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 Breathing in Helium will make your voice very high pitched  Breathing in any of the noble gases below neon will make your voice become very low… but.

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Presentation on theme: " Breathing in Helium will make your voice very high pitched  Breathing in any of the noble gases below neon will make your voice become very low… but."— Presentation transcript:

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2  Breathing in Helium will make your voice very high pitched  Breathing in any of the noble gases below neon will make your voice become very low… but can also make you suffocate!

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5  Indications of a chemical rxn: 1. Production of heat and light. 2. Production of a gas. 3. Formation of a precipitate 4. Color change. Precipitate: a solid that is produced as a result of a chemical rxn in solution and that separates from the solution 8-3

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7 Eggs+ Milk +Yeast Cake Reactants Products

8 2Na ( l ) + Cl 2 (g) 2NaCl Reactants Products

9 2Na ( l ) + Cl 2 (g) 2NaCl (s) element coefficient Diatomic molecule Phase of matter compound S= Solid l = Liquid G = Gas aq = aqueous (dissolved in water) The arrow = “Yields” or “Forms”

10 H O N Cl Br I F -All come in pairs

11 Sodium = Na (Element) Oxygen= O 2 (Diatomic) Sodium Chloride= NaCl (Compound)

12  You must have the same number of each element on both sides of the equation! IT’S A BACK AND FORTH GAME!!!  Let’s see what that looks like with our first example

13 Na + Cl 2 NaCl USE COEFFICIENTS!!!!! -DON’T TOUCH SUBSCRIPTS

14 Na + Cl 2 NaCl Na = 1 Cl = 2 Na = 1 Cl = 1 PROBLEM!!! Number of atoms We need to balance this reaction to make sure the number of chlorine atoms and sodium atoms on each side are equal… Using coefficients!

15 2Na + Cl 2 2NaCl Na = 2 Cl = 2 Na = 2 Cl = 2 Number of atoms Notice that I have the same number of atoms on both sides

16 Let’s do that again! H 2 + O 2 H2OH2O

17 H2OH2O H = 2 O = 2 H = 2 O = 1 Number of atoms PROBLEM!!! We know that we need at least 2 O in the products…

18 H 2 + O 2 2H2O2H2O

19 2H2O2H2O H = 2 O = 2 H = 4 O = 2 Number of atoms Great now what do we do?! Need more Hydrogen in the reactants PROBLEM!!!

20 2H 2 + O 2 2H2O2H2O H = 4 O = 2 H = 4 O = 2 Number of atoms Success

21 Mg + O 2 MgO

22 CaCl 2 + PbO 2 PbCl 4 + CaO

23 CaCl 2 + PbO 2 PbCl 4 + CaO Ca = 1 Cl = 2 Pb = 1 O = 2 Ca = 1 Cl = 4 Pb = 1 O = 1 Go through one at a time and figure out the problems Need 2 more Need 1 more

24 2CaCl 2 + PbO 2 PbCl 4 +2CaO Ca = 2 Cl = 4 Pb = 1 O = 2 Ca = 2 Cl = 4 Pb = 1 O = 2 Go through one at a time and figure out the problems Need 2 more

25 HgO Hg + O 2 Fe + O 2 Fe 2 O 3 CaO + HCl CaCl 2 + H 2 O

26 ◦ Polyatomics  If you have polyatomic ions, treat them as a group, not individual elements  IT MAKES THINGS CONCISE!

27 Al 2 (SO 4 ) 3 + Ca(NO 3 ) 2 Al(NO 3 ) 3 + Ca(SO 4 )

28 Al 2 (SO 4 ) 3 + Ca(NO 3 ) 2 Al(NO 3 ) 3 + Ca(SO 4 ) Al = 2 (SO 4 ) = 3 Ca = 1 (NO 3 ) = 2 Al = 1 (SO 4 ) = 1 Ca = 1 (NO 3 ) = 3

29 Al 2 (SO 4 ) 3 + Ca(NO 3 ) 2 2Al(NO 3 ) 3 + Ca(SO 4 ) Al = 2 (SO 4 ) = 3 Ca = 1 (NO 3 ) = 2 Al = 2 (SO 4 ) = 1 Ca = 1 (NO 3 ) = 6

30 Al 2 (SO 4 ) 3 + Ca(NO 3 ) 2 2Al(NO 3 ) 3 + 3Ca(SO 4 ) Al = 2 (SO 4 ) = 3 Ca = 1 (NO 3 ) = 2 Al = 2 (SO 4 ) = 3 Ca = 3 (NO 3 ) = 6

