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RIP W. E. B. DuBois (1963) Read for Tuesday Read for Tuesday Chapter 3: Sections 1-2 Chapter 3:...

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RIP W. E. B. DuBois (1963) Read for Tuesday Chapter 3: Sections 1-2 HOMEWORK – DUE Tuesday 9/1/15 HW-BW 1 (Homework Bookwork) #’s 4-9 all, 21, 26, 30, 36, 39, 44, 47, 49, 53-69 odd, 78, 80, 81 HW-WS 1 (Homework Worksheet) (from course website) HOMEWORK – DUE Thursday 9/3/15 HW-BW 2 (Homework Bookwork) #’s 1-7 all, 14, 19, 21, 23, 36, 47-50 all, 57, 59-66 all, 81, 82, 113, 114, 116, 127, 132 HW-WS 2 (Homework Worksheet) (from course website) Lab Monday/Tuesday
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Page 1: RIP W. E. B. DuBois (1963) Read for Tuesday Read for Tuesday Chapter 3: Sections 1-2 Chapter 3: Sections 1-2 HOMEWORK – DUE Tuesday 9/1/15 HOMEWORK – DUE.

RIP W. E. B. DuBois (1963) Read for Tuesday

Chapter 3: Sections 1-2 HOMEWORK – DUE Tuesday 9/1/15

HW-BW 1 (Homework Bookwork) #’s 4-9 all, 21, 26, 30, 36, 39, 44, 47, 49, 53-69 odd, 78, 80, 81

HW-WS 1 (Homework Worksheet) (from course website) HOMEWORK – DUE Thursday 9/3/15

HW-BW 2 (Homework Bookwork) #’s 1-7 all, 14, 19, 21, 23, 36, 47-50 all, 57, 59-66 all, 81, 82, 113, 114, 116, 127, 132

HW-WS 2 (Homework Worksheet) (from course website) Lab Monday/Tuesday

EXP #1- PRELAB!!!!

Page 2: RIP W. E. B. DuBois (1963) Read for Tuesday Read for Tuesday Chapter 3: Sections 1-2 Chapter 3: Sections 1-2 HOMEWORK – DUE Tuesday 9/1/15 HOMEWORK – DUE.

Classification of Matter

Pure Substance: Matter that is constant in its chemical composition and properties.

Mixture: A blend of two or more pure substances in any ratio each retaining their identity.

Physical changes can separate mixtures into one or more pure substances.

Page 3: RIP W. E. B. DuBois (1963) Read for Tuesday Read for Tuesday Chapter 3: Sections 1-2 Chapter 3: Sections 1-2 HOMEWORK – DUE Tuesday 9/1/15 HOMEWORK – DUE.

Matter

Pure Substance

Mixture

Are properties andcomposition constant?

YESNO

Physical change

Page 4: RIP W. E. B. DuBois (1963) Read for Tuesday Read for Tuesday Chapter 3: Sections 1-2 Chapter 3: Sections 1-2 HOMEWORK – DUE Tuesday 9/1/15 HOMEWORK – DUE.

Matter

Pure Substance

Mixture

Chemical Compound

Element

Are properties andcomposition constant?

YESNO

Physical change

Is separation by chemical reaction into simpler substances possible?

NO YES

Chemical change

Page 5: RIP W. E. B. DuBois (1963) Read for Tuesday Read for Tuesday Chapter 3: Sections 1-2 Chapter 3: Sections 1-2 HOMEWORK – DUE Tuesday 9/1/15 HOMEWORK – DUE.

ElementElements are substances that cannot be broken

down chemically into simpler substances.

All of the atoms that make up a particular element are chemically the same as all of the others

All of the atoms that make up a particular element contain the same number of protons as all other atoms of that element.

Page 6: RIP W. E. B. DuBois (1963) Read for Tuesday Read for Tuesday Chapter 3: Sections 1-2 Chapter 3: Sections 1-2 HOMEWORK – DUE Tuesday 9/1/15 HOMEWORK – DUE.

Pure SubstancesChemical Compounds: Two or more elements combined chemically in specific ratios to form a pure substance.

Water = H2O

Methanol = CH3OH

Nitroglycerine = C3H5(NO3)3

Page 7: RIP W. E. B. DuBois (1963) Read for Tuesday Read for Tuesday Chapter 3: Sections 1-2 Chapter 3: Sections 1-2 HOMEWORK – DUE Tuesday 9/1/15 HOMEWORK – DUE.

Matter

Pure Substance

Mixture

Chemical Compound

ElementHeterogeneous

Mixture Homogeneous

Mixture

Are properties andcomposition constant?

YESNO

Physical change

Is separation by chemical reaction into simpler substances possible?

NO YES

Are properties consistant from sample to sample?

