King henry died by drinking chocolate milk worksheet

Presentation on theme: “Metric Conversions king henry died by drinking chocolate milk”— Presentation transcript:

1 Metric Conversions king henry died by drinking chocolate milkk h da b d c m king henry died by drinking chocolate milk k h da base d c m

2 Metric System ConversionsExample: cm= _____________ km

3 TRY THIS: 1000 mg = ______ g 1

4 TRY THIS: 0.15 m = __________ mm 150

5 TRY THIS: kg = __________ cg 6400

6 TRY THIS: 89 ds = __________ das 0.89

7 What is density? m ÷ ALWAYS REMEMBER UNITS! × D V

8 Let’s try a density problem togetherFrank’s paper clip has a mass of 9 g and a volume of 3 mL. What is its density? Density = Mass  Volume Density = 9 g  3 mL = 3 g/mL 9 g = 3 g 3 mL mL m ÷ × D V

9 Let’s try a density problem togetherFrank’s eraser has a density of 6 g/mL and a volume of 2 mL. What is its mass? Mass = Density × Volume Mass = 6g/mL × 2mL = 12g 6 g x 2 mL = 12 g mL m ÷ × D V

10 Let’s try a density problem togetherFrank’s shot-put has a mass of 500 g and a density of 100 g/mL. What is its volume? Volume = Mass  Density Volume = 500 g  100 g/mL = 5 mL m ÷ × D V

11 Significant Figures 56,400,000 Is there a decimal? PACIFIC (Present)ATLANTIC (Absent) Start from LEFT & count all #s from first nonzero Start from RIGHT & count all #s from first nonzero 56,400,000 4 sig figs 3 sig figs

12 You Try! How many sig figs in the following: Examples: a) 1001 kmb) m c) 129,870 m d) km e) L f) 6.02 x 1023 atoms g) 20,000 cm h) g Number of Significant Figures: a) 4 b) 4 c) 5 d) 1 e) 4 f) 3 g) 1 h) 2

13 CALCULATIONS WITH SIG FIGS SummaryMultiplying or dividing: round to the measurement with the smallest number of significant figures. 2.6 cm x 3.78 cm = cm2 Adding or subtracting: round the answer to the smallest number of decimal places. 165.5 cm + 8 cm cm = cm = 9.8 cm2 = 178 cm

14 PARTS OF THE ATOM Nucleus: Contains protons and neutrons VERY tinyContains all the atom’s mass Electron Cloud: Contains electrons Makes up the atom’s volume No mass

15 Isotope Naming

16 Electron Configurations7p

17 Electron ConfigurationsElectrons fill sublevels in the order on the chart. Read it like a book – from left to right. Every sublevel to the left and above is filled. Boron (B) 1s22s22p1 Oxygen (O) 1s22s22p4 7p

18 Electron ConfigurationsElectrons fill sublevels in the order on the chart. Read it like a book – from left to right. Every sublevel to the left and above is filled. Silicon (Si) 1s22s22p63s23p2 Gallium (Ga) 1s22s22p63s23p64s23d104p1 7p

19 2A: Alkaline Earth MetalsPeriodic Table ROWS: Periods COLUMNS: Families or Groups 8A: Noble Gases 1A: Alkali Metals 7A: Halogens 2A: Alkaline Earth Metals

20 Trends in ElectronegativityAbility of an atom to attract electrons in another atom * Fluorine has the highest electronegativity, Noble gases have ZERO electronegativity

21 Trends in Ionization EnergyThe energy required to remove an electron from an atom

22 Trends in Atomic Size Atomic Size The distance between the nucleus and the outer edge of the electron cloud

23 Ionic Bonds – electrons transferredMetals lose electrons and Nonmetals gain electrons to form an ionic bond. Properties of Ionic Compounds Crystal lattice structure High melting points Conducts electricity when melted or dissolved in water

24 Predicting Ionic ChargesGroup 1A: Lose 1 electron to form 1+ ions H+ Li+ Na+ K+ Rb+

25 Predicting Ionic ChargesGroup 2A: Loses 2 electrons to form 2+ ions Be2+ Mg2+ Ca2+ Sr2+ Ba2+

26 Predicting Ionic ChargesLoses 3 electrons to form 3+ ions Group 3A: B3+ Al3+ Ga3+

27 Predicting Ionic ChargesNeither! Group 4A elements rarely form ions EXCEPTION: Sn and Pb!! Treat like transition metals Do they lose 4 electrons or gain 4 electrons? Group 4A:

