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서론 - 화학 공부의 요점

Chapter 1

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1.1 Chemistry and some Fundamental Definitions

1.2 The Scientific Approach: Developing a Model

1.3 Chemical Problem Solving

1.4 Measurement in Scientific Study

1.5 Uncertainty in Measurement: Significant Figures

Chapter 1 : Keys to the Study of Chemistry

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제 1 장 - 화학 공부의 요점

1. 화학이란 ?2. 과학적 방법 : 모형3. 화학문제 풀기 : 단위와 환산인자4. 과학적 연구에서 측정 , SI 단위5. 측정의 불확실성 : 정밀도 , 정확도 , 유효숫자

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불을 다루는 학문 ( 火學 ) 이 아니라

숭례문 火災

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화학이란 ?

변화 ( 變化 )

학문 ( 學問 )Chemistry:

The Study of Change

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the study of matter,

its properties,

the changes that matter undergoes,

and

the energy associated with these changes.

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Chemistry: A Science for the 21st Century

• Health and Medicine

• Sanitation systems

• Surgery with anesthesia

• Vaccines and antibiotics

• Energy and the Environment

• Fossil fuels

• Solar energy

• Nuclear energy

한국인의 평균 수명

1930 년 2009 년30 세

80 세

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Chemistry: A Science for the 21st Century

• Materials and Technology

• Polymers, ceramics, liquid crystals

• Room-temperature superconductors?

• Molecular computing?

• Battery

• Nano materials and technology

• Food and Agriculture

• Genetically modified crops

• “Natural” pesticides

• Specialized fertilizers

하버 법에 의한 질소비료 생산량 1 억톤 / 년연간 세계 천연가스 수요의 3-5%전체 에너지 수요의 1-2%세계인구 1/3 의 식량을 감당

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Chemical Industry

우리나라 GDP 의 30% 감당

자동차 에너지

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Changes in

물질 (Matter)

조성 (Composition)types and amounts of substances in a sample of matter

에너지 (Energy)

성질(Properties)some intrinsic or extrinsic quality of

a sample of matterphysical ,chemical

pure substance (element, compound),mixture (homogeneous, inhomogeneous)

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물질의 성질

Chemical Properties those which the substance shows as it interacts with, or transforms into, other substances.

Examples:flammability, corrosiveness

Physical Properties those which the substance shows by itself without interacting with another substance.

Examples:color, melting point, boiling point, density

Physical and chemical properties

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B Chemical change

Physical and chemical change.

A Physical change

hydrogen burns in air to form water

ice melts to form water

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An extensive property of a material depends upon how much matter is being considered.

An intensive property of a material does not depend upon how much matter is being considered.

크기 성질과 세기 성질

Pressure, temperature, color, density, …

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Physical States of Matter

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PROBLEM: Decide whether each of the following process is primarily a physical or a chemical change, and explain briefly:

PLAN: “Does the substance change composition or just change form?”

(a) Frost forms as the temperature drops on a humid winter night.

(b) A cornstalk grows from a seed that is watered and fertilized.

(c) Dynamite explodes to form a mixture of gases.

(d) Perspiration evaporates when you relax after jogging.

(e) A silver fork tarnishes slowly in air.

SOLUTION:

(a) physical change (b) chemical change (c) chemical change

(d) physical change (e) chemical change

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energy due to the position of the object or energy from a chemical reactionPotential Energy

Kinetic Energy energy due to the motion of the object

Energy is the capacity to do work.

Potential and kinetic energy can be interconverted.

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Energy is the capacity to do work. Figure 1.3A

less stable

more stable

change in potential energy EQUALSkinetic energy

A gravitational system. The potential energy gained when a lifted weight is converted to kinetic energy as the weight falls.

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Energy is the capacity to do work. Figure 1.3B

less stable

more stable

change in potential energy EQUALSkinetic energy

A system of two balls attached by a spring. The potential energy gained by a stretched spring is converted to kinetic energy when the moving balls are released.

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Energy is the capacity to do work. Figure 1.3C

less stable

more stable

change in potential energy EQUALSkinetic energy

A system of oppositely charged particles. The potential energy gained when the charges are separated is converted to kinetic energy as the attraction pulls these charges together.

