CH 101 FINAL EXAM GUIDE
CH 101 FINAL EXAM GUIDE CH101 Rupar
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CH 101 FINAL EXAM STUDY GUIDE Formulas and Things to Memorize Avogadros number 602 X 1023 thingsmolecules Atomic number number of protons and electrons Atomic mass sum of protons and neutrons in the nucleus Metric System 0 Kilo103 o Centi10392 o Milli10393 0 Micro 106 o Nano 10399 Speed of light c 300 x 108 Planck s Constant 6626 X 103934 Js Order of electromagnetic Spectrum wavelength frequency 0 AM radio TVFM Microwaves Infrared ltVisible lightgt Ultraviolet X ray Gamma Rays OOOOOOO Energy of a photon Ehv Frequency c M L is wavelength Energy of a photon emitted o 218 x103918 x1nf2 1n3 When given electronegativities you can determine polarity o If the difference is between 01 9 04 NONPOLAR o If the difference is between 05 9 19 POLAR o If the difference is gt 20 IONIC Molar mass conversions 0 Mass of moles X molar mass 0 Molar mass mass of moles o Moles massmolar mass Percent yield actualtheoretical 100 Change in energy AE q w Heat q mCsAT Work pAV qsurroundings qcalorimeter 39qsystem for calorimeters Standard Enthalpy of Formation Boyle s Law 39 P1V1 P2V2 Charles s Law 39 T1V1T2V2 Avogadro s Law 39 V1T1 VzTz The Ideal Gas Law I PV nRT nmoles R ideal gas constant 008206 Dalton s Law of partial pressure I ltotal Iltotal Avg kinetic energy of gasses 0 KB 12mV2 Rate of effusion o AH 2n AHfO products 2n AHf reactants RateA Molar MaSSB RateB Molar MassA Common Polyatomic Ions ion name ion name NH 4 ammoni um 003339 carbonate NO 2 nitrite HO O 3 hydrogen carbonate N03 nitrate ClO39 hypochlorite 8032 sulfite C102quot chlorite S 043 sulfate Cl 03 chlorate HS 04 hydrogen sulfate 0104 perchlorate OH hydroxide C2H303 acetate C Nquot cyanide Mnoq permanganate P043 phosphate Crz 073 dichromate HP 043 39 hydrogen phosphate CrOa2 chromate H2 P04 dihydrogen phosphate 023 peroxide To memorize order of orbitals Compounds Containing the Following Ions are Generally Soluble Excep ons when combined with ions on the left the compound is insoluble Li Na K NH4 none N03 C2H302 none Cl Br I Ag Hg22 Pb2 SO42 Ag Ca2 Sr2 Ba Pb2 Compounds Containing the Following Ions are Generally Insoluble Excep ons when combined with ions on the left the compound is soluble or slightly soluble OH Li Na K NH4 Ca2 Sr2 Ba2 82 U M W NH4 0032 P043 Li Na K NH4 7d is Start here and move along the arrows one by one 6f 69 339 7f 79 7117 Pure Substances made of a single type of atom or molecule all samples have the same characteristics Elements cannot be decomposed simpler by reactions single type of atom may or may not be combined can combine for compounds Ex He Compounds one moleculearray of ions 2 or more atoms same compounds behave the same way Ex H20 Mixture matter varies 2 types of atoms in variable proportions Homogeneous mixture uniform composition every sample identical could have different characteristics of other same with same components distribution Heterogeneous mixture non uniform portions have different composition and properties Neon gas pure substance element Co2 pure substance compound Soup heterogeneous mixture Coke homogeneous mixture Signi cant Figures All non placeholding numbers are significant Interior OS are significant Trailing 0s after decimal place always significant Trailing 0s before decimal place ambiguous Leading OS are not significant Exact numbers have unlimited sig figs X amp result has lowest number of sig figs of given number amp result has same number of decimal places as the lowest Round at the end keep track of sig figs during calculations Atomic Experiments Millikans experiment charges were multiples of l6 x lOAl9 o Deducted the charge is discrete found the mass with Thomson using cathode rays Thomson Plum Pudding Model 0 First time atom was shown to be divisible has inner structure Proved incorrect with Gold foil experiment Rutherford 0 Because some alpha particles were de ected must have been some positive charge in the center most of mass is in nucleus 0 Found protons Atomic mass average of all isotopes Sum of fractional abundance x mass Spectrometry moves ions past magnets to measure how much ions are de ected Definition The capacity to do work Kinetic energy is recognized as temperature THERMAL ENERGY Potential due to composition and relative position Chemical Potential electrostatic forces between atoms electrons Spontaneous process things with lots of potential energy want to get rid of it ie things that burn The Mole Avogadro s number 6022 X 1023 things The Nature of Light The way light interacts with matter helps us describe it Law of Quantum Mechanics molecular scale behaves as a wave Waves Amplitude height of wave Wavelength lamba distance covered by wave crest to crest Frequency Mu of waves that pass through a point in a given time Number of waves number of cycles Units Hz or sAl Light Electromagnetic radiation All waves move through space at same speed ie speed of light o C 300 X 10 A8 Since we know speed we can nd wavelength or frequency 0 V sAl cms Wavelength m Light behaves in photos energy determined by frequency and consequently wavelength 0 E hv o h Planck s constant 6626 X IDA34 J X s 0 Color depends on wavelength I White light mixture of visible I When object absorbs all black Electromagnetic spectrum 0 Visible light is only a small part of all wavelengths o Shorter wavelength higher frequency 0 Higher energy electromagnetic radiation can be damaging to biological molecules ie UV rays Called ionizing radiation 0 The lowest we encounter is I AM radio I TVFM I Microwaves I Infrared I ltVisible lightgt I Ultraviolet I X ray I Gamma Rays To calculate number of photons released 0 Ehv then Total energy energy per photon Constructive interference In phase Destructive interference out of phase Diffraction waves encounter obstacles in barriers about same size as wavelength and bend around it Diffraction through slits separated show interference patterns Particles to not diffract waves do Photoelectric