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Chapter 4: The Major Classes of Chemical Reactions

Chapter 4: The Major Classes of Chemical Reactions. 4.1 The Role of Water as a Solvent 4.2 Precipitation Reactions and Acid-Base Reactions. The Role of Water as a Solvent: The Solubility of Ionic Compounds.

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Chapter 4: The Major Classes of Chemical Reactions

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  1. Chapter 4: The Major Classes of Chemical Reactions 4.1 The Role of Water as a Solvent 4.2 Precipitation Reactions and Acid-Base Reactions

  2. The Role of Water as a Solvent: The Solubility of Ionic Compounds Electrical conductivity - The flow of electrical current in a solution is a measure of the solubility of ionic compounds or a measurement of the presence of ions in solution. Electrolyte - A substance that conducts a current when dissolved in water. Soluble ionic compound dissociate completely and may conduct a large current, and are called strong Eeectrolytes. NaCl(s) + H2O(l) Na+(aq) + Cl -(aq) When sodium chloride dissolves into water the ions become solvated, and are surrounded by water molecules. These ions are called “aqueous” and are free to move through out the solution, and are conducting electricity, or helping electrons to move through out the solution

  3. The Electrical Conductivity of Ionic Solutions Fig. 4.1

  4. Determining Moles of Ions in Aqueous Solutions of Ionic Compounds - I Problem: How many moles of each ion are in each of the following: a) 4.0 moles of sodium carbonate dissolved in water b) 46.5 g of rubidium fluoride dissolved in water c) 5.14 x 1021 formula units of iron (III) chloride dissolved in water d) 75.0 ml of 0.56M scandium bromide dissolved in water e) 7.8 moles of ammonium sulfate dissolved in water a) Na2CO3 (s) 2 Na+(aq) + CO3-2(aq) H2O

  5. Determining Moles of Ions in Aqueous Solutions of Ionic Compounds - I Problem: How many moles of each ion are in each of the following: a) 4.0 moles of sodium carbonate dissolved in water b) 46.5 g of rubidium fluoride dissolved in water c) 5.14 x 1021 formula units of iron (III) chloride dissolved in water d) 75.0 ml of 0.56M scandium bromide dissolved in water e) 7.8 moles of ammonium sulfate dissolved in water a) Na2CO3 (s) 2 Na+(aq) + CO3-2(aq) moles of Na+ = 4.0 moles Na2CO3 x = 8.0 moles Na+ and 4.0 moles of CO3-2 are present H2O 2 mol Na+ 1 mol Na2CO3

  6. Determining Moles of Ions in Aqueous Solutions of Ionic Compounds - II H2O b) RbF(s) Rb+(aq) + F -(aq)

  7. Determining Moles of Ions in Aqueous Solutions of Ionic Compounds - II H2O b) RbF(s) Rb+(aq) + F -(aq) 1 mol RbF 104.47 g RbF moles of RbF = 46.5 g RbF x = 0.445 moles RbF thus, 0.445 mol Rb+and 0.445 mol F -are present

  8. Determining Moles of Ions in Aqueous Solutions of Ionic Compounds - II H2O b) RbF(s) Rb+(aq) + F -(aq) 1 mol RbF 104.47 g RbF moles of RbF = 46.5 g RbF x = 0.445 moles RbF thus, 0.445 mol Rb+and 0.445 mol F -are present H2O c) FeCl3 (s) Fe+3(aq) + 3 Cl -(aq) moles of FeCl3 = 9.32 x 1021 formula units 1 mol FeCl3 6.022 x 1023 formula units FeCl3 x = 0.0155 mol FeCl3 3 mol Cl - 1 mol FeCl3 moles of Cl - = 0.0155 mol FeCl3 x = 0.0465 mol Cl - and 0.0155 mol Fe+3 are also present.

  9. Determining Moles of Ions in Aqueous Solutions of Ionic Compounds - III H2O d) ScBr3 (s) Sc+3(aq) + 3 Br -(aq) Converting from volume to moles: 1 L 103 ml 0.56 mol ScBr3 1 L Moles of ScBr3 = 75.0 ml x x = 0.042 mol ScBr3 3 mol Br - 1 mol ScBr3 Moles of Br - = 0.042 mol ScBr3 x = 0.126 mol Br - 0.042 mol Sc+3 are also present H2O e) (NH4)2SO4 (s) 2 NH4+(aq) + SO4- 2(aq) 2 mol NH4+ 1 mol(NH4)2SO4 Moles of NH4+ = 7.8 moles (NH4)2SO4 x = 15.6 mol NH4+ and 7.8 mol SO4- 2are also present.

