C h e m i s t r y 1 A : C h a p t e r 4 P a g e 1 Chapter 4: Chemical Quantities and Aqueous Reactions: Homework: Read Chapters 4. Work out sample/practice exercises Keep up with assignments in Lab Check MasteringChemistry.com assignment and complete before due date Reaction Stoichiometry: Chemical arithmetic. Stoichiometry is the study of numerical relationships between chemical quantities in a chemical reaction. Using ratios found in balanced chemical reactions to solve quantitative problems. Convert to moles, mole to mole ratio, convert to desired result. Practice: 1. a) Balance the following unbalanced equation: BaO (s) + HCl (aq) BaCl 2 (aq) + H 2 O (l) b) There are 2 moles of BaO and 2 moles of HCl, which is the limiting reactant? 2. Balance the equation below and calculate the mass in grams of water vapor produced when 56.0 g of pure hydrogen gas burns in an unlimited supply of oxygen. H 2 (g) + O 2 (g) H 2 O (g)
C h e m i s t r y 1 A : C h a p t e r 4 P a g e 2 3. How many mg of solid iron (III) chloride (FeCl 3 ) result when 15.5 milligrams of solid iron (Fe) is reacted with an excess of chlorine gas (Cl 2 ). Start by balancing the equation. Yields: Actual or Experimental Yield = that produced in a lab environment. Theoretical Yield = the highest (maximum) possible yield assuming 100% perfection using up all of the limiting reactant. Percent Yield = Limiting reactants: Actual Yield Theoretical Yield x 100% The limiting reactant is completely consumed producing the least amount of product. It runs out first and limits the amount of product. It produces the lowest (maximum) theoretical yield. Limiting reactant problems occur when two values for reactants are given. Excess reactants: Practice: Excess reactants are not completely consumed in the reaction. 4. a) Balance the following unbalanced equation: Ca (s) + H 2 O (l) Ca(OH) 2 (aq) + H 2 (g) b) How many grams of H2O would be required to react with 8.00 g of calcium? c) Solve for the theoretical mass of hydrogen gas produced. d) When the reaction described above was carried out in the lab, 0.346 grams of hydrogen gas was obtained. Calculate the percent yield for this reaction.
C h e m i s t r y 1 A : C h a p t e r 4 P a g e 3 5. a) Balance the following unbalanced equation: P 4 (s) + F 2 (g) PF 5 (g) b) Calculate the theoretical mass in grams of phosphorus pentafluoride produced from 12.6 grams F 2 and 9.00 g phosphorus. c) What is the limiting reactant from the reaction above? d) What is the % yield of the reaction above if 15.8g of PF 5 were experimentally produced in a lab starting with the given amounts above? Solution Concentrations: Homogenous mixtures are called solutions. The major component that keeps its same state is called the solvent. The minor component, the one that may have a change in state, is the solute. Dilute solutions have a small amount of solute, concentrated solutions have a large amount of solute compared to the solvent. Molarity = (moles of solute)/(liter of solution) Molarity can be used as a conversion factor to solve for moles in stoichiometry problems Solution Dilution: M 1 V 1 = M 2 V 2 When diluting solutions it is best to use a volumetric flask and safest to start with some solvent, then add the concentrated amount, followed by more solvent to reach the desired total volume. Note: Volumes are not always additive: for example 50 ml of water plus 50 ml of ethanol only produce 95 ml volume total due to some differences in density and intermolecular attractions Other concentrations for solutions include: mass percent, volume percent, mass/volume percent, ppm, ppb, normality, molality, mole fraction.
C h e m i s t r y 1 A : C h a p t e r 4 P a g e 4 Practice: 6. Calculate the Molarity of a solution made from 30.0 grams of sucrose (C 12 H 22 O 11 ) added to enough water to make 300 ml total solution. 7. Describe quantitatively and qualitatively how one should safely prepare 600 ml of 1.0 M H 2 SO 4 (aq) solution from 18.0 M H 2 SO 4 stock solution. 8. How many grams of hydrogen gas will be theoretically formed by the reaction of 10.0 ml of 6.00 M HCl with excess zinc? Zn (s) + HCl (aq) ZnCl 2 (aq) + H 2 (g) unbalanced
C h e m i s t r y 1 A : C h a p t e r 4 P a g e 5 Titrations: In a titration, a substance in a solution of known concentration is reacted with another of unknown concentration. A burette is used to add a known volume to a flask with an indicator that will change (endpoint) when the reactants reach their stoichiometric ratio (equivalence point). Practice: 9. Solve for the molarity concentration in a 10.0 ml portion of acetic acid solution (HC 2 H 3 O 2 ) that required 39.6 ml of 0.202 M NaOH solution to reach the equivalence point in a titration.