effect electrons will be emitted when metals are impacted with high enough frequency Observed not sum of photons energy but energy of individual photons Source of light is quantized not continuous comes in chunks of photons Threshold frequency has to equal binding energy of the emitted electron Hv si When matter absorbs energy it is often released as light ie glowsticks light bulbs used in ame tests Bohr Model to show structure when atom undergoes energy transitions Quantized energy of atom related to electron position discrete l electrons travel in orbits at xed distance 2 lower orbit transition emit photon 3 higher orbit transition obtain photon 4 distance between orbits determines E of produced photon Quantum Mechanics Predicts the wave particle duality of matter every piece of matter has a wave deBroglie predicted that wavelength of a particle is inversely related to its momentum wvlength hmv Proven in the two slit experiment If electrons behaved like particles it would only be two light spots and diffraction wouldn t exist Interference is caused by e with itself not multiple e s Wave and particle nature are complementary properties You cannot observe both at the same time Uncertainty principle Heisenberg product of uncertainties in position and speed of particle is inversely proportional to mass cannot know position and velocity exactly at same time Because of this indeterminacy can only predict probability Properties of atoms are related to electron s energies Electron energy depends on which orbital it s in Wave function2 is probability distribution map aka an orbital Multiple solutions Quantum Numbers N principal quantum number Depicts the energy of the electron in particular orbital Larger value means larger energy the e has The closer the e gets to the nucleus the less energy it has because of interactions with the nucleus but the more energy that is released when the electron jumps closer o 218 x IDA18 J 1112 L describes the shape of the orbital Every value of n has n different L values 1 values from 0 to nl 0 Ex lfNis 2 L 01 L0 S Ll P L2 D o L3 F ML speci es the orientation of the orbital values are integers from 1 to 1 If LO ML0 If L2 ML 2 l 0 l 2 Each wavelength in the spectrum corresponds to electron transitions between orbitals For an excited electron it can relax to nl orbitals If n5 can relax to 4 3 2 1 Energy of photon released 218 x IDA18 lnf2 lniz OOO Radial distribution function represents the total probability of nding an electron within a thin spherical shell at distance r probability decreases with increasing r but volume of shell increases Number of nodes n l N is principal quantum number N principal quantum number denotes energy of the electron in the orbital L angular momentum Nl L 0 S orbital I Spherical lowest energy each level has one S orbital Ll P orbital I For any value of n gt 1 you have 3 p orbitals I Two lobed look like two balloons tied together L2 D orbital I For any value of ngt2 you have 5 p orbitals I Situated in planes I Shape is like 4 balloons tied together L3 F orbital I For any value of ngt3 you have 7 f orbitals I 8 lobed Why are atoms spherical Average of orbitals approximate a sphere Periodic Table Designed by Mendeleev based on elemental reactivity with oxygen Periodic law when elements are arranged in order of increasing atomic mass patterns appear periodically QUANTUM MECHANICS EXPLAIN WHY Electronic spin I Intrinsic property spin to create magnetic eld I MS l2 or l2 I Each orbital holds two electrons of opposite spin Sublevel Energy splitting in multi electron atoms I If you put an electron in an orbital you can calculate its energy IF it s a hydrogen atom by using 218 x 103918 lnz I Only reliant on energy I For non hydrogen atoms L and M1 are important Interaction of electrons I Feel attraction to positive charge of the nucleus and negative repulsion of other electrons Electrons farther away feel both forces the closer electrons repel the outsite electrons with shielding force EX the 2p orbital is more shielded with respect to the 2s orbital Order of multielectron system MEMORIZE I ls 2s 2p 3s 3p 4s 3d 4s 4p 5s 4d Pauli exclusion principle No two electrons can have the same 4 quantum numbers 0 ie only two electrons with opposite spin allowed per orbital Aufbau principle Electrons ll the lowest energy orbital possible Hund s rule when filling orbitals with same energy place on electron in each orbital before pairing them correct incorrect 1 1 1 1 1 1 1 q J 152 252 2p 15392 253 2p 3 0 Valence electrons electrons in sublevels with highest principal energy levels inner electrons that ARENT valence core electrons o In other words the amount of electrons in the outermost energy shell 0 Determine chemical properties Electron configuration Atomic Element Electron Number of Number 333111301 Con guration 1Valence Electrons 151 Is2 153251 15552 18228pr 1322832p2 1332532 13239232 2p4 1322532 Ne 1332822S DNDOOQOxMhLQMO e mOZOUJUJgfIEIE D b OOQQMbLAJMO MH H o Noble gas notation Use previous noble gas in group 18 closest to atom you are describing then add the additional electrons Periodic Trends I Alkali Metals Group 1 One valence electron Highly reactive I Alkali Earth metals Group 2 Two valence electrons Want to lose both electrons to gain noble gas con guration I Chalogens Group 16 I Halogens Group 17 1 electron short of noble gas notation Tend to gain one electron to have an octet Metals VS Elements that border on the amphoteric line shown in green are metalloids They Nonmetals have characteristics of both metals and nonmetals Aluminum Al however definitely has mostly metallic characteristics and boron B is mostly nonmetalllc H l m Md r Metals Nonmetals 1 Have luster 1 Are dull 2 Are malleable and ductile 2 Are brittle 3 Conduct heat and electricity 3 Do not conduct heat or electricity very well 4 Tend to lose electrons 4 Tend to gain electrons Zeff effective nuclear charge If valence electrons do not experience the full strength of a nuclear charge it is less strongly held