31 Al 2 (SO 4 ) 3 + Ca(NO 3 ) 2 2Al(NO 3 ) 3 + 3Ca(SO 4 ) Al = 2 (SO 4 ) = 3 Ca = 1 (NO 3 ) = 2 Al = 2 (SO 4 ) = 3 Ca = 3 (NO 3 ) = 6 good Still don’t match

32 Al 2 (SO 4 ) 3 +3Ca(NO 3 ) 2 2Al(NO 3 ) 3 + 3Ca(SO 4 ) Al = 2 (SO 4 ) = 3 Ca = 3 (NO 3 ) = 6 Al = 2 (SO 4 ) = 3 Ca = 3 (NO 3 ) = 6

33  There are five main types of reactions ◦ Synthesis ◦ Decomposition ◦ Single Displacement ◦ Double Displacement ◦ Combustion

34  Also known as “Addition”  Is a reaction where two or more substances come together to form a new compound A + B AB 2H 2 + O 2 2H 2 O Think of it like two people starting to date

35  A reaction where a compound will separate into two or more different elements or compounds  ONE REACTANT!!! AB A + B H 2 CO 3 H 2 O + CO 2 Think of a couple splitting up

36  When one metal (not bonded) replaces another metal that is bonded to an anion  THE METALS (CATIONS) ARE MOVING! A + BX AX +B Mg+ Cu(SO 4 ) Mg(SO 4 ) + Cu Based off of the activity series

37  When two metals are bonded to two different anions, and then switch places. A compound that is insoluble must be formed Ba(NO 3 ) 2 + Na 2 (SO 4 ) Ba(SO 4 ) + 2Na(NO 3 ) AX + BY BX + AY Think of this like the couple switching dates

38  CARBOHYDRATES with O 2 that produces H 2 O ◦ Think of anything BURNING Carbohydrate + O 2 CO 2 + 2H 2 O C 8 H 18 + O 2 + House

39  CARBOHYDRATES with O 2 that produces H 2 O ◦ Think of anything BURNING ◦ Ex: Carbohydrate + O 2 CO 2 + H 2 O CH 4 + O 2 CO 2 + H 2 O C 3 H 8 + O 2

40 5) What type of rxns are the following a) CO 2 + H 2 O → H 2 CO 3 ______________________ b) Cl 2 + 2KI → 2KCl + I 2 _____________________ c) CaCO 3 → CaO + CO 2 _____________________ d) 2HCl + CaCO 3 → H 2 CO 3 + CaCl 2 ____________ e) Fe + CuSO 4 → FeSO 4 + Cu_________________ 8-22

41 5) What type of rxns are the following a) CO 2 + H 2 O → H 2 CO 3 __ Synthesis_______ b) Cl 2 + 2KI → 2KCl + I 2 ____Single Dis.____ c) CaCO 3 → CaO + CO 2 ____Decomp_______ d) 2HCl + CaCO 3 → H 2 CO 3 + CaCl 2 ___Double Dis_________ e) Fe + CuSO 4 → FeSO 4 + Cu___Single Dis_______ 8-22

42 ◦ Formulas ◦ Words (TWO WAYS) 2) Write the reaction for silver (I) nitrate reacting with copper to form copper (II) nitrate and silver Cu(SO 4 ) + 2Na Na 2 (SO 4 ) + Cu 1) Copper (II) sulfate + SodiumSodium sulfate+ Copper

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44 Iron + SulfurIron (II) sulfide

45 Iron + SulfurIron (II) sulfide This is why it is so important to remember how to name compounds! Fe +SFeS

46 Nickel + Copper (II) Nitrate Nickel (I) Nitrate + Copper Lead (II) iodide +potassium nitrate Potassium iodide + lead (II) nitrate

47 Try writing balanced equations for the following reactions: 2) Aluminum reacts with oxygen to form aluminum (III) oxide. 3) Chlorine reacts with lithium bromide to form lithium chloride and bromine. 4) Sodium phosphate reacts with iron (II) chloride to form sodium chloride and iron (II) phosphate. 8-15

48  We have gone over the types of reactions, balancing them, and learning how to write equations when given the reaction in words.  Today we are going to be learning how to predict the outcome of a SINGLE DISPLACEMENT REACTION

49  When one metal (not bonded) replaces another metal that is bonded to an anion AX + BBX + A Cu(SO 4 ) + Mg Mg(SO 4 ) + Cu Based off of the activity series

50 Remember Super Stud!!!! -the super stud (or the higher Element on the table will kick Out the one lower than it Reactivity

51  Always check the reactivity series to see if any reaction will occur… Which one of these will cause reaction? AgCl + Mg MgCl 2 + Ag

52  Always check the reactivity series to see if any reaction will occur… Which one of these will cause reaction? AgCl + Mg MgCl 2 + Ag

53  Will a single displacement reaction occur? Al + Cu(SO 4 ) ?

54  Will a single displacement reaction occur? ◦ Yes because Al is more reactive than Cu  What will happen in this reaction Al + Cu(SO 4 ) ?

