chemistry 9th ed. 2009 whitten, davis, peck, stanley thomson books/cole 代理商 ...

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Chemistry 9th ed. 2009

Whitten, Davis, Peck, StanleyThomson Books/Cole

代理商偉明圖書有限公司02-2363-8586范先生

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Chapter 1Chapter 1

The Foundations of The Foundations of ChemistryChemistry

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Chapter OutlineChapter Outline• Matter and Energy 物質與能量• Chemistry – A Molecular View of Matter• States of Matter • Chemical and Physical Properties 化學與物理性質• Chemical and Physical Changes 化學與物理變化• Mixtures, Substances, Compounds, and Elements• Measurements in Chemistry• Units of Measurement• Use of Numbers• The Unit Factor Method (Dimensional Analysis)• Percentage 百分比• Density and Specific Gravity 密度與比重• Heat and Temperature• Heat Transfer and the Measurement of Heat

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Chemistry is everywhereTent → plasticSteel cooking grate → alloy of iron and carbonTrees → Photochemical reaction → CO2 + H2O → carbohydrateOur bodies → inorganic and bioorganic compounds …...

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ChemistryChemistry• the science that describes matter—

– properties,– the changes it undergoes,– the energy changes that accompany those

processes 研究物質的特性 , 組成 , 結構和變化的科學• Organic Chemistry 有機化學 Inorganic Chemistry 無機化學 Analytical Chemistry 分析化學…• Physical Chemistry 物理化學 Biochemistry 生物化學…

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1-1 Matter and Energy1-1 Matter and Energy

• Matter– Mass– Occupies space

• Energy: The capacity to do work or transfer heat Kinetic energy & Potential energy 動能位能

Exothermic & Endothermic 放熱吸熱

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Fig. 1-1, p. 4

The Law of Conservation of Matter物質不滅定律

There is no observable change in the quantity of matter during a chemical reaction or during a physical change.

無論物質經過任何化學變化或物理變化 , 其反應物的總質量和產物的總質量相同

鎂燃燒生成氧化鎂

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The Law of Conservation of Matter物質不滅定律The law of Conservation of Energy能量不滅定律

• Energy cannot be created or destroyed in a chemical reaction or in a physical change. It can only be converted from one form to another

The Law of Conservation of Matter and Energy 質能不滅定律

• The combined amount of matter and energy in the universe is fixed

• Einstein’s Relativity• E=mc2

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Dalton’s Atomic TheoryDalton’s Atomic Theory道耳吞原子學說道耳吞原子學說 (in 1808)(in 1808)

Fig1-2 Relative size for atoms of the noble gasesFig. 1-2, p. 6

1-2 Chemistry –A molecular View of Matter

•所有的物質都是由極小的微粒所組成 , 稱之為原子 (atom)•不同元素 (element) 的原子不同 , 但同一元素所含的原子均相同

•原子是組成物質的最小粒子 , 不能再分割•化學反應是物質中原子的重新排列或組合•不同元素的原子形成化合物 (compound) 時 , 原子間的比例為一固定的簡單整數比

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Table 1-1, p. 6

電子質子中子

物質的基本粒子

The basic building blocks of atoms

11p. 7

Atomic number 原子序 the number of protons in the nucleus

Silver

12

Fig. 1-3, p. 7

A molecule 分子 is the smallest particle of an element or compound that can have a stable independent existence

Fig 1-3 Models of diatomic molecules of some elements, approximately to scale. There are called space-filling models ( 比例模型 ) because they show the atoms with their approximate relative sizeOthers: N2, Cl2, Br2, I2

13Fig. 1-4, p. 7

Fig 1-4 (a) A model of the P4 molecule of white phosphorus. (b) A model of the S8 ring found in rhombic sulfur (c) Top view of the S8 ring in rhombic sulfur

polyatomic molecules

14Fig. 1-5, p. 8

Fig 1-5 Formulas and ball-and-stick models for molecules of some compounds. Ball-and-stick ( 球棍模型 ) models represent the atoms as a smaller spheres than in space-filling models, in order to show the chemical bonds between the atoms as “stick”

甲烷乙醇

Compound 化合物

15Fig. 1-6a, p. 8

National Institute of Science and Technology logoCobalt atoms on a copper surface

Using the Scanning tunneling Microscope

34 iron atoms arranged on a copper surface

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Atom 原子the smallest particle of an element that maintains its chemical identity through all chemical and physical change

