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First Semester Exam Review. Physical Science. 2013 Exam Schedule: Tuesday, Dec. 17: 1 st & 3 rd Wednesday, December 18: 5 th & 7 th Thursday, December 19: 2 nd & 4 th Friday, December 20: 6 th & 8 th. Exams are 25% of the semester grade!

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First Semester Exam Review


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    1. First Semester Exam Review Physical Science 2013 Exam Schedule: Tuesday, Dec. 17: 1st & 3rd Wednesday, December 18: 5th & 7th Thursday, December 19: 2nd & 4th Friday, December 20: 6th & 8th Exams are 25% of the semester grade! Calculation: Sem. Grade = (0.75)(18-wk.avg.) + (0.25)(Exam) Exam = Desired Sem. Grade – (0.75)(18-wk. avg.) 0.25

    2. Experimental DesignYou will be given an experiment in which you must identify the following components. • Independent Variable – “You change it.” • Dependent Variable - “It changed.” • Constants – variables that remain the same • Control – does not receive the treatment

    3. Determining # of Sig. Fig. examples: Sig. Figs in Calculations: 1) 10000 = 1 sig. fig. 2) 10001 = 5 sig. figs 3) 10000. = 5 sig. figs 4) 10000.0 = 6 sig. figs 5) 0.0001 = 1 sig. fig. 6) 0.00010 = 2 sig. figs 7) 0.000101 = 3 sig. figs 8) 10.00010 = 7 sig. figs Addition and Subtraction: 5.687 g -2.31g = 3.377 g = 3.38 g (rounded to 2 decimal places) Multiplication and Division: (12.500 cm)(2.00 cm) = 25.0cm2 Significant Figures

    4. Scientific Notation • Write 59000000 m in scientific notation. 5.9 x 107 m This shows 2 sig. figs. If you needed to round to 1 sig. fig. : 6 x 107 m • Write 0.0000256 m in scientific notation. 2.56 x 10-5 m (rounded to 3 sig. figs) If you need to round to 2 sig figs: 2.6 x 10-5 m If you need to round 1 sig. figs: 3 x 10-5 m

    5. Scientific Notation Convert the following to Scientific Notation: • 1,870,000 m • 320,000,000 • 0.000 000 005 6 kg • 0.000 987

    6. Scientific Notation Convert the following to Scientific Notation: • 1,870,000 m = 1.87 x 106 m • 320,000,000 = 3.2 x 108 • 0.000 000 005 6 kg = 5.6 x 10-9 kg • 0.000 987 = 9.87 x 10-4

    7. Dimensional Analysis Kilo- Hecto- Deca- Basic unit (m, l, g) deci- centi- milli- Convert 45.000 km to m 45.000 km x 1000m = 45,000 m 1 km Convert 2000 cm to m 2000 cm x 1m = 20 m 100 cm Convert 500000 mm to km 500000 mm x _1 km = 0.5 km 106 mm

    8. Dimensional Analysis Problems • 3.50 kg to g • 0.50 cg to mg • 120.50 g to kg • 0.0500 kg to mg

    9. Check your work: a) 3.50 kg x 1000 g = 3500 g 1 kg b) 0.50 cg x 10 mg = 5 mg c) 120.50 g x _1kg___ = 0.1205 kg 1000 g d) 0.0500 kg x 106 mg = 5.00 x 104 mg 1 kg

    10. Volume Solid rectangular object:

    11. The volume of an irregularly-shaped object can be determined through the water-displacement method: What is the volume of the object?

    12. The volume of an irregularly-shaped object can be determined through the water-displacement method: What is the volume of the object? 75 mL – 50 mL = 25 mL The volume of the object is 25 mL. The object occupies 25 mL of volume.

    13. Density D = m v V = m D M = D x v Objects with a density greater than water’s density, which is 1.00 g/mL, will sink in water, while objects with a density less than water’s will float in water. Temperature affects density. Since matter tends to expand when heated and contract when cooled, the density decreases with increasing temperature, and increases with decreasing temperature.

    14. Density Problem: a) The density of silver (Ag) is 10.5 g/cm3. Find the mass of Ag that occupies 965 cm3 of space.

    15. Density Problem Solution: The density of silver (Ag) is 10.5 g/cm3. Find the mass of Ag that occupies 965 cm3 of space. D = m (solve for m) v D = 10.5g/cm3 m = D v m= ? m = (10.5 g/cm3)(965 cm3) v = 965 cm3 m = 10,132.5 g

    16. Density Problem: b) A 2.75 kg sample of a substance occupies a volume of 250.0 cm3. Find its density in g/cm3.

    17. Density Problem Solution: b) A 2.75 kg sample of a substance occupies a volume of 250.0 cm3. Find its density in g/cm3. D = m D = ? v m = 2.75 kg D = 2750 g 2.75 kg x 1000g = 2750 g 250.0 cm3 1 kg v = 250.0 cm3 D = 11g/cm3

    18. Density Problem: c) Under certain conditions, oxygen gas (O2) has a density of 0.00134 g/mL. Find the volume occupied by 250.0 g of O2under the same conditions.

