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HN ADV CHEMISTRY CHAPTERS 1- 4

HN ADV CHEMISTRY CHAPTERS 1- 4. Chapter 1: Chemical Foundations. 2 fundamental concepts of chemistry 1. Matter is composed of various types of atoms 2. One subs changes into another by reorganizing the way the atom are attached to each other The SI System pg 8 & 9

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HN ADV CHEMISTRY CHAPTERS 1- 4

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  1. HN ADV CHEMISTRY CHAPTERS 1- 4

  2. Chapter 1: Chemical Foundations • 2 fundamental concepts of chemistry • 1. Matter is composed of various types of atoms • 2. One subs changes into another by reorganizing the way the atom are attached to each other • The SI System pg 8 & 9 • Table 1.1  base units • Table 1.2  prefixes • Uncertainty in Measurement • numbers obtained by using some measuring device • #’s that are the same regardless of who makes the measurement certain digits • last digit is estimated and varies w/the person who makes the measure uncertain digits

  3. all #’s have some degree of uncertainty  amt depends on precision of device • significant figures/digits – all certain digits and the first uncertain digit • Pg 11 Sample exercise 1.1 • Significant Figures and calculations • pg. 13 rules for counting sig figs • pg. 14 rules for sig figs in calculations and rounding • Precision & Accuracy • accuracy – agreement of a value w/the true value • precision – degree of agreement among several measurements of the same quantity – how close #’s are to each other • Types of error • 1. random error – (indeterminate error) – measurement has an equal probability of being high or low • 2. systemic error – (determinate error) – occurs in same direction each time

  4. Dimensional Analysis • used to convert one unit to another • uses conversion factors in ratio (want/have) – multiply by the #’s on top of ratio and divide by #’s on the bottom – make sure all units cancel until the last un-cancelled unit matches the unit of the answer • Density- ratio of a subs mass to its volume • D = m v

  5. Chapter 2: Atoms, Molecules and Ions • Chemical Laws • 1. Law of Conservation of Mass – mass is neither created nor destroyed it only changes form • 2. Law of Definite Proportions – a given cmpd always contains exactly the same proportion of elements by mass  proposed by Proust – Dalton proposed that matter is composed of atoms so a cmpd contains atoms in the same proportions • 3. Law of Multiple Proportions – when 2 elements form a series of cmpds, the ratios of the masses of the 2nd element that combine w/1 g of the 1st can always be reduced to small whole #’s • Pg 42 Sample exercise 2.1 • 4. Avogadro’s hypothesis – at the same temp & pressure, equal volumes of different gases contain the same # of particles • Subatomic Particles Mass Charge • Electron 9.11 x 10-31 kg -1 • Proton 1.67 x 10-27 kg +1 • Neutron 1.67 x 10-27 kg none One atom turns to another and says, “Darn it. I’ve lost an electron!” The other atom says, “Are you sure?” and the first says, “I’m positive!”

  6. Molecules and Ions • - chemical bonds – forces that hold atoms together • - covalent bonds – e- shared – particle formed called a molecule – • molecules represented by chemical formula or structural formula • - ionic bonds – particles exchange e- to form charged particles ions (+) ion = cation (-) ion = anion Chapter 3: Stoichiometry • Atomic masses - based on 612C as std – mass = 12 amu • - mass spectrometer is used to compare masses of atoms • - given on periodic table  avg atomic mass of all naturally occurring isotopes • - can use a mass spec to determine the isotopic composition of a natural element • Pg 81 Sample exercise 3.1

  7. The Mole • - the # of C atoms in exactly 12 g of pure 612C = 6.02214 x 1023 6.022 x 1023 = Avogadro’s # • *doesn’t matter what the subs is it is still = Avogadro’s # • Pg 83 Sample exercise 3.2 • Pg 84 Sample exercise 3.3 • Pg 85 Sample exercise 3.4, 3.5 • Molar mass • - the mass in grams of one mole of a cmpd – the sum of all the atomic masses multiplied by the subscripts • Pg 86 Sample exercise 3.6, 3.7 • Pg 88 Sample exercise 3.8

  8. Percent composition • - compare the mass of each element to the total mass of the cmpd mass percent • Pg 89 Sample exercise 3.9 • Determining the formula of a cmpd • - need to determine the % mass of each element in the cmpd • - empirical formula – smallest whole # ratio of elements in a cmpd • - molecular formula – some multiple of the emp. formula based on the ratio Molecular molar mass = multiple of Empirical molar mass empirical • Pg 93 Sample exercise 3.11 • Chemical equations • - depict chem rxns • - Reactants  Products