NO YES

Chemical change

Page 8: RIP W. E. B. DuBois (1963) Read for Tuesday Read for Tuesday Chapter 3: Sections 1-2 Chapter 3: Sections 1-2 HOMEWORK – DUE Tuesday 9/1/15 HOMEWORK – DUE.

MixturesHeterogeneous Mixture: A mixture of matter in whichthe properties change from sample to sample.Hetero

change

Sand and sugar

Quartz and gold

Page 9: RIP W. E. B. DuBois (1963) Read for Tuesday Read for Tuesday Chapter 3: Sections 1-2 Chapter 3: Sections 1-2 HOMEWORK – DUE Tuesday 9/1/15 HOMEWORK – DUE.

Matter

Pure Substance

Mixture

Chemical Compound

ElementHeterogeneous

Mixture Homogeneous

Mixture

Are properties andcomposition constant?

YESNO

Physical change

Is separation by chemical reaction into simpler substances possible?

NO YES

Are properties consistant from sample to sample?

NO YES

Chemical change

Page 10: RIP W. E. B. DuBois (1963) Read for Tuesday Read for Tuesday Chapter 3: Sections 1-2 Chapter 3: Sections 1-2 HOMEWORK – DUE Tuesday 9/1/15 HOMEWORK – DUE.

Mixtures

Homogeneous Mixture: A mixture of matter in whichthe properties remain constant from sample to sample.Homo

constant

Salt water

Kool-AidBrass

Page 11: RIP W. E. B. DuBois (1963) Read for Tuesday Read for Tuesday Chapter 3: Sections 1-2 Chapter 3: Sections 1-2 HOMEWORK – DUE Tuesday 9/1/15 HOMEWORK – DUE.

Classification of MatterLabel each of the following as either: element, compound, homogenous mixture or heterogeneous mixture

Carbon dioxide

Tea brewed from tea bags

Water, distilled and deionized

Oxygen gas

An egg

Apple Juice

compound

homogenous mixture

compound

element

heterogeneous mixture

homogenous mixture

Page 12: RIP W. E. B. DuBois (1963) Read for Tuesday Read for Tuesday Chapter 3: Sections 1-2 Chapter 3: Sections 1-2 HOMEWORK – DUE Tuesday 9/1/15 HOMEWORK – DUE.

Before There Were Atoms…Three laws that lead to the atomic view of the atom:

1) Law of Conservation of Mass

Total mass must be same before and after a reaction

2) Law of Definite Proportions

No matter the source, a compound is always made of the same elements in the same mass ratio

3) Law of Multiple Proportions

If two elements combine to form more than one type of compound with each other, the masses of one element that combined with a fixed mass of the other element are in ratios of small whole numbers

Page 13: RIP W. E. B. DuBois (1963) Read for Tuesday Read for Tuesday Chapter 3: Sections 1-2 Chapter 3: Sections 1-2 HOMEWORK – DUE Tuesday 9/1/15 HOMEWORK – DUE.

Before There Were Atoms…Three laws that lead to the atomic view of the atom:

1) Law of Conservation of Mass

Total mass must be same before and after a reactionI am BBQing and start with 20 pounds of charcoal. When I am done, there is only 4 pounds of ash left. How can we explain the apparent failure of the Law of Mass Conservation?

48.0 g of carbon react with 128 g of oxygen, how much CO2 should be formed?

176 g CO2(g)

48.0 g of carbon react with 148 g of oxygen. After the reaction is complete, there is still only 176 g of CO2 formed. Has the Law of Conservation of Mass failed?

20.0 g O2(g) left over

Page 14: RIP W. E. B. DuBois (1963) Read for Tuesday Read for Tuesday Chapter 3: Sections 1-2 Chapter 3: Sections 1-2 HOMEWORK – DUE Tuesday 9/1/15 HOMEWORK – DUE.

48.0 g of carbon react with 128 g of oxygen, forming 176 g CO2, how much CO2 should be formed from 72.0 g of carbon?48.0 g of carbon react with 128 g of oxygen, forming 176 g CO2, how much CO2 should be formed from 72.0 g of carbon?

Before There Were Atoms…Three laws that lead to the atomic view of the atom:

2) Law of Definite Proportions

No matter the source, a compound is always made of the same elements in the same mass ratio

X = 264 g CO2(g)

2 2

48.0 g C

g CO17

7=

2

6 g

.0

C

C

O

g

X 227

482.0

.0 g

176

C=g g CO

CO C

g X

Page 15: RIP W. E. B. DuBois (1963) Read for Tuesday Read for Tuesday Chapter 3: Sections 1-2 Chapter 3: Sections 1-2 HOMEWORK – DUE Tuesday 9/1/15 HOMEWORK – DUE.

48.0 g of carbon react with 128 g of oxygen, forming 176 g CO2, how much O2 should be react with 72.0 g of carbon?

