2 A solution is a homogenous mixture of 2 or more substances The solute is(are) the substance(s) present in the smaller amount(s) The solvent is the substance present in the larger amount SolutionSolventSolute Soft drink(l) Air(g) Soft Solder(s) H2OH2O N2N2 Pb Sugar, CO 2 O 2, Ar, CH 4 Sn aqueous solutions of KMnO 4
3 An electrolyte is a substance that, when dissolved in water, results in a solution that can conduct electricity. A nonelectrolyte is a substance that, when dissolved, results in a solution that does not conduct electricity. nonelectrolyte weak electrolyte strong electrolyte
4 Strong Electrolyte – 100% dissociation NaCl(s) Na + (aq) + Cl - (aq) H2OH2O Weak Electrolyte – not completely dissociated CH 3 COOH CH 3 COO - (aq) + H + (aq) Conduct electricity in solution? Cations (+) and Anions (-)
5 Ionization of acetic acid CH 3 COOH CH 3 COO - (aq) + H + (aq) A reversible reaction. The reaction can occur in both directions. Acetic acid is a weak electrolyte because its ionization in water is incomplete.
6 Hydration is the process in which an ion is surrounded by water molecules arranged in a specific manner. H2OH2O
7 Nonelectrolyte does not conduct electricity? No cations (+) and anions (-) in solution C 6 H 12 O 6 (s) C 6 H 12 O 6 (aq) H2OH2O
8 Precipitation Reactions Precipitate – insoluble solid that separates from solution molecular equation ionic equation net ionic equation Pb 2+ + 2NO 3 - + 2Na + + 2I - PbI 2 (s) + 2Na + + 2NO 3 - Na + and NO 3 - are spectator ions PbI 2 Pb(NO 3 ) 2 (aq) + 2NaI (aq) PbI 2 (s) + 2NaNO 3 (aq) precipitate Pb 2+ + 2I - PbI 2 (s)
9 Solubility is the maximum amount of solute that will dissolve in a given quantity of solvent at a specific temperature.
10 Writing Net Ionic Equations 1.Write the balanced molecular equation. 2.Write the ionic equation showing the strong electrolytes completely dissociated into cations and anions. 3.Cancel the spectator ions on both sides of the ionic equation 4.Check that charges and number of atoms are balanced in the net ionic equation AgNO 3 (aq) + NaCl(aq) AgCl(s) + NaNO 3 (aq) Ag + + NO 3 - + Na + + Cl - AgCl(s) + Na + + NO 3 - Ag + + Cl - AgCl(s) Write the net ionic equation for the reaction of silver nitrate with sodium chloride.
11 Properties of Acids Have a sour taste. Vinegar owes its taste to acetic acid. Citrus fruits contain citric acid. React with certain metals to produce hydrogen gas. React with carbonates and bicarbonates to produce carbon dioxide gas Cause color changes in plant dyes. 2HCl(aq) + Mg(s) MgCl 2 (aq) + H 2 (g) 2HCl(aq) + CaCO 3 (s) CaCl 2 (aq) + CO 2 (g) + H 2 O(l) Aqueous acid solutions conduct electricity.
12 Have a bitter taste. Feel slippery. Many soaps contain bases. Properties of Bases Cause color changes in plant dyes. Aqueous base solutions conduct electricity. Examples:
13 Arrhenius acid is a substance that produces H + (H 3 O + ) in water Arrhenius base is a substance that produces OH - in water
14 A Brønsted acid is a proton donor A Brønsted base is a proton acceptor acidbaseacidbase A Brønsted acid must contain at least one ionizable proton!