  9. - bonds have been broken and reformed to produce new subs • - must show that all atoms in reactants are accounted for in products by balancing equation (law of conservation of matter) • - shows states of matter • - shows # of atoms/molecules in rxn • - shows # of moles of R and P in rxn • Pg 100 Sample exercise 3.14 • Stoichiometric Calculations: • 1. Balanced equation • 2. State given • 3. State what is wanted • 4. Convert given to moles • 5. Mole to Mole ratio (from balanced equation) • 6. Convert to units of answer

  10. Pg 105 Sample exercise 3.16, 3.17 • **any step not needed just SKIP • LimitingReactants • - the reactant consumed firsts limits the amt of product formed • - to determine the LR do a stoichiometry problem starting w/1 R and end w/the other - compare to amts given • - to determine amt of product start w/LR and end w/product  • Pg 110 Sample exercise 3.18 • Percent yield • - amt of product formed when a reactant is completely consumed is called the theoretical yield • - amt of product formed in reality is called the actual yield • - these can be used to determine the % yield • Actual x 100 = % yield • Theoretical • Pg 111 Sample exercise 3.19

  11. Chapter 4: Types of Chemical Reactions and Sol’nStoichiometry • - Water H H • - able to dissolve many things 105o • - bent • - O-H bonds are covalent bonds O • - e- shared unequally – oxygen has a greater pull on e- so they are closer to it  H b/c e- are farther away from it polar molecule • Hydration – ionic solid dissolving in H2O • - (+) of H2O is attracted to anions of cmpd • - (-) of H2O is attracted to cations of cmpd • - salt “falls apart” – breaks into ions and is indicated w/the (aq) • Some non-ionic cmpds are soluble in H2O  polar dissolves polar and nonpolar dissolves nonpolar

  12. Sol’ns • - made up or solutes and solvents • - some can conduct electricity  electrolytes • - strong electrolytes – conduct efficiently • - weak electrolytes – conduct poorly • - non-electrolytes – do not conduct • - conductivity is a factor of how many ions are present in a sol’n • - strong electrolytes ionize completely • - soluble salts are easily hydrated • - strong acids produce H+ and other ions • - all form (aq) sol’n • - completely dissociate into ions • - exception: H2SO4 , only 1 H+ dissociates • - strong bases release OH- from cations

  13. - weak electrolytes have only a small amt of ionization • - weak acids contain an acidic H+ and nonacidic H’s – ex: H3PO4 • - weak bases partially lose OH- in sol’n • - nonelectrolytes – dissolve in H2O but do not produce any ions • Composition of sol’ns • - need to know 2 things to do Stoichiometriccalcs • 1. nature of reactants • 2. amts of chem in sol’n • - Molarity – (M) • - expression of sol’n concentration • M = moles of solute • L of sol’n • Pg 134 Sample exercise 4.1, 4.2, 4.3 • Pg 135 Sample exercise 4.4, 4.5

  14. Std sol’n – sol’n whose conc. Is accurately known • Pg 136 Sample exercise 4.6 • Std sol’n – sol’n whose conc. Is accurately known • Pg 136 Sample exercise 4.6 • Dilutions • - sol’ns are purchased in conc. form and H2O is added to desired molarity • M1V1 = M2V2 • - when solving for volume of conc. sol’n, remember that H2O must be added to reach the total volume needed • Pg 138 Sample exercise 4.7

  15. Types of ChemRxns • - Precipitation Rxns – 2 sol’ns are mixed and a solid forms and separates from sol’n(precipitate, ppt) • - products must have a net charge of 0 – (+) = (-) • - rules for determining solubilities on pg. 144 Table 4.1 • - slightly soluble – a tiny amt of solid dissolves (not noticeable)  sometimes insoluble • Pg 144 Sample exercise 4.8 • - equations so far have been molecular equations – do not show what is happening b/n particles in the sol’n • Ex: K2CrO4(aq) + Ba(NO3)2(aq) BaCrO4(s) + 2KNO3(aq) • - complete ionic equations – all subs that are strong electrolytes are represented as ions • Ex: 2K+ (aq) + CrO42-(aq) + Ba2+(aq) + 2NO3-(aq) BaCrO4(s) + 2K+(aq) + 2NO3- (aq) • *not all ions participate in the rxnspectator ions, e.g. K+ NO3-