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Energy is the capacity to do work. Figure 1.3D

less stable

more stable

change in potential energy EQUALSkinetic energy

A system of fuel and exhaust. A fuel is higher in chemical potential energy than the exhaust. As the fuel burns, some of its potential energy is converted to the kinetic energy of the moving car.

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Scientific Approach: Developing a Model

Observations : Natural phenomena and measured events; universally consistent ones can be stated as a natural law.

Hypothesis: Tentative proposal that explains observations.

Experiment: Procedure to test hypothesis; measures one variable at a time.

Model (Theory):Set of conceptual assumptions that explains data from accumulated experiments; predicts related phenomena.

Further Experiment: Tests predictions based on model.

revised if experiments do not support it

altered if predictions do not support it

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Measurement and Error

물체와 연결된 어떤 성질을 측정

기준 량→도량형도 ( 度 ) 는 길이 또는 길이를 측정하기 위한 자 , 양 ( 量 ) 은 부피 및 되 , 형 ( 衡 ) 은 무게 및 저울을 뜻한다 .

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Measurements in the wall of the city hall in Re-gensburg, Germany. The measurements are the

foot, the yard, and the fathom.

광무 6 년 (1905 년 ) 의 1 두 표준 원기 .

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Converting Units of Length

PROBLEM: To wire your stereo equipment, you need 325 centimeters (cm) of speaker wire that sells for $0.15/ft. What is the price of the wire?

PLAN: Known - length (in cm) of wire and cost per length ($/ft)

We have to convert cm to inches and inches to ft followed by finding the cost for the length in ft.

SOLUTION:length (cm) of wire

length (ft) of wire

length (in) of wire

Price ($) of wire

2.54 cm = 1 in

12 in = 1 ft

1 ft = $0.15

Length (in) = length (cm) x conversion factor

= 325 cm x in2.54 cm

= 128 in

Length (ft) = length (in) x conversion factor

= 128 in x ft12 in

= 10.7 ft

Price ($) = length (ft) x conversion factor

= 10.7 ft x $0.15ft

= $1.60

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Quantity Measured Unit Symbol

Length meter m

Mass kilogram kg

Time second s

Thermodynamic temperature kelvin K

Amount of substance mole mol

Electric current ampere A

Luminous intensity candela Cd

SI Base Units

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Prefixes Used with SI UnitsPrefix Symbo

lMeanin

g Length(m) Time(s)

peta- P 1015 Ly(9x1015)

tera- T 1012 Sun-Jupiter

giga- G 109 　 조선시대 평균수명 (32 년 )

mega- M 106 한국 길이 10 일

kilo- k 103 　 Time to run 5km

deci- d 10-1 　 심장박동

centi- c 10-2 　

milli- m 10-3 빗방울 반지름

micro- μ 10-6 cell

nano- n 10-9 big molecules

pico- p 10-12 　 빛이 300μm 를 지나는 시간

femto- f 10-15 Nucleus radius molecular vibration

atto- a 10-18 　

zepto z 10-21 　

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Physical Quantity Unit Symbol Formula

Frequency hertz Hz (cycles) s-1

Force newton N kg·m·s-2

Pressure pascal Pa N·m-2

Energy joule J N·m

Power watt W J·s-1

Electric potential difference volt V W·A-1

Electric charge coulomb C A·s

Electric resistance ohm Ω V·A-1

Electric capacitance farad F C·V-1

Electric conductance siemens S A·V-1

Magnetic flux weber Wb V·s

Magnetic flux density tesla T Wb·m-2

Inductance henry H Wb·A-1

Luminous flux lumen lm cd·sr

Illuminance lux lx lm·m-2

Activity(radionuclide) becquerel Bq (disintegration) s-1

Absorbed dose(radiation) gray Gy m2·s-2

SI-Derived units with special names

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Quantity SI Non-SI Unit Symbol Formula

Energy J calorie cal 1 cal = 4.184 *J

erg erg 1 erg = 10-7 J

electronvolt eV 1 eV = 1.60219×10-19 J

Force N dyne dyn 1 dyn = 10-5 N

Length m Angstrom Å 1 Å = 10-10 m

Mass kg gram g 1 g = 10-3 kg

atomic mass unit amu 1 amu = 1.66057×10-27 kg

Pressure Pa atmosphere atm 1 atm = 1.013×105 Pa

torr torr 1 torr = 1 mm Hg = 133 Pa

Temperature K Celsius °C 1°C = 1 K

Volume m3 liter L 1 L = 10-3 m3

Non-SI units used in chemistry with con-version factors to SI

*exact value

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JET'S FUEL RAN OUT AFTER METRIC CONVERSION ERRORS

By RICHARD WITKIN (The New York Times); National DeskJuly 30, 1983, SaturdayLate City Final Edition, Section 1, Page 7, Column 4, 819 words