to an atom This value is equal to the amount of valence electrons Z atomic number number of core electrons Arrows indicate increase gt A IrrIqm J 1 2 H He 11wquot hum Ilmml manquot 6 nthn quotUuu HIM AH nqu 3 4 10 LI Be Ne v 1 I 11 quot l V W 11 12 15 18 Na Mg P Ar 9 19 20 21 22 23 33 34 35 36 K Ca Sc TI V As Se Br Kr 1 37 38 39 40 41 51 52 53 54 Rb Sr Y Zr Nb Sb Te I Xe 119 1w mm WINK Wu I um w mm 55 56 5770 71 72 73 83 84 85 86 Cs Ba Lu Hf Ta Bl Po 1 n 87 88 89102 103 104 105 F a Lr Rf Db l l l 1 1m 1 l l l Lanthanide series 57 58 59 60 61 62 63 64 65 66 67 68 69 70 La Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb mu 1m mu um w w 1 w 1 nuH Wm 14m w 1va 11111 Actinide series 89 90 91 92 93 94 95 9e 97 9a 99 100 101 102 c Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No 111 law 1 HI 1 v l quot UH i l 1v l 1W Ixquot l ll iu39l I39H39 13quotquot 3 1 Ionization energy energy needed to remove an electron from an atom valence electrons are easiest to remove The larger the Zeff the more energy required to ionize gt A mm J 1 2 H He 11w 1mm Ilhunl lu lllnu n milm quot11qu wquot 1H1 3 4 5 6 7 8 9 10 LI Be B C N O F Ne 1 y 1 11 12 13 14 15 16 17 18 Na Mg Al SI P S Cl Ar 9 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 K Ca Sc TI V Cr Mn Fe Co NI Cu Zn Ga Ge As Se Br Kr mm nuts 1 14 1mm nui wr mun unu w u an an m 1 l hm I w I 1 w my 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te Xe m 49 w w u 01 x h 7 a m m u wn Inn 4 m w um um um I 1391 391 mm 11 m m my mum llH 1 mun ulvm n m39nun HHI mm 111 Nn 11M Inn mumm lllIHH Hmva 1501 H mm 111le 1v luxlmm nkih III nlatww mm 55 56 5770 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 Cs Ba Lu Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bl t 1391 11 w u m A rm 5 lw m mm m vm In39x I r v w w lmtw ml 3 w mws l 1 if 1 l 11 mmm Lrhulv mun wu muln 1 11mm muuw uv Hnrumv murmle mumumu unnluln mun lmlum 87 88 89102 103 104 105 106 107 108 109 110 111 112 114 a Lr Rf Db Sg Bh Hs Mt Uun Uuu Uub Uuq 13 4 Uri n J l 1 L39I 1 1311 14011 7v 1 n gt1 1 1 NJ L r 71 q mmmmm Lanthanide series 57 58 59 60 61 62 63 64 65 66 67 68 69 70 La Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er m Yb 11x14 m 1 H1 u um w 1 19 1 pm mm mm 11 I J 139u 11111 l Actinide series 39 90 91 92 93 94 95 9e 97 9a 99 100 101 102 c Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No 111 1m 1 HI 1 v l quot Ni i l 1v l 1W Ixquot l ll iz 1 13quotquot 3 1 Electron af nity How much energy it takes to remove an electron from a system group 17 has the highest magnitude Highest EA in any period is the halogen or the one in group 17 A 1 M J 1 2 H He luvquot hum Ilhunl mam n 1 nitnu 1 I M M nqu 3 4 5 10 LI Be B Ne 11 12 13 18 Na Mg Al Ar 19 20 21 22 23 24 25 26 27 28 29 30 31 36 K Ca Sc TI V Cr Mn Fe Co NI Cu Zn Ga Kr we Wis 11 M 13944 Hul wr mun unu w u an an m 1 l 4 1 mm 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te Xe m 49 w m u 01 x h 7 a m m u wn Inn 4 m w um um um I 1391 391 mm 11 m m m mm llH 1 mun ulvm n mum HHI mm ln v Nn 11M Inn nmmm lIlmn MIHHIH 1501 mm 11111111 in 1 hmmm 111111 II nlatww 111m 55 56 5770 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 Cs Ba Lu Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bl t um w u m A rm 5 lw m mm m vm In39x I r v w w lmtw ml 3 w mws l 1 if 1 l 11 mmm mhulv mun wil 4111quot 1 11mm muuw up quotllr39llllll munmum mumulmv umuum mur 111mm 87 88 89102 103 104 105 106 107 108 109 110 111 112 114 a Lr Rf Db Sg Bh Hs Mt Uun Uuu Uub Uuq 13 4 Uri n J v 1 39I 1 13111 14011 7v 1 n 1 11 NJ L r 71 Aquot Lanthanide series 57 58 59 60 61 62 63 64 65 66 67 68 69 70 La Ce r Nd Pm Sm Eu Gd Tb Dy 0 Er Tm Yb xvl mu mu um up in im 114 I mm mm My 1mm mu Metallic character Metals tend to have lower Ionization energy and non metals tend to have large electron af nity lt 1 lwznh H He 3 hailquot 9 llii II LI Be F Ne m n 1811 Na Mg Cl Ar 4 2 57 I 439 m 31 I 32mquot 33 55 32 K Ca Sc TI V Cr Mn Fe Co NI Cu Zn Ga Ge As Se Br Kr I 96 Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te Xe 39 In NI H4 H II x km W 4 ltI 111 Hi Iw II Inn 1 Iv M II II IIII I H HI J I m r nI I I1 urgslH IIIsmelh 5770 IIII II II71r2JH IIII7gIIII III I7IXI IIV397I5III39I my II7IIIII II 495w quot18396qu rhgqvm IBIZI my I u 398 III 3518 l e W Cs Ba Lu Hf Ta W Re Os Ir Pt Au Hg Tl Pb BI Po At Rn yaw w I711 m 2 m w mm vw I M 4quot lmw ml a w39 mm m1 I111 1 1 II Iggnn I gnaw 104 II315IIII NI39II1ampIIIEI 139BI7II IIlealam m Igr39 IIIII1IIISIIII IIIII1H1IIIIII 11413114 InII 1x1IEIIIIII Fr Ra Lr Rf Db Sg Bh Hs Mt Uun Uuu Uub Uuq I l I J I I I I I I39JI Iquot 1 I I I I I I I I m Ilmnmm mmm quot dylVW39 mumm umwmmu lIIi II39II mum In rmn hmmm quotIlium m mm 57 58 59 60 61 62 63 67 68 69 70 La 9e Pr Nd Pm 3 E9 99 Th Dy H9 Er Tm Yb 39Lanthanide series 1 r I I A I y 39 I II Actinide series 96 97 98 Ac Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No 139 1 HI w M 1m v I 39l Inn 2 I In Iquot II I Ilw39l l I 4 lt 3 5 6 I 9 w LI Be B C N O F Ne 1113 quothigh JIJJI 113411 Jii 39 I 57 3173 11 12 13 14 15 16 17 18 Na Mg Al SI P S Cl Ar 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 K Ca Sc TI V Cr Mn Fe Co NI Cu Zn Ga Ge As Se Br Kr 123 I ZCI IJI 3175 31112 7131 911 311 ITIIL39 2343 III I Chili 25133 4221113 Mini III ZITI 37 38 39 40 49 50 51 52 53 54 Rb Sr Y Zr In Sn Sb Te Xe u M an w mun M I m J Lem I395III 4erng 5770 IIII39II In3971r2JH V III IlIIII III In II I I 81 82 Exam 4 I81I4I a5 86 Cs Ba Lu Hf Tl Pb BI Po Rn Iv Iggnn I gnaw 104 3lsIllI I39I1ampIIIEI L113 I IIlealam m quoglur IIIII1IIiISIIII IIIII1H1IIIIII vnx1n1tglu tnII 1lIIIIIII Fr Ra Lr Rf Db Sg Bh Hs Mt Uun Uuu Uub Uuq I l I J l I I I I I II Iquot 1 I I I I I I I I 1 57 58 59 063quot 61 62 63 m W16 65 66 gym quot3539quot quotquot631quot w 70H 39Lanthanide series La Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb IIIII In v 1 r I I A I y 39 I I q Actimde series Ac Th Pa Np Pu Am Cm Bk Cf Es Fm Md No I I I l I IF39I I I I Itquot I I quot 39I I Naming Ionic Compounds Forces that in uence bonding O O O Nucleus repulsion