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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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AP Chapter 1Some 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
Conservation of Matter and Energy
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.
Chemical Property
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.
Physical Property
Physical Property
Chemical Change
Physical Change
Mixture
Heterogeneous Homogeneous
Mixture
Heterogeneous Solution
Mixtures
• Can be separated by physical means.• Can vary in composition
Pure Substance
• Does not vary in composition.• Elements or Compounds.
Element
• a pure substance consisting of one type of atom.
• Cannot be decomposed chemically.
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
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.
What is it?
Na
H2O2
S8
Democritus and John Dalton
HONClFIBr
Metric Prefixes
••
•••••
Scientific Notation
• (7.75 x 10-6 m) x (2.07 x 1019 m) =1.60425 x 1014
1.60 x 1014 m2
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.
Convert 152 mL to dL
Convert 152 mL to dL
1.52 dL
Convert 1.38kg to cg
Convert 1.38kg to cg
138,000 cg
Convert 0.56 cm to m
Convert 0.56 cm to m
5600 m
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?
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
A geologist measures an advancing glacier at a rate of 2.5 mm/hr. What
is this rate in km/year?
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
Significant Digits
Rules for Significant Digits
Rules for Significant Digits
• All non-zero digits are significant.
Rules for Significant Digits
• All non-zero digits are significant.• “Trailing” zeros after the decimal point
ARE significant.
Rules for Significant Digits
• All non-zero digits are significant.• “Trailing” zeros after the decimal point
ARE significant.• Zeros between significant digits are
significant.
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.
How to Determine Significant Digits
How to Determine Significant Digits
•Underline the leftmost nonzero digit.
How to Determine Significant Digits
•Underline the leftmost nonzero digit.
•Use the rules for significant digits to determine the rightmost significant digit.
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.
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.
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
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
Math Operations with Significant Digits
• When multiplying and/or dividing your answer must reflect the smallest number of significant digits.
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
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
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 =
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
Addition and/or Subtraction reflects the fewest decimal places
decimal places.
• 24.6 cm − 17.01 cm = 7.59 cm
Addition and/or Subtraction reflects the fewest decimal places
decimal places.
• 24.6 cm − 17.01 cm = 7.59 cm = 7.6 cm
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 =
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
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
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.
Density• Density is the mass per unit volume.• D = Mass ÷ Volume
÷
Solving For Variables
Common Volume Units
Illegal ivory is often detected on the basis of density. What is the density of the elephant tusk?
Mass = 2680g
Volume = 1450cm3
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
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?
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
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
The density of methanol is 0.788g/cm3.
What is the minimum volume of a tank that can hold 795kg of methanol.
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
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
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
• Temperature is a measurement of the “average kinetic energy” of the particles in a material.
Celsius and Kelvin Conversion +/- 273