Chem final study guide PDF

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Chemistry Final Study Guide CH 1 1. Solid o Potent. energy > Kin. energy (rigid) 2. Liquid o Kin. energy > moves around, no def shape o ↑ Temp = ↑ kin. energy 3. Gas o Compressable (becomes lighter) 4. Atom: smallest unit of elem. 5. Mole. – combo of 2+ atoms 6. Classifying composition- MATTER is. . . o Pure substance = uniform composition/intensive property t/o 1 type - Cannot be separated!!  Elements -> can’t be broken down into simpler subst.  Conducts electricity? o YES- Metal (hard solid) o NO- NM (s, l, or g) o OR Metalloids (semi conductors)  Compounds -> can be simplif. into ele. o Mixture = Varied comp. w/ 2 + pure subst. / comp./ele. - Can be seperated  Homogeneous (solution) -> uniform t/o [ie. Salt water]  Hetero -> nonuniform distrib. [physical seeing]  Det. Hetero/ Homo? SIGNIFICANCE

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- Lake water- homo, sand water- hetero, oil water- hetero, tap water- homo, airhomo, brass- homo, soil- hetero {not uniform} 7. Classifying element, compound, or mixture w/ image??~??? ch1notes review 8. SIG FIGS Zero in b/w is sig. (405, 40.5) Zeros to right of sig fig is sig (5.00) Lead in zeros NOT sig. ( 0.151, 0.0000405) Trailing Zeros SIG. ( 1500000.) Multi./ Div. sig fig = fewest # (sf)  Ex: 3.489 (3.3) = 12 ( 2 sig fig) o Add./Subt. Sig fig = depends on fewest dec. places (dp)  Ex: 5.67 – 2.3 = 3.3 (1 dp)

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9. Scientific Not. = Left -> positive power ; Right -> neg. power 10. 1ml = 1 cm^3 1 in = 2.54 cm 11. Density: higher = sink ; low = float o Ex: Mass 11.2 g. Density? Given 2.0 cm box o D = m/v  V= 2x2x2= 8 cm ^3  D= 11.2g/8cm^3= 1.4 g/cm^3 o Al vs Au = which denser?  GOLD b/c > mass (same vol)

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Ch 2 1. Lavosier’s Law of Conservation of Mass a. Mass not gained/lost 2. Proust’s Law of Definite Proportions 3. Dalton’s Atomic Theory a. matter = indivisible part. Atoms b. all atoms of an ele. = same mass + properties c. atoms not created/destroyed d. atoms combine in simple whole # ratios 4. ATOM a. IS div. (of subatom. part. ) 5. J.J. Thomson’s Electron discovery a. Cathode ray tubes test 6. Milikan’s oil drop a. Det. Electron charge and mass 7. ELECTRON = MASS of atom a. Atom mostly empty space + pos. sphere outside < plum pudding model 8. Which more abundant? a. Given 107.9 amu. B/w Ag-107 or Ag-109; Ag-107 bc close to amu 9. Nat. abundance??/ Avg atom mass= fractional abundance x atm. Mass + ~~~ 10.

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Ch 3 1. Electrolytes- conduct electricity, release ions a. Strong: complete dissociation; ex: HCl, nitric acid b. Weak: acetic acid 2. Nonelectrolytes- no electricity conduct., doesn’t dissociate in h2o 3. Hydrates a. Cobalt (II) Chloride Hexahydrate: CoCl2 – 6H2O b. Calcium Sulfate hemihydrate: CaSO4- ½ H2O c. Magnesium Sulfate Heptahydrate: MgSO4 – 7 H2O d. Nickel (II) Chloride Hexahydrate: NiCl2 – 6H2O 4. Types of Compounds a. Ionic aka salts – electrolytes i. Crystal lattice arrange. ii. Hard, brittle solid, high mp, high bp, high d, strong electro. b. Molecular- nonelectro. i. No overall charge ii. soft solid, low mp, low bp, low d, weak electro/nonelect. c. Acid + Bases- electro. 5. ***CH 3 polyatomic ions chart *** 6. Polyatomic atom = -ate 7. Monoatomic atom= -ide 4