between nearby atoms Electron to electron repulsion Nucleus to electron attraction Lewis bonding theory 0 O Emphasizes valence electrons Can predict stability shape size polarity Ionic compounds 0 O O O The metal tends to lose an electron and form a CATION Nonmental tends to gain the electron and form an ANION Dissolved in water an ionic solution will conduct an electric eld Why you keep plugs and electronics from water because most water is not pure and has ionic compounds in it Ionic solids are hard to melt crystals and release a lot of energy when broken down Octet rule 0 Atoms want to have 8 valence electrons to obtain noble gas con guration o Exceptions H Li Be and B loses l 2 or 3 valence electrons to obtain He con guration with 2 valence electrons in the ls shell To write a compound 1 Write the metal cation and positive charge 2 Write the nonmetal anion and charge 3 Switch the charges to the subscript of the opposite atom 4 Reduce subscripts to lowest ratio 5 Check the total charge by multiplying the charge by the subscript a Example Al 3 0 2 A1203 Naming 0 Name of cation is the same 0 Anions have ide suf x I NaO sodium oxide I CsBr cesium bromide 0 Some transition metals can form more than one cation 0 MEMORIZE I Chromium 2 or 3 I Iron 2 or 3 I Tin 2 or 4 I Copper 1 or 2 0 Also memorize polyatomic ions These are composed of covalently bonded atoms that posses a charge Most are oxyanions Tips to memorize polyatomic ions Memorize the ates I ate 1 oxygen 9 PER pre x I ate 1 oxygen 9 ITE suf x I ate 2 oxygen 9 HYPO pre x ITE suf x Hydrate ionic compounds add pre xhydrate I Magnesium Sulfate Pentrahydrate MgSO4 SHZO Covalent Compounds 0 Nonmentals have high ionization energy and there is dif culty to remove electrons 0 Because of this when nonmentals bond to each other they SHARE electrons sharing holds them together 0 The more atoms the electrons share the shorter and stronger the bonds 0 Covalent compounds do not form lattices like ionic compounds they are direction and this is why they have speci c geometries Mass VIolar Mass and Moles conversions 0 When doing these calculations use the formula mass of moles molar mass 0 Every mole contains 6022 X 1023 formula units 0 One mole of a compound or element is equal to its molar mass Ex One mole of Oxygen 1600 g one mole of water 1800 g So 6022 X 1023 units of water is 18 g just like 6022 X 1023 units of oxygen is 16 g Chemical Formulas 0 Empirical Give the relative number of atoms ie the smallest atomic ratio 0 Molecular the actual number of atoms I Ex Glucose molecular is C6H1206 empirical would be CH20 o composition mass of element in one mol mass of 1 mol of compound 0 100 I Ex composition of C in C02 12011 gmol C 44066 gmol C02 273 I Ex How much oxygen is present in 32 g of H20 Mass of H20 is 1802 g Dimensional analysis 32 g 1802 we have 177 moles of H20 Since there is only one molecule of 0 per molecule we also have 177 moles of oxygen 177 moles of oxygen 2832 g 2832 32 g total 100 885 0xygen o Calculating Emprical Formulas I Convert given to grams I Convert grams to moles I Create a pseudo formula using moles I Divide each molecular ratio by the smallest number of moles I Multiply all ratios of moles to make whole numbers Look at example 0 Ex C 60 H 448 03553 C 60g H 448 g 0 3553 g divide by each molar mass C 4996 mol H 443 mol 02221 mol Psuedo mol formula C4996H44302221 Divide all subs by smallest 2221 C2249H199701 H rounds to 2 0 is 1 but we need to x the carbon 0 If the nal ratio has a decimal of 33 or 67 multiply all subs by 3 o If the nal ratio has a decimal of 25 or 75 multiply all subs by 4 o If the nal ratio has a decimal of 5 multiply all subs by 2 Because carbon is 2249 mole we will multiply by 4 The empirical formula is C9H804 Electronegatvitiy and Bond Polarity Polar covalent bonds occur between unlike nonmetals resulting in an unequal pairing causing a partial positive charge and partial negative charge Think of water 0 has a partial negative charge H has a partial positive charge The partial negative charge is because it attracts electrons more readily this is called electronegativity Electronegativity is the ability of an atom to attract bonding electrons to itself the larger the difference in electronegativity the more polar the bond The trend increases upwards and to the right ie Flourine is the most electronegative atom noble gasses are stable and therefore hardly electronegative and Francium is the least electronegative When given electronegativities you can determine polarity o If the difference is between 01 9 04 NONPOLAR o If the difference is between 05 9 19 POLAR o If the difference is gt 20 IONIC Dipole moments are a measure of bond polarity the more electrons shared the larger the moments are Dipole moments are directly proportional to partial charges Drawing Skeleton Structures Hydrogens are always terminal and the least electronegative element is in the center Calculate the total number of electrons add one for each negative ion and subtract for each positive ion Place two bonding electrons between each bonded atom first then try to ll the valence shells starting with the terminal atoms until you run out of given electrons Form double or triple bonds if octets are not satisfied Formal Charges Some atoms during bonding have more or fewer electrons in the bonded form than it brought to the reaction ITO calculate Formal charges on neutral atomsmolecules is 0 Charged ions will have a formal charge EX Ammonium NH4 Resonance Structures Lewis theory suggests there is some degree of delocalization to stabilize molecules ie sometimes there is more than one lewis structure for each molecule these are resonance structures The true nature of the molecule eXists somewhere in between 0 Resonance structures only differ in position of electrons and must have the same number of electrons Remember only 3rd row and below can break the octet rule Overall charge of the molecule cannot change 0 The BEST resonance structure minimizes formal charges has