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# INTROGENERALCHEM CHEM100

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This 19 page Class Notes was uploaded by Miss Tyrell Rippin on Tuesday October 20, 2015. The Class Notes belongs to CHEM100 at San Diego State University taught by McNamaraSchroeder in Fall. Since its upload, it has received 34 views. For similar materials see /class/225326/chem100-san-diego-state-university in Chemistry at San Diego State University.

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Date Created: 10/20/15

11 13 15 Chapter 1 The Language of Chemistry A physical property is one that a substance can undergo without changing its identity A chemical property is one that describes how a substance can react with another substance to form a new substance 21 Oxygen loses its identity therefore this is a chemical property b When oxygen boils it transforms from a liquid into a gas but does not change identity therefore this is a physical property c When oxygen dissolves in water it mixes with water on a molecular level but does not change its identity therefore this is a physical property 1 When oxygen reacts with hydrogen to form water oxygen loses its identity therefore this is a chemical property a Air is a mixture of oxygen and nitrogen 13 Mercury is a pure substance found on the periodic table c Aluminum foil is made only of aluminum therefore it is a pure substance d Table salt is NaCl a pure compound that cannot be separated Length is measured in meters mass in kilograms temperature in Kelvins time in seconds amount of substance in moles a630m xi m i 0630km lOOOm f 15 b 1440 ms XL 1440 s lOOOms 1000 1000m c 0000065 kg x i x 65 mg 1kg 1s d 1300 pg lg OW 1300 mg IOOOtzg k 111 113 11 Chapter 1 a 00945 I 945 x 1072 therefore there are three signi cant gures b 8322 8322 x 10 therefore there are four signi cant gures 0 106 z 106 x 102 therefore there are three signi cant gures 1 0000130 130 x 10 therefore there are three signi cant gures 3 259000 g because last three zeros are shown all digits are signi cant therefore there are six signi cant gures b 102 cm 2 102 gtlt 102 therefore there are three signi cant gures c 0002 m 2 x 1039 therefore there is one signi cant gure d 2001 kg 2001 x 103 therefore there are four signi cant gures 6 00605 s 605 x 10quot2 therefore there are three signi cant gures a 000839 move the decimal place to the right three places to get 839 therefore 2 839 x 104 b 83264 move the decimal place to the left four places to get 83264 therefore a 83264 x 104 c 372 move the decimal place to the left two places to get 372 therefore 2 372 x 102 1 00000208 move the decimal place to the right ve places to get 208 therefore a 208 x 104 a 936800 move the decimal place to the left ve places to get 9368 therefore 2 9368 x 105 b 1638 move the decimal place to the left three places to get 1638 therefore 2 1638 x 103 c 00000568 move the decimal place to the right five places to get 568 therefore E 568 x 10quot 1 000917 move the decimal place to the right tluee places to get 917 therefore 2 917 X to 3 a 1n multiplication the number of signi cant gures in the answer is equal to the number of signi cant gures in the least exact number In this example 321 has three signi cant gures therefore the answer must have three signi cant gures 2 484 cmz 121 125 12 129 The Language of Chemistry 3 b 11 has two signi cant figure therefore the answer must have two signi cant gures E 84 In the addition to two numbers the result should have only as many digits to the right of the decimal point as the starting number with the least number of digits to the tight ofthe decimal point a 123 has only two digits after the decimal point therefore 123 12786 1402 b 3961 has only three digits after the decimal point therefore 3961 246543 2 28615 Each answer will have three signi cant gures because 0800 miles has three signi cant figures which is the minimum a 0800 miles 19933513 2129km 1mile 39 i b 0800 miles XiW1X 1WJ1Z9 x 103 m 39 1mile j 1km 1mile 1m 1 16093ka 1000111 1000 mm X JX 1m d 0800 miles x c 0800 miles x 1 6093mlx11000m x1100le 129 x 105 m 1km 1m 1 I 1 129x106m mie 375 gal X1M3397853ij 19 31 1L Jl42x104mL lgal 10km xii lmlle r w 1 6214 miles H6903km From Table 15 1 oz 2835 g a 00011 x 16535935 Wm gel 027 kg 11b 1000 