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KF Chemistry 1st 6 weeks Exam Review Week 6 CCA September 26, 2014 Pure Substances vs. Mixtures • Pure substance: matter that has a fixed (constant) composition and unique properties. Contains only 1 type element or compound; homogeneous Mixture: Contains at least 2 PHYSICALLY combined compounds; can be homogeneous or heterogeneous Elements • pure substance that cannot be separated into simpler substance by physical or chemical means. Compounds Pure substance composed of two or more different elements joined by chemical bonds. – Made of elements in a specific ratio that is always the same – Has a chemical formula – Can only be separated by chemical means, not physically Mixtures • A combination of two or more pure substances that are not chemically combined. Zn + Cu • substances held together by physical forces, not chemical • No chemical change takes place • Each item retains its properties in the mixture • They can be separated physically 3 classes of MIXTURES Solution Colloid homogenous Examples Particle Type homogenous Suspension heterogenous salt water, Soot, fog, Muddy water, air mayonnaise Italian dressing ions, atoms Small Clusters Large Clusters small medium large Scatter Light? (TYNDALL EFFECT) No yes yes Settle while standing? No No yes Separate by filtration? No No yes Particle Size Mixtures vs. Compounds http://www.bbc.co.uk/schools/ks3bitesize/science/chemistry/elements_com_mix_6.shtml Properties of Matter (4D, Level 3) • • • Identify each of the following items as a mixture or a pure substance. - If a mixture, identify as homogeneous or heterogeneous. - If a pure substance, identify as a compound or an element. Sample Gold Water Italian Dressing Milk Cobalt Coffee Calcium Carbonate Cake Batter Type of Sample Type of Mixture or Pure Substance Physical Properties • Physical Property: Can be observed without changing the substance composition – Ex: Hardness, Color, Conductivity, Malleability, Melting Point, Boiling Point • Physical Change: Properties of the material may change, but the COMPOSITION does not Water – MP is 0°C Gallium – MP is 30°C Chemical Properties • Chemical Property: Can ONLY be observed by changing the substance composition – Ex: Burn, rot, rust, decompose, ferment, explode, corrode • Chemical Change: The COMPOSITION of matter ALWAYS changes • Also called a chemical reaction Properties of Matter (4A, Level 2) • Identify the following properties as physical or chemical. Property Temperature Bond Strength Calorie Content Mass Density Reactivity Length Melting Point Physical Property Chemical Property Properties of Matter (4A, level 3) • Identify the following changes as physical or chemical. Property A rock is crushed Condensation of water vapor Pancakes cook Salt is dissolved in water An apple is cut Food is digested Alcohol evaporates Ice melts Physical Change Chemical Change Extensive vs. Intensive Properties • Extensive Properties: Depends on amount • Ex: Volume • Intensive Properties: Depends on type of matter • Ex: Hardness Properties of Matter (4B, Level 3) • Identify the following properties as extensive or intensive. Property Temperature Bond Strength Calorie Content Mass Density Reactivity Length Melting Point Extensive Property Intensive Property States of Matter • Solid: • Definite Shape • Definite Volume • Incompressible • Liquid: • Indefinite Shape • Definite Volume • Not Easily Compressed • Gas: • Indefinite Shape • Indefinite Volume • Easily compressed States of Matter Solid Density Viscosity Compressibility Structure Shape Volume Movement Draw a PictureNanoscopic Eyes Liquid Gases The Law of Conservation of Mass Law of Conservation of Mass During a chemical change, matter is neither created nor destroyed. According to the law of conservation of mass, how much zinc was present in the zinc carbonate? A 40 g B 88 g C 104 g D 256 g (All practice problems are from TEA released TAKS Tests) In the procedure shown above, a calcium chloride solution is mixed with a sodium sulfate solution to create the products shown. Which of the following is illustrated by this activity? (A) The law of conservation of mass (B) The theory of thermal equilibrium (C) The law of conservation of momentum (D) The theory of covalent bonding Use SIGNIFICANT FIGURES to answer • • • • • All non-zero numbers are significant EX: 23.4 is 3 sig.figs., 19 is 2 sig.figs., etc. All zeros between numbers are significant— EX: 2003 is 4s.f., 2.07 is 3 s.f., 1.009 is 4 s.f. (captive zeros) All zeros to the right of numbers aren’t significant if there’s no decimal EX: 1200 is 2s.f., 1000 is 1. (trailing zeros) All zeros to the left of numbers aren’t significant if there’s no number in front EX: 0.0045 is 2s.f. (Leading zeros) All zeros after a number and a decimal are significant EX: 1.00 is 3 s.f. 230.0 is 4 s.f. When multiplying and dividing, use the least number of significant figures in your answer. When adding and subtracting, use the least number of decimal places in your answer. LOOK AT YOUR FORMULA SHEET FOR THE RULES! ! ! SIG FIGS-NOW YOU TRY IT 1. 22.4 x 1.2= 5. 561.25 + 105.2= 2. 1036 x 2.00= 6. 124.12 - 104.121= 3. 0.00686/19.00= 7. 343.2 x 510= 4. 1.200/13.686= 8. 100.25 + 68.750= Scientific Notation (6.02E23) Open ( Enter 6.02 2nd (Blue or Yellow button) EE ( , button above 7) Type in the exponent 23) Your screen looks like: (6.02E23) 0 6 2 3 . change these numbers from Standard Notation to Scientific Notation 1) 9872432 5) 356890 2) .0000345 6) 80345 3) .08376 7) 0.00000075 4) 5673 8) 1000 Use this example to help you learn how to find neutrons Example: Carbon (C) Atomic Number = 6 Mass Number = 12 #Protons = 6 #Electrons = 6 #Neutrons = (mass number) – (atomic number) = (12) – (6) =6 Element Name aluminum Symbol Atomic Number Al 13 Mass Number Number of Electrons Number of Protons Number of Neutrons 13 13 14 bismuth calcium 20 copper iodine 129 82 8 50 30 Isotopes of Carbon Naturally occurring carbon consists of three isotopes, 12C, 13C, and 14C. State the number of protons, neutrons, and electrons in each of these carbon atoms. 12C 6 13C 14C 6 6 #P _______ _______ _______ #N _______ _______ _______ #E _______ _______ _______ Average Atomic Mass Use this worked to example to practice the upcoming problems! 35Cl has atomic mass 34.97 amu (75.76%) and 37C has atomic mass 36.97 amu (24.24%). • Use atomic mass and relative abundance of each isotope to calculate the contribution of each isotope to the weighted average. 34.97 x 36.97 x 75.76 100 24.24 100 = = • Sum is atomic mass of Cl 26.49 amu + 8.961 amu 35.45 amu Average Atomic Mass of Magnesium Isotopes Mass of Isotope Abundance 24Mg = 24.0 amu 78.70% = ______ 25Mg 26Mg = 25.0 amu 10.13% = ______ = 26.0 amu 11.17% = ______ Find the Average Atomic mass of Magnesium 28 Average Atomic Mass of Boron Isotopes Mass of Isotope Abundance 10B = 10.013 amu 19.8% = ______ 11B = 11.009 amu 80.02% = ______ Find the Average Atomic mass of Boron 29 Contrast & Compare Fission vs. Fusion Fission Fusion ALPHA & BETA DECAY (12B, Level 3) What is the alpha particle?__________________________________________ What is the beta particle?___________________________________________ Define gamma rays________________________________________________