Air Canada said yesterday that its Boeing 767 jet ran out of fuel in midflight last week be-cause of two mistakes in figuring the fuel supply of the airline's first aircraft to use metric measurements. After both engines lost their power, the pilots made what is now thought to ...

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On 23 July 1983, Air Canada Flight 143 ran completely out of fuel about halfway through its flight from Montreal to Edmonton. Fuel loading was miscalculated through misunderstanding of the recently adopted metric system. For the trip, the pilot calculated a fuel requirement of 22,300 kg. There were 7,682 liters al-ready in the tanks.

Problem 3 - If a liter of jet fuel has a mass of 0.803 kg, how much fuel needed to be added for the trip?

Answer:In order to calculate how much more fuel had to be added, the crew needed to convert the quantity in the tanks, 7,682 L, to a weight, subtract that figure from 22,300 kg, and convert the result back into a volume (L). 7,682 L x (0.803 kg/ 1 L)= 6,169 kg22,300 kg – 6,169 kg = 16,131 kg 16,131 kg x ( 1 L / 0.803 kg) = 20,088 L of jet fuel. <= volume required.

mass/unit volume = 0.803 kg/L

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Between the ground crew and flight crew, however, they arrived at an incorrect conversion factor of 1.77, the weight of a liter of jet fuel in pounds. This was the conversion factor provided on the refueller's paperwork and which had always been used for the rest of the airline's fleet.

Their calculation produced: 7,682 liters x (1.77 pounds/liter) = 13,597 which they interpreted as kilograms but was actually the fuel mass in pounds! Then they continued the calculation:

22,300 kg – 13,597 'kg' = 8,703 kg 8,703 kg ÷ 1.77 = 4,916 L <= actual volume fueled.

….so they were actually 15,172 L short of fuel! (20,088 L-4,916 L)

mass/unit volume = 1.77 lb/L

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Launch• Delta 7425• Launch 12/11/98• 629 kg launch mass

Cruise• 4 midcourse maneuvers• 10–Month Cruise

Mars Orbit Insertion and Aerobraking• Arrival 9/23/99• MOI is the only use of the main engine. The 16- minute burn depletes oxidizer and captures vehicle into 13–14 hour orbit.• Subsequent burn using hydrazine thrusters reduce orbit period further.• Aerobraking to be completed prior to MPL ar-rival [12/3/99]

Mars Climate Orbiter

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Mars Surveyor '98 program

On 9/23/99, $125 million Mars Climate Orbiter en-tered Mar’s atmosphere 100 km (62 miles) lower than planned and was destroyed by heat.

1 lb force = 1 N

1 lb force = 4.45 N

NASA, JPL, Lockheed Martin

Actual trajectoryand AMD ΔV’s

Estimated trajectoryand AMD ΔV’s

Mars

To Earth

Schematic MCO En-counter Diagram

Not to scale

57 km

226km

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On September 23, 1999 NASA lost the $125 million Mars Climate Orbiter spacecraft after a 286-day journey to Mars. Miscalculations due to the use of English units instead of metric units apparently sent the craft slowly off course -- 60 miles in all. Thrusters used to help point the spacecraft had, over the course of months, been fired incorrectly because data used to control the wheels were calculated in incorrect units. Lockheed Martin, which was performing the calcula-tions, was sending thruster data in English units (pounds) to NASA, while NASA's navigation team was expecting metric units (Newtons).

Answer: 10,000,000 'Newtons' x ( 1 pound / 4.448 Newtons) = 2,200,000 pounds instead of 10 million pounds so the error is a 'missing' 7,800,000 pounds of thrust…an error that would definitely be noticed at launch!!!