smaller formal charges if necessary and keeps negative formal charges on more electronegative atoms VSEPR Molecules Geometry depends on electron geometry which is derived from the amount of electron groups around the central atom An electron group is a bond whether single double or triple it equates to one electron group or a lone pair Example NH3 has 3 bonds and l lone pair so it has 4 electron groups We use this to create AXE notation A central atom X bonding pairs E lone pairs 0 Example CF4 has four bonding pairs and no lone pairs AX4 0 N02 has a double bond a single bond and one lone pair AX2E VSEPR says that electrons around the central atom want to be as far away as possible and the result is molecular geometry 0 Remember that lone pairs have more electron density so they distort the ideal bond angles Clnclron Bonding Lone Elnclron Molncular Approximate Groups I Groups Pairs Gnomctry Geometry Bond Anglns Example 2 2 0 Linear Lmeac 18039 039 c 1 3 3 o T 39 I mum 120 T ngona planar puns u 39f 3 K 3 2 1 Trigoml plans Ben 12039 l 4 4 0 Tettahedral Tetrahodtal 109 5 H C H l H Tngonal quot N 4 3 1 Tetrahedtal pwammm 1539 I b 4 2 2 Tetrahedtal 89m lt109539 H ZOI H A 39c 3quot s 5 o Trigonal Ttigonal 120 lequaloriall 39 39 39 L 39 bipvtamidal bipwamidal 90 axial a 9 39 139 TngonaI lt120 quuato all 5 upmmmaa See quot lt9o39 anal F 3 F 39g F F quot3rquot F Tngonal 5 3 2 Damtamde Tampa 90 s 2 3 9 L 13039 Fquotxc P Q uwmidu quot9339 39 6 6 D octahedral Octahcdral 90quot Square 6 5 1 Octahcdral mammal lt90 Squaw 6 4 2 Octahodrai pun 90 Valence Bond Theory More accurate that VSEPR Bonds occur when orbitals overlapinteract For example a H atom has one V electron in its ls orbital when it combines with another the ls orbital is full and it lowers the energy of the system VBT de nes valence electrons residing in quantum orbitals and hybrid orbitals Explains why carbon can have four bonds It has 2 2s electrons and 2 2p electrons however that only allows for two bonds But when hybridizing the s and p orbitals there become 4 orbitals each with one electron that can welcome another ie four bonds ie tetravalent Memorization for VBT spElectron groupsl o 4 bonds sp3 hybridization tetrahedral there are four of them 0 3 electron groups sp2 hybridization trigonal planar there are three of them 0 2 electron groups sp hybridization linear shape there are two of them seen in atoms that form triple bonds Sigma Bonds and Pi bonds Sigma bonds mean orientation lSt bond formed parallel gives free rotation Pi bonds have perpendicular orientation 0 Sigma bond is just a single bond When you have a double bond you have 1 sigma bond and 1 pi bond When you have triple bond you have 1 sigma bond and 2 pi bonds Stoichiometry Molar mass conversions 0 Mass of moles X molar mass 0 Molar mass mass of moles o Moles massmolar mass Mass to mass conversions o SiCl4 2H2O 9 4HCl SiO2 I With 76 grams of SiCl4 how many grams of HCl are formed I 76g 1701 gmol moles of SiCl4 I 4 moles HCl Moles of SiCl4 So moles of SiCl4 X 4 because of ratio moles of HCl I Moles of HCl X 365 gmole mass of products 652 g HCl Mass A 9 moles A 9 moles B 9 mass B 2H2 02 9 2H2O What mass of O2 needed to react with 4 g of H2 How much water is produced 0 l 4 g H2 1 mol H2 1 mol 02 320 g 32 g of O2 needed 20 g 2 mol H2 1 mol O2 0 2 4 g H2 1 mol H2 2 mol H20 18 g 36 g H2O produced 20 g 2 mol H2 1 mol H2O Limiting Agents and Theoretical Yield The limiting reagent is the reactant that produces the smaller amount of product you can use either product for setting up ratios 0 How much CaCl produced from 3 mol Ca and 5 mol C1 Which is limiting 39 Ca 1 C12 9 CaC12 I 3 mol Ca 11 ratio 3 mol CaCl2 I 5 mol C12 11 ratio 5 mol CaCl2 I Ca is limiting This is like saying 3 mol Ca mol C12 will produce 3 mol CaCl2 with 2 mol C12 leftover because there is not enough Ca left to react with 0 CH4 202 9 CO2 2 H2O If we have 5 mols CH4 and 8 mols O2 which is limiting I 5 mol CH4 11 ratio 5 mol CO2 I 8 mol O2 21 ratio 4 mol CO2 I Oxygen is limiting because 4 lt 5 This is like saying when 5 and 8 moles react 4 mol C02 will be made with 1 mol CH4 leftover Theoretical yield is the maximum amount of product that can be formed based upon the limiting reagent identify the limiting reagent and the of moles it can producethis is your theoretical yield 0 What is the maximum amount of water that can be formed from 4 mol H2 2H2 O2 9 2H2O I The answer is 2 mol H2O because for every 2 mol of H2 and 1 mol of O2 we can form 2 molecules of water If we do not have a second mol of O2 another set of water molecules can not be formed 0 How many moles of Si3N4 can be formed from 12 mol of Si and 1 mole of N2 381 2 N2 9 Sl3N4 I SteIi 1 identi limitini reaient 1 mol N2 1 mol Si3N4 2 mol N2 0500 mol I Step 2 The theoretical yield is 0400 mol because silicon is the limiting reagent Actual and Percent Yield Actual the amount produced in a reaction Percent yield actualtheoretical 100 If the problem indicates that element x is in excess the other reagent is limiting Use the limiting reagent to calculate actual yield with a mass 9 mol 9 mol 9 mass equation Solutions homogenous mixtures Solvent major component Solute minor component 0 Aqueous solutions are solutions with water as the solvent 0 Concentration is given in units of massvolume o Molarity amt of solute in moles amt of solvent in Liters 0 We can rearrange this equation to solve for moles or volume I Moles molarity volume I Volume Molarity moles 0 Important note 1 mol of NaCl 1 L of water DOES NOT EQUAL a 1M solution because adding NaCl will increase the volume to greater than 1 L 0 To make a l M solution add 1 mol of solute add enough water to dissolve it dilute with more H20 until volume is 1L 0 How would you prepare a 2500 mL of 0150 M CaC12 solution I 0250 L 0150 MolL 0375 mol of CaC12 needed I Place 037 5 mol 416 g if calculated with molar mass to a container dissolve in minimum amount of water then dilute to 250 mL Diluting Solutions 0 M1V12M2V2 o M is the concentration molarity