  16. - net ionic equations – includes only those ions that participate in rxn • Ex: Ba2+(aq) + CrO4 2-(aq) BaCrO4(s) • Pg 146 Sample exercise 4.9 • Stoichiometry of pptrxns • - must decide what will occur when 2 sol’ns are mixed – write down species (ions) in sol’n • - to obtain moles of reactants use volume of sol’n and Molarity • Pg 147 Sample exercise 4.10 • *Steps (1-6) pg 148 • Pg 148 Sample exercise 4.11 • Acid-base rxns • - according to Bronstead-Lowry • - acid is a proton (H+) donor • - base is a proton (H+) acceptor

  17. *OH- is such a strong base that for stoichiometrycalcs. It can be assumed to react completely w/any weak acid •   - rxn is called a neutralization rxn • Pg 150 Sample exercise 4.12 • Pg 151 Sample exercise 4.13 • Acid-base titrations • - volumetric analysis – technique for determining the amt of a certain subs by doing a titration - titration – delivery (from a buret) of a measured volume of a sol’n a known conc. (the titrant) into a sol’n containing the sub being analyzed (the analyte) - when exactly enough titrant has been added to react w/the analyte called the equivalence point or stoichiometric point *marked by an indicator – a subs that changes color according to pH • - point at which indicator changes color called the endpoint • *you need to choose an indicator that changes color at the equivalence point

  18. Requirements for successful titration: • 1. Exact rxn b/n titrant and analyte must be known • 2. Stoichiometric/equivalence point must be marked accurately • 3. Volume of titrant required to reach Stoichiometric point must be known • *Use strong acids/bases for titrant • Standardizing the sol’n • Pg 153 Sample exercise 4.14, 4.15  • Redoxrxns • - one or more e- are transferred b/n subs • - most energy producing rxns are redoxrxns • - oxidation states/oxidation #’s – help track e- in redoxrxns • *Rules for assigning oxidation #’s pg 156 Table 4.2 • Pg 157 Sample exercise 4.16

  19. **Exception: +2/3 -2 • Fe O4 - b/c it’s an “avg” • +8 -8 - some Fe’s are 2+ and some are 3+ • *occurs in other cmpds DON’T FREAK!! • -4+1 0 +4-2 +1-2 • Ex: CH4 + 2O2 CO2 + 2H2O • Oxidation loss 8e- RA • Reduction gained 2e- OA • *if a subs is reduced it ------------------------------------------- *if a subs is oxidized it causes oxidation in the -7-6-5-4-3-2-10+1+2+3+4+5+6+7 causes reduction in the other oxidizing agent other reducing agent gain of e- loss of e- reduction oxidation • pg 160 Sample exercise 4.17 • Pg 161 Sample exercise 4.18

  20. Balancing RedoxRxns • - use half-rxn method • - separate into one for oxidation and one for reduction • Ex: Ce4+ + Sn2+ Ce3+ + Sn4+ • ReductionOxidation • Ce4+ Ce3+ Sn2+ Sn4+ • 2[1e- + Ce4+ Ce3+] Sn2+ Sn4+ + 2e- • 2e- + 2Ce4+ 2Ce3+ Sn2+ Sn4+ + 2e- • 2Ce4+ + Sn2+ 2Ce3+ + Sn4+ • 2(4+) + 2  2(3+) + 4 • +10  +10 • *pg 162 Steps for balancing Redoxrxns using half rxns

  21. +7 -2 +2 +3 +2 • Ex 2: MnO4- + Fe2+ Fe3+ + Mn2+ • ReductionOxidation • MnO4- Mn2+ Fe2+ Fe3+ • 8H+ + MnO4-  Mn2+ + 4H20 • 5e- + 8H+ + MnO4-  Mn2+ + 4H2O 5[Fe2+ Fe3+ +1e-] • 5Fe2+ 5Fe3+ + 5e- • 8H+ + MnO4- + 5Fe2+ Mn2+ + 4H2O + 5Fe3+ • +17 +17 • Pg 165 Sample exercise 4.19 • - redoxrxns can also occur in basic sol’ns • Pg 167 Sample exercise 4.20 (in basic sol’n)

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