C h e m i s t r y 1 A : C h a p t e r 4 P a g e 6 Gas Stoichiometry: Practice: Molar volume at STP for any ideal gas is 22.4 L/mol STP = standard temperature (0 C) and pressure (1 atmosphere) 10. Pure iron may be prepared from iron(iii) oxide and carbon monoxide: Fe 2 O 3 (s) + CO (g) Fe (s) + CO 2 (g) a) Balance the unbalanced equation above. b) Calculate the maximum theoretical mass in grams of Fe produced from 40.0 L of carbon monoxide at STP from the reaction above. Dissolving a Solute When mixing a solute with the solvent, attractions between the like substances must break while attractions between solute-solvent are formed. If the new solute-solvent attractions are strong enough, the solute will dissolve. Water is a polar molecule with uneven distribution of electrons: (more on the oxygen -, less on the hydrogen + ) When ionic compounds dissolve in water, each ion is attracted to the polar water and surrounded by a solvent cage of water. The result is a solution with free moving charged particles that may conduct electricity. Materials that dissolve in water to form a solution that will conduct electricity are called electrolytes.
C h e m i s t r y 1 A : C h a p t e r 4 P a g e 7 Electrolytes in Aqueous Solution: Strong Electrolytes will largely dissociate into its ions in an aqueous solution and are written as separated ions in the ionic reactions. Examples: Strong acids, Strong Bases, Soluble salts. Strong Acids: HCl, HBr, HI, HNO 3, H 2 SO 4, HClO 4, HClO 3 Strong Bases: soluble hydroxides from Group 1A (not including H) and Group 11A, not including the top two) LiOH, NaOH, KOH, RbOH, Ca(OH) 2, Sr(OH) 2, Ba(OH) 2 Soluble Salts: Ionic compounds that contain the cations from Group 1A; Li +, Na +, K +, Rb +, Cs +, or ammonium ion, NH 4 +. A compound is probably soluble if it has the anion; Cl -1, Br -1, I -1 except with Ag +, Hg 2 +2, or Pb +2, and most compounds that include NO 3-1, ClO 4-1, C 2 H 3 O 2-1, Soluble with most SO 4-2 except Ba +2, Hg 2 +2, or Pb +2. Weak Electrolytes will partially dissociate into its ions in an aqueous solution, but are written as compounds in an ionic equation. Weak electrolytes are the weak acids and weak bases such as HC 2 H 3 O 2 or NH 3 Nonelectrolytes will stay in their formula in an aqueous solution. Nonelectrolytes are the insoluble salts, gases, liquids, and other solids. Writing Molecular, Ionic, and Net ionic equations: Example: For the double displacement reaction of lead (II) nitrate reacting with aluminum chloride. Whole equation: Unbalanced Pb(NO 3 ) 2 (aq) + AlCl 3 (aq) PbCl 2 (s) + Al(NO 3 ) 3 (aq) Total ionic equation: Unbalanced Pb +2 (aq) + NO 3-1 (aq) + Al +3 (aq) + Cl -1 (aq) PbCl 2 (s) + Al +3 (aq) + NO 3-1 (aq) Identify the spectator ions and leave them out when writing the net ionic eq. Net ionic equation: Unbalanced Pb +2 (aq) + Cl -1 (aq) PbCl 2 (s)
C h e m i s t r y 1 A : C h a p t e r 4 P a g e 8 Chemical Reactions: Evidence of chemical reactions: Examples formation of a solid, or formation of gas, light given off, change in temperature, etc. Driving forces of chemical reactions: 1) Transfer of electrons occurs in redox (oxidation-reduction) reactions 2) Formation of solid 3) Formation of gas 4) Formation of water or a weak electrolyte (weak acid or weak base) 5) Other driving forces not listed may occur Classifying Chemical Reactions: Types of Chemical Reactions: (D-D) double displacement: AB + CD AD + CB (P) precipitation: a solid is formed (N) neutralization or acid-base: water is formed (G) gas evolution: gas is formed (Syn) synthesis or combination: A + B AB (D) decomposition: AB A + B (R) redox or oxidation-reduction: electrons are transferred (C) combustion: add O 2 gas, possible products include: CO 2, H 2 O, NO 2, SO 2. (SR) single replacement: A + BC AC + B or BA + C The more active metal will have a partner, less active is alone. Activity series: Au<Hg<Ag<Cu<H2<Pb<Sn<Ni<Cd<Fe<Cr< Zn<Mn<Al< Mg< Na<Ca< K<Li