15 Monoprotic acids HCl H + + Cl - HNO 3 H + + NO 3 - CH 3 COOH H + + CH 3 COO - Strong electrolyte, strong acid Weak electrolyte, weak acid Diprotic acids H 2 SO 4 H + + HSO 4 - HSO 4 - H + + SO 4 2- Strong electrolyte, strong acid Weak electrolyte, weak acid Triprotic acids H 3 PO 4 H + + H 2 PO 4 - H 2 PO 4 - H + + HPO 4 2- HPO 4 2- H + + PO 4 3- Weak electrolyte, weak acid
16 Identify each of the following species as a Brønsted acid, base, or both. (a) HI, (b) CH 3 COO -, (c) H 2 PO 4 - HI (aq) H + (aq) + I - (aq)Brønsted acid CH 3 COO - (aq) + H + (aq) CH 3 COOH (aq)Brønsted base H 2 PO 4 - (aq) H + (aq) + HPO 4 2- (aq) H 2 PO 4 - (aq) + H + (aq) H 3 PO 4 (aq) Brønsted acid Brønsted base
17 Neutralization Reaction acid + base salt + water HCl(aq) + NaOH(aq) NaCl(aq) + H 2 O H + + Cl - + Na + + OH - Na + + Cl - + H 2 O H + + OH - H 2 O
18 Neutralization Reaction Involving a Weak Electrolyte weak acid + base salt + water HCN(aq) + NaOH(aq) NaCN(aq) + H 2 O HCN + Na + + OH - Na + + CN - + H 2 O HCN + OH - CN - + H 2 O
19 Neutralization Reaction Producing a Gas acid + base salt + water + CO 2 2HCl(aq) + Na 2 CO 3 (aq) 2NaCl(aq) + H 2 O +CO 2 2H + + 2Cl - + 2Na + + CO 3 2- 2Na + + 2Cl - + H 2 O + CO 2 2H + + CO 3 2- H 2 O + CO 2
20 Oxidation-Reduction Reactions (electron transfer reactions) 2Mg 2Mg 2+ + 4e - O 2 + 4e - 2O 2- Oxidation half-reaction (lose e - ) Reduction half-reaction (gain e - ) 2Mg + O 2 + 4e - 2Mg 2+ + 2O 2- + 4e - 2Mg + O 2 2MgO
21 Zn(s) + CuSO 4 (aq) ZnSO 4 (aq) + Cu(s) Zn is oxidizedZn Zn 2+ + 2e - Cu 2+ is reducedCu 2+ + 2e - Cu Zn is the reducing agent Cu 2+ is the oxidizing agent Copper wire reacts with silver nitrate to form silver metal. What is the oxidizing agent in the reaction? Cu(s) + 2AgNO 3 (aq) Cu(NO 3 ) 2 (aq) + 2Ag(s) Cu Cu 2+ + 2e - Ag + + 1e - AgAg + is reducedAg + is the oxidizing agent
22 Oxidation number The charge the atom would have in a molecule (or an ionic compound) if electrons were completely transferred. 1.Free elements (uncombined state) have an oxidation number of zero. Na, Be, K, Pb, H 2, O 2, P 4 = 0 2.In monatomic ions, the oxidation number is equal to the charge on the ion. Li +, Li = +1; Fe 3+, Fe = +3; O 2-, O = -2 3.The oxidation number of oxygen is usually –2. In H 2 O 2 and O 2 2- it is –1. 4.4
23 4.The oxidation number of hydrogen is +1 except when it is bonded to metals in binary compounds. In these cases, its oxidation number is –1. 6. The sum of the oxidation numbers of all the atoms in a molecule or ion is equal to the charge on the molecule or ion. 5.Group IA metals are +1, IIA metals are +2 and fluorine is always –1. HCO 3 - O = – 2H = +1 3x( – 2) + 1 + ? = – 1 C = +4 What are the oxidation numbers of all the elements in HCO 3 - ? 7. Oxidation numbers do not have to be integers. Oxidation number of oxygen in the superoxide ion, O 2 -, is –½.
