Chapter 1 review

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Chapter 1 review

  1. 1. Chapter 1:Matter – Its Properties and Measurement
  2. 2. The Scientific MethodObservation Hypothesis Experiment Established Theory modify modify Experiment Theory CHM1311 Matter 2
  3. 3. Properties of Matter• Matter: occupies space and displays mass and inertia• Composition: relative proportions of the components of a sample of matter ex. water is 11.19% H and 88.81% O by massCHM1311 Matter 3
  4. 4. Properties of Matter•  Physical property: –  a property that can be measured or observed without changing the matter’s composition•  Chemical property: –  a property that comes with observing a change in chemical composition•  Extensive property: depends on the quantity of matter present•  Intensive property: does NOT depend on the quantity of matter presentCHM1311 Matter 4
  5. 5. Elements and Compounds•  Element: –  cannot be decomposed into a simpler substance through chemical processes; distinguished by the unit of the atom•  Compound: –  a substance made from the atoms of two or more elements bonded chemically in defined proportions•  Compounds can only be decomposed into their respective elements via chemical processesCHM1311 Matter 5
  6. 6. Pure Substances and Mixtures•  A pure substance –  A substance with a fixed and uniform composition and distinct properties (ex: pure water)•  A mixture: –  A combination of two or more pure substances which can vary in composition and properties a) homogeneous: ex: salt water b) heterogeneous: ex: oil and water•  It is possible to separate mixtures through physical porcessesCHM1311 Matter 6
  7. 7. Pure Substances and MixturesCHM1311 Matter 7
  8. 8. Measuring Matteran observed measurement not followed by a unit is meaningless! The seven base SI units are:CHM1311 Matter 8
  9. 9. Who cares about units anyway? •  Mars Climate Orbiter •  probe sent by NASA to Mars to study its weather •  the $168 million probe was destroyed in 1999 after entering the Martian atmosphere •  desired altitude: 140-150 km •  altitude attained: 57 km •  investigation revealed that the on board computer used SI units, while the computers on Earth were using BE unitsCHM1311 Matter 9
  10. 10. SI Prefixes Value Prefix Symbol 1012 tera- T 109 giga- G 106 mega- M 103 kilo- k 102 hecto- h 101 deca- da 10-1 deci- d 10-2 centi- c 10-3 milli- m 10-6 micro- µ 10-9 nano- n 10-12 pico- p CHM1311 Matter 10
  11. 11. Mass versus Weight •  mass: –  measures the quantity of matter in an object •  weight: –  the force of gravity on an objectThe kilogram (kg) is the official SI unit, but we will most oftenuse the gram (g): 1 kg = 1000 g CHM1311 Matter 11
  12. 12. Volume•  volume: the size of a cube (i.e., m3)•  we will most often use the litre (L) for measuring volumes 1000 mL = 1 L 1000 L = 1 m3CHM1311 Matter 12
  13. 13. Temperature•  the SI unit is the kelvin (K)•  absolute zero temperature is 0 K or -273.15oC•  the freezing point of water is 273.15 K or 0oC•  the boiling point of water is 373.15 K or 100oC always use the temperature in K in your calculations!CHM1311 Matter 13
  14. 14. Accuracy and Precision• Accuracy: –  indicates how close a measured value is to the actual (or accepted) value• Precision: –  indicates the degree of reproducibility of a measured quantityCHM1311 Matter 14
  15. 15. Accuracy and Precision accurate not accurate, and precise but precise not precise, neither accurate but accurate nor precise• accurate measurements are usually precise, but a systematic error will produce values which are precise but not accurate CHM1311 Matter 15
  16. 16. Scientific measurements•  Scientific notation: N x 10n 6.022 045 x 1023 instead of 602 204 500 000 000 000 000 000 N=6.022 045 and n=23•  Significant figures –  digits considered to be significant in the calculation or measurement of a quantity this balance is precise to ±0.01 kg an object that has a mass of 6.732 kg will give a measurement of 6.73 ± 0.01 kg __________CHM1311 Matter 16
  17. 17. Rules for sig figs…•  all non zero digits are significant 4 6.732 kg has __ significant figures•  zeros between two sig figs are also significant 5 6.0061 kg has __ significant figures•  zeros to the left of a sig fig are not significant 3 0.0502 kg has __ significant figuresCHM1311 Matter 17