C h e m i s t r y 1 A : C h a p t e r 4 P a g e 9 Classifying and Predicting Products and Chemical Reactions: The following examples will help you classify the type of reaction and predict the expected products. Refer to the solubility rules to determine if solid forms. (D-D) double displacement: AB + CD AD + CB A double displacement reaction starts with two ionic compounds in which the ions exchange to produce new balanced ionic compounds. Always write the cation before the anion and verify that the net charge of a compound is zero. Ionic compounds may be acid, base, or salt. Oxidation charges do not alter in double displacement reactions so this type will never be a redox reaction. (NR) When all ionic products are strong electrolytes, there is NO REACTION. FeBr 3 (aq) + 3 LiNO 3 (aq) 3LiBr (aq) + Fe(NO 3 ) 3 (aq) NR Driving forces for double displacement reactions include (P) (N) (P) Precipitation, formation of solid (N) Neutralization or Acid-Base Reaction, formation of water (G) gas evolution, formation of a gas (WE) formation of a weak electrolyte, a soluble weak acid or weak base Formation of a new solid in a double displacement reaction when there is no solid on the reactant side is called PRECIPITATION. Ni(NO 3 ) 2 (aq) + Na 2 S (aq) NiS (s) + 2 NaNO 3 (aq) DD,P Formation of water in a double displacement reaction is called NEUTRALIZATION (also known as Acid-Base reaction) 3 KOH (aq) + H 3 PO 4 (aq) K 3 PO 4 (aq) + 3 H 2 O (l) DD,N (WE) Formation of a WEAK ELECTROLYTE, generally a weak acid, produced from a double displacement reaction Al(C 2 H 3 O 2 ) 3 (aq) + 3 HCl (aq) AlCl 3 (aq) + 3 HC 2 H 3 O 2 (aq) DD,WE (G) Gas Evolution: Several common products are in the gas phase. You should recognize the following as gas products. CO 2, SO 2, NH 3, H 2 S, H 2, N 2, O 2, F 2, Cl 2, NO 2
C h e m i s t r y 1 A : C h a p t e r 4 P a g e 10 Special cases of (G) Gas Evolution in (DD) Double Displacement: There are three common products that further decompose leading to three products in place of just the two products in a double displacement reaction. They are as follows. NH 4 OH (aq) H 2 SO 3 (aq) H 2 CO 3 (aq) NH 3 (g) + H 2 O (l) SO 2 (g) + H 2 O (l) CO 2 (g) + H 2 O (l) The double displacement reactions below show examples of the decomposition of one of the ionic products. The driving forces include both formation of water and a gas, because water is formed these can be labeled DD, N, G reactions. The last one is DD, N, G and P. K 2 CO 3 (aq) + 2HCl (aq) H 2 CO 3 (aq) + 2 KCl (aq) K 2 CO 3 (aq) + 2HCl (aq) CO 2 (g) + H 2 O (l) + 2 KCl (aq) Na 2 SO 3 (aq) + 2HCl (aq) H 2 SO 3 (aq) + 2 NaCl (aq) Na 2 SO 3 (aq) + 2HCl (aq) SO 2 (g) + H 2 O (l) + 2 NaCl (aq) (NH 4 ) 2 SO 4 (aq) + Ba(OH) 2 (aq) 2 NH 4 OH (aq) + BaSO 4 (s) (NH 4 ) 2 SO 4 (aq) + Ba(OH) 2 (aq) 2 NH 3 (g) + 2 H 2 O (l) + BaSO 4 (s) (Syn) Synthesis or combination: A + B AB A synthesis reaction starts with two or more substances and creates a single compound product. The reverse of a synthesis reaction is Decomposition. Remember which elements are diatomic elements: H 2, N 2, O 2, F 2, Cl 2, Br 2, I 2. Know the elements in their pure form: gas, solid, and liquid. a) two elements (metal plus nonmetal) ionic compound. K (s) + Cl 2 (g) KCl (s) b) metal oxide and water base (metal hydroxide) BaO (g) + H 2 O (l) Ba(OH) 2 (aq) c) nonmetal oxide and water acid CO 2 (g) + H 2 O (l) H 2 CO 3 (aq)