V is volume 0 Most important note make sure ALL units of given values match 0 If an ionic compound dissolves in water it is an electrolyte and will conduct electricity in a solution 0 A strong electrolyte dissociates 100 o A weak electrolyte has lt 100 dissociation o Acids are covalent examples of electrolytes Example HCl 0 A strong acid completely dissociates into its ions 0 A weak acid does not completely dissociate Ex CH3C02H 9 CH3C02 H o Nonelectrolytes most molecular compounds They may make bonds with the water molecules but do not conduct electricity 0 Solubility 0 Some ionic compounds dissolve in water and some do not Use solubility tables to gure these out Excep ons Compounds Containing when combined with ions on the Following Ions are the left the compound is Generally Soluble insoluble Li Na K NH4 none N031 C2H302 none Cl Br I Ag Hg22 Pb2 SO42 Ag Ca2 Sr2 Ba2 Pb2 o Excep ons Compounds Containing when combined with ions on the Following Ions are the left the compound is Generally Insoluble soluble or slightly soluble OH Li Na K NH4 Ca2 Sr2 Ba2 32 Li Na K NH 0032 P043 Li Na K NH4 o Predicting Precipitation Reactions 0 Determine which ions each aqeous reactant has 0 Determine the formulas of all possible products 0 Use solubility tables to predict if possible products are soluble or insoluble o If all possible products are aqueousinsoluble no reaction occurs If one or more of the products forms a solid a reaction occurs 0 Example CaClz Na2CO3 9 Ca2 2C1139 2Na1 CO3239 Possible products CaClz aqueous soluble CaC03 SOLIDHNSOLUBLE NaCl aqueoussoluble NaC03 aqeoussoluble CaClz Na2C03 9 CaCO3 s 2NaC1 aq Molecular Ionic Net ionic Equations 0 Molecular equations describe chemicals put into water and the product molecules formed Ex 2KOH aq MgN032 aq 9 KN03 aq MgOH2 s Complete Ionic Equations 0 Strong electrolytes are written as ions insoluble substances weak electrolytes and nonelectrolytes are written in molecular form 0 Equation written as all ions 0 Ions that are the same in reactants and products are spectator ions Net ionic equation Ionic equation spectator Ions 0 Example 2K aq on39 aq Mg2 aq 2N03391aq e K aq N03391 aq MgOH2 s Net ionic 2 OH39 aq Mg2 aq 9 MgOH2 s Acidmase Reactions 0 Acids ionize in water to form H ions H is the same as H30 hydronium ion 0 Bases dissociate in water to form OH Hydroxide ions Have OH groups NH4 or NaC03 o In a reaction of an acid and a base h pulls an OH to form a water molecule Cation from the base and anion from the acid form a salt I Acid base reactions make Salt amp water 6 this can help you remember I Ex NaOH HCl 9 NaCl H20 0 The number of H molecules in the reactants lets you know this I HzXYZ Diprotic I H3XYZ Triprotic I A tripotic acid means you need 3 hydroxide molecules to balance the equation diprotic means you need 2 hydroxide molecules to balance the equation etc always balance the equation rst and make sure units for V are liters 0 When you have equal amounts of both an acid and a base they react Salt amp Water and neutralize each other form a pink color change This is called the endpoint I Remember moles of H moles of OH 0 Example 10mL NaOH unknown concentration NaOH requires 1254 mL of 100 M HCl to neutralize What is the concentration I Step 1 Balance the reaction NaOH HCl 9 NaCl H20 This tells us that the H to OH ratio is 11 I Step 2 Write what you know 0 10mL NaOH V1 0 1245 mL HCl V2 HCl 0100 M M2 CONVERT TO LITERS because molarity is moles per LITER Conceptual equation 001L NaOHM NaOH 01254 L HCl0100M HCl 0 This is M V which nds number of moles Reference molarity equation M MolNolume I If our acid to base ratio is 11 nd the number of moles in the mass of HCl given The moles of HCl moles of NaOH Use the of moles to calculate the molarity by dividing by volume Plug back into equation to check To nd moles use Mmolvolume Manipulate to solve Moles M V 0 001254 L 010M 0001254 moles of HCl 0 Now that we know moles divide the moles by the volume of NaOH I 0001254 mol 001L NaOH 01254 mol m or M Thermochemistry 0 Energy anything that has the capacity to do work 0 Kinetic energy is manifested as heat on an atomic level 0 Potential energy is chemical potential for this class Work force acting over a distance Heat ow of energy caused by a difference in temperature The Law of Conservation of Energy says that the total energy of the universe is constant To nd kinetic energy 12 mv2 units Joules o A joule is the amount of energy to move 1kg of a mass at a speed of 1 ms A calorie is the amount of energy needed to raise a temperature of one g of water 1 degree C o kCal is the energy needed to raise 1000 g of water 1 degree C I food calories are kcals o A kwh 36 x 106 J kilowatt hour 0 De ning the system and the surroundings 0 System material or process we are studying o Surroundings everything else and the exchange of energy 0 Concepts to understand 0 Energy of the universe is constant so AEuniverse0 because there is no change I Think of the universe as system surroudings I AEsystem AEsurroundings AEuniverse 0 0 When energy ows OUT of a system AE system lt 0 AE surroundings gt0 0 When energy ows INTO a system AE system gt 0 AE surroundings lt0 I AE surroundings AE system or vice versa 0 Example a glass of water cools The water is the system Because it is cooling off it is losing heat energy into its surroundings Therefore AESystem lt 0 AEsurmundings gt 0 o lntemal energy state function sum of the potential and kinetic energy of the system The CHANGE of internal energy depends on the amount of internal energy at the beginning and end of the situationreaction O reaCtion Eproducts Ereactants 0 Note 1 mol of C and 1 mol of 02 has a greater amount of thermal energy than 1 mol of C02 because some of it is lost in the reaction process exothermic The reactants in molecular phase have a higher energy and combustion to form COz gives off some of this energy I The reverse forming C and 02 fro COz would require putting energy in this is like what plants do 0000 O 0 Energy changes 0 Energy can