g W453 59gl 27 X102 g b 0601b xi is 060 lb xtiee 227 x 105 mg 135 137 143 Chapter 1 10L 10 gtlt103 mL 1000 cm3 quot1x10310x10x10cm Therefore each side ofthe cube whose volume is 10 L is 10 cm cm 1700 mm x 11 J 1700 cm mm Volume L x W x H 1700 cm x 650 cm x 325 cm 359 cm3 g 3 CITI Mass Density gtlt Volume 22 x 359 em3 79 g C OF 4 3201 gtlt 5 C 9 F 5 C 9 F C 68 F 32 F gtlt J 20 C K C27327 C273300K 39 2 CpJoules 201 JZOD69 J J g x C 32g gtlt28 C 18 C g x C mass of solution J densit p y volumeof solution Volume of solution 1984 mL Mass of solution mass of ask and solution mass of ask 22419 g 54381 g 31962 g 22419g 1130 mL 1984ij g p density Use the unit conversion method 84 gtlt103 k X1 g 501 g 1674 k1 145 147 155 157 159 The Language of Chemistry quot 11b 1 1 O f WWW z11011 5C ltL4536g t A mass j41242 28463gl 1278811 04 mL DenSIy ivolume L 123 mL l 12393 mL 1 g I p 2 density 1932 gr cm3 1 2416 kg 2416 g x I a Water changes into hydrogen gas and loses its identity therefore this is a chemical property 13 Lead does not change its identity therefore this is a physical property 0 These properties of lead do not change its identity therefore this is a physical property d Lead changes its identity and properties therefore this is a chemical prOperty 3 9620000 kg 962 x 106 kg b 54870 days 5487 x 104 days 18 i 0253 s2253 x1045 c 253 milliseconds x M ms d 0000274 kilometers x Llooom 274 X 1041 m 1 km 458 mg X 162 X 10quot3 02 28330 mg a 82 x 102 375 X104 820 37500 383 gtlt104 b 521 x 102 274 x 10quot 00521 000274 548 x 10 2 c 101 x 104 723 x 10 0000101 000323 733 x 10quot3 10Lgtlt 1000 mL 10 x103 cm3 Heat lost by man 2 heat gained by water 161 163 171 13 Chapter 1 x i f V L 3 10 Mass ofwater 1000 gallons x 3 7853EJX 1000m Jx 5 i 3783300 g 1 gal 1L 1m 1 Cp x mass gtlt AT 4184 i iC x 3785300 g X 249OC 394 x 107 J in 15 h g ii x i X 24 hours 63 x to8 J per day V 15 h The unspeci ed solution contains two components NaCl and water You could obtain pure NaCl by evaporating all the water from the mixture Five grams of pure NaCl could then easily be weighed out in a laboratory and the speci ed mixture of salt and water could be easily prepared You must modify your original hypothesis and retest it The uk location stands for United Kingdom England a country that uses the SI metric system In the United States the Fahrenheit system is still in popular use so converting 175 Celsius degrees to Fahrenheit degrees should solve the problem 175 C x 95 F0C 32 34701 a 360 m x 100 cmm x 1 inf254 cm x 1 ft12 in x l yd3 ft 394 yd b 260 in x 254 cmfin 660 cm c 299 x 108 ms gtlt kin1000111 x 1 mile161 km x 60 sx min x 60 minhr x 24 hrday x 365 dayyr 590 x 109 milesyr d This is a uid measure so we use the density of water to convert mL to ounces of water Assume waters density to 100 gmL 640 mL x 100 gmL gtlt 12835 ozg 226 oz The price per mL of the active ingredient is Recommended product 650 185 mL cream x 1 mL of cream0005 mL active ingredient 703 mL of active ingredient Generic product 00165 mL of cream X 1 mL of cream0007 mL active ingredient 606mL of active ingredient The generic product is more cost effective 1622000 810 Volume ofeach side ofthe box 3 100 cm x 100 cm x 0205 em 250 x 10339 cm3 The Language of Chemistry Mass of each side ofthe box I 250 x 103 cm3 X 250 gzcm3 525 X 103 g Mass ot box 2 600 X 625 x1031quot 375 X104 g Density of box 2 mass of boxx volume of box 375 x 101 9100 x 10 cm3 00375 gmm3 Since the volume of the box is 100 x 106 cm the total amount of water to be added is 375 gtlt104 mLX1iOOx106mL X963x105mL 175 You may have rounded off your gures before you nished the calculation 7 The Language of Chemistry 3 Hints for Faster Coverage Mass volume and temperature can be given a cursory overview as most students will have an intuitive understanding of these concepts F or students with a scienti c background the difference between mixtures and pure compounds can be eliminated Additionally for this audience the introduction to the metric system can be scaled back Exercise Overview EYK Expand Your Knowledge exercise The Composition ofMatter 11 14 149 163 EYK 164 EYK Measurements Metric System Uncertainty and Signi cant Figures 15 112 150 152 157 165 EYK 167 EYK 168 EYK 170 EYK 175 EYK Scienti c Notation and Calcu1ations Using Scienti c Notation 113 120 151 Units and