  10. Determining Moles of Ions in Aqueous Solutions of Ionic Compounds - III H2O d) ScBr3 (s) Sc+3(aq) + 3 Br -(aq) Converting from volume to moles: 1 L 103 ml 0.56 mol ScBr3 1 L Moles of ScBr3 = 75.0 ml x x = 0.042 mol ScBr3 3 mol Br - 1 mol ScBr3 Moles of Br - = 0.042 mol ScBr3 x = 0.126 mol Br - 0.042 mol Sc+3 are also present H2O e) (NH4)2SO4 (s) 2 NH4+(aq) + SO4- 2(aq) 2 mol NH4+ 1 mol(NH4)2SO4 Moles of NH4+ = 7.8 moles (NH4)2SO4 x = 15.6 mol NH4+ and 7.8 mol SO4- 2are also present.

  11. Fig. 4.2

  12. Fig. 4.3

  13. The Solubility of Ionic Compounds in Water The solubility of ionic compounds in water depends upon the relative strengths of the electrostatic forces between ions in the ionic compound and the attractive forces between the ions and water molecules in the solvent. There is a tremendous range in the solubility of ionic compounds in water! The solubility of so called “insoluble” compounds may be several orders of magnitude less than ones that are called “soluble” in water, for example: Solubility of NaCl in water at 20oC = 365 g/L Solubility of MgCl2 in water at 20oC = 542.5 g/L Solubility of AlCl3 in water at 20oC = 699 g/L Solubility of PbCl2 in water at 20oC = 9.9 g/L Solubility of AgCl in water at 20oC = 0.009 g/L Solubility of CuCl in water at 20oC = 0.0062 g/L

  14. The Solubility of Covalent Compounds in Water The covalent compounds that are very soluble in water are the ones with -OH group in them and are called “polar” and can have strong polar (electrostatic)interactions with water. Examples are compound such as table sugar, sucrose (C12H22O11); beverage alcohol, ethanol (C2H5-OH); and ethylene glycol (C2H6O2) in antifreeze. H Methanol = Methyl Alcohol H C O H H Other covalent compounds that do not contain a polar center, or the -OH group are considered “nonpolar” , and have little or no interactions with water molecules. Examples are the hydrocarbons in gasoline and oil. This leads to the obvious problems in oil spills, where the oil will not mix with the water and forms a layer on the surface! Octane = C8H18 and / or Benzene = C6H6

  15. Acids - A Group of Covalent Molecules Which Lose Hydrogen Ions to Water Molecules in Solution When gaseous hydrogen iodide dissolves in water, the attraction of the oxygen atom of the water molecule for the hydrogen atom in HI is greater that the attraction of the of the iodide ion for the hydrogen atom, and it is lost to the water molecule to form an hydronium ion and an iodide ion in solution. We can write the hydrogen atom in solution as either H+(aq) or as H3O+(aq) they mean the same thing in solution. The presence of a hydrogen atom that is easily lost in solution is an “Acid” and is called an “acidic” solution. The water (H2O) could also be written above the arrow indicating that the solvent was water in which the HI was dissolved. HI(g) + H2O(L) H+(aq) + I -(aq) HI(g) + H2O(L) H3O+(aq) + I -(aq) H2O HI(g) H+(aq) + I -(aq)

  16. Fig. 4.4

  17. Strong Acids and the Molarity of H+ Ions in Aqueous Solutions of Acids Problem: In aqueous solutions, each molecule of sulfuric acid will loose two protons to yield two hydronium ions, and one sulfate ion. What is the molarity of the sulfate and hydronium ions in a solution prepared by dissolving 155g of concentrate sulfuric acid into sufficient water to produce 2.30 Liters of acid solution? Plan: Determine the number of moles of sulfuric acid, divide the moles by the volume to get the molarity of the acid and the sulfate ion. The hydronium ions concentration will be twice the acid molarity. Solution: Two moles of H+ are released for every mole of acid: H2SO4 (l) + 2 H2O(l) 2 H3O+(aq) + SO4- 2(aq)