28 Predicting Ionic ChargesNitride Gains 3 electrons to form 3- ions Group 5A: P3- Phosphide As3- Arsenide

29 Predicting Ionic ChargesOxide Gains 2 electrons to form 2- ions Group 6A: S2- Sulfide Se2- Selenide

30 Predicting Ionic ChargesGains 1 electron to form 1- ions Group 7A: F1- Fluoride Br1- Bromide Cl1- Chloride I1- Iodide

31 Predicting Ionic ChargesStable noble gases do not form ions! Group 8A:

32 Covalent Bonds – electrons sharedBond between two nonmetals. Properties of Covalent Compounds Lower melting points Does not conduct electricity when dissolved in water

33 Metallic Bonds – “sea” of electronsBond between two metals. Bond is the attraction between electrons and positive nucleus. Properties of Metallic Compounds Malleable (pounded into shapes) Ductile (pulled into wires) Conducts electricity as a solid.

34 NAMING COMPOUNDS Use roman numerals for transition metalsIONIC COMPOUNDS: Name the CATION then the ANION Use roman numerals for transition metals Example: Cu(NO3)2 copper (II) nitrate COVALENT COMPOUNDS (molecules): Use prefixes to match the subscript. Second atom name ends in “ide” Example: N2O4 dinitrogen tetroxide

35 WRITING FORMULAS Example: iron (II) chloride Fe2+ Cl- Fe 1 Cl 2 FeCl2IONIC COMPOUNDS: USE THE CROSS RULE: Example: iron (II) chloride Fe2+ Cl- Fe 1 Cl 2 FeCl2 COVALENT COMPOUNDS (molecules): Use prefixes to identify the subscripts. Example: trichloro tetrafluoride Cl3F4

36 Lewis Structure of MoleculesAdd up all the valence electrons. This is the number of dots in the final structure. Place least electronegative element in center. (Never H) Place two electrons between central and outer atoms Fill up outer atom octets. Fill up central atom octet. Make double or triple bonds if necessary so all atoms meet the octet rule. Br2 2(7) = 14

37 Lewis Structure of MoleculesAdd up all the valence electrons. This is the number of dots in the final structure. Place least electronegative element in center. (Never H) Place two electrons between central and outer atoms Fill up outer atom octets. Fill up central atom octet. Make double or triple bonds if necessary so all atoms meet the octet rule. H2O 6 + 2(1) = 8

38 Lewis Structure of MoleculesAdd up all the valence electrons. This is the number of dots in the final structure. Place least electronegative element in center. (Never H) Place two electrons between central and outer atoms Fill up outer atom octets. Fill up central atom octet. Make double or triple bonds if necessary so all atoms meet the octet rule. CO2 4 + 2(6) = 16

39 Balancing Reactions Balance reactions so that the same number of atoms are on both sides. ONLY change coefficients, NOT subscripts. _____C3H ____O2  ____CO ____H2O ____ Pb(NO3) ____ HCl  ____ PbCl ____ HNO3 Balance in the following order Polyatomic ions (as a group, only if they appear on both sides) Metals Nonmetals Hydrogen 5.Oxygen

40 Calculating Percent Composition ExampleDetermine the percent composition of each element in Mg(NO3)2 Determine the contribution of each element Molar mass On a per mol basis

41 Mole Conversions 1 mole = 6.02  1023 particles 1 mole = _______ g*particles could be molecules, ions, atoms, formula units 1 mole = _______ g *1 mole equals the molar mass of the atom or compound 1 mole = 22.4 L * for gases only at STP (Standard Temperature (0˚C and Standard Pressure (1 atm))

42 Mole Conversions a. What is the number of moles of CaS in 120 g?1 mole =72.2 g CaS = 1.7 moles 120 g CaS 1 mole 72.2 g CaS b. What is the mass in grams of 1.81 x 1023 molecules of CO2? 1 mole =6.02  1023 particles, 1 mole =44.0 g CO2 = 13.2 g 1.81 x molecules CO2 1 mole 72.2 g CO2 6.02 x molecules

43 Mole Conversions c. How many molecules of CO2 are in 0.50 moles?1 mole = 6.02  1023 particles = 3.01×1023 molecules 0.50 moles CO2 1 mole 6.02 x molecules d. How many liters do 7.87×1023 molecules of H2S occupy at STP? 1 mole =6.02  1023 particles, 1 mole =22.4 L at STP = 29.3 L 7.87 x molecules CO2 1 mole 22.4 L 6.02 x molecules

