DAWSON COLLEGE DEPARTMENT OF CHEMISTRY & CHEMICAL TECHNOLOGY FINAL EXAMINATION CHEMISTRY 202-NYB-05 May 2, 200 9:30 2:30 Print your Name: Student Number: MARK DISTRIBUTION. / 8 INSTRUCTORS: Please circle the name of your instructor: 2. / 5 3. / 3 INSTRUCTIONS: J. Ali I. Dionne M. Haniff D. Baril M. Di Stefano S. Holden O. Behar N. Duxin / Y-S. Uh S. Mutic This exam set consists of 6 questions. Please ensure that your copy of this examination is complete. Answer all questions in the space provided.. Calculators may not be shared. Programmable calculators are not permitted. 2. No books or extra paper are permitted. 3. In order to obtain full credit, you must show the method used to solve all problems involving calculations and express your answers to the correct number of significant figures. 4. Your attention is drawn to the College policy on cheating. 5. A Periodic Table is provided. (last page). 6. If a mathematical equation is used to solve a problem, the equation should be clearly written. 7. Write your answer in the appropriate space when required. USEFUL DATA: Avogadro s Number N A 6.022 x 0 23 mol Gas Constant 0.08206 L atm K mol R 8.34 L kpa K mol 8.34 J K mol atm 0.3 kpa 760 mmhg 760 torr J kg m 2 s 2 0.3 J L atm 4. / 6 5. / 5 6. / 8 7. / 6 8. / 6 9. / 9 0. / 7. / 6 2. / 7 3. / 7 4. / 6 5. / 6 6. / 5 TOTAL /00
Question Ethanol is the common alcohol with molecular formula C 2 H 5 OH. An alcohol-water solution is prepared by dissolving 0.00 cm 3 of ethanol, with density d ethanol 0.789 g/cm 3, in a sufficient volume of water to produce 00.00 cm 3 of solution. Density of solution is d soln 0.932 g/cm 3. For a given solution calculate the following for ethanol: - 2 - a. the mass percent For 00.00 cm 3 of solution: 00.00 cm 3 x 0.932 g cm 3 93.2 g solution In this solution 0.00 cm 3 ethanol are present: 0.00 cm 3 x 0.789 g cm 3 7.89 g ethanol Mass percent 7.89 g 93.2 g b. the molarity x 00% 8.46% ethanol For.00 L of solution:.00 L x 0.932 g ml x 000 ml L 932 g solution Ans. Mass%: In this solution 8.46% is ethanol then: 932 g solution x 7.80 g ethanol 00. g solution 78.8g ethanol Molarity mol solute L solution 78.8 g x mol ethanol 46.07 g ethanol.00 L.7 M ethanol c. the molality Ans. molarity: Mole of ethanol for.00l solution 78.8 g x mol 46.07 g.7 mol Number of kilogram of solvent in one liter of solution: 932 g solution - 78.8 g ethanol 853 g solvent Molality mol solute kg solvent 853 g x.7 mol kg 000 g solvent 2.00 molal ethanol Ans. molality: d. the mole fraction. Mole of ethanol for.00 L solution 78.8 g x mol 46.07 g.7 mol Mole of solvent for.00 L solution 853 g solvent mol 8.05 g 47.3 mol Mole fraction mol solute total mol solution.7 mol.7 mol + 47.3 mol 3.49x0-2 ethanol Ans. Mole fraction:
Question 2 Toluene, C 7 H 8 is a component of gasoline (octane, C 8 H 8 ). It is present in gasoline as an octane booster at concentrations between 3 to 5% by mass (25% in racing cars gasoline). Consider a solution of octane with 20.% by mass of toluene at 20 C - 3 - a. Calculate the total vapor pressure of this solution Data: P octane 0.5 mm Hg at 20 C, T b 26 C P toluene 22 mm Hg at 20 C, T b C (3 marks) for 00 g of solution you have : (keep one extra sig.fig. for calculations) 00. g solution x 20 % toluene 20. g toluene 00. g solution - 20. g toluene 80. g octane n toluene 20 g x mol toluene 92.4 g 0.27 mol n octane 80 g x mol octane 4.23 g 0.700 mol χ toluene 0.27 mol 0.27 mol + 0.700 mol 0.237 χ octane - 0.237 0.763 (only 2 components) o Total vapor pressure P toluene o x χ toluene + P octane x χ octane (22 mm Hg x 0.237) + (0.5 mm Hg x 0.763) 3 mm Hg ans. total vapor pressure: ( mark) b. Calculate the mole ratio of toluene to octane in the vapor phase above the solution Molar ratio n toluene Vap. n octane Vap. Since PV nrt then P n therefore: n toluene Vap. n octane Vap. P toluene P octane 22 mm Hg x 0.237 0.5 mm Hg x 0.763 5.2 mm Hg 8.0 mm Hg 0.65 ans. mole ratio: toluene/octane: c. If the actual vapor pressure measured is 5.2 mm Hg, will the boiling point of this solution be higher or lower than the one expected from Raoult s law? Explain. ( mark) The actual vapor pressure is higher than the ideal one (calculated) therefore, the boiling point will be lower than the one expected. The intermolecular forces between toluene - octane are weaker than the one of the pure compounds.