  18. Strong Acids and the Molarity of H+ Ions in Aqueous Solutions of Acids Problem: In aqueous solutions, each molecule of sulfuric acid will loose two protons to yield two hydronium ions, and one sulfate ion. What is the molarity of the sulfate and hydronium ions in a solution prepared by dissolving 155g of concentrate sulfuric acid into sufficient water to produce 2.30 Liters of acid solution? Plan: Determine the number of moles of sulfuric acid, divide the moles by the volume to get the molarity of the acid and the sulfate ion. The hydronium ions concentration will be twice the acid molarity. Solution: Two moles of H+ are released for every mole of acid: H2SO4 (l) + 2 H2O(l) 2 H3O+(aq) + SO4- 2(aq) 1 mole H2SO4 98.09 g H2SO4 Moles H2SO4 = = 1.58 moles H2SO4 155 g H2SO4 x

  19. Strong Acids and the Molarity of H+ Ions in Aqueous Solutions of Acids Problem: In aqueous solutions, each molecule of sulfuric acid will loose two protons to yield two hydronium ions, and one sulfate ion. What is the molarity of the sulfate and hydronium ions in a solution prepared by dissolving 155g of concentrate sulfuric acid into sufficient water to produce 2.30 Liters of acid solution? Plan: Determine the number of moles of sulfuric acid, divide the moles by the volume to get the molarity of the acid and the sulfate ion. The hydronium ions concentration will be twice the acid molarity. Solution: Two moles of H+ are released for every mole of acid: H2SO4 (l) + 2 H2O(l) 2 H3O+(aq) + SO4- 2(aq) 1 mole H2SO4 98.09 g H2SO4 Moles H2SO4 = = 1.58 moles H2SO4 155 g H2SO4 x 1.58 mol SO4-2 2.30 l solution Molarity of SO4- 2 = = 0.687 Molar in SO4- 2

  20. Strong Acids and the Molarity of H+ Ions in Aqueous Solutions of Acids Problem: In aqueous solutions, each molecule of sulfuric acid will loose two protons to yield two hydronium ions, and one sulfate ion. What is the molarity of the sulfate and hydronium ions in a solution prepared by dissolving 155g of concentrate sulfuric acid into sufficient water to produce 2.30 Liters of acid solution? Plan: Determine the number of moles of sulfuric acid, divide the moles by the volume to get the molarity of the acid and the sulfate ion. The hydronium ions concentration will be twice the acid molarity. Solution: Two moles of H+ are released for every mole of acid: H2SO4 (l) + 2 H2O(l) 2 H3O+(aq) + SO4- 2(aq) 1 mole H2SO4 98.09 g H2SO4 Moles H2SO4 = = 1.58 moles H2SO4 155 g H2SO4 x 1.58 mol SO4-2 2.30 l solution Molarity of SO4- 2 = = 0.687 Molar in SO4- 2 Molarity of H+ = 2 x 1.58 mol H+ / 2.30 liters = 1.38 Molar in H+

  21. Fig. 4.5

  22. Precipitation Reactions: A Solid Product is Formed When ever two aqueous solutions are mixed, there is the possibility of forming an insoluble compound. Let us look at some examples to see how we can predict the result of adding two different solutions together. Pb(NO3) (aq) + NaI(aq) Pb+2(aq) + 2 NO3-(aq) + Na+(aq) + I-(aq) When we add these two solutions together, the ions can combine in the way they came into the solution, or they can exchange partners. In this case we could have lead nitrate and sodium iodide, or lead iodide and sodium nitrate formed, to determine which will happen we must look at the solubility table (p.141) to determine what could form. The table indicates that lead iodide will be insoluble, so a precipitate will form. Pb(NO3)2 (aq) + 2 NaI(aq) PbI2 (s)+ 2 NaNO3 (aq)

  23. Precipitation Reactions: Will a Precipitate Form? If we add a solution containing potassium chloride to a solution containing ammonium nitrate, will we get a precipitate? KCl(aq) + NH4NO3 (aq) = K+(aq) + Cl-(aq) + NH4+(aq) + NO3-(aq) By exchanging cations and anions we see that we could have potassium chloride and ammonium nitrate, or potassium nitrate and ammonium chloride. In looking at the solubility table it shows all possible products as soluble, so there is no net reaction! KCl(aq) + NH4NO3 (aq) = No Reaction! If we mix a solution of sodium sulfate with a solution of barium nitrate, will we get a precipitate? From the solubility table it shows that barium sulfate is insoluble, therefore we will get a precipitate! Na2SO4 (aq) + Ba(NO3)2 (aq) BaSO4 (s) + 2 NaNO3 (aq)