55  Will a single displacement reaction occur? ◦ Yes because Al is more reactive than Cu  What will happen in this reaction ◦ Cu will no longer be bonded to the SO 4, as Al will take its place  What will the formula be when Al bonds with SO 4 ? Al + Cu(SO 4 ) ? + ?

56 Al(SO 4 )

57 Al 2 (SO 4 ) 3 + Cu Al + Cu(SO 4 ) Balance!

58 Al 2 (SO 4 ) 3 + 3 Cu2 Al + 3 Cu(SO 4 ) DONE!

59 Ca + FeBr 2 1. Will a reaction occur? ?

60 Ca + FeBr 2 1.Will a reaction occur? YES 2.What will happen in this reaction? ?

61 Ca + FeBr 2 1.Will a reaction occur? YES 2. What will happen in this reaction? Calcium will kick out the iron (calcium will now be bonded to Bromine) 3. What will the formula of calcium and bromine be? ?

62 CaBr

63 Ca + FeBr 2 Fe + CaBr 2

64 Ca + FeBr 2 Fe + CaBr 2

65 Ca + Co(ClO 3 ) 2 ? Co + Ca(ClO 3 ) 2

66 Sr + PbCl 4 ? Pb + SrCl 2 Sr + PbCl 4

67 Ba + MgCl 2 Rb + SnI 4 Fe + HCl

68 Platinum (II) chloride + Nickel Lead (IV) sulfate + Zinc

69  What reactions have we talked about so far? ◦ Synthesis ◦ Decomposition ◦ Single Displacement ◦ Combustion  We need to be able to predict the products of ◦ Single Displacement ◦ Double Displacement… but why?

70 2Na ( l ) + Cl 2 (g) 2NaCl (s) Phase of matter ∆H S= Solid l = Liquid g = Gas aq = aqueous (dissolved in water)

71  From now on, we will be writing the phases of matter when we write equations  To help with this, there are a couple of things that can help

72  From now on, we will be writing the phases of matter when we write equations  To help with this, there are a couple of things that can help when we are at room temperature ◦ Diatomics are always gases ◦ Metals by themselves are usually solids (except Hg!) ◦ For our purposes, assume that most of the compounds (that we talk about here) are in water (aq)

73  Look at your handy dandy solubility table ◦ Rows are the cation ◦ Columns are anion  Everything that is labelled S- means it is soluble in water  Everything else just means insoluble in water

74  Aluminum and Hydroxide  Calcium and Chloride  Strontium and Sulfate

75  When Double Displacement reactions occur, a compound that is INSOLUBLE in water MUST be formed

76 CaCl 2 + Na 2 (SO 4 )?

77 CaCl 2 (aq) + Na 2 (SO 4 ) (aq) ?

78 CaCl 2 + Na 2 (SO 4 )NaCl + Ca(SO 4 )

79 NaCl (aq) + Ca(SO 4 ) (s) CaCl 2 (aq) +Na 2 (SO 4 ) (aq)

80  Check to see if you form insoluble compound ◦ If so, Double Displacement Reaction occurs  Determine the formula of products by crisscross  Write down the phases of you products ◦ At least one needs to be (s)  Balance

81 KCl (aq) +Al(NO 3 ) 3 (aq)

82  Since both of the products that would have formed are soluble, no reaction occurs! KCl (aq) +Al(NO 3 ) 3 (aq) No Reaction

83 PbBr 2 (aq) +Na 2 (CO 3 ) (aq)

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85 NaBr + Pb(CO 3 )

86 (NH 4 )I (aq) +Hg(NO 3 ) (aq)

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88 (NH 4 )(NO 3 ) +HgI

89 Silver Chlorate + Aluminum Acetate

90 AgCl (aq) +Al(C 2 H 3 O 2 ) 3(aq) AlCl 3 + Ag(C 2 H 3 O 2 ) Phase and Balance

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