Before There Were Atoms…Three laws that lead to the atomic view of the atom:

2) Law of Definite Proportions

No matter the source, a compound is always made of the same elements in the same mass ratio

48.0 g of carbon react with 128 g of oxygen, forming 176 g CO2, how much O2 should be react with 72.0 g of carbon?

X = 192 g O2(g)

2 2

48.0 g C

g 12

7

8

2

g

.

O=

C

O

0 g

X 227

482.0

.0 g

128

C=g

g C O

O g X

Page 16: RIP W. E. B. DuBois (1963) Read for Tuesday Read for Tuesday Chapter 3: Sections 1-2 Chapter 3: Sections 1-2 HOMEWORK – DUE Tuesday 9/1/15 HOMEWORK – DUE.

Before There Were Atoms…Three laws that lead to the atomic view of the atom:

2) Law of Definite Proportions

No matter the source, a compound is always made of the same elements in the same mass ratio

A 219.6 kg sample of NaCl from The Great Salt Lake contains 86.4 kg of sodium, what mass of chloride would be present in a 76.8 kg sample of NaCl from the Dead Sea?

X = 46.8 kg Cl

219.6 kg Na

g 133.2 kg

76.8

Cl

kg=

NaCll

l

C

CX 133.2 kg C

219.6 k76.8 kg

g NaCl=

l NaC Cll g X

219.6 kg NaCl – 86.4 kg Na = 133.2 kg Cl

A 219.6 kg sample of NaCl from The Great Salt Lake contains 86.4 kg of sodium, what mass of chloride would be present in a 76.8 kg sample of NaCl from the Dead Sea?

219.6 kg NaCl – 86.4 kg Na = 133.2 kg Cl

Page 17: RIP W. E. B. DuBois (1963) Read for Tuesday Read for Tuesday Chapter 3: Sections 1-2 Chapter 3: Sections 1-2 HOMEWORK – DUE Tuesday 9/1/15 HOMEWORK – DUE.

Before There Were Atoms…Three laws that lead to the atomic view of the atom:

3) Law of Multiple Proportions

If two elements combine to form more than one type of compound with each other, the masses of one element that combined with a fixed mass of the other element are in ratios of small whole numbers

16.0 g of oxygen react with 2.0 g of hydrogen

64.0 g of oxygen react with 4.0 g of hydrogen

416.0 g of oxygen + 1.0 g of hydrogen

16.0 g of oxygen + 2.0 g of hydrogen

64.0 g O4.0 g H

16.0 g O2.0 g H

264.0 g O4.0 g H

16.0 g O2.0 g H

Page 18: RIP W. E. B. DuBois (1963) Read for Tuesday Read for Tuesday Chapter 3: Sections 1-2 Chapter 3: Sections 1-2 HOMEWORK – DUE Tuesday 9/1/15 HOMEWORK – DUE.

Before There Were Atoms…Three laws that lead to the atomic view of the atom:

3) Law of Multiple Proportions

If two elements combine to form more than one type of compound with each other, the masses of one element that combined with a fixed mass of the other element are in ratios of small whole numbers

89.4 g of oxygen react with 33.6 g of carbon

50.5 g of oxygen react with 37.9 g of carbon

89.4 g O33.6 g C

50.5 g O37.9 g C

289.4 g O33.6 g C

50.5 g O37.9 g C

Page 19: RIP W. E. B. DuBois (1963) Read for Tuesday Read for Tuesday Chapter 3: Sections 1-2 Chapter 3: Sections 1-2 HOMEWORK – DUE Tuesday 9/1/15 HOMEWORK – DUE.

The Atom

REALLY early atomic theory…Democritus ~350 BCAtomos - Greek meaning indivisible

Modern Definition: Smallest piece that matter can be broken up

into and still maintain the properties of an element

Page 20: RIP W. E. B. DuBois (1963) Read for Tuesday Read for Tuesday Chapter 3: Sections 1-2 Chapter 3: Sections 1-2 HOMEWORK – DUE Tuesday 9/1/15 HOMEWORK – DUE.

My Atom Broke

Subatomic ParticlesNucleus

Protons – p+

Carry a single positive chargeNumber of p+ = ATOMIC NUMBER1.673x10-24 g

Page 21: RIP W. E. B. DuBois (1963) Read for Tuesday Read for Tuesday Chapter 3: Sections 1-2 Chapter 3: Sections 1-2 HOMEWORK – DUE Tuesday 9/1/15 HOMEWORK – DUE.