  24. Fig. 4.6

  25. Solubility Rules for Ionic Compounds in Water Soluble Ionic Compounds Insoluble Ionic Compounds 1) All common compounds of Group 1A(I) ions (Na+,K+, etc.) and ammonium(NH4+) are soluble. 2) All common nitrates (NO3-), acetates (CH3COO-), and most perchlorates (ClO4-) are soluble. 3) All common chlorides (Cl-), bromides (Br -), and iodides (I-) are soluble, except those of Ag+, Pb2+, Cu2+, and Hg22+. 4) All common sulfates (SO42-) are soluble, except those of Ca+2, Sr2+, Ba2+, and Pb2+. 1) All common metal hydroxides are insoluble, except those of Group 1A (1) and the larger members of Group 2A(2) (beginning with Ca2+). 2) All common carbonates (CO32-) and phosphates (PO43-) are insoluble, except those of Group 1A(1) and NH4+. 3) All common sulfides are insoluble, except those of Group 1A(1), Group 2A(2), and NH4+.

  26. Predicting Whether a Precipitation Reaction Occurs; Writing Equations: a) Calcium Nitrate and Sodium Sulfate solutions are added together. Molecular Equation Total Ionic Equation Net Ionic Equation - b) Ammonium Sulfate and Magnesium Chloride are added together.

  27. Predicting Whether a Precipitation Reaction Occurs; Writing Equations: a) Calcium Nitrate and Sodium Sulfate solutions are added together. Molecular Equation Ca(NO3)2 (aq) + Na2SO4 (aq) CaSO4 (s) + NaNO3 (aq) Total Ionic Equation Net Ionic Equation b) Ammonium Sulfate and Magnesium Chloride are added together.

  28. Predicting Whether a Precipitation Reaction Occurs; Writing Equations: a) Calcium Nitrate and Sodium Sulfate solutions are added together. Molecular Equation Ca(NO3)2 (aq) + Na2SO4 (aq) CaSO4 (s) + NaNO3 (aq) Total Ionic Equation Ca2+(aq)+2 NO3-(aq) + 2 Na+(aq)+ SO4-2(aq) CaSO4 (s) + 2 Na+(aq+) 2 NO3-(aq) Net Ionic Equation b) Ammonium Sulfate and Magnesium Chloride are added together.

  29. Predicting Whether a Precipitation Reaction Occurs; Writing Equations: a) Calcium Nitrate and Sodium Sulfate solutions are added together. Molecular Equation Ca(NO3)2 (aq) + Na2SO4 (aq) CaSO4 (s) + NaNO3 (aq) Total Ionic Equation Ca2+(aq)+2 NO3-(aq) + 2 Na+(aq)+ SO4-2(aq) CaSO4 (s) + 2 Na+(aq+) 2 NO3-(aq) Net Ionic Equation Ca2+(aq) + SO4-2(aq) CaSO4 (s) Spectator Ions are Na+ and NO3- b) Ammonium Sulfate and Magnesium Chloride are added together.

  30. Predicting Whether a Precipitation Reaction Occurs; Writing Equations: a) Calcium Nitrate and Sodium Sulfate solutions are added together. Molecular Equation Ca(NO3)2 (aq) + Na2SO4 (aq) CaSO4 (s) + NaNO3 (aq) Total Ionic Equation Ca2+(aq)+2 NO3-(aq) + 2 Na+(aq)+ SO4-2(aq) CaSO4 (s) + 2 Na+(aq+) 2 NO3-(aq) Net Ionic Equation Ca2+(aq) + SO-4(aq) CaSO4 (s) Spectator Ions are Na+ and NO3- b) Ammonium Sulfate and Magnesium Chloride are added together. In exchanging ions, no precipitates will be formed, so there will be no chemical reactions occurring! All ions are spectator ions!