44 Mole Conversions e. What is the number of liters that 27 g of oxygen gas occupies at STP? 1 mole =32.0 g O2, 1 mole =22.4 L at STP = 18.9 L 27 g O2 1 mole 22.4 L 32 g O2

45 Stoichiometry CalculationsZn(s) + 2 HCl(aq)  ZnCl2(aq) + H2(g) 38a. How many moles of H2 if 4 moles of HCl react? CONVERSION FACTOR: mole ratio: ___ mole _____ = ___ mole ______ 2 mol HCl = 1 mol H2 4 mol HCl 1 mol H2 = 2 mol H2 2 mol HCl

46 Stoichiometry CalculationsZn(s) + 2 HCl(aq)  ZnCl2(aq) + H2(g) 38b. How many grams of Zn if 3 moles of ZnCl2 are formed? CONVERSION FACTORS: mole ratio: ___ mole _____ = ___ mole ______ molar mass: 1 mole ____ = _____ g ______ 1 mol Zn = 1 mol ZnCl2 1 mol Zn = g Zn 3 mol ZnCl2 1 mol Zn 65.34 g Zn = g Zn 1 mol ZnCl2

47 Stoichiometry CalculationsZn(s) + 2 HCl(aq)  ZnCl2(aq) + H2(g) 38c. How many grams of HCl if 6.87 g Zn react? CONVERSION FACTORS: mole ratio: ___ mole _____ = ___ mole ______ molar mass: 1 mole ____ = _____ g ______ 1 mol Zn = 2 mol HCl 1 mol Zn = g Zn 1 mol HCl = g HCl 6.87 g Zn 1 mol Zn 2 mol HCl 36.45 g HCl = 7.7 g HCl 65.34 g Zn 1 mol HCl

48 Stoichiometry CalculationsMg + 2 HNO3  Mg(NO3)2 + H2 39a. If 40 g Mg react, how many grams H2 form? CONVERSION FACTORS: mole ratio: ___ mole _____ = ___ mole ______ molar mass: 1 mole ____ = _____ g ______ 1 mol Mg = 1 mol H2 1 mol Mg = 24.3 g Zn 1 mol H2 = 2.0 g H2 40 g Mg 1 mol Mg 1 mol H2 2.0 g H2 = 3.3 g H2 24.3 g Mg

49 Stoichiometry CalculationsMg + 2 HNO3  Mg(NO3)2 + H2 39b. If 1.7 g H2 was ACTUALLY produced, what was the percent yield? Actual Yield: Given in problem (1.7 g H2) Theoretical Yield: Calculated in part a (3.3 g H2)

50 Stoichiometry CalculationsMg + 2 HCl  ZnCl2 + H2 40a. If 40 g Zn and 40 g HCl react, how many grams H2 form? This is a limiting reactant problem. Do 2 calculations. CONVERSION FACTORS: mole ratio: ___ mole _____ = ___ mole ______ molar mass: 1 mole ____ = _____ g ______ First Calculation: 1 mol Zn = 1 mol H2 1 mol Zn = 65.4 g Zn 1 mol H2 = 2.0 g H2 40 g Zn 1 mol Zn 1 mol H2 2.0 g H2 = g H2 65.4 g Zn

51 Stoichiometry CalculationsMg + 2 HCl  ZnCl2 + H2 40a. If 40 g Zn and 40 g HCl react, how many grams H2 form? This is a limiting reactant problem. Do 2 calculations. CONVERSION FACTORS: mole ratio: ___ mole _____ = ___ mole ______ molar mass: 1 mole ____ = _____ g ______ Second Calculation: 2 mol HCl = 1 mol H2 1 mol HCl = 35.5 g HCl 1 mol H2 = 2.0 g H2 40 g HCl 1 mol HCl 1 mol H2 2.0 g H2 = g H2 36.5 g HCl

52 Stoichiometry CalculationsMg + 2 HCl  ZnCl2 + H2 40b. If 40 g Zn and 40 g HCl react, how many grams H2 form? What is the limiting reactant? First Calculation (Zn is reactant): Produced 1.22 g H2. Second Calculation (HCl is reactant): Produced 1.10 g H2. HCl produced the LEAST amount of product so it is the limiting reactant. You can only produce 1.10 g H2.

Top 4 king henry died by drinking chocolate milk worksheet