Molecule 分子the smallest particles of an element or compound that can have a stable independent existence

Element 元素a substance that cannot be decomposed into simpler substance by chemical means

Compound 化合物composed two or more different elements in fixed proportions can be decomposed into their constituent elements

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p. 8

i. Krypton 氪

Example 1-1 Models

ii. ethane 乙烷 iii. Nitrogen 氮

iv. Aspirin 阿斯匹靈v. Sulfur dioxide 二氧化硫 vi. Copper 銅

atommoleculeelementcompound

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1-3 Sates of Matter

Iodine 碘 Bromine 溴 Chlorine 氯

Fig 1-7 p. 9

(s) (l) (g)

19

p. 9

Solid (s) Gas (g) Liquid (l)

Example 1-2 Models

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1-4 Chemical and Physical Properties•Chemical properties •Chemical change changes in composition 鎂燃燒生成氧化鎂

•Physical properties•Absence of any change of composition

color, density, hardness, melting point, boiling point, and electrical and thermal conductivities

•State change (ice-water-steam)Extensive Properties 外延性質

• Depends on the amount of material in a sample•與尺度有關之性質 , 如重量、體積、熱 ... ..

Intensive Properties 內涵性質 • Independent of the amount of material in a sample•與尺寸無關之性質 , 如密度、顏色、黏度、壓力、溫度、濃度 … ..

No two different substances have identical sets of chemical and physical properties

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Fig. 1-8, p. 11

Endothermic(absorb heat)

Exothermic(release heat)

蒸發 / 凝結

冷凍 / 溶化

昇華 / 沉澱

Physical Changes Occur among the three states of matter

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Fig. 1-9a, p. 12

Physical properties:(a)melts at 0oC(b)boils at 100oC(c)dissolves a wide range of substances

Chemical properties:(d) water reacts violently with sodium to form hydrogen gas and sodium hydroxide

Water

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Table 1-2, p. 12

醋酸苯溴鐵

甲烷氧氣

氯化鈉

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Fig. 1-10, p. 13

1-5 Chemical and physical changes•Chemical change•One or more substances are used up•One or more new substances are formed•Energy is absorbed or released

•Physical change•No change in chemical composition•Change in energy

solid liquid liquid gas

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p. 13

1-6 Mixtures, Substances, Compounds, and ElementsMixture•have variable composition•May be separated by physical methods• components retain its properties•Mixture of different components may have

widely different propertiesHeterogeneous mixture 非均勻混合物• Do not have same composition throughout• Components are distinguishable• Salt and charcoal, foggy air, vegetable soup

Homogeneous mixture (solution)• Have same composition throughout• Components are indistinguishable• Salt water, alloys, air Galena( 方鉛礦 ) and quartz ( 石英 )

26Fig. 1-11a, p. 15

A mixture of iron and sulfur is a heterogeneous mixture

It ca be separated by physical means, such as removing the iron with magnet

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Fig. 1-12, p. 15

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Electrolysis apparatus for small-scale chemical decomposition of water by electric energy

(88.9%)

(11.1%)

Fig. 1-13, p. 16

Chemical decomposition of liquid water

Water as a compound!!

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Fig. 1-14, p. 17

Compound

CompoundCalcium Oxide

elementsFig 1-14 Diagram of the decomposition of calcium carbonate

elements

A compound is a pure substance consisting of two or more different elements in a fixed ratioLaw of definite Proportions 定比定律 (Law of Constant Composition)

Heat CompoundCarbon dioxide

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Fig. 1-15, p. 17

The physical and chemical properties of a compound are different from the properties of its constituent elements Sodium Chloride (NaCl)

•White solid•This compound is formed by the

combination of sodium and chloride•Sodium

•A soft, silvery white metal•Chloride

•A pale green, corrosive, poisonous gas•Na + Cl NaCl

Light and Heat Produce

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Table 1-3, p. 18

硼鋇鉍溴

鎘鈣碳

氯鈷鉻銅

金鋁銀

氦汞碘鉀

鎂鋰氪

錳氮鈉氖

氫鐵氟

鉛鉑硫鍗

鍶錫矽

鈦鈾鎢鋅

磷氧鎳

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Table 1-4, p. 18

10/88 naturally occurring elements make up more than 99% by mass

Living matter

No biological role

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Table 1-5, p. 19

1-7 Measurements in Chemistry

發光強度

電流

長度質量

燭光絕對溫標

莫耳數克分子量

時間

溫度

物質數量

The international System of Units (SI) metric system 公制

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Table 1-6, p. 19

SI: the international system of units

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1-8 Units of MeasurementMass ( 質量 ): doesn’t vary as its position

changeWeight ( 重量 ): is a measure of the

gravitational attraction of the earth for the body, and this varies with distance from the center of the earth