Signi cant Figures in Ca1cu1ations 121426 153 456 160 EYK 162 EYK 169 EYK Mass Volume Density and Temperature 127 138 141 142 145 148 161 EYK 166 EYK 173 Heat and Ca10rirnetry 139 140 143 144 158 159 EYK 171 EYK 172 EYK Boxes 174 EYK Solutions to EvenNumbered Exercises 12 A physical property is one that a substance can undergo without changing its identity A chemical property is one that describes how a substance can react with another substance to form a new substance a Magnesium must react with hydrochloric acid to form Mng to produce hydrogen gas Thus magnesium loses its identity therefore this is a chemical property b When magnesium melts it transforms from a solid into a liquid but does not change identity therefore this is a physical property c Magnesium being shiny and able to conduct electricity does not change its identity therefore these are physical properties Chapter I 1 Magnesium changes from a shiny metal to a white solid unable to conduct electricity because it reacts with oxygen to form MgO Thus it changes its identity and this is a chemical property a Seawater is a mixture of salt water and other substances that can be separated by separation techniques b Table sugar is a pure substance that cannot be separated without losing its identity c Smoke is a mixture of small particles dispersed in air that can be separated by ltration d Blood is a mixture ofproteins water ions and other chemicals that can be separated by various techniques 3 3 w Area m veloc1ty ms volume r m densin legm 30813km 1000 81 m 1km I 1000ms b 036 S X 1 J367 ms S 03140111 x rrrrrrrrr 7140 g lOOOmg g 100 d 0095 mg x0 g z 96 Mg lmg a 160 em 160 x 101 cm last zero is included therefore three signi cant gures b 00063 m 63 x 10 3 m therefore two significant gures c 100 km 2 l x 102 km therefore one signi cant gure d 29374 g 29374 x 100 g therefore ve significant gures e 107 lbin2 107 lb inz therefore three signi cant gures a 2 2 m 3 212 gtlt 101 In therefore three signi cant gures b 0023 kg 23 x 10393 kg therefore two signi cant gures c 4694 em 4694 x 10 cm therefore four signi cant gures a 45359 g 45359 x 102 g therefore ve signi cant gures li39t 116 ll8 The Language of Chemistry e 16030 km 2 16030 x 100 km therefore ve signi cant gures Move the decimal point over the number of places given by the exponent lling in spaces with zeros lithe exponent is negative move the decimal point to the left if the exponent is positive ineve the decimai place to the right a 23 x 103 00023 019 x i02 290 e 392 x 10 0000392 0173 gtlt104 17300 a 276000 kilogram 275 x 105 kg 13 87000 years 87 x 104 years c 0097 second 97 x 1033 s a in division the number ofsigni cant figures in the answor is equal to the number of significant gures in the least exact starting number In this example 123 has tinee signi cant gures therefore the answer must have three signi cant figures 2 l 13 b 0095 has two signi cant gures therefore the answer must have only two signi cant gures 5 0067 In the subtraction of two numbers the result should have only as many digits to the right of the decimal point as the starting number with the least number of digits to the right of the decimal point a 391 has only two digits after the decimal point therefore 391 763 115 b 892 has only one digit after the decimal point therefore 892 54832 1 344 375 miles x 68 h 55 miles 2 35 10 295 oz gtlt jw g x mgqgg g 836 x104mg 102 lg 3600 L093 x S x me x 11quot 246 milesh 910 3 1h 16093 km 1000 m 233 xii 5gxl90 3 2835 x 104 mg a 130 132 140 Chapter 1 9k 21 0820 miles gtlt1 g7 mx 132 km 1 mlle k 1 1000 1 13108201111165 l gtlt IE 132 x 10quot m 1111116 1km 100 1000 100 308201 1111123 x A m 1x m x 111 132 x 105 111 1111116 114ml 1m 1 3111111101113 10 L 1000 c1113 gtlt iii 1x Jill v x 131 00010 m3 100 cm 100 cm 100 cm A 1 Volume 2 wit DEE 3165 cm 8115113 0 7899quin cm 0 QC 450F 32 F 720C 9 F C FT OC 9 F 32 F 5 C F 125 C 39139 32 F 257 F 5 C Heat C39p gtlt mass x AT 0750 kg x 9995 750 g 1 kg ic x 350 g x 680 C 130 C 908 x 103 J g Heat 0220 p 2 density mass of urine volume of urine Mass of urine 25853 g Volume of urine volume of water 1 Volume of water mass of water x 23718 01113 densny of water 144 146 148 154 156 T110 Language ot Chcmistry p density J x 10900 glam 2 10900 gme Heat 1 Q X mass x Alquot 084 113 x 100 g x 20 C 17 gtlt102 1 z 