24 The Oxidation Numbers of Elements in their Compounds
25 NaIO 3 Na = +1O = -2 3x(-2) + 1 + ? = 0 I = +5 IF 7 F = -1 7x(-1) + ? = 0 I = +7 K 2 Cr 2 O 7 O = -2K = +1 7x(-2) + 2x(+1) + 2x(?) = 0 Cr = +6 What are the oxidation numbers of all the elements in each of these compounds? NaIO 3 IF 7 K 2 Cr 2 O 7
26 Types of Oxidation-Reduction Reactions Combination Reaction A + B C 3Mg + N 2 MgN 2 Decomposition Reaction 2KClO 3 2KCl + 3O 2 C A + B 00+2-3 +1+5-2+10
27 Types of Oxidation-Reduction Reactions Combustion Reaction A + O 2 B S + O 2 SO 2 00 +4-2 2Mg + O 2 2MgO 00 +2-2
28 Displacement Reaction A + BC AC + B Sr + 2H 2 O Sr(OH) 2 + H 2 TiCl 4 + 2Mg Ti + 2MgCl 2 Cl 2 + 2KBr 2KCl + Br 2 Hydrogen Displacement Metal Displacement Halogen Displacement Types of Oxidation-Reduction Reactions 0 +1+20 0+40+2 0 0
29 The Activity Series for Metals M + BC MC + B Hydrogen Displacement Reaction M is metal BC is acid or H 2 O B is H 2 Ca + 2H 2 O Ca(OH) 2 + H 2 Pb + 2H 2 O Pb(OH) 2 + H 2
30 The Activity Series for Halogens Halogen Displacement Reaction Cl 2 + 2KBr 2KCl + Br 2 0 0 F 2 > Cl 2 > Br 2 > I 2 I 2 + 2KBr 2KI + Br 2
31 Ca 2+ + CO 3 2- CaCO 3 NH 3 + H + NH 4 + Zn + 2HCl ZnCl 2 + H 2 Ca + F 2 CaF 2 Precipitation Acid-Base Redox (H 2 Displacement) Redox (Combination) Classify each of the following reactions.
32 What type of reaction is shown below? 2Mg + O 2 2MgO a. combination b.single replacement c. neutralization d.decomposition
33 What type of reaction is shown below? 2Na + Cl 2 2NaCl a. combination b.single replacement c. neutralization d.decomposition
34 Solution Stoichiometry The concentration of a solution is the amount of solute present in a given quantity of solvent or solution. M = molarity = moles of solute liters of solution What mass of KI is required to make 500. mL of a 2.80 M KI solution? volume of KI solutionmoles KIgrams KI M KI 500. mL= 232 g KI 166 g KI 1 mol KI x 2.80 mol KI 1 L soln x 1 L 1000 mL x
36 Dilution is the procedure for preparing a less concentrated solution from a more concentrated solution. Dilution Add Solvent Moles of solute before dilution (i) Moles of solute after dilution (f) = MiViMiVi MfVfMfVf =
37 How would you prepare 60.0 mL of 0.200 M HNO 3 from a stock solution of 4.00 M HNO 3 ? M i V i = M f V f M i = 4.00 M M f = 0.200 MV f = 0.0600 L V i = ? L V i = MfVfMfVf MiMi = 0.200 M x 0.0600 L 4.00 M = 0.00300 L = 3.00 mL Dilute 3.00 mL of acid with water to a total volume of 60.0 mL.
38 Titrations In a titration a solution of accurately known concentration is added gradually added to another solution of unknown concentration until the chemical reaction between the two solutions is complete. Equivalence point – the point at which the reaction is complete Indicator – substance that changes color at (or near) the equivalence point Slowly add base to unknown acid UNTIL the indicator changes color
39 Titrations can be used in the analysis of Acid-base reactions Redox reactions H 2 SO 4 + 2NaOH 2H 2 O + Na 2 SO 4 5Fe 2+ + MnO 4 - + 8H + Mn 2+ + 5Fe 3+ + 4H 2 O