Triple-beam balance 0.01g三樑天平

Electronic top-load balance0.001g 電子秤

Analytical balance0.0001g 電子分析天平

Fig. 1-16a, p. 20

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Fig. 1-17 1-18, p. 21

Length

inch

centimeter

Volume

Beaker150ml 燒杯

buret25ml 玻璃量管

volumetric flask1000ml 定量瓶

Graduated cylinder100ml 量筒

volumetric pipet10ml 定量吸管

Flask錐形瓶

Liters=1000cm3 milliliter=1cm3= 1cc

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Table 1-8, p. 21

Appendix C Common Units, Equivalences, and Conversion Factors

Page A-8 to A-10 in textbook

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Fig. 1-19, p. 23

1-9 Use of Numbers

Scientific Notation 科學記數法4,300,000=4.3x106

0.000348=3.48x10-4

Significant Figures 有效位數 are digits believed to be correct by the person who makes a measurement

Numbers obtained by countingexact numbers

Numbers obtained measurementsare not exact estimate numbers

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• The volume is read at the bottom of the liquid curve (meniscus).

• Meniscus of the liquid occurs at about 20.15 mL. Certain digits: 20.15 Uncertain digit: 20.15

Measurement of Volume Using a BuretMeasurement of Volume Using a Buret

• A digit that must be estimated is called uncertain. • A measurement always has some degree of

uncertainty.• Record the certain digits and the first uncertain

digit (the estimated number).

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Fig. 1-20, p. 24

Significant Figures 有效位數

Precision 精確•Degree of agreement among several measurements of the same quantity.

Accuracy 準確 •Agreement of a particular value with the true value.

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Neither accurate nor precise Precise but not accurate Both precise and accurate

Precision and AccuracyPrecision and Accuracy

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38.57 4 significant figure

288g 3 significant figure


288g 3 significant figure

1. Nonzero digits are always significant .

Some simple rules govern the use of significant figures

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2. Zeros are sometimes significant , and sometimes they are nota. Zeroes at the beginning of a number are never significant

b. Zeroes between nonzero digits are always significant c. Zeroes at the end of a number that contains a decimal

point are always significant d. Zeroes are at the end of a number that does not contain a

decimal point may or may not be significant a. 0.052 2 significant figure

5.2x10-2


3.64x10-3

b. 2007 4 significant figure


a. 0.052 2 significant figure

5.2x10-2


3.64x10-3

b. 2007 4 significant figure


c. 38.0 3 significant figure


d. 24,300km 3,4,5 significant figure

2.43x104 2.430x104 2.4300x104

c. 38.0 3 significant figure


d. 24,300km 3,4,5 significant figure

2.43x104 2.430x104 2.4300x104

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3. Exact numbers can be considered as having an unlimited number of significant figures. This applies to defined quantities.

1 yard = 3 feet1,3 are exactDo no apply the rules of significant figures

1 inch= 2.54cm

1 yard = 3 feet1,3 are exactDo no apply the rules of significant figures

1 inch= 2.54cm

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Example Significant FiguresExample Significant Figures

Give the number of significant figures for each of the following results.

a. A student’s extraction procedure on tea yields 0.0105 g of caffeineb. A chemist records a mass of 0.050080 g in an analysisc. In an experiment a span of time is determined to be 8.050x10-3 sec

a. 0.0105g 1.05x10-2g 3 significant figures

b. 0.050080g 5.0080x10-2 g 5 significant figuresc. 8.050x10-3 sec 4 significant figures

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4. In addition and subtraction, the last digit retained in the sum or difference is determined by the position of the first doubtful digit

(a) Add 37.24ml and 10.3ml

37.24ml +10.3 ml 47.54 ml is reported as 47.5ml

(b) Subtract 21.2342 from27.81

27.87 g - 21.2342g 6.6358 g is reported as 6.64g

(a) Add 37.24ml and 10.3ml

37.24ml +10.3 ml 47.54 ml is reported as 47.5ml

(b) Subtract 21.2342 from27.81

27.87 g - 21.2342g 6.6358 g is reported as 6.64g

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5. In multiplication and division, an answer contains no more significant figures than the least number of significant figures used in the operation.

