 12th Edition | ISBN: 9780321696724 | Authors: Theodore E. Brown; H. Eugene LeMay; Bruce E. Bursten; Catherine Murphy; Patrick Woodward ISBN: 9780321696724 27

Solution for problem 24E Chapter 1

Chemistry: The Central Science | 12th Edition

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Chemistry: The Central Science | 12th Edition | ISBN: 9780321696724 | Authors: Theodore E. Brown; H. Eugene LeMay; Bruce E. Bursten; Catherine Murphy; Patrick Woodward

Chemistry: The Central Science | 12th Edition

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Problem 24E

Problem 24E

Units and Measurement (Section)

Use appropriate metric prefixes to write the following measurements without use of exponents: (a) 2.3 × 10-10 L, (b) 4.7 × 10-6 g, (c) 1.85 × 10-12 m, (d) 16.7 × 106 s, (e) 15.7 × 103 g, (f)1.34 × 10-3 m, (g) 1.84 × 102 cm.

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General Chemistry I Study Guide Exam 3 Chapter 7  Valence Shell Electron Pair Repulsion (VSEPR) – predicting molecular shape; basic idea is that electrons repel each other; electrons are found in different domains (lone pairs/single bonds/double bonds/triple bonds) o electrons will arrange themselves as far as possible o arrangements minimize repulsive interactions o make sure to know the different types and bond angles of molecular geometry  Electron domain geometry- arrangement of electron domains around central atom  Bond angle- angle between two adjacent A-B bonds  4 Steps To determine geometry- o Draw lewis structure o Count the number of electron domains around the central atom o Determine electron-domain geometry by applying VSEPR model o Determine molecular geometry by considering the positions of the atoms only  Lone pairs take up more space than bonded pairs of electrons  Van der Waal’s Forces o London forces – natural attraction between all molecules  Increases with molar mass  Size of electron cloud determines strength  Similar molecular weight = same London force strength  Weakest of all forces  Only type between nonpolar molecules o Dipole-dipole forces- attraction between oppositely charged portions of 2 or more polar molecules  The more polar, the stronger the attraction forces (higher melting and boiling point)  Stronger the London forces  Bigger the dipole moment, the more polar the molecule  Hydrogen bonds – special type of dipole attraction where a hydrogen gets trapped between two highly electronegative elements (F, O, N) o Both molecules MUST have a hydrogen directly attached to a F, O, or N. o Strongest attraction of the 3 intermolecular forces  Dispersion forces – result from Coulombic attractions between instantaneous dipoles of non-polar molecules  Valence Bond Theory – atoms share electrons when atomic orbitals overlap o the H-H bond in H2 forms when the singly occupied 1s orbitals of the 2 H atoms overlap o A bond forms when single occupied atomic orbitals on 2 atoms overlap o The 2 electrons shared in the region of orbital overlap must be opposite spin o Formation of a bond results in a lower potential energy for the system  Hybridization- accounts for observed bond angles in molecules that could not be described by the direct overlap of atomic orbitals o If the s orbital and 3 p orbitals hybridize, then it is sp3 hybridization  Example – methane CH4  Steps to determine hybridization- o Draw lewis structure o Count electron domains on the central atom. This will be equal to the number of hybrid orbitals o Draw the ground state orbital diagram for the central atom o Maximize number of unpaired valence electrons by promotion o Combine the necessary number of atomic orbitals to generate required number of hybrid orbitals o Place electrons in hybrid orbitals, putting one electron in each orbital before pairing any electrons  Sigma bond – forms when sp2 hybrid orbitals overlap st o 1 bond between ANY 2 atoms is ALWAYS a sigma bond  Pi bond – when the overlap of the orbitals does NOT lie on a line drawn between the 2 nuclei o Only p orbitals can form pi bonds nd rd o The 2 or 3 bonds of a double or triple bond are always pi bonds o Pi bonds are not as strong as sigma bonds because the orbitals do not overlap as much in a pi bond Chapter 8  For chemical reactions, each species on the left is a reactant  Each species on the right is a product  (g) – gas; (l) – liquid; (aq) – aqueous; (s) – solid  Equations must be balanced so the law of conservation of mass is obeyed o Achieved by writing stoichiometric coefficients to the left of the chemical formulas o Steps for balancing  Change coefficients of compounds before changing the coefficients of elements  Treat polyatomic ions that appear on both sides of the equation as units  Count atoms or polyatomic ions carefully  Combination reaction – 2 or more reactants combine to form a single product  Decomposition – 2 or more products form from a single reactant  Combustion – substance burns in presence of oxygen o ALWAYS produces CO2 and H2O o Incomplete if CO or C is produced  A 1.50 g sample of hydrocarbon undergoes complete combustion to produce 4.40 g of CO 2nd 2.70 g of H O2 What is the empirical formula of this compound  A 0.250 g sample of hydrocarbon undergoes complete combustion to produce 0.845 g of CO a2d 0.173 g of H O.2What is the empirical formula of this compound  Know how to use the mole ratios the determine how much product will form from a balanced equation  Limiting reactant – the reactant that is used up first o Excess reactants are present in quantities greater than necessary to react with the quantity of the limiting reactant o A 2.00 g sample of ammonia is mixed with 4.00 g of oxygen. Which is the limiting reactant and how much excess reactant remains after the reaction has stopped o try example 8.7 in chemistry book  Theoretical yield – amount of product that forms when all of the limiting reactant reacts to form the desired product  Actual yield – amount of product actually determined from the reaction  Percent yield – tells what percentage the actual yield is of the theoretical yield Chapter 9  Solution – homogenous mixture of 2 or more substances  Solvent – substance present in largest amount  Solute – other substances present  Electrolyte – a substance that dissolves in water to yield a solution that conducts electricity  Dissociation – electrolyte breaks apart into its constituent ions  Ionization – a molecular compound forms ions when it dissolves  Nonelectrolyte – a substance that dissolves in water to yield a solution that does not conduct electricity  Strong electrolyte – dissociates completely o Example- strong acids (HCl), strong bases (NaOH)  Weak electrolyte – a compound that produces ions upon dissolving but exists in solution predominantly as molecules that are not ionized  Precipitate – insoluble product that separates from a solution  Hydration – occurs when water molecules remove the individual ions from an ionic solid surrounding them so the substances dissolves  Solubility – maximum amount of solute that will dissolve in a given quantity of solvent at a specific temperature  Double replacement/metathesis – reactions in which cations in 2 ionic compounds exchange anions  Ionic equation – compounds that exist completely as ions in solution are represented as those ions  Net ionic equation – an equation that includes only the species that are involved in the reaction o Ions that appear on both sides are called spectator ions and are not included in the overall reaction  Oxidation/reduction – chemical reaction in which electrons are transferred from one reactant to another (also called redox) o Oxidation = LOSS o Reduction = GAIN o OIL RIG or LEO goes GER o (oxidation Is loss, reduction is gain); (loss equals oxidation, gain equals reduction)  redox – sum of an oxidation half-reaction and a reduction half reaction  oxidation number – charge an atom would have if electrons were transferred completely o elements have an oxidation number of zero o steps to determine oxidation number  start with the ones you know  the total contribution to charge must sum to zero if it is a neutral compound; if not, it must sum to the charge of the compound.  Displacement reaction – an atom or an ion in a compound is replaced by an atom of another element  Molarity (M) – molar concentration; defined as the number of moles of solute per liter of solution o M = mol/L  Dilution – process of preparing a less concentrated solution from a more concentrated one  Moles of solute before dilution = moles of solute after dilution  McVc = MdVd Chapter 10  System – a part of the universe that is of specific interest  Surroundings – rest of the universe outside of the system  Thermochemistry – study of heat in chemical reactions  Heat – transfer of thermal energy o Either absorbed or released o Si unit is a Joule, J  Exothermic process – occurs when heat is transferred from the system to the surroundings  Endothermic process – occurs when heat is transferred from the surroundings to the system  Thermodynamics – study of the interconversion of heat and other kinds of energy  3 types of systems o open system – can exchange mass and energy with the surroundings o closed system – allows the transfer of energy but not mass o isolated system – does not exchange either mass or energy with its surroundings  state functions – properties that are determined by the state of the system, regardless of how that condition was achieved o pressure o volume o energy o temperature  First law of thermodynamics – energy can be converted from one form to another, but not created nor destroyed  Pressure increases when volume is constant  Volume increases when pressure is constant  Enthalpy of reaction – the difference between the enthalpies of the products and the enthalpies of reactants I have notes that may help with thermodynamics, however, they go further than where we are in class, so just ignore that part. I think we may be learning it on Tuesday, but if not, ignore it. Thanks! Also, I recommend looking in the book for practice problems. She usually picks problems straight from the book or the internet! Good Luck!