40 What volume of a 1.420 M NaOH solution is required to titrate 25.00 mL of a 4.50 M H 2 SO 4 solution? WRITE THE CHEMICAL EQUATION! volume acidmoles redmoles basevolume base H 2 SO 4 + 2NaOH 2H 2 O + Na 2 SO 4 4.50 mol H 2 SO 4 1000 mL soln x 2 mol NaOH 1 mol H 2 SO 4 x 1000 ml soln 1.420 mol NaOH x 25.00 mL = 158 mL M acid rxn coef. M base
41 WRITE THE CHEMICAL EQUATION! volume redmoles redmoles oxidM oxid 0.1327 mol KMnO 4 1 L x 5 mol Fe 2+ 1 mol KMnO 4 x 1 0.02500 L Fe 2+ x 0.01642 L = 0.4358 M M red rxn coef. V oxid 5Fe 2+ + MnO 4 - + 8H + Mn 2+ + 5Fe 3+ + 4H 2 O 16.42 mL of 0.1327 M KMnO 4 solution is needed to oxidize 25.00 mL of an acidic FeSO 4 solution. What is the molarity of the iron solution? 16.42 mL = 0.01642 L25.00 mL = 0.02500 L
46 Gases assume the volume and shape of their containers. Gases are the most compressible state of matter. Gases will mix evenly and completely when confined to the same container. Gases have much lower densities than liquids and solids. Physical Characteristics of Gases NO 2 gas
47 Units of Pressure 1 pascal (Pa) = 1 N/m 2 1 atm = 760 mmHg = 760 torr Pressure = Force Area (force = mass x acceleration)
48 P x V = constant P 1 x V 1 = P 2 x V 2 Boyle’s Law Constant temperature Constant amount of gas P 1/V For a fixed amount of an ideal gas kept at a fixed temperature, P and V are inversely proportional
49 A sample of chlorine gas occupies a volume of 946 mL at a pressure of 726 mmHg. What is the pressure of the gas (in mmHg) if the volume is reduced at constant temperature to 154 mL? P 1 x V 1 = P 2 x V 2 P 1 = 726 mmHg V 1 = 946 mL P 2 = ? V 2 = 154 mL P 2 = P 1 x V 1 V2V2 726 mmHg x 946 mL 154 mL = = 4460 mmHg P x V = constant
50 As T increasesV increases Variation in Gas Volume with Temperature at Constant Pressure V TV T
51 Variation of Gas Volume with Temperature at Constant Pressure V TV T V/T = constant V 1 /T 1 = V 2 /T 2 T (K) = t ( ° C) + 273.15 Charles’ & Gay-Lussac’s Law Temperature must be in Kelvin
52 A sample of carbon monoxide gas occupies 3.20 L at 125 ° C. At what temperature will the gas occupy a volume of 1.54 L if the pressure remains constant? V 1 = 3.20 L T 1 = 398.15 K V 2 = 1.54 L T 2 = ? T 2 = V 2 x T 1 V1V1 1.54 L x 398.15 K 3.20 L = = 192 K V 1 /T 1 = V 2 /T 2 T 1 = 125 ( ° C) + 273.15 (K) = 398.15 K
53 Avogadro’s Law V number of moles (n) V = constant x n V 1 / n 1 = V 2 / n 2 Constant temperature Constant pressure
54 Ammonia burns in oxygen to form nitric oxide (NO) and water vapor. How many volumes of NO are obtained from one volume of ammonia at the same temperature and pressure? 4NH 3 + 5O 2 4NO + 6H 2 O 1 mole NH 3 1 mole NO At constant T and P 1 volume NH 3 1 volume NO
58 Ideal Gas Equation Charles’ law: V T (at constant n and P) Avogadro’s law: V n (at constant P and T) Boyle’s law: P (at constant n and T) 1 V V V nT P V = constant x = R nT P P R is the gas constant PV = nRT
59 The conditions 0 ° C and 1 atm are called standard temperature and pressure (STP). PV = nRT R = PV nT = (1 atm)(22.414L) (1 mol)(273.15 K) R = 0.082057 L atm / (mol K) Experiments show that at STP, 1 mole of an ideal gas occupies 22.414 L.