4.56 x 1.4 = 6.4 two significant figures

What is the area of a rectangle 1.23 cm wide and 12.34cm long?

A= l x w =(12.34cm) x (1.23cm) =15.2cm2

(calculator result =15.1782)

4.56 x 1.4 = 6.4 two significant figures

What is the area of a rectangle 1.23 cm wide and 12.34cm long?

A= l x w =(12.34cm) x (1.23cm) =15.2cm2

(calculator result =15.1782) Exercise 32

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Example: Significant Figures in Mathematical OperationsExample: Significant Figures in Mathematical OperationsCarry out the following mathematical operations, and give each

result with the correct number of significant figures.a. 1.05x10-3 ÷ 6.135b. 21-13.8c. (2.560)x(8.8) ÷ 275.15

a. 1.05x10-3 ÷ 6.135 1.71x10-4 3 significant figures

b. 21-13.8 7 The number with the least number decimal places (21)

has nonec. (2.560)(8.8) ÷ 275.15 0.0835908 The least precise measurement has 2 significant figures 8.4x10-2

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1-10 The Unit Factor Method (Dimensional Analysis 因次分析 )

Example 1-5 Unit FactorExpress 1.47 mi. in inches.Miles feet inches 1 mile = 5,280 ft 1 ft = 12 inch

Example 1-5 Unit FactorExpress 1.47 mi. in inches.Miles feet inches 1 mile = 5,280 ft 1 ft = 12 inch

Unit Factor: a factor in which the numerator and denominator are expressed in different units but represent the same or equivalent amounts.

1 ft = 12 in 1=12 in/ft 1=1 ft/12inThe reciprocal of any unit factor is also a unit factor

Unit Factor: a factor in which the numerator and denominator are expressed in different units but represent the same or equivalent amounts.

1 ft = 12 in 1=12 in/ft 1=1 ft/12inThe reciprocal of any unit factor is also a unit factor

?in. =1.47 mi x 5280ftmi

x 12in.ft

=9.31x104 in.(93139.2)

Unit Conversion

Volume Calculation

Mass Conversion

Volume Conversion

Energy Conversion

English-Metric Conversion

Factor-label Method

Exercise 36

Unit Conversion

50

Example 1-6 Unit Conversion The Ångestrom (Å) is a unit of length 1x10-10m, that provides a convenient scale on which to express the radii of atoms. Radii of atoms are often expressed in nanometers. The radius of a phosphorus atom is 1.10 (Å). what is the distance expressed in centimeters and nanometers?

Example 1-6 Unit Conversion The Ångestrom (Å) is a unit of length 1x10-10m, that provides a convenient scale on which to express the radii of atoms. Radii of atoms are often expressed in nanometers. The radius of a phosphorus atom is 1.10 (Å). what is the distance expressed in centimeters and nanometers?

Unit Conversion

Å m cm

Å m nm

= 1.1 Å =1.10x10-8cm? cm x 1cm1x10-2m

1x10-10m1 Å

x

= 1.1 Å =1.10x10-1nm? nm x 1nm1x10-9m

1x10-10m1 Å

x

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Example : Express 9. 32 yards in millimeters.Example : Express 9. 32 yards in millimeters.

Example : Express 627 milliliters in gallons.Example : Express 627 milliliters in gallons.

= 9.32yd

=8.52x103mm

? mm x

= ? mm9.32 yd3ft1yd x 12in

1ft x 2.54cm1in x 10mm

1cm

=0.166155 gal 0.166 gal

? mm = 627ml x = 627 ml? gal

1L1000mLx 1.06qt

1L x 1gal4qt

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Example : Express 2.61 x 104 cm2 in ft2. Example : Express 2.61 x 104 cm2 in ft2.

Example : Express 2.61 ft3 in cm3

Example : Express 2.61 ft3 in cm3

=28.09380619 ft2

28.1 ft2

? ft2 = 2.61x104cm2 x 1in2.54cm

x2 1ft12in

2

=73906.9696 cm3

7.39x104 cm3

? cm3 = 2.61 ft3 x 2.54cm1in

3x12in1ft

3

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p. 31

1-11 Percentage

U.S. pennies made since 1982 consist of 97.6% zinc and 2.4% copper. The mass of a particular penny is measures to be 1.494 grams. How many grams of zinc does in this penny contain?