  18. 18. Rules for sig figs…•  if the value is greater than 1, all zeros to the right of the decimal point are significant 4 6.000 kg has __ significant figures•  when converting to scientific notation, it may sometimes be ambiguous whether hanging zeros are significant or not4500 kg could be 4.5 x 103, 4.50 x 103, or 4.500 x 103 kg therefore 4500 kg could have 2, 3, or 4 sig figs!CHM1311 Matter 18
  19. 19. Rules for sig figs •  a whole number with perfect precision has an infinite number of significant figuresif we determine the average of 3 trials, we can assume it s 3.000 000 000 … trials CHM1311 Matter 19
  20. 20. Rules for sig figs…•  addition/subtraction: –  the answer must have the same number of sig figs after the decimal as the element of the calculation with the least number of sig figs after the decimal point + 0.2225 + 2.73 + 2.06 + 0.321 ! 1.1 3.27rounded to ______ 1.0 rounded to ______ + 3.2735 + 0.96 CHM1311 Matter 20
  21. 21. Rules for sig figs…•  multiplication/division: –  the answer must have the same number of sig figs as the element of the calculation with the least number of sig figs2.2 x 3.7845 = 8.32590 8.3 rounded to ______3.76 / 4.236 = 0.8876298 0.888 rounded to ______(2.27 x 7.324) / 3.3 = 5.0380 5.0 rounded to ______ CHM1311 Matter 21
  22. 22. Rules for sig figs…•  Logarithms –  the answer must have the same number of sig figs as the log elementlog(957) = 2.980911... = 2.98 ??= log(9.57 x 102)= log(9.57) + log(102)= 0.980911... + 2.00000... = 2.981 CHM1311 Matter 22
  23. 23. In summary…• on tests and the final exam,• on homework assignments,CHM1311 Matter 23
  24. 24. Conversion Factors •  to convert a quantity from one unit to another, we need to use a conversion factor Dimensional AnalysisQuantity with Quantity with Conversion desired unit = given unit X factor CHM1311 Matter 24
  25. 25. Example 1: Conversion factors Convert 345.3 cm into metres.Solution 100 cm = 1 m 1m ? m = 345.3 cm x = 3.453 m 100 cm •  N.B. the number of sig figs in the conversion factor is infinite ! CHM1311 Matter 25
  26. 26. Example 2: Conversion factorsThe density of the lightest metal, lithium (Li) is 5.34 x 102 kg/m3.Convert this value to g/cm3.Solution 1000 g = 1 kg 100 cm = 1 m 3 2 kg 1000 g 1 m %3 ? g/cm = 5.34 x 10 3 • •$ = m 1 kg # 100 cm 2kg 1000 g 1 m3 5.34 x 10 • • 6 = 0.534 g/cm3 m3 1 kg 10 cm3 CHM1311 Matter 26
  27. 27. Density mass•  density = ρ = volume•  density is a intensive property and is a very useful conversion factor !•  the SI unit is kg/m3, but we will most often use g/cm3 for solids and liquids and g/L for gases 1 g/cm3 = 1 g/mL = 1000 kg/m3 1 g/L = 0.001 g/mLCHM1311 Matter 27
  28. 28. Example : Using density A piece of platinum has a density of 21.5 g/cm3 and a volume of 4.49 cm3. What is its mass?Solution m = ! m = •V V 3 21.5 g Pt ? g Pt = 4.49 cm Pt • 3 = 96.5 g Pt cm Pt CHM1311 Matter 28
  29. 29. Percent Composition• number of parts of a component in 100 parts of the whole –  ex. 10% means 10 parts x per 100 parts of the whole• IMPORTANT: must be defined by a unit! –  ex. a rock contains 3.5% gold by mass means 3.5 g of gold per 100 g of rock –  ex. a bottle of wine contains 10.7% alcohol by volume means 10.7 mL of alcohol per 100 mL of wineCHM1311 Matter 29
  30. 30. Percent Composition• when expressed as a conversion factor, the numerator and denominator must have the SAME UNITS –  ex. a rock contains 3.5% gold by mass 3.5 g gold 3.5 kg gold 3.5 oz gold = = 100 g rock 100 kg rock 100 oz rock –  ex. a bottle of wine contains 10.7% alcohol by volume ! 10.7 mL EtOH 10.7 L EtOH 10.7 tbsp EtOH = = 100 mL wine 100 L wine 100 tbsp wineCHM1311 Matter 30
  31. 31. Example : Using percent composition A solution of sucrose in water is 28.0% sucrose by mass and has a density of 1.118 g/mL. What mass of sucrose (in grams) is in 3.50 L of this solution? Solution 1000 mL 1.118 g solution 28.0 g sucrose? g sucrose = 3.50 L solution • • • 1L mL solution 100 g solution= 1.10x10 3 g sucrose CHM1311 Matter 31
  32. 32. Chapter 1: Key Concepts1.  the forms and properties of matter2.  SI units and prefixes3.  accuracy vs. precision4.  significant figures5.  scientific notation6.  conversion factorsCHM1311 Matter 32
  33. 33. Chapter 1: Suggested Problems 13, 14, 19, 20, 23, 27, 31, 45, 47, 51, 63, 65, 80, 81, 89CHM1311 Matter 33

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