60 What is the volume (in liters) occupied by 49.8 g of HCl at STP? PV = nRT V = nRT P T = 0 ° C = 273.15 K P = 1 atm n = 49.8 g x 1 mol HCl 36.45 g HCl = 1.37 mol V = 1 atm 1.37 mol x 0.0821 x 273.15 K Latm molK V = 30.7 L
61 Density (d) Calculations d = m V = PMPM RT m is the mass of the gas in g M is the molar mass of the gas Molar Mass ( M ) of a Gaseous Substance dRT P M = d is the density of the gas in g/L
62 A 2.10-L vessel contains 4.65 g of a gas at 1.00 atm and 27.0 ° C. What is the molar mass of the gas? dRT P M = d = m V 4.65 g 2.10 L = = 2.21 g L M = 2.21 g L 1 atm x 0.0821 x 300.15 K Latm molK M = 54.5 g/mol
63 Gas Stoichiometry What is the volume of CO 2 produced at 37 0 C and 1.00 atm when 5.60 g of glucose are used up in the reaction: C 6 H 12 O 6 (s) + 6O 2 (g) 6CO 2 (g) + 6H 2 O(l) g C 6 H 12 O 6 mol C 6 H 12 O 6 mol CO 2 V CO 2 5.60 g C 6 H 12 O 6 1 mol C 6 H 12 O 6 180 g C 6 H 12 O 6 x 6 mol CO 2 1 mol C 6 H 12 O 6 x = 0.187 mol CO 2 V = nRT P 0.187 mol x 0.0821 x 310.15 K Latm molK 1.00 atm = = 4.76 L
64 Dalton’s Law of Partial Pressures V and T are constant P1P1 P2P2 P total = P 1 + P 2
65 Consider a case in which two gases, A and B, are in a container of volume V. P A = n A RT V P B = n B RT V n A is the number of moles of A n B is the number of moles of B P T = P A + P B X A = nAnA n A + n B X B = nBnB n A + n B P A = X A P T P B = X B P T P i = X i P T mole fraction (X i ) = nini nTnT
66 A sample of natural gas contains 8.24 moles of CH 4, 0.421 moles of C 2 H 6, and 0.116 moles of C 3 H 8. If the total pressure of the gases is 1.37 atm, what is the partial pressure of propane (C 3 H 8 )? P i = X i P T X propane = 0.116 8.24 + 0.421 + 0.116 P T = 1.37 atm = 0.0132 P propane = 0.0132 x 1.37 atm= 0.0181 atm
68 Kinetic Molecular Theory of Gases 1.A gas is composed of molecules that are separated from each other by distances far greater than their own dimensions. The molecules can be considered to be points; that is, they possess mass but have negligible volume. 2.Gas molecules are in constant motion in random directions, and they frequently collide with one another. Collisions among molecules are perfectly elastic. 3.Gas molecules exert neither attractive nor repulsive forces on one another. 4.The average kinetic energy of the molecules is proportional to the temperature of the gas in kelvins. Any two gases at the same temperature will have the same average kinetic energy
69 Kinetic theory of gases and … Compressibility of Gases Boyle’s Law P collision rate with wall Collision rate number density Number density 1/V P 1/V Charles’ Law P collision rate with wall Collision rate average kinetic energy of gas molecules Average kinetic energy T P TP T
70 Kinetic theory of gases and … Avogadro’s Law P collision rate with wall Collision rate number density Number density n P nP n Dalton’s Law of Partial Pressures Molecules do not attract or repel one another P exerted by one type of molecule is unaffected by the presence of another gas P total = P i
71 The distribution of speeds for nitrogen gas molecules at three different temperatures The distribution of speeds of three different gases at the same temperature
72 Gas diffusion is the gradual mixing of molecules of one gas with molecules of another by virtue of their kinetic properties. NH 3 17 g/mol HCl 36 g/mol NH 4 Cl r1r1 r2r2 M2M2 M1M1 = molecular path
73 Deviations from Ideal Behavior 1 mole of ideal gas PV = nRT n = PV RT = 1.0 Repulsive Forces Attractive Forces
74 Effect of intermolecular forces on the pressure exerted by a gas.
75 Van der Waals equation nonideal gas P + (V – nb) = nRT an 2 V2V2 () } corrected pressure } corrected volume
76 A 13 gram gaseous sample of an unknown hydrocarbon occupies a volume of 11.2 L at STP. What is the hydrocarbon a. CH b. C 2 H 4 c. C 2 H 2 d. C 3 H 3
77 A force is applied o a container of gas reducing its volume by half. The temperature of the gas: a. decreases b. increases c. remains constant d. the temperature change depends upon the amount of force used
78 Ammonia burns in air to form nitrogen dioxide and water. 4NH 3 +7O 2 4NO 2 + 6H 2 O If 8 moles of NH3 are reacted with 14 moles of O 2 in a rigid container with an initial pressure of 11 atm, what is the partial pressure of NO 2 in the container when the reaction runs to completion? ( Assume constant temperature) a. 4 atm b. 6 atm c. 11 atm d. 12 atm