1.494g sample x 100g sample97.6g zinc =1.46g zinc

Exercise 63, 64

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p. 31, Tabl1 1-9, p.32

1-12 Density and Specific Gravity 密度與比重Density=

MassVolume D = m

Vg/cm3 or g/ml for solids and liquids,g/L for gases

gasoline

water

cork

oak

mercurybrass

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Example 1-13A 47.3-ml sample of ethyl alcohol has a mass of 37.32g. What is the density?

Example 1-13A 47.3-ml sample of ethyl alcohol has a mass of 37.32g. What is the density?

p. 32

D = V37.32 g47.3 ml = = 0.789g/ml

Exercise 40

Example 1-14If 116g of ethanol is need for a chemical reaction, what volume of liquid would you use?

Example 1-14If 116g of ethanol is need for a chemical reaction, what volume of liquid would you use?

V =

m

mD

116 g0.789g/ml

= = 147mlExercise 42

Example 1-15Express the density of mercury in lb/ft3?Example 1-15Express the density of mercury in lb/ft3?

Mercury density:13.59g/cm3

= 13.59

=848.4lb/ft3

lbft3?

g cm3 x 2.54cm

1in x 12in1ft

1lb453.6g x

3 3

D = Vm

m = D x V

V = Dm

56

p. 33

Specific gravity 比重The ratio of its density to the density of water, both at the same temperature

Sp. Gr = Dsubstance

Dwater

Dice<Dwater Dsolid ethanol > Dliquid ethanol

Example 1-16The density of table salt is 2.16 g/ml at 20oC. What is its specific gravity?

Example 1-16The density of table salt is 2.16 g/ml at 20oC. What is its specific gravity?

Sp. Gr =Dsubstance

Dwater

2.16g/ml1.00g/ml=

= 2.16

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Example 1-17 Specific Gravity, Volume, Percentage by MassBattery acid is 40.0% sulfuric acid, H2SO4, and 60% water by mass. Its specific gravity is 1.31. Calculate the mass of pure H2SO4 in 100ml of battery acid.

Example 1-17 Specific Gravity, Volume, Percentage by MassBattery acid is 40.0% sulfuric acid, H2SO4, and 60% water by mass. Its specific gravity is 1.31. Calculate the mass of pure H2SO4 in 100ml of battery acid.

40g H2SO4 100g sol

= 100.0ml sol

=52.4g H2SO4

? x40.0g H2SO4

100g sol1.31g sol1ml sol xH2SO4

Exercise 48

40% H2SO4 =

In this case, the specific gravity equal to the density =1.31g/ml

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Fig 1-22, p. 33

1-13 Heat and Temperature

Heat is a form of energy that always flows spontaneously from a hotter body to a colder body — it never flows the reverse direction. Hotter Body Colder Body

Temperature Measure: Mercury thermometer Centigrade temperature scale 0oC 100oC Fahrenheit temperature scale 32oF 212oF Kelvin (absolute) temperature scale ?K=oC+273.15o or ? oC=K-273.15o

Fahrenheit and Centigrade Relationships ?oF=1.8x?oC+32oF

Fahrenheit and Centigrade Relationships ?oF=1.8x?oC+32oF ?oC= 1.8

oF-32

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Example 1-18 Temperature ConversionWhen the temperature reaches “100oF” in the shade”, it’s hot. What is this temperature on the Celsius scale?

Example 1-18 Temperature ConversionWhen the temperature reaches “100oF” in the shade”, it’s hot. What is this temperature on the Celsius scale?

?oC= 1.8oF-32

1.8100-32

= = 38oC

Example 1-19 Temperature ConversionWhen the absolute temperature is 400K, what is this Fahrenheit temperature?

Example 1-19 Temperature ConversionWhen the absolute temperature is 400K, what is this Fahrenheit temperature?