C h e m i s t r y 1 A : C h a p t e r 4 P a g e 11 (D) Decomposition: AB A + B A decomposition reaction starts with one reactant and breaks up into two or more substances. The reverse of a decomposition reaction is synthesis. Remember the diatomic elements: H 2, N 2, O 2, F 2, Cl 2, Br 2, I 2. a) ionic compound two elements (metal plus nonmetal) 2 Fe 2 O 3 (s) 4 Fe (s) + 3 O 2 (g) b) metal carbonate metal oxide and carbon dioxide CaCO 3 (s) CaO (s) + CO 2 (g) c) metal bicarbonate metal carbonate + water + carbon dioxide 2 Al(HCO 3 ) 3 (s) Al 2 (CO 3 ) 3 (s) + 3 H 2 O (l) + 3 CO 2 (g) d) metal connected to halogen with oxygen metal halide + oxygen Ca(BrO 3 ) 2 (s) CaBr 2 (s) + 3 O 2 (g) e) hydrated salt anhydrous salt plus water CuSO 4. 5H 2 O (s) CuSO 4 (s) + 5 H 2 O (l) (R) Redox or oxidation-reduction: Redox reactions have a transfer of electrons. An easy what to tell that a redox reaction takes place is to check for element charges and see if they change from reactant to product. Some synthesis and decomposition reactions are redox such as the synthesis (a), and decomposition (a), (d) reactions. All single replacement and combustion reactions are redox reactions. (C) Combustion: elements burn with O 2 gas to produce oxides, Common products include: CO 2, H 2 O, NO 2, SO 2. NH 3 (g) + O 2 (g) C 4 H 10 (s) + O 2 (g) C 17 H 24 S (s) + O 2 (g) C 12 H 22 O 11 (s) + O 2 (g) (SR) Single Replacement: A + BC AC + B or BA + C The more active metal will have a partner, less active is alone. Activity series: Au<Hg<Ag<Cu<H2<Pb<Sn<Ni<Cd<Fe<Cr< Zn<Mn<Al< Mg< Na<Ca< K<Li Do NOT memorize the activity series. The necessary information will be given to you as needed. Most of the transition metals commonly have a +2 charge, except Ag +1, Cr +3, and Au +3
C h e m i s t r y 1 A : C h a p t e r 4 P a g e 12 Predict the products if a reaction occurs or write NR for no reaction Zn(NO 3 ) 2 (aq) + Cu (s) Cu(NO 3 ) 2 (aq) + Ag (s) AgNO 3 (aq) + Cu (s) The nonmetal activity levels increase as you go from the bottom to the top of the periodic table: Given that F>Cl>Br>I. Predict the products or write NR Redox Reactions: NaBr (s) + Cl 2 (g) NaBr (s) + I 2 (s) Redox reactions exchange electrons. When balancing a reaction (1) electrons lost = electrons gained (2) atoms on either side of the reaction must balance (3) total charge on either side of the reaction must balance. Oxidation Reduction increases oxidation number reduces oxidation number lose electrons gain electrons 2 Cl -1 Cl 2 + 2 e - K + + 1 e - K Cl -1 is oxidized K + is reduced Cl -1 is a reducing agent K + is an oxidizing agent Rules for Assigning Oxidation Numbers 1) elements zero 2) monatomic ions charge on ion 3) oxygen mostly (-2) Exceptions: peroxide, O -2 2, (-1) and superoxides, O -1 2, (-1/2) 4) hydrogen mostly (+1) exceptions: hydrides (-1) 5) halogens mostly (-1) Exceptions: (+1, +3, +5, +7) in a compound with lighter halogens or oxygen 6) Compounds sum of oxidation numbers equal zero 7) polyatomic ions sum of oxidation numbers equal charge of ion
C h e m i s t r y 1 A : C h a p t e r 4 P a g e 13 Balancing Complicated Redox Rxns using the Half Rxn Method (Ch 18): The half reaction method of balancing for the electrochemistry reactions splits apart the oxidation half reaction from the reduction half reaction and then adds the two together for an overall reaction. 1) Assign oxidation numbers to each atom. 2) Determine what is being oxidized and what is reduced. 3) Split the skeleton equation into two half reactions label appropriately as oxidation or reduction reactions. 