8. Naming molecular comp. = -ide (name by numbering) a. CCl4 carbon tetrachloride b. SO3 sulfur trioxide c. N2O4 dinitrogen teroxide d. PF5 phosphorus pentafluoride 9. PREFIXES a. Tetra- 4 b. Penta- 5 c. Hexa- 6 d. Hepta- 7 e. Octa- 8 f. Nona- 9 g. Deca- 10 10. Special cases of molec. formulas a. Hydrogen peroxide H2O2 b. H2O water c. NH3 Ammonia 11. OXOACIDS a. HNO3 nitric acid b. H2SO4 sulfuric acid c. H2CO3 carbonic acid 12. Naming Binary Acids a. –ate to – ic or –ite to –ous for polyatomic ion b. If not polyatomic, use hydro- and –ic c. H3PO3 phosphorous acid d. HNO2 nitrous acid e. H2SO3 Sulfurous acid

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Ch 4

1. Mass % element = Mass ele/ mass entire comp x 100 2. 1 mol = 6.022 x 10^23 particles (atoms/mole/units) 3. Dilution = Mi (Vi) = Mf (Vf )

Ch 5

1. Combustion of Hydrocarbons skeleton- CH + O2 -> CO2 + H2O 2. Acid a. H2S (g) b. H2CO3 -> CO2 + H2O c. H2SO3 -> SO2 + H2O 3. Base a. NH4OH -> NH3 + H2O

4. Metal + Water -> H2 + Metal Hydroxide 6

a. 2Na + H2O -> H2 + NaOH 5. Balancing Reaction a. **MUST check activity series to see if metal more reactive then the one in compound 6. Halogens- increasing activity displacing in reaction F Cl Br I

Ch 6 ~ 1. Heat a. H2 + I2 -> 2HI b. Q = +53 kj/mol I2 c. Endo or exo? ENDO b/c + KJ d. Energy change? When 2.5 mol I2 reacts i. 2.5 mol I2 x 53 kJ = 133 kJ 1 mol I2 2. CALORIMETRY coffee cup a. 5 g methanol, temp ↑ from 20-> 35 celsius, heat capacity= 7.7 kJ/cels. b. Heat of combustion? i. 2CH3OH + 3 O2-> 2CO2 + 4 H2O ii. T = Tf- Ti = 35 – 20= 15 celsius iii. Q rxn= -Q calorimetry 1. -15 x 7.7 kJ/Celsius = -116 kJ/ 5g CH3OH= 23.1 kj/g methanol c. Per mol? 7

23.1 kj/g (32.05 g/1mol)= 739 kJ/mol 3. Mass of ele/comp consumed when given mass of other comp/ele? a. Mass -> MM = mol-> mol/mol ratio = mol-> MM = Mass 4. **Law of Conserv. Of Mass= Mass comsumed react = mass formed prod 5. Finding limiting reactant a. Given reaction equation. Using # of mole. In picture b. Calc B > actual B i. B is limiting reactant c. Calc B< actual B i. A is limiting reactant d. Calculated value = required amt; given amt = what’s present e. Choose to find mol of the one that requires more mol according to reaction Ex: 0.25 mol P4, 0.75 mol O2 P4 + 5O2 -> 2P2O5 i.

0.25 mol P4 needs to react with 5 mol O2 vs .75 mol O2 needs 1 mol P4

6. How many mol of NH3 (prod) can be produced then? a. BASED on using limiting reactant i. 5 mol H2 (2 mol NH3/3 mol H2)= 3.3 mol NH3 ii. 3.1 mol N2 (2 mol NH3/1mol N2)= 6.2 mol NH3 7. GIVEN grams instead of mols a. Use MM to convert g -> mol for both given masses b. Then for one of the mols calculated, do mol/mol ratio to see which is limiting react. 8. Percent yield = actual/ theoretical yield x 100 9. **Quantities of Heat ** a. 1 cal = 4.184 J b. 1 Cal = 1000 cal =1 kcal c. Calorie = amt of ener it takes to raise temp of 1 g of water by 1 cels. 8

d. 190 calories (Cal). Energy in J? i. 190 Cal x (1000 cal/1Cal) (4.184 J/1 cal) 10. Specific Heat a. Q = m (specific heat) (final temp in cels) b. Specific heat special and given for all solutions c. Q system + Q surr = 0 i. Exo into surr : q< O neg. ii. Endo into syst: q> 0 pos. d. Smallest specific heat = greatest temp change

Ch 7 1. 2. 3. 4.

C= 3.0 x 10^8 m/s speed of light C = wavelength x frequency (v) If wavelength ^, frequency V Wavelength? Given frequency(v) 100.7 MHz a. 100.7 Mhz (10^6 s-1/ 1MHz) i. Wavelength = 3.0 x 10^8 / a. = 2.98 m ii. Nm = 10^-9 m b. Electromagnetic spectrum i. Gamma ray < xray< ultraviolet 10^-8...