? oC=K-273.15o

=400-273 =127oC

?oF=1.8x?oC+32oF =1.8x127+32oF =261oF Exercise 50

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1-14 Heat Transfer and the Measurement of Heat

Joule (J): the SI unit of energy and work job 1 J = 1kg m2/s2

Kinetic energy ( 動能 ) =1/2 mv2

Calorie: 4.184J

Joule (J): the SI unit of energy and work job 1 J = 1kg m2/s2

Kinetic energy ( 動能 ) =1/2 mv2

Calorie: 4.184J

Specific heat ( 比熱 ): • the amount of heat required to raised the temperature of one

gram of the substance one degree Celsius with no change in phase • A physical property, is different for the solid, liquid and gaseous

phase of the substance Appendix E

Specific heat ( 比熱 ): • the amount of heat required to raised the temperature of one

gram of the substance one degree Celsius with no change in phase • A physical property, is different for the solid, liquid and gaseous

phase of the substance Appendix E

Specific Heat=(amount of heat in J)

(mass of substance in g)(temperature change in oC)

Unit: J/goC

61

Example 1-20 Specific HeatHow much heat s, in joules, is required to raise the temperature of 205g of water from 21.2oC to 91.4oC?

amount of heat = (mass)(specific heat)(temperature change)

amount of heat = (205g)(4.18J/goC)(91.4oC-21.2oC)

= 6.02x104J (60.2kJ)

Example 1-20 Specific HeatHow much heat s, in joules, is required to raise the temperature of 205g of water from 21.2oC to 91.4oC?


amount of heat = (205g)(4.18J/goC)(91.4oC-21.2oC)

= 6.02x104J (60.2kJ)

Specific Heat= (amount of heat )(mass)(temperature change )

Exercise 58,59

62

Thermal equilibriumwhen two objects at different temperature are brought into contact, heat flow from the hotter to the colder body Reach to the same temperature

endothermic = exothermic 吸熱 = 放熱

Fig 1-23, p. 38

63

Example 1-21 Specific HeatA 588-gram chunk of iron is heated to 97.5oC. Then it is immersed in 247grams of water originally at 20.7oC. When thermal equilibrum has been reached, the water and iron are both at 36.2oC. Calculate the specific heat of iron. Temperature change of water = 36.2-20.7 = 15.5oC Temperature change of iron = 97.5-36.2 = 61.3oC Joules gained by water = Joules lost by iron Let x = specific heat of iron (247g)(4.18 J/goC)(15.5oC)=(588g)(x J/goC)(61.3oC) 16003.13 = 36044.4 x x = 0.444J/goC

Example 1-21 Specific HeatA 588-gram chunk of iron is heated to 97.5oC. Then it is immersed in 247grams of water originally at 20.7oC. When thermal equilibrum has been reached, the water and iron are both at 36.2oC. Calculate the specific heat of iron. Temperature change of water = 36.2-20.7 = 15.5oC Temperature change of iron = 97.5-36.2 = 61.3oC Joules gained by water = Joules lost by iron Let x = specific heat of iron (247g)(4.18 J/goC)(15.5oC)=(588g)(x J/goC)(61.3oC) 16003.13 = 36044.4 x x = 0.444J/goCExercise 62

64

Example 1-22 Comparing Specific HeatsWe add the same amount of heat to 10.0grams of each of the following substances staring at 20oC: H2O(l); Hg(l); liquid Benzene,C6H6(l); and solid aluminum, Al(s). Rank the samples from lowest to highest final temperature. Refer to Appendix E fro required data. Substance specific heat H2O(l) 4.18 Hg(l) 0.138 C6H6(l) 1.74 Al(s) 0.90


specific heat : H2O(l) > C6H6(l) > Al(s) >Hg(l)

The ranking from the lowest to highest final temperature is

H2O(l) < C6H6(l) < Al(s) < Hg(l)

Example 1-22 Comparing Specific HeatsWe add the same amount of heat to 10.0grams of each of the following substances staring at 20oC: H2O(l); Hg(l); liquid Benzene,C6H6(l); and solid aluminum, Al(s). Rank the samples from lowest to highest final temperature. Refer to Appendix E fro required data. Substance specific heat H2O(l) 4.18 Hg(l) 0.138 C6H6(l) 1.74 Al(s) 0.90


specific heat : H2O(l) > C6H6(l) > Al(s) >Hg(l)

The ranking from the lowest to highest final temperature is

H2O(l) < C6H6(l) < Al(s) < Hg(l)Exercise 71

chemistry 9th ed. 2009 whitten, davis, peck, stanley thomson books/cole 代理商 ...

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