2284 s Densityx 3 33 m 2 8 10200gm1 v0111mc 22394 mL 3 115311 138 smquot138mL 109g 150gx quot1000 a 00045 L x 5 451m 1L q b 287x 10quot8 l 993 287 ns a 00057 km 50 1 57 m 39 1 f U 100000036 kg XL19 X100103 J 36 mg 9 3 000256 256 x 10 3 three signi cant gures b 128009 128009 x 102 six signi cant gures 1 300730 200730 x 100 six signi cant gures 1 201 201 x 102 three signi cant gures 0 009864 9864 x 104 four signi cant gures 21 32 x 10331 x 10 99 x 10 2 two signi cant gures b 947 x 10quot0 232 x 10 z 408 x 10 three signi cant gures s 56 X 1039 3 23 x 10392 29 x 1040010 signi cant gures 1 36 X 10 3 w 14 x 1051 r 35 x 10 3 two signi cant gures 21 27 0111 108 cm x 2247 cm 85 x 10I cm two signi cant gures from 2 l64 168 Chapter l b 254 cm 0541 cm 22 cm I 091 cm two signi cant gures from 223 cl 2163 g 4284 g x 00372 g 0964 g three significant gures from 00372 111837 mL 3857 n391L 213 niL 68l mL three significant tiguresli o111 213quot Heat Cquot x mass gtlt AT 7 1 J A g1 116 M O a 20252 T gOC massx AT 127gx 250 C 02 mg leapsule gtlt 75 kg 15 mg x e 3 capsules per dose Both students measurements contained two signi cant figures and both correctly expressed their results in exponential notation rounded to two significant figures Student B got his result by converting inches to centimeters and then calculating the area 85 in x 254 einfin 2159 cm rounded to 216 cm ll in x 254 cmfin 2794 cm rounded to 279 cm 216 cm x 279 cm 6026 cmz rounded to 603 cm3 He should have rounded the results ofhis length and width conversions to two signi cant gures before he calculated the area The conversion factor does not control the number of signi cant figures the measurements do A scienti c law is a concise statement of a natural phenomenon to which no exception is known A theory is a hypothesis that is proposed to account for or explain a natural phenomenon We know that 0C FF 32 x SEQ Substitute 40 for both C and 0F and calculate the result 40 72 x 5940 quotJ A graphical solution is equally valid There are 3785 Li39gal so 1 gallon ofyour friend s gasoline would cost 3785 Ltgar x sownL 250 As the gasoline costs 300 per gallon at your favorite stations keep buying it their 170 173 174 The Language 0f Chemigtry 39 1 39mm a 53 kg gtlt lfi xL m g 7 77 lb v 42339 1kg 1quot 53 cm x m l En 20 9 1n 254 cm 162calx Jimmy X 1113 1141 day I year 4200ca1 The density depends on only twe variables mass and volume Because the mass of the sperm whale cannot change it must be able to change its volume 17 Chapter 2 Hints for Faster Coverage It is possible for students to understand electronic con guration and the importance of electron in the property ofatoms without discussing the concepts of quantum numbers This may be advisable on a less rigorous coursc in freshman chemistry Exercise Overview EYK Expand Your Knowledge exercise Daiton s Atomic Theory 21 28 265 EYK The Structure of Atoms 29 214 247 255 258 262 EYK 264 EYK Isotopes 215 2l8 263 EYK The Periodic Table 219 228 Electron Organization Within the Atom 229 242 248 251 259 261 266 EYK 26 27 l EYK 238 276 EYK Atomic Structure Periodicity Chemical Reactivity 243 246 252 254 272 EYK Solutions to EvenNumbered Exercises 22 24 26 28 The law of de nite composition states that the mass proportions of elements in a chemical compound are xed and invariant 37685 gc 5214 x10quot x100 5214 5150g 0 W1312 x10 2 x100l312 5150g W 3474 x 10quot x 100 3434 515034 Atomic mass S 2 x 160000 amu 320000 amu Atomic mass Be 075 x 120000 amu 90 amu Atomic number 16 element is sulfur t t 0 218 220 Atomic Structure Protons Neutrons Electrons Mass amu Element 19 20 19 39 K 34 45 34 79 Se 20 20 20 40 Ca 11 12 11 23 Na It would be an anion with a charge 1 Total percent 100 Isotope at 26 mm X Isotope at 25 amu 100 X 100 3 100 IOU X 26 100 25 256 26X 25X 2500 2650 X60100 X40 Isotope at 26 amu 60 Isotope at 25 amu 40 is is 32m in ion 32m 1 Helium b Magnesium c Potassium d Nitrogen e F luorine i3 14 228 230 232 240 Chapter 2 Element Group Period Si IV 3 Ge IV 4 As V 4 Sb V 5 Si Ge As and Sb are elements that have properties of both metals and nonmetals and are called metalloids Group I Group 11 True Atomic spectra consist of emission of light at frequencies that re ect energy differences between ground and excited atomic states Electron Shell Subshells n z 1 s n 2 5 p n 2 3 