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AP Chapter 1 Some Fundamental Concepts

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AP Chapter 1. Some Fundamental Concepts. “A Fine Lady is Grossly Misunderstood”. MATTER. Mass. Mass. Measurement of the amount of matter. ENERGY. Kinetic Energy. Potential Energy. POSITION. CONDITION. COMPOSITION. Law of Conservation of Matter. Law of Conservation of Energy. - PowerPoint PPT Presentation

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Page 1: AP Chapter 1

AP Chapter 1Some Fundamental

Concepts

Page 2: AP Chapter 1

“A Fine Lady is Grossly Misunderstood”

Page 3: AP Chapter 1

MATTER

Page 4: AP Chapter 1

Mass

Page 5: AP Chapter 1

Mass

• Measurement of the amount of matter.

Page 6: AP Chapter 1
Page 7: AP Chapter 1

ENERGY

Page 8: AP Chapter 1

Kinetic Energy

Page 9: AP Chapter 1

Potential Energy

Page 10: AP Chapter 1

POSITION

Page 11: AP Chapter 1

CONDITION

Page 12: AP Chapter 1

COMPOSITION

Page 13: AP Chapter 1

Law of Conservation of Matter

Page 14: AP Chapter 1

Law of Conservation of Energy

Page 15: AP Chapter 1

Conservation of Matter and Energy

Page 16: AP Chapter 1

Chemical Property

• Does it react and if so how?• A chemical property is any of a material's

properties that becomes evident during a chemical reaction; that is, any quality that can be established only by changing a substance's chemical identity.

Page 17: AP Chapter 1

Chemical Property

Page 18: AP Chapter 1

Physical Property

• Any property other than a description of reactivity or non-reactivity.

• A physical property is any aspect of an object or substance that can be measured or perceived without changing its identity.

Page 19: AP Chapter 1

Physical Property

Page 20: AP Chapter 1

Physical Property

Page 21: AP Chapter 1

Chemical Change

Page 22: AP Chapter 1

Physical Change

Page 23: AP Chapter 1

Mixture

Heterogeneous Homogeneous

Page 24: AP Chapter 1

Mixture

Heterogeneous Solution

Page 25: AP Chapter 1

Mixtures

• Can be separated by physical means.• Can vary in composition

Page 26: AP Chapter 1

Pure Substance

• Does not vary in composition.• Elements or Compounds.

Page 27: AP Chapter 1

Element

• a pure substance consisting of one type of atom.

• Cannot be decomposed chemically.

Page 28: AP Chapter 1

Compound

• A pure substance consisting of two or more different elements that can be separated into simpler substances by chemical reactions.

• Chemical compounds consist of a fixed ratio of atoms.

• H2O, NaCl

Page 29: AP Chapter 1

Atoms and Molecules

• The atom is a basic unit of matter consisting of a dense, central nucleus surrounded by a cloud of negatively charged electrons.

• The atom is the smallest particle of an element.

• A molecule consists of two or more atoms bonded together.

Page 30: AP Chapter 1

What is it?

Na

H2O2

S8

Page 31: AP Chapter 1

Democritus and John Dalton

Page 32: AP Chapter 1

HONClFIBr

Page 33: AP Chapter 1
Page 34: AP Chapter 1

Metric Prefixes

••

•••••

Page 35: AP Chapter 1

Scientific Notation

Page 36: AP Chapter 1

• (7.75 x 10-6 m) x (2.07 x 1019 m) =1.60425 x 1014

1.60 x 1014 m2

Page 37: AP Chapter 1

Factor – Label Method

• I do not plan on putting factor – label problems on this particular test.

• However you will be using factor – label method on many tests this year.

Page 38: AP Chapter 1

Convert 152 mL to dL

Page 39: AP Chapter 1

Convert 152 mL to dL

1.52 dL

Page 40: AP Chapter 1

Convert 1.38kg to cg

Page 41: AP Chapter 1

Convert 1.38kg to cg

138,000 cg

Page 42: AP Chapter 1

Convert 0.56 cm to m

Page 43: AP Chapter 1

Convert 0.56 cm to m

5600 m

Page 44: AP Chapter 1

The maximum dosage of a medication is 50,000 μg/day. An IV drip can be set in

mg/hr. What should the IV drip be set at?