be transferred through heat q or work W q Gains heat Loses heat W Gains energy in work Releases E by doing work AE Gains internal E Loses internal E 0 Example A car traveling down a road performs 30 kJ of work and releases 100 kj of work 0 Q 100 kJ released 0 W 30 kJ work performed 0 AEsystem 39130 AEsurroundings Finding AE Ileat O 0 AB q w 0 Example A reaction produces 31 kJ of heat and a decrease in volume It requires the surroundings to do 76 J of work on the products What is AE o ststem 3l kJ o Wsystem 76 it is positive because work is done on it by surroundings o AE 3176 45 kJ Temperature increases proportionally to the amount of heat absorbed This proportionality constant is heat capacity or C 0 Heat qC AT Specif1c heat is the instrinsic ability to absorb heat The amount of energy needed to raise the temperature of lg of the substance 1 OC CS units JgOC Relationship between C heat capacity and CS speci c heat CS C X lm C is proportional to mass and speci c heat We can calculate quantity of heat absorbed if we know the mass speci c heat and raise in temperature Remember the MCAT equation for all you medicine people out there 0 q mCSAT 0 Example how much heat absorbed by copper penny that is 31 grams when the temperature raises from 8 C to 37 C Copper CS 0385 I Q 31 0385 45 537 J 0 Double check if the object losesreleases heat Q should be negative Solving problems when two objects of different temperatures come in contact with one another 0 Since we know that AEsmoundmgs AEsystem the heat lost by the hotter object will be the energy gained by the water or surroundings in the problem Since we are measuring heat set these problems up by setting q of the hot q of the cold Or I Mhot X Cs hot X mcoldX Cs cold X I You can nd any of these variables However if you solving for nal temperature it is important to place the unknown variable in the correct location In the equation AT for the hot object will be represented by temperature given x AT for the colder object will be x temperature given Then you algebraically solve for x Work 0 Pressure volume work is caused by volume change against external pressure When a gas expands AV the system does work on the surroundings W is negative It is using work to push against the air think of helium pushing against a balloon to expand it o If volume decreases the systems work is negative because the surroundings are pushing on the system 0 Work pAV p is usually given in atm and AV in L Multiplying these two will give you atm L but to convert to joules you will multiply the answer by 1013 0 Ex if a balloon is in ated from 0100 L to 185 L against external pressure of 100 atm how much work is done I W 100175 175 atm L 175 1013 177 J ofwork performed Enthalpy 0 When solving for AE it is dif cult to measure temperature changes of every chemical in the system so we can use the temperature change of the surroundings ie bomb calorimeter O qsurroundings qcalorimeter 39qsystem o Calorimeters have a constant volume and because AV is 0 we can manipulate AEsystemq 0 q is negative because q is the change in temp of the surroundings 0 Heat capacity of the calorimeter is the amount of heat absorbed by the calorimeter for each 10 change Measured in Coal or kJOC 0 Example 2 mol of H20 are formed from 2 mol H2 and 1 mol O2 in the calorimeter the temperature raises 5 degrees Cc 20kJOC Find AE for burning one mole of H2 I AEsystem 39qcalorimeter I To nd q Cc AT 20 kJOC 5 C 10kJ I However this is for 2 mol of H20 so to nd AE per mole divide by the number of moles of H2 10kJ2 mol H2 5kJmol o Enthalpy is H the sum of internal energy of the system and the product of pressure and volume H E pV A change in enthalpy indicates heat lost or gained When AH is negative heat is lost EXOTHERMIC products have less energy When AH is positive heat is gained ENDOTHERMIC products have more B The more reactants the larger the enthalpy change Ex C3H8 502 9 3CO2 4H2O Aern 2044 kJ How much heat evolved from complete combustion of 132 kg of C3H8 We know that with one mol of 331 1892044 kJ is released Find the number of moles of C3H8 after converting kg to g then multiplying 2044 by the number of moles If it was asking for oxygen nd moles of oxygen then multiply by 2044 and then divide by 5 mol because of the reaction ratio about 30 of the exam OOOOO Hess s Law When a reaction is multiplied by a factor AHIxn is increased by that factor also If the reaction is reversed the sign of AH is inverted If the reaction can be expressed as steps then Aern for the overall reaction is the some of the reaction heats for each step 0 Example I Given Cus C12 g 9 CuCl2s Aern 206 kJ 2Cu s C12 g 9 2CuCl Aern 36 kJ Calculate Aern for Cu CuCl2 9 2CuCl I First step is to nd one reaction that we can reverse to ip the sign of AH Technically both reactions are reversible but we want to nd the reaction where only one constituent appears on one side only In this case CuCl2 ONLY appears on the right hand side amongst the given equations so we are going to ip that equation We cant ip the other one because we want the 2CuCl on the right side since that is our product in the equation we are calculating I Once the reaction is ipped we ip the sign of the AH CuCl2 9 Cu s C12 g AH 206 M 0 2Cu s C12 g 9 2CuCl AH 36 M I Then we essentially add all the reactants and all of the products AND the enthalpies CuCl2 2Cu C12 9 2CuCl Cu C12 206 36 I Solve like you re solving for algebra CuCl2 Cu 9 2CuCl AH 170 endothermic Standard Enthalpy of Formation The standard state is the state of the material at standard conditions For formation these are pure gas at latm 25 degrees Celsius substances are concentrated at 1M AH enthalpy change at standard state AHf enthalpy change to form one mol of a pure compound at standard state units are in kJmol AHfO is 0 if the molecule is in its standard state When calculating these equations the products always have a coef cient of one but in these equations the reactants can have fractions to equal the 1 mole of product Example write the