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Chapter 1, Problem 24E is Solved
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Textbook: Chemistry: The Central Science
Edition: 12
Author: Theodore E. Brown; H. Eugene LeMay; Bruce E. Bursten; Catherine Murphy; Patrick Woodward
ISBN: 9780321696724

The full step-by-step solution to problem: 24E from chapter: 1 was answered by , our top Chemistry solution expert on 04/03/17, 07:58AM. This full solution covers the following key subjects: use, prefixes, measurement, measurements, metric. This expansive textbook survival guide covers 49 chapters, and 5471 solutions. Chemistry: The Central Science was written by and is associated to the ISBN: 9780321696724. This textbook survival guide was created for the textbook: Chemistry: The Central Science, edition: 12. Since the solution to 24E from 1 chapter was answered, more than 399 students have viewed the full step-by-step answer. The answer to “Units and Measurement (Section)Use appropriate metric prefixes to write the following measurements without use of exponents: (a) 2.3 × 10-10 L, (b) 4.7 × 10-6 g, (c) 1.85 × 10-12 m, (d) 16.7 × 106 s, (e) 15.7 × 103 g, (f)1.34 × 10-3 m, (g) 1.84 × 102 cm.” is broken down into a number of easy to follow steps, and 50 words.

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Units and Measurement (Section)Use appropriate metric