4) Complete and balance each half reaction (This method does not need to use the oxidation numbers) a) balance atoms that change their oxidation numbers b) balance O by adding H 2 O c) balance H by adding H + d) balance charges by including the electrons as products in the oxidation reaction and electrons as reactants in the reduction reaction e) If the solution is basic: add equal OH -1 ions on both sides of the equation to counter all H + ions, every H + and OH -1 become H 2 O essentially removing all H + ions from the reaction. 5) Combine the two half reactions to obtain overall equation a) Multiply each half reaction by the factor needed to have equivalent electrons lost and gained. b) Simplify and reduce to the lowest whole number ratio 6) Check balancing a) Do atoms balance? b) Are the sum of charges of reactants equal to those of products? c) Are the same number on electrons lost and gained? Redox: Balance the following. a) Al (s) + SO 4-2 (aq) Al +3 (aq) + SO 2 (g) (acidic conditions) b) A breathalyzer detects ethanol in suspected drunk drivers using the reaction below -2 Cr 2 O 7 (aq) + C 2 H 5 OH (l) Cr +3 (aq) + C 2 H 4 O (l) (acidic conditions)
C h e m i s t r y 1 A : C h a p t e r 4 P a g e 14 c) MnO 4-1 + Mn +2 MnO 2 (acidic conditions) d) MnO 4-1 + IO 3-1 Mn +2 + IO 4-1 (basic conditions) e) Br 2 BrO 3-1 + Br -1 (basic conditions) Redox Applications: Combustion, Bleaching, Batteries, Metallurgy, Corrosion, Respiration, Titration Redox Titration: Electrons transfer. Titrate with burette as liquid is slowly added to a flask. Color changes to indicate endpoint just like acid/base titrations. Example : The indicator, MnO 4-1 has an intense dark purple color. For the example below, MnO 4-1 is a dark purple reactant in the burette. It will colorlessly react with oxalic acid in the flask until the oxalic acid runs out. At the endpoint, excess MnO 4-1 turns the flask solution purple. Calcium levels in blood may be determined by adding oxalate ion to precipitate the calcium oxalate, CaC 2 O 4 Ca +2 (in blood) + C 2 O 4-2 CaC 2 O 4 (s) (solid is separated out of the blood) followed by dissolving the precipitate in acid CaC 2 O 4 (s) + 2H +1 (aq) H 2 C 2 O 4 (aq) + Ca +2 and titrating the resulting oxalic acid with KMnO 4 (K + is a spectator ion which is left out of the reaction.) 5H 2 C 2 O 4 (aq) + 2MnO 4-1 (aq) + 6H +1 (aq) 10CO 2 (g) + 2Mn +2 (aq) +8H 2 O(l) How many milligrams of Ca +2 are present in 10.0 ml of blood if 21.08 ml of 0.00988M KMnO 4 solution is needed?
C h e m i s t r y 1 A : C h a p t e r 4 P a g e 15 Memorize the Soluble Solubility Rules: Compounds containing these ions are generally soluble Li +, Na +, K +, NH 4 + NO 3 -, C 2 H 3 O 2 - Exceptions: (when combined with ions on left, compound is insoluble) none none Cl -, Br -, I - Ag +, Hg 2 +2, Pb +2 SO 4-2 Ag +, Ca +2, Sr +2, Ba +2, Pb +2 Be able to use the Insoluble Solubility Rules: Compounds containing these ions are generally insoluble OH - Li +, Na +, K +, NH 4 + Exceptions: (when combined with ions on left, compound is soluble or slightly soluble Ca +2, Sr +2, Ba +2 S -2 Li +, Na +, K +, NH 4 +, Ca +2, Sr +2, Ba +2 CO 3-2, PO 4-3 Li +, Na +, K +, NH 4 + Practice: 11. List elements that are diatomic. 12. Name several driving forces for a chemical reaction. 13. Name several signs of evidence that would indicate a chemical reaction happened. 14. What information can one obtain from a balanced chemical equation? 15. a) Complete and balance the double-displacement whole equation. Include the physical states. Na 2 SO 4 (aq) + AgNO 3 (aq) b) Write the balanced total ionic equation for the above reaction. Include charges on ions and physical states. Circle the spectator ions. c) Write the balanced net ionic equation for this reaction. d) What is/are the driving force(s) for this reaction?