1H

1.013

Li6.9411

Na22.99

19K

39.1037

Rb85.47

55Cs

132.9187Fr

(223)

4Be9.0112

Mg24.31

20Ca40.08

38Sr

87.6256Ba

137.3388

Ra(226)

21Sc

44.9639Y

88.9157La

138.9189Ac(227)

22Ti

47.8840Zr

91.2272Hf

178.49104Rf

(261)

23V

50.9441

Nb92.91

73Ta

180.95

24Cr52.00

42Mo95.94

74W

183.85

25Mn54.94

43Tc(99)75Re

186.21

26Fe

55.8544

Ru101.07

76Os190.2

27Co58.93

45Rh

102.9177Ir

192.22

28Ni

58.6946Pd

106.4278Pt

195.08

29Cu63.55

47Ag

107.8779

Au196.97

30Zn65.39

48Cd

112.4180

Hg200.59

5B

10.8113Al

26.9831

Ga69.72

49In

114.8281Tl

204.38

6C

12.0114Si

28.0932

Ge72.61

50Sn

118.7182Pb207.2

7N

14.0115P

30.9733As

74.9251Sb

121.7583Bi

208.98

8O

16.0016S

32.0734Se

78.9652Te

127.6084Po(209)

9F

19.0017Cl

35.4535Br

79.9053I

126.90

10Ne20.18

18Ar39.95

36Kr83.80

54Xe

131.29

2He4.00

58Ce

140.12

59Pr

140.91

60Nd

144.24

61Pm(147)

62Sm

150.36

63Eu

151.97

64Gd

157.25

65Tb

158.93

66Dy

162.50

67Ho

164.93

68Er

167.26

69Tm168.93

70Yb

173.04

71Lu

174.97

85At

(210)

86Rn(222)

90Th(232)

91Pa(231)

92U

(238)

93Np(237)

94Pu(244)

95Am(243)

96Cm(247)

97Bk(247)

98Cf

(251)

99Es

(252)

100Fm(257)

101Md(258)

102No(259)

103Lr(260)

105Db(262)

106Sg

(263)

107Bh(262)

108Hs(265)

109Mt(266)

110Ds(271)

111Rg(272)

112--

(277)

114--

(285)

116--

(289)

1IA

2IIA

3IIIB

4IVB

5VB

6VIB

7VIIB

8VIII

9VIII

10VIII

11IB

12IIB

13IIIA

14IVA

15VA

16VIA

17VIIA

18VIIIA

Page 22: RIP W. E. B. DuBois (1963) Read for Tuesday Read for Tuesday Chapter 3: Sections 1-2 Chapter 3: Sections 1-2 HOMEWORK – DUE Tuesday 9/1/15 HOMEWORK – DUE.

My Atom BrokeSubatomic Particles

NucleusNeutrons – no

No charge# can vary from atom to atom

1.675x10-24 g (roughly the same mass as p+)

Page 23: RIP W. E. B. DuBois (1963) Read for Tuesday Read for Tuesday Chapter 3: Sections 1-2 Chapter 3: Sections 1-2 HOMEWORK – DUE Tuesday 9/1/15 HOMEWORK – DUE.

My Atom Broke

Subatomic ParticlesOutside the nucleus

Electrons – e-

Carry a single negative charge~1830 e- = mass of p+ or no

9.11x10-28gResponsible for most of the chemistry that

ever happens

Page 24: RIP W. E. B. DuBois (1963) Read for Tuesday Read for Tuesday Chapter 3: Sections 1-2 Chapter 3: Sections 1-2 HOMEWORK – DUE Tuesday 9/1/15 HOMEWORK – DUE.

Atoms vs. IonsAtoms are NEUTRAL!!!!!

This means that they have zero charge#p+ = #e-

Page 25: RIP W. E. B. DuBois (1963) Read for Tuesday Read for Tuesday Chapter 3: Sections 1-2 Chapter 3: Sections 1-2 HOMEWORK – DUE Tuesday 9/1/15 HOMEWORK – DUE.

Charge!!!

1 p+

1 e – –0

47 p+

47 e– +0

When #p+ = #e-, the atom has no charge and is neutral

Page 26: RIP W. E. B. DuBois (1963) Read for Tuesday Read for Tuesday Chapter 3: Sections 1-2 Chapter 3: Sections 1-2 HOMEWORK – DUE Tuesday 9/1/15 HOMEWORK – DUE.

Atoms vs. IonsAtoms can gain or lose e- to form IONS

ANY charged particle is called an ionLosing e- gives POSITIVELY charge

Page 27: RIP W. E. B. DuBois (1963) Read for Tuesday Read for Tuesday Chapter 3: Sections 1-2 Chapter 3: Sections 1-2 HOMEWORK – DUE Tuesday 9/1/15 HOMEWORK – DUE.

10 e- =11 e- =

Charge!!!

When an atom LOSES electrons 10 e–

11 p+ = 11 p+

–+1

NaNa Na+Na Na+ + e-

Before After


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