  31. Acid - Base Reactions: Neutralization Rxns. An Acid is a substance that produces H+ (H3O+) ions when dissolved in water. A Base is a substance that produces OH - ions when dissolved in water. Acids and bases are electrolytes, and their strength is categorized in terms of their degree of dissociation in water to make hydronium or hydroxide ions. Strong acids and bases dissociate completely, and are strong electrolytes. Weak acids and bases dissociate weakly and are weak electrolytes. The generalized reaction between an Acid and a Base is: HX(aq) + MOH(aq) MX(aq) + H2O(L) Acid + Base = Salt + Water

  32. Fig. 4.7

  33. Selected Acids and Bases Acids Bases Strong Strong Hydrochloric acid, HCl Sodium hydroxide, NaOH Hydrobromic acid, HBr Potassium hydroxide, KOH Hydriodoic acid, HI Calcium hydroxide, Ca(OH)2 Nitric acid, HNO3 Strontium hydroxide, Sr(OH)2 Sulfuric acid, H2SO4 Barium hydroxide, Ba(OH)2 Perchloric acid, HClO4 Weak Weak Hydrofluoric acid, HF Ammonia, NH3 Phosphoric acid, H3PO4 Acidic acid, CH3COOH (or HC2H3O2) Table 4.2

  34. Writing Balanced Equations forNeutralization Reactions - I Problem:Write balanced chemical reactions (molecular, total ionic, and net ionic) for the following chemical reactions: a) calcium hydroxide(aq) and hydroiodic acid(aq) b) lithium hydroxide(aq) and nitric acid(aq) c) barium hydroxide(aq) and sulfuric acid(aq) Plan: These are all strong acids and bases, therefore they will make water and the corresponding salts. Solution:

  35. Writing Balanced Equations forNeutralization Reactions - I Problem:Write balanced chemical reactions (molecular, total ionic, and net ionic) for the following chemical reactions: a) calcium hydroxide(aq) and hydroiodic acid(aq) b) lithium hydroxide(aq) and nitric acid(aq) c) barium hydroxide(aq) and sulfuric acid(aq) Plan: These are all strong acids and bases, therefore they will make water and the corresponding salts. Solution: a) Ca(OH)2 (aq) + 2HI(aq) CaI2 (aq) + 2H2O(l)

  36. Writing Balanced Equations forNeutralization Reactions - I Problem:Write balanced chemical reactions (molecular, total ionic, and net ionic) for the following chemical reactions: a) calcium hydroxide(aq) and hydroiodic acid(aq) b) lithium hydroxide(aq) and nitric acid(aq) c) barium hydroxide(aq) and sulfuric acid(aq) Plan: These are all strong acids and bases, therefore they will make water and the corresponding salts. Solution: a) Ca(OH)2 (aq) + 2HI(aq) CaI2 (aq) + 2H2O(l) Ca2+(aq) + 2 OH -(aq) + 2 H+(aq) + 2 I -(aq) Ca2+(aq) + 2 I -(aq) + 2 H2O(l)

  37. Writing Balanced Equations forNeutralization Reactions - I Problem:Write balanced chemical reactions (molecular, total ionic, and net ionic) for the following chemical reactions: a) calcium hydroxide(aq) and hydroiodic acid(aq) b) lithium hydroxide(aq) and nitric acid(aq) c) barium hydroxide(aq) and sulfuric acid(aq) Plan: These are all strong acids and bases, therefore they will make water and the corresponding salts. Solution: a) Ca(OH)2 (aq) + 2HI(aq) CaI2 (aq) + 2H2O(l) Ca2+(aq) + 2 OH -(aq) + 2 H+(aq) + 2 I -(aq) Ca2+(aq) + 2 I -(aq) + 2 H2O(l) 2 OH -(aq) + 2 H+(aq) 2 H2O(l)

  38. Writing Balanced Equations for Neutralization Reactions - II b) LiOH(aq) + HNO3 (aq) LiNO3 (aq) + H2O(l)

  39. Writing Balanced Equations for Neutralization Reactions - II b) LiOH(aq) + HNO3 (aq) LiNO3 (aq) + H2O(l) Li+(aq) + OH -(aq) + H+(aq) + NO3-(aq) Li+(aq) + NO3-(aq) + H2O(l) OH -(aq) + H+(aq)H2O(l)

  40. Writing Balanced Equations for Neutralization Reactions - II b) LiOH(aq) + HNO3 (aq) LiNO3 (aq) + H2O(l) Li+(aq) + OH -(aq) + H+(aq) + NO3-(aq) Li+(aq) + NO3-(aq) + H2O(l) OH -(aq) + H+(aq)H2O(l) c) Ba(OH)2 (aq) + H2SO4 (aq) BaSO4 (s) + 2H2O(l)