Problem 1 - A solid rocket booster is ordered with the specification that it is to produce a total of 10 million pounds of thrust. If this number is mistaken for the thrust in Newtons, by how much, in pounds, will the thrust be in error? (1 pound = 4.45 Newtons)

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Converting Units of Volume

PROBLEM: The volume of an irregularly shaped solid can be determinedfrom the volume of water it displaces. A graduated cylinder contains 19.9mL of water. When a small piece of galena, an ore of lead, is submerged in the water, the volume increases to24.5mL. What is the volume of the piece of galena in cm3 and in L?

PLAN: The volume of galena is equal to the change in the water volume before and after submerging the solid.

volume (mL) before and after addition

volume (mL) of galena

volume (cm3) of galena

subtract

volume (L) of galena

1 mL = 1 cm3 1 mL = 10-3 L

SOLUTION:

(24.5 - 19.9)mL = volume of galena

4.6 mL×mL

1 cm3

= 4.6 cm3

4.6 mL×mL

10-3 L = 4.6x10-3 L

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Converting Units of Mass

PROBLEM: International computer communications are often carried by optical fibers in cables laid along the ocean floor. If one strand of optical fiber weighs 1.19 x 10-3lbs/m, what is the total mass (in kg) of a cable made of six strands of optical fiber, each long enough to link New York and Paris (8.84 x 103km)?

PLAN: The sequence of steps may vary but essentially you have to find the length of the entire cable and convert it to mass.

length (km) of fiber

mass (lb) of fiber

length (m) of fiber

1 km = 103 m

mass (kg) of cablemass (lb) of cable

6 fibers = 1 cable

1 m = 1.19×10-3 lb

2.205 lb = 1 kg

SOLUTION:

8.84×103km×km

103m

8.84×106m × m

1.19×10 -3lbs

1.05 × 104lb ×cable

6 fibers

= 1.05×104lb

= 8.84×106m

2.205 lb

1kg×

6.30×104lb

cable

6.30×104lb =cable

2.86×104 kgcable

=

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Calculating Density from Mass and Length

PROBLEM: Lithium (Li) is a soft, gray solid that has the lowest densityof any metal. If a slab of Li weighs 1.49 x 103 mg and has sides that measure 20.9 mm by 11.1 mm by 11.9 mm, whatis the density of Li in g/cm3 ?

PLAN: Density is expressed in g/cm3 so we need the mass in grams and the volume in cm3.

mass (mg) of Li

lengths (mm) of sides

mass (g) of Li

density (g/cm3) of Li

103 mg = 1 g

10 mm = 1 cm

lengths (cm) of sides

volume (cm3)

multiply lengths

SOLUTION:

20.9mm×

1mg10-3g

= 1.49g1.49×103mg×

10mm1cm

= 2.09cm

Similarly the other sides will be 1.11 cm and 1.19 cm, respectively.

2.09×1.11×1.20 = 2.76cm3

density of Li =1.49g

2.76 cm3= 0.540 g/cm3

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Figure 1.8 The freezing and boiling points of water.

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Temperature Scales and Interconversions

Kelvin ( K ) - The “Absolute temperature scale” begins at absolute zero and only has positive values.

Celsius ( oC ) - The temperature scale used by science, formally called centigrade, most commonly used scale around the world; water freezes at 0oC, and boils at 100oC.

Fahrenheit ( oF ) - Commonly used scale in the U.S. for our weather reports; water freezes at 32oF and boils at 212oF.

T (in K) = T (in oC) + 273.15 T (in oC) = T (in K) - 273.15

T (in oF) = 9/5 T (in oC) + 32T (in oC) = [ T (in oF) - 32 ] 5/9

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Converting Units of Temperature

PROBLEM: A child has a body temperature of 38.70C.

PLAN: We have to convert 0C to 0F to find out if the child has a fever and we use the 0C to kelvin relationship to find the temperature in kelvins.

(a) If normal body temperature is 98.60F, does the child have a fever?

(b) What is the child’s temperature in kelvins?

SOLUTION:

(a) Converting from 0C to 0F9

5(38.70C) + 32 = 101.70F

(b) Converting from 0C to K 38.70C + 273.15 = 311.8K

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The number of significant figures in a measurement depends upon the measuring device.