    19. Density Problem Solution: c) Under certain conditions, oxygen gas (O2) has a density of 0.00134 g/mL. Find the volume occupied by 250.0 g of O2under the same conditions. v = m D D = 0.00134 g/mL m = 250.0 g v = 250.0 g v = ? 0.00134 g/mL v = 18,656 mL

    20. Kinetic Theory of Matter and States of Matter Kinetic Theory: all matter is made up of tiny particles in constant motion. More motion among particles (increase in KE), temperature increases, and the changes in state of matter occur. Temperature is the measure of the average kinetic energy (KE) of the particles that make up matter.

    21. Boyle’s Law When temperature is held constant, volume and pressure are inversely proportional. P1V1 = P2V2 At constant temperature, as the volume decreases, the particle speed increases, so the number of collisions with the walls of the container increases, therefore the pressure also increases. As volume increases, particle motion slows, collisions with the walls of the container decrease, & pressure decreases.

    22. Charles’ Law • When pressure is held constant, volume and temperature and are directly proportional. • V1 = V2 T1 T2 As particle speed increases, temperature increases, and the particles expand, so the volume increases.

    23. Chemical or Physical? • boiling of water • bursting of a balloon • crumpling paper • burning of gasoline • rotting of an egg • exploding fireworks

    24. Chemical or Physical? • boiling of water P • bursting of a balloon P • crumpling paper P • burning of gasoline C • rotting of an egg C • exploding fireworks C

    25. Chemical or Physical? • freezing of water • evaporation of gasoline • rusting a nail • sawing of wood • crushing a can • toasting a marshmallow

    26. Chemical or Physical? • freezing of water P • evaporation of gasoline P • rusting a nail C • sawing of wood P • crushing a can P • toasting a marshmallow C

    27. According to the law of conservation of mass, how does the mass of the products in a chemical reaction compare to the mass of the reactants? a. There is no relationship. b. The mass of the products is greater. c. The mass of the reactants is greater. d. The masses are equal.

    28. Ten Signs of Chemical Change • Bubbles of gas appear. (double replacement rxn.) • A precipitate forms. (a solid falls out of solution when 2 aqueous solutions react – usually in double replacement reactions) • A color change occurs. • The temperature changes. • Light is emitted.

    29. Ten Signs of Chemical Change (cont.) • A change in volume occurs. • A change in electrical conductivity occurs. • A change in melting point or boiling point occurs. • A change in odor or taste occurs. • A change in any distinctive chemical or physical property occurs.

    30. Determine the number of protons, neutrons and electrons in a given element. Ex. Uranium-238

    31. Determine the number of protons, neutrons and electrons in a given element. Ex. Uranium-238 92 protons, 146 neutrons, 92 electrons

    32. Calculate the average atomic mass of silver if 51.48% of its atoms have a mass of 106.905 amu and 48.17% of its atoms have a mass of 108.905 amu. (0.5148)(106.905 amu) = 55.034 amu (0.4817)(108.905 amu) = +52.460 amu 107.494 amu

    33. At room temperature, most metals are ____. • radioactive • gases • liquids • solids

    34. Group 1 elements are called the ________________________. What makes this group highly reactive?

    35. Group 1 elements are called the alkali metals What makes this group highly reactive? The alkali metals are highly reactive because they all have one valence electron. An atom needs a full valence shell, like a Noble Gas, to be stable and unreactive.

    36. Why is hydrogen is grouped with the alkali metals?

    37. Why is hydrogen is grouped with the alkali metals? Hydrogen has one valence electron, as do the alkali metals (group 1).

    38. What do elements in the same group have in common? What do elements in the same period have in common?

    39. What do elements in the same group have in common? same # of valence electrons What do elements in the same period have in common? same # of energy levels

    40. What is the relationship between the elements sodium and potassium?

    41. What is the relationship between the elements sodium and potassium? They are both alkali metals in group 1 and they both have 1 valence electron.

    42. Which statement is true of the elements sodium and potassium? a. They are polyatomic ions. b. They are compounds. c. They are nonmetals. d. They are metals.

    43. The properties of sodium and potassium are: a.quite different because they have different electron arrangements b.quite different because they have similar electron arrangements c.quite similar because they have different electron arrangements d.quite similar because they have similar electron arrangements

    44. 20. Which of the statements below is true of the elements sodium and potassium? a. they form the compound NaK b. they have very different chemical properties c. belong to the same period d. belong to the same group (family)

    45. Why are the atomic masses of the elements not whole numbers?

    46. Why are the atomic masses of the elements not whole numbers? Each element has various isotopes that are present in nature, and the atomic mass is a weighted average of all of the isotopes that occur naturally. Atomic masses are calculated using the percent abundance and mass of each of its isotopes.

    47. Isotopes • Atoms of the same element that have different numbers of neutrons, and thus, different masses. • All Isotopes of carbon all have 6 protons. • Carbon -12 = 6 protons + 6 neutrons • Carbon -13 = 6 protons + 7 neutrons • Carbon -14 = 6 protons + 8 neutrons

    48. Bohr Model K = 2 e- L = 8 e- M = 18 e- N = 32 e- Identify this element. A: Boron Know the Bohr models of elements 1-20.

    49. Lewis Dot Diagrams – show only valence electrons Be able to determine the correct Lewis dot diagram for a given element. (Look @ group # for a hint).