  41. Writing Balanced Equations for Neutralization Reactions - II b) LiOH(aq) + HNO3 (aq) LiNO3 (aq) + H2O(l) Li+(aq) + OH -(aq) + H+(aq) + NO3-(aq) Li+(aq) + NO3-(aq) + H2O(l) OH -(aq) + H+(aq)H2O(l) c) Ba(OH)2 (aq) + H2SO4 (aq) BaSO4 (s) + 2H2O(l) Ba2+(aq) + 2OH -(aq) + 2 H+(aq) + SO42-(aq) BaSO4 (s) + 2 H2O(l) Ba2+(aq) + 2 OH -(aq) + 2 H+(aq) + SO42-(aq) BaSO4 (s) + 2 H2O(l)

  42. An Acid-Base Titration Fig. 4.8

  43. Finding the Concentration of Acid from an Acid - Base Titration Volume (L) of base (difference in buret readings) M (mol/L) of base Moles of base molar ratio Moles of acid volume (L) of acid M (mol/L) of acid

  44. Finding the Concentration of Base from an Acid - Base Titration - I Problem: A titration is performed between sodium hydroxide and potassium hydrogenphthalate (KHP) to standardize the base solution, by placing 50.00 mg of solid potassium hydrogenphthalate in a flask with a few drops of an indicator. A buret is filled with the base, and the initial buret reading is 0.55 ml; at the end of the titration the buret reading is 33.87 ml. What is the concentration of the base? Plan: Use the molar mass of KHP (204.2 g/mol) to calculate the number of moles of the acid, from the balanced chemical equation, the reaction is equal molar, so we know the moles of base, and from the difference in the buret readings, we can calculate the molarity of the base. Solution: HKC8H4O4 (aq) + OH -(aq) KC8H4O4-(aq) + H2O(aq)

  45. O C O K+ O H C O Potassium Hydrogenphthalate KHC8H4O4 O C K+ O O C H+ O

  46. Finding the Concentration of Base from an Acid - Base Titration - II 50.00 mg KHP 204.2 g KHP 1 mol KHP 1.00 g 1000 mg moles KHP = x = 0.00024486 mol KHP Volume of base = Final buret reading - Initial buret reading = 33.87 ml - 0.55 ml = 33.32 ml of base one mole of acid = one mole of base; therefore 0.00024486 moles of acid will yield 0.00024486 moles of base in a volume of 33.32 ml. 0.00024486 moles 0.03332 L molarity of base = = 0.07348679 moles per liter molarity of base = 0.07349 M

  47. An Aqueous Strong Acid-Strong BaseReaction on the Atomic Scale Acid + Base Salt + H2O Fig. 4.9

  48. An Acid-Base Reaction That Formsa Gaseous Product The reaction of acid with carbonates or bicarbonates will produce carbon dioxide gas that is released from solution as a gas in the form of bubbles that leave the solution. Fig. 4.10

  49. Types of Chemical Reactions - I I) Combination reactions that are redox reactions a) A metal and a nonmetal form an Ionic compound b) Two nonmetals form a covalent compound c) Combination of an element and a compound II) Combination reactions that are not redox reactions a) A metal oxide and a nonmetal form an ionic compound with a polyatomic anion b) Metal oxides and water form bases c) Nonmetal oxides and water form acids III) Decomposition reactions a) Thermal decomposition i) Many ionic compounds with oxoanions form a metal oxide and a gaseous nonmetal ii) Many metal oxides, chlorates, and perchlorates release oxygen b) Electrolytic decomposition

  50. Types of Chemical Reactions - II IV) Displacement Reactions a) Single-Displacement Reactions - Activity Series of the Metals i) A metal displaces hydrogen from water or an acid ii) A metal displaces another metal ion from solution iii) A halogen displaces a halide ion from solution b) Double-Displacement Reactions - i) In precipitation reactions: A precipitate forms ii) In acid-base reactions: Acid + Base form a salt & water V) Combustion Reactions - All are Redox Processes a) Combustion of an element with oxygen to form oxides b) Combustion of hydrocarbons to yield water & carbon dioxide Reactants Products

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