더 정밀한 측정장치 → 더 많은 유효숫자

32.30C32.330C

측정 , 측정의 불확실성 ( 오차 ) 과 유효숫자

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a measured value of 12.53 means the true value lies between 12.525 and 12.535

=> 1, 2, 5 are assured numbers, and the last digit 3 is meaningful, but has a little doubtful. 1, 2, 5, 3 are significant figures.   another measured value of 1.2 means

the true value lies between 1.15 and 1.25   

Significant figures

example

Case 1: addition/subtraction 12.53+1.2

12.525 < 12.53 < 12.535

1.15 < 1.2 < 1.25

13.675 < 13.73 < 13. 785

the last digit of 3 is meaningless. So, we put the answer

13.7.

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Significant figures

Case 2: multiplication/division 12.53×1.2

12.525 < 12.53 < 12.535

1.15 < 1.2 < 1.25

14.40375 < 15.036 < 15.66875

the last 3 digits of 0, 3, 6 are meaningless. So, we put the answer

15.

Case 3: scientific notation 12.53×12

12.525 < 12.53 < 12.535

11.5 < 12 < 12.5

144.0375 < 150.36 < 156.6875

the last 3 digits of 0, 3, 6 are meaningless. How can we put the answer in correct way?

1.5 × 102

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Rules for Determining Which Digits are Significant

All digits are significant

• Make sure that the measured quantity has a decimal point.• Start at the left of the number and move right until you reach the

first nonzero digit.• Count that digit and every digit to it’s right as significant.

Numbers such as 5300 L are assumed to only have 2 significant figures. A terminal decimal point is often used to clarify the situation, but scientific notation is the best!

except zeros that are used only to position the decimal point.

Zeros that end a number and lie either after or before the decimal point are significant; thus 1.030 ml has four significant figures, and 5300. L has four significant figures also.

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Determining the Number of Significant Figures

PROBLEM: For each of the following quantities,

underline the zeros that are significant figures(sf), and

determine the number of significant figures in each quantity.

For (d) to (f), express each in exponential notation first.

(b) 0.1044 g(a) 0.0030 L (c) 53,069 mL

(e) 57,600. s(d) 0.00004715 m (f) 0.0000007160 cm3

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Rules for Rounding Off Numbers1. If the digit removed is more than 5, the preceding number increases by 1. 5.379 rounds to 5.38 if three significant figures are retained and to 5.4 if two significant figures are retained.

2. If the digit removed is less than 5, the preceding number is unchanged. 0.2413 rounds to 0.241 if three significant figures are retained and to 0.24 if two significant figures are retained.

3.If the digit removed is 5, the preceding number increases by 1 if it is odd and remains unchanged if it is even.17.75 rounds to 17.8, but 17.65 rounds to 17.6. If the 5 is followed only by zeros, rule 3 is followed; if the 5 is followed by nonzeros, rule 1 is followed: 17.6500 rounds to 17.6, but 17.6513 rounds to 17.7

4. Be sure to carry two or more additional significant figures through a multistep calculation and round off only the final answer.

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Issues Concerning Significant Figures

graduated cylinder < buret ≤ pipet

numbers with no uncertainty

1000 mg = 1 g

60 min = 1 hr

These have as many significant digits as the calculation requires.

be sure to correlate with the problem

FIX function on some calculators

Electronic Calculators

Choice of Measuring Device

Exact Numbers

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Significant Figures and Rounding

PROBLEM: Perform the following calculations and round the answer to the correct number of significant figures:

PLAN: In (a) we subtract before we divide; for (b) we are using an exact number.

SOLUTION:

7.085 cm

16.3521 cm2 - 1.448 cm2(a)

11.55 cm3

4.80×104 mg(b)

1 g

1000 mg

7.085 cm

16.3521 cm2 - 1.448 cm2(a) =

7.085 cm

14.904 cm2

= 2.104 cm

11.55 cm3

4.80×104 mg

(b)

1 g

1000 mg=

48.0 g

11.55 cm3

= 4.16 g/ cm3

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정밀도와 정확도 Errors in Scientific Measurements

우연오차 (Random Error) - In the absence of systematic error, some values that are higher and some that are lower than the actual value.

정밀도 (Precision) -Refers to reproducibility or how close the measurements are to each other.

정확도 (Accuracy) -Refers to how close a measurement is to the real value.

계통오차 (Systematic Error) - Values that are either all higher or all lower than the actual value.

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정밀도와 정확도

4.917

Ran

dom

Err

or

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정밀도와 정확도

4.803

Sys

tem

atic

Err

or

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정밀도와 정확도

precise and accurate

precise but not accurate