formation reaction of C O 9 CO 0 We imply that the product has a coef cient of l we need one mol of standard state C and one mol of standard state oxygen 0 Csngralgh e 02 9 CO This is not balanced So we put a one half in front of the 02 0 Answer CSDgraphiteJr 12 O2 9 CO g Any reaction can be written as a sum of formation reactions so we can nd AHO from the sum of the AH of the products AHfO of the reactants 0 Formula AH 2n AHf products 2n AHf reactants I N is the coef cient or number of moles 0 Example CaCO3 9 Ca0 02 at standard conditions given I AHfO CaC03 12076 I AHfO CaO 6349 I Though we are not given it we know AHfO of 02 is 0 because it is in its standard form and the reaction is at standard conditions latm 25 C I lmol6349 1 mol 0 1 mol l2076 I 5277 kJ Gasses Properties to understand 0 Gases are mostly empty space particles travel in straight lines until they encounter something 0 When the particles bounce off of something that is considered pressure Boyle s Law 0 Used a U shaped tube to deduct that in a closed system when mass is constant pressure and volume are inversely proportional 39 P1V1 P2V2 Charles s Law 0 Volume is directly proportional to temperature under constant pressure and amount of gass Graph of V amp T is a straight line TEMPERATURE IN KELVIN 39 T1V1 TzV 2 Avogadro s Law 0 Volume is directly proportional to the number of gas molecules P and T are constant 39 V1T1 V2T2 The Ideal Gas Law This law can hel ou determine behavior of gasses at standard conditions STP for gasses is lso know that Volume of one mol of gas at STP 224 L I PV nRT nmoles R ideal gas constant 008206 0 We can use this equation rearranged to nd any of the variables and if you have both the number of moles and the volume you can convert to mass in grams using mm calculation and nd the density IN gL of the gas 0 Ex calculate density of N2 at STP I We know volume is 224 L and we know the mass because 1 mol of N2 is the molar mass ofN2 280g I 28224 125 gL Partial Pressure 0 Logic to understand Two moles of the same gas or one mole of one one mole of another behave the same way and have a pressure of 2atm at standard conditions I When gasses mix their molecules behave independently and a container of mixed gasses occupy the same volume Since each gas has the same temperature this is why we can think of mixtures as one gas Ex Air 0 The pressure of a single gas in a mixture is called partial pressure O Ptotal 11total I Ntotal is the same as Ngls 1Ngas 2 NGas 3 0 To nd a partial pressure example I Find partial pressure of neon in a mixture of neon and xenon V 87 L Ptotal 39 atm T 598 K 017 mol Xe I Ptotal P Ne PXe rearrange to solve for PNe I PNe Ptotal PXe We need to nd PXe using ideal gas law 0958 atm I PNe 39 0958 29 atm Kinetic Molecular Theory I Gas particles are constantly moving and there is a lot of empty space I When the temperature gets higher the speed of the particles gets faster I AVERAGE Kinetic energy of the gas particles is directly proportional to the temperature in Kelvin This is conceptual 0 Avg KE dependent on average mass and velocity 0 KB 12mv2 0 However in air KB of Oxygen KE nitrogen KE Argon To have the same Kinetic energy they must vary in velocity and have different mass ie different m and v values to reach the same KE value I Think about it if oxygens molar mass is smaller than argon s oxygen is going to travel faster Small mass faster v Higher mass slower v Heavier molecules must have a slower average speed Density and Pressure I More gas molecules higher density higher pressure Effusion I Diffusion is when molecules move from high 9 concentration I Effusion is when a collection of molecules escapes through a small hole into a vacuum We can use this to find molar mass of a gas because different gasses escape at different rates RateA Molar M 1333 Rate is in molesec of LSec RateB Molar M assA I Example A mole of neon effuses from a container at 76 sec The same amount of an unknown gas requires 155 seconds Identify the gas 0 First we have to change the given to rates Neon 17 6 X1155 0 Rate A 17 6 Rate B 1155 but in the equation these look like 15576 mm B mm A 155 76 meB 2018 204 mmB2018 I 416 mmB 2018 I 839 mmB molar mass of Krypton Deviations from the Ideal Gas Law An important assumption of the IDL is gas particles are not attracted to each other and the particles have 0 volume Factors that cause deviation from the ideal gas law 0 In real gasses molecules have volume and the molecules have weak attraction to each other PVnRT STP is 273 K and latm o If you solve for volume you nd that at STP 1 mol of gas has a volume of 224 L However in an example with Argon at high pressure the volume is higher than an ideal gas Why Argon molecules have the same shape but at high pressure there are a lot of these molecules and the volume becomes important 0 At low pressure small of atoms occupying small volume 0 At high pressure many molecules occupying the same volume because there are so many the sum of the volume of all the atoms becomes important and is slightly larger total volume 0 More important with molecules that are much bigger for instance Xenon versus Helium Same concept occurs with very low temperatures pressure is lower in real gasses than ideal gasses The proportion of temperature to pressure is different 0 Why Real molecules are slightly attracted to each other 0 When molecules move slower the collisions don t allow molecules to bounce off of each other as vigorously they kind of seem to stick to each other Connection Keep going lower in temperature and the molecules do stick together to form a liquid PVRT in an ideal gas is l but other gasses have deviations Real gasses behave close to ideal at low pressures but as you get higher they have strong deviations IDL is correct at higher temp and lower pressure so this is why we use it and it has predictive power SUMMARY AT MUCH HIGHER PRESSURE LARGER REAL VOLUME AT MUCH LOWER TEMPERATURE LOWER PRESSURE GOOD LUCK
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