Page 45: AP Chapter 1

The maximum dosage of a medication is 50,000 μg/day. An IV drip can be set in

mg/hr. What should the IV drip be set at?2.08 mg/hr

Page 46: AP Chapter 1

A geologist measures an advancing glacier at a rate of 2.5 mm/hr. What

is this rate in km/year?

Page 47: AP Chapter 1

A geologist measures an advancing glacier at a rate of 2.5 mm/hr. What

is this rate in km/year? 0.0219 km/year

Page 48: AP Chapter 1

Significant Digits

Page 49: AP Chapter 1

Rules for Significant Digits

Page 50: AP Chapter 1

Rules for Significant Digits

• All non-zero digits are significant.

Page 51: AP Chapter 1

Rules for Significant Digits

• All non-zero digits are significant.• “Trailing” zeros after the decimal point

ARE significant.

Page 52: AP Chapter 1

Rules for Significant Digits

• All non-zero digits are significant.• “Trailing” zeros after the decimal point

ARE significant.• Zeros between significant digits are

significant.

Page 53: AP Chapter 1

Rules for Significant Digits

• All non-zero digits are significant.• “Trailing” zeros after the decimal point

ARE significant.• Zeros between significant digits are

significant.• All other zeros are NOT significant

unless indicated to be so by having a bar placed over them.

Page 54: AP Chapter 1

How to Determine Significant Digits

Page 55: AP Chapter 1

How to Determine Significant Digits

•Underline the leftmost nonzero digit.

Page 56: AP Chapter 1

How to Determine Significant Digits

•Underline the leftmost nonzero digit.

•Use the rules for significant digits to determine the rightmost significant digit.

Page 57: AP Chapter 1

How to Determine Significant Digits

•Underline the leftmost nonzero digit.

•Use the rules for significant digits to determine the rightmost significant digit.

•Every digit in between the leftmost and rightmost significant digits are significant as well.

Page 58: AP Chapter 1

Counting or Exact NumbersCounting numbers: If there are 10 people in a

room there are not 9.5 or 10.76 people in the room. Counting numbers are exact.

Ones in Conversion Factors: 1 kilometer = 1000 meters.  Exactly 1 km is equal to exactly 1000m. The 1 is considered to be an exact number and so is the 1000.

Since Counting numbers are exact they have an infinite number of significant digits.

Page 59: AP Chapter 1

Determine the Significant Digits(Examples in Notebook)

• 70.12 L • 0.000800 mg • 82.003 µm • 27.0 km • 50 people • 1.002 cm • _ • 200 kg

• -270.8 ºC • 1000 mL • 42,729.00 cm • 225 beans • 99.294 dm • 0.06900 m • _• 3,200,000 µL

Page 60: AP Chapter 1

Determine the Significant Digits (Examples in Notebook)

• 70.12 L 4• 0.000800 mg 3• 82.003 µm 5• 27.0 km 3• 50 people infinite• 1.002 cm 4• _ • 200 kg 2

• -270.8 ºC 4• 1000 mL 1• 42,729.00 cm 7• 225 beans infinite• 99.294 dm 5• 0.06900 m 4• _• 3,200,000 µL 5

Page 61: AP Chapter 1

Math Operations with Significant Digits

• When multiplying and/or dividing your answer must reflect the smallest number of significant digits.

Page 62: AP Chapter 1

Math Operations with Significant Digits

• When multiplying and/or dividing your answer must reflect the smallest number of significant digits.

• (17.3 cm)(28 cm) = 484.4 cm2

Page 63: AP Chapter 1

Math Operations with Significant Digits

• When multiplying and/or dividing your answer must reflect the smallest number of significant digits.

• (17.3 cm)(28 cm) = 484.4 cm2 = 480 cm2

Page 64: AP Chapter 1

Math Operations with Significant Digits

• When multiplying and/or dividing your answer must reflect the smallest number of significant digits.

• (17.3 cm)(28 cm) = 484.4 cm2 = 480 cm2

• 708g ÷ 4.700ml = 150.63829 g/ml =

Page 65: AP Chapter 1

Math Operations with Significant Digits

• When multiplying and/or dividing your answer must reflect the smallest number of significant digits.