79 At STP, one liter of which of the following gases contains the most molecules? a. H 2 b. He 2 c. N 2 d. Each gas contains the same number of molecules at STP
81 Energy is the capacity to do work. Radiant energy comes from the sun and is earth’s primary energy source Thermal energy is the energy associated with the random motion of atoms and molecules Chemical energy is the energy stored within the bonds of chemical substances Nuclear energy is the energy stored within the collection of neutrons and protons in the atom Potential energy is the energy available by virtue of an object’s position
82 Heat is the transfer of thermal energy between two bodies that are at different temperatures. Energy Changes in Chemical Reactions Temperature is a measure of the thermal energy. Temperature = Thermal Energy
83 Thermochemistry is the study of heat change in chemical reactions. The system is the specific part of the universe that is of interest in the study. open mass & energyExchange: closed energy isolated nothing
84 Exothermic process is any process that gives off heat – transfers thermal energy from the system to the surroundings. Endothermic process is any process in which heat has to be supplied to the system from the surroundings. 2H 2 (g) + O 2 (g) 2H 2 O (l) + energy H 2 O (g) H 2 O (l) + energy energy + 2HgO (s) 2Hg (l) + O 2 (g) energy + H 2 O (s) H 2 O (l)
85 Thermodynamics is the scientific study of the interconversion of heat and other kinds of energy. State functions are properties that are determined by the state of the system, regardless of how that condition was achieved. Potential energy of hiker 1 and hiker 2 is the same even though they took different paths. energy, pressure, volume, temperature U = U final - U initial P = P final - P initial V = V final - V initial T = T final - T initial
86 First law of thermodynamics – energy can be converted from one form to another, but cannot be created or destroyed. U system + U surroundings = 0 or U system = - U surroundings C 3 H 8 + 5O 2 3CO 2 + 4H 2 O Exothermic chemical reaction! Chemical energy lost by combustion = Energy gained by the surroundings system surroundings U is the change in internal energy of a system
87 Enthalpy and the First Law of Thermodynamics U = q + w U = H - P V H = U + P V q = H and w = -P V At constant pressure: U is the change in internal energy of a system
88 Enthalpy (H) is used to quantify the heat flow into or out of a system in a process that occurs at constant pressure. H = H (products) – H (reactants) H = heat given off or absorbed during a reaction at constant pressure H products < H reactants H < 0 H products > H reactants H > 0
89 Thermochemical Equations H 2 O (s) H 2 O (l) H = 6.01 kJ/mol Is H negative or positive? System absorbs heat Endothermic H > 0 6.01 kJ are absorbed for every 1 mole of ice that melts at 0 0 C and 1 atm.
90 Thermochemical Equations CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O (l) H = -890.4 kJ/mol Is H negative or positive? System gives off heat Exothermic H < 0 890.4 kJ are released for every 1 mole of methane that is combusted at 25 0 C and 1 atm.
91 H 2 O (s) H 2 O (l) H = 6.01 kJ/mol The stoichiometric coefficients always refer to the number of moles of a substance Thermochemical Equations If you reverse a reaction, the sign of H changes H 2 O (l) H 2 O (s) H = - 6.01 kJ/mol If you multiply both sides of the equation by a factor n, then H must change by the same factor n. 2H 2 O (s) 2H 2 O (l) H = 2 x 6.01 = 12.0 kJ
92 H 2 O (s) H 2 O (l) H = 6.01 kJ/mol The physical states of all reactants and products must be specified in thermochemical equations. Thermochemical Equations H 2 O (l) H 2 O (g) H = 44.0 kJ/mol How much heat is evolved when 266 g of white phosphorus (P 4 ) burn in air? P 4 (s) + 5O 2 (g) P 4 O 10 (s) H = -3013 kJ/mol 266 g P 4 1 mol P 4 123.9 g P 4 x 3013 kJ 1 mol P 4 x = 6470 kJ
93 A Comparison of H and U 2Na(s) + 2H 2 O(l) 2NaOH(aq) + H 2 (g) H = -367.5 kJ/mol U = H - P V At 25 ° C, 1 mole H 2 = 24.5 L at 1 atm P V = 1 atm x 24.5 L = 2.5 kJ U = -367.5 kJ/mol – 2.5 kJ/mol = -370.0 kJ/mol
94 The specific heat (s) of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius. The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius. C = m x s Heat (q) absorbed or released: q = m x s x t q = C x t t = t final - t initial
95 How much heat is given off when an 869 g iron bar cools from 94 o C to 5 o C? s of Fe = 0.444 J/g ° C t = t final – t initial = 5 ° C – 94 ° C = -89 ° C q = ms t = 869 g x 0.444 J/g ° C x –89 ° C= -34,000 J