C h e m i s t r y 1 A : C h a p t e r 4 P a g e 16 16. Crude gunpowders often contain a mixture of solid potassium nitrate and charcoal (solid carbon). When such a mixture is heated until a reaction occurs, a solid residue of potassium carbonate is produced. The explosive force of the gunpowder comes from the fact that two gases are also produced, carbon monoxide and nitrogen, which increase in volume with great force and speed. Write the balanced chemical equation for the reaction. 17. Balance and Classify in as many ways possible the following reactions. Possible classifications: (R) redox or oxidation-reduction (D) decomposition (N) neutralization (C) combustion (G) gas evolution (D-D) double displacement (SR) single replacement (Syn)synthesis or combination (P) precipitation a) NaI (s) + F 2 (g) NaF (s) + I 2 (s) b) NH 3 (g) + O 2 (g) NO 2 (g) + H 2 O (l) c) N 2 (g) + O 2 (g) N 2 O 5 (g) d) C 4 H 10 (l) + O 2 (g) CO 2 (g) + H 2 O (g) e) As(OH) 3 (aq) + H 2 SO 4 (aq) As 2 (SO 4 ) 3 (aq) + H 2 O (l) f) aqueous silver nitrate and solid copper reacts to form aqueous copper (II) nitrate and solid silver. g) solid magnesium chlorate heated produces oxygen gas and solid magnesium chloride
C h e m i s t r y 1 A : C h a p t e r 4 P a g e 17 h) A chemical process that uses electricity is called an electrolysis reaction. When an electrical current passes through a concentrated aqueous sodium chloride solution, several useful products can be isolated: aqueous sodium hydroxide, chlorine gas, and hydrogen gas. Write the balanced chemical equation for this process. i) Fe (s) + Br2 (l) FeBr3 (s) j) H2SO4(aq) + Na2CO3(aq) Na2SO4(aq) + H2O(l) + CO2(g) k) Al 2 (CO 3 ) 3 (s) Al 2 O 3 (s) + CO 2 (g) l) HC 2 H 3 O 2 (aq) + Mg (s) H 2 (g) + Mg(C 2 H 3 O 2 ) 2 (aq) m) Ni(NO 3)2 (aq) + K 2 S (aq) NiS (s) + KNO 3 (aq) n) KOH (aq) + (NH 4 ) 2 SO 4 (aq) NH 3 (g) + K 2 SO 4 (aq) + H 2 O (l) 18. For the following reactions below: Complete and balance the reactions. Indicate what the driving force(s) of the reaction will be. If there are no driving forces write NR for no reaction. a & b are double displacement; c is synthesis. DRIVING FORCE(S) PRODUCTS a) HBr (aq) + Li 2 CO 3 (aq) + + b) Pb(NO 3 ) 2 (aq) + (NH 4 ) 2 S (aq) + c) Cl 2 (g) + Al (s)
C h e m i s t r y 1 A : C h a p t e r 4 P a g e 18 19. Complete and Balance the following chemical reactions. Include the state that the substances are in, (s), (l), (g), or (aq). If no reaction in a double displacement then write No reaction, but balance it anyway. a) Combustion reaction: C 8 H 18 (l) + O 2 (g) b) Single replacement reaction: (Zn > Cu) Solid zinc is added to aqueous copper (II) chloride c) Decomposition reaction: Fe(HCO 3 ) 3 (s) d) Double Displacement reaction: aqueous sulfuric acid is added to aqueous ammonium carbonate e) Synthesis reaction: N 2 O 5 (g) + H 2 O (l) f) Double displacement reaction: KNO 3 (aq) + (NH 4 ) 3 PO 4 (aq) g) Double displacement reaction: aqueous acetic acid + aqueous barium hydroxide h) Decomposition: CaCO 3 (s) j) Neutralization: HCl (aq) + K 2 SO 3 (aq)