• (17.3 cm)(28 cm) = 484.4 cm2 = 480 cm2

• 708g ÷ 4.700ml = 150.63829 g/ml = 151g/ml

Page 66: AP Chapter 1

Addition and/or Subtraction reflects the fewest decimal places

decimal places.

• 24.6 cm − 17.01 cm = 7.59 cm

Page 67: AP Chapter 1

Addition and/or Subtraction reflects the fewest decimal places

decimal places.

• 24.6 cm − 17.01 cm = 7.59 cm = 7.6 cm

Page 68: AP Chapter 1

Addition and/or Subtraction reflects the fewest decimal places

decimal places.

• 24.6 cm − 17.01 cm = 7.59 cm = 7.6 cm

• 8.5g + 1.32g + 0.18g =

Page 69: AP Chapter 1

Addition and/or Subtraction reflects the fewest decimal places

decimal places.

• 24.6 cm − 17.01 cm = 7.59 cm = 7.6 cm

• 8.5g + 1.32g + 0.18g = 10g

Page 70: AP Chapter 1

Addition and/or Subtraction reflects the fewest decimal places

decimal places.

• 24.6 cm − 17.01 cm = 7.59 cm = 7.6 cm

• 8.5g + 1.32g + 0.18g = 10g = 10.0g

Page 71: AP Chapter 1

MASS WEIGHT

Meaning Measurement of the amount of matter in an object

Measurement of the gravitational force

Variable/constant constant variable

Measured with balance Scale

Units Grams, kilograms, etc.

Pounds, ounces, etc.

Page 72: AP Chapter 1

Density• Density is the mass per unit volume.• D = Mass ÷ Volume

÷

Page 73: AP Chapter 1

Solving For Variables

Page 74: AP Chapter 1

Common Volume Units

Page 75: AP Chapter 1

Illegal ivory is often detected on the basis of density. What is the density of the elephant tusk?

Mass = 2680g

Volume = 1450cm3

Page 76: AP Chapter 1

Illegal ivory is often detected on the basis of density. What is the density of the elephant tusk?

Mass = 2680g

Volume = 1450cm3

D = 2680g ÷ 1450cm3 = 1.8482g/cm3 = 1.85g/cm3

Page 77: AP Chapter 1

Calcium chloride is used as a deicer on roads in winter. Its density is 2.50g/cm3. What mass of calcium chloride will the truck hold if its volume is 4560L?

Page 78: AP Chapter 1

Calcium chloride is used as a deicer on roads in winter. Its density is 2.50g/cm3. What mass of calcium chloride will the truck hold if its volume is 4560L?4560L = 4,560,000cm3

Page 79: AP Chapter 1

Calcium chloride is used as a deicer on roads in winter. Its density is 2.50g/cm3. What mass of calcium chloride will the truck hold if its volume is 4560L?4560L = 4,560,000cm3

M = D x V = 2.50g/cm3 x 4,560,000cm3 = 11,400,000g

Page 80: AP Chapter 1

The density of methanol is 0.788g/cm3.

What is the minimum volume of a tank that can hold 795kg of methanol.

Page 81: AP Chapter 1

The density of methanol is 0.788g/cm3. What is the minimum volume of a tank that can hold 795kg of methanol.

795kg = 795,000g

Page 82: AP Chapter 1

The density of methanol is 0.788g/cm3. What is the minimum volume of a tank that can hold 795kg of methanol.

795kg = 795,000g

795,000g ÷ 0.788g/cm3 = 1,008,883cm3 = 1,010,000cm3

Page 83: AP Chapter 1

The density of methanol is 0.788g/cm3. What is the minimum volume of a tank that can hold 795kg of methanol.

795kg = 795,000g

795,000g ÷ 0.788g/cm3 = 1,008,883cm3 = 1,010,000cm3

= 1010L

Page 84: AP Chapter 1

• Temperature is a measurement of the “average kinetic energy” of the particles in a material.

Page 85: AP Chapter 1

Celsius and Kelvin Conversion +/- 273