519 d There are two electrons per orbital There are three orbitals in a p subshell and each orbital can hold two electrons Therefore a p subshell can hold six electrons The p snbshell consists of three bi lobed teardropshaped orbitals perpendicular to each other The noble gas notation is used to shorten the electronic con guration of higher atomic number elements a He 3 192 2 electrons He2522p1 2 electrons 3 electrons 5 electrons The element with five protons is boron 242 244 246 248 250 260 262 Atomic Structure l5 1 He 2 152 quot 2 electrons Hel2522p5 2 electrons 7 electrons I 9 electrons The element with nine protons is fluorine c Ar 1532922p63533p6 18 electrons Arl452 18 electrons 2 electrons 20 electrons The element with 20 protons is calcium d Kr 1522522p63323p64323dm4p6 2 36 electrons Krl s2 36 electrons 2 electrons 38 electrons The element with 38 protons is strontium The negative charge indicates that there is one more electron than proton 1332532106 10 electrons Since there must be nine protons the anion is F1quot Group VII All elements possess seven outer electrons nsznps Group VIII All elements possess eight outer electrons nsznp except helium which is 152 Atomic emission spectra For the electron shell n 3 there are 5 p and d subshells The s subshell can have a maximum of two electrons thep subshell can have a maximum of six electrons the d subshell can have a maximum often electrons Therefore the maximum number of electrons in the n 3 shell is 18 One can also employ the Zn2 rule for the maximum number of electrons in a shell 232 2 x 9 18 Sulfur is a nonmetal They are inert and do not readily react to form compounds because they have a full outer electron shell In chemical reactions atoms tend to attain eight electrons in its outer shell con guration which is the outer shell con guration of the noble gases except helium The chemical properties are identical but the nucleus of Carbon12 contains six neutrons and that of Carbonl3 contains seven neutrons No The emission spectra of an atom will contain only those wavelengths of light corresponding to transitions between allowed energy levels In example 21 the mass percents were calculated as follows carbon C 4095 hydrogen Ii 452 oxygen 0 2 5452 Therefore 8200 g Vitamin C x 04095 3358 g C in Vitamin C 8200 g Vitamin C x 00452 3 0371 g H in Vitamin C 8200 g Vitamin C x 05452 4471 g 0 in Vitamin C 274 Chapter 2 The statement is incorrect there are no more than eight elements that have equal numbers of protons electrons and neutrons l Ie C N 0 Ne Si 8 and Ca Yes Observing the spectral lines when body uid is incinerated in a ame would be one method ofdetecting various elements a Two unpaired electrons h The negative sign means that the atom has an extra electron more than the neutral atom thus this atom has three unpaired electrons c The positive sign means that the atom has one fewer electron than the neutral atom thus this atom has one unpaired electron 1 Four unpaired electrons a This is an excited state of Boron where the ground state is 152 25 b This is the ground state of neon is2 252 2p c This is an excited state of uorine where the ground state is Arllls2 3615 Elements that appear in the same group in the periodic table will exhibit similar chemical properties a From Table 28 nitrogen has three electrons in the 2p subshell b From Table 28 neon has its outer shell completely full c From Table 28 sodium s outer subshell is the 3s and this atom has one atom in this s subshell which makes it half full 1 From Table 28 beryllium s outer subshell is the 25 which is full a From Table 28 F has one unpaired electron The positive sign means that this atom has one less electron than the neutral atom and thus has two unpaired electrons b Sn has the electron configuration Krj4d10532p2 The 2 charge means that this atom has two less electrons than the neutral atom and thus has zero unpaired electrons 0 Bi has the electron configuration Xel4f45d10653p3 The 3 charge means that this atom has three less electrons than the neutral atom and thus has zero unpaired electrons zero unpaired electrons d Ar has the electron con guration Nel352p65 The positive charge means that this atom has one less electron than the neutral atom and thus has one unpaired electron

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