Question 3 A 0.46 g sample of cumene, a non-volatile non-ionic compound, is dissolved in 0.0 g cyclohexane (C 6 H 2 ), producing a solution that freezes at -.25 C. Cyclohexane has a normal freezing point of 6.50 C and a freezing point depression constant of 20.2 C/m. What is the molar mass of cumene? - 4 - (3 marks) Colligative properties involving the freezing point depression of a non ionic solute: ΔT f K f x m where ΔT f 6.50 C - (-.25 C) 7.25 C The concentration of cumene can be found: ΔT f K f m 7.75 C 20.2 C.m - 0.384 m The number of mole of cumene is: molality mole solute kg solvent Mole solute molality x kg solvent 0.384 m x 0.0 g x kg 000 g 3.84x0-3 mol Finally, the molar mass of cumene mass cumene mole cumene 0.46 g 0.00384 mol 20. g mol (or.20x0 2 g mol ) Ans. Mol. mass cumene:
Question 4 Hydrofluoric acid, (HF) is a weak acid that can be used in the fluoridation of water. An aqueous solution of 0.00 M HF has an osmotic pressure of 2.64 atm at 25 C. - 5 - a. Calculate the van t Hoff factor for HF at this concentration Colligative properties involving the osmotic pressure of an ionic compound in solution: i M R T i Π M R T 2.64 atm (0.00 mol L L.atm )(0.082 )(25 + 273)K K.mol.08 Ans. van t Hoff factor: b. Does it differ from the maximum van t Hoff factor expected for a monoprotic acid? If so, explain why. (2 mark) Yes it differs from the expected van't Hoff factor of 2 for a monoprotic acid. Therefore. HF is a week acid since its dissociation is partial because of ion pairing at this concentration corresponding to an incomplete dissociation of the acid (weak acid). c. What is the percent ionization of HF at this concentration? HF à H + + F - C-x x x i mole of particle in solution mole solute added 2x + (c - x) mole solute added x + C C Since I.08 and C 0.00 M à.08 x + 0.00 0.00 then x 0.008 % ionization x C x 00% 0.008 M x 00% 8% 0.00 M Ans. % ionization:
Question 5 Iodide ion is oxidized in acidic solution to triiodide ion I 3 by hydrogen peroxide. - 6 - H 2 O 2 (aq) + 3 I (aq) + 2 H + (aq) à I 3 (aq) + 2H 2 O(l) A series of four experiments was run at different concentrations, and the initial rates of I 3 formation were determined (see table). Initial concentration (mol L - ) Initial concentration (mol L - ) Initial concentration (mol L - ) Initial rate (mol L - s - ) H 2 O 2 I H + Exp 0.00 0.00 0.00050.5x0-6 Exp 2 0.020 0.00 0.00050 2.30x0-6 Exp 3 0.00 0.020 0.00050 2.30x0-6 Exp 4 0.00 0.00 0.0000.5x0-6 a. From the table above, obtain the reaction orders with respect to each of the following species: H 2 O 2, I, H +. (3 marks) The reaction rate is given by: rate k [H 2 O 2 ] α [I - ] β [H + ] γ For α : For β : For γ : Exp.2 Exp. 2.30x0-6 k(0.020) α (0.00) β (0.00050) γ -6.5x0 k(0.00) α (0.00) β (0.00050) γ à 2 2 α then α Exp.3 Exp. 2.30x0-6 k(0.00) α (0.020) β (0.00050) γ -6.5x0 k(0.00) α (0.00) β (0.00050) γ à 2 2 β then β Exp.4 Exp..5x0-6 k(0.00) α (0.00) β (0.0000) γ -6.5x0 k(0.00) α (0.00) β (0.00050) γ à 2 γ then γ 0 The reaction orders are: H 2 O 2 : I _ : H + : 0 Ans. Reaction order: H 2 O 2 : I : H + : b. Find the rate constant with its units. The reaction rate is given by: rate k [H 2 O 2 ] [I - ] Using Exp.4 :.6x0-6 M.s - k (0.00 M) (0.00 M) k.6x0-6 M.s - (.0x0-2 M) 2.2x0-2 - - M s Ans. rate constant:
Question 6 The reaction below was monitored as a function of time at a temperature of 400 K: - 7-2NOCl (g) 2NO(g) + Cl 2 (g) A plot of /[NOCl] against time yielded a straight line with slope of 6.7x0-4 L mol - s -. a. Write the rate law for the reaction. rate 6.7x0-4 L mol - s - [NOCl] 2 (second order) b. What is the half-life for the reaction if the initial concentration of NOCl is 0.20 M? t 2 k[a] o t 2 6.7x0-4 L.mol - s - [0.20 mol.l - ] 7.5x0 3 s. Ans. half-life: c. If the initial concentration of NOCl is 0.35 M, what is the concentration of NOCl after 5.0 min? [NOC_] k t + [NOC_] o [NOC_] ( 6.7x0-4 L.mol - s - ) x 5.0 min x 60 s min + 0.35 M 3.06 M- or [NOCl] 0.33 M Ans. [NOCl] after 5.0 min: d. If the initial concentration of NOCl is 0.35 M, How long will it take for the concentration to drop to 20% of its original value? [NOC_] k t + [NOC_] o t k x ( [NOC_] - ) [NOC_] o 6.7x0-4 L.mol - s - x ( 0.35 M x 20/00-0.35 M ).7x04 s. Ans. time after 20% drop:
Question 7 a. Consider the potential energy profiles for three different chemical reactions. - 8 - Indicate which reaction is the slowest one. Explain your choice Reaction is the slowest one since its activation energy is the highest. b. Consider the potential energy profiles for a chemical reaction. Mechanism 2A + B D 2A + B C D C Mechanism 2 2A B + C C D Mechanism 3 A A + B + C C 2D Circle the proposed mechanism that is consistent with the reaction profile shown and explain your choice. The curve is consistent with mechanism 2 since the slow step is the first one (highest E a ). The rate law will be rate k [A] 2 c. Beside concentration and pressure, give two parameters you can change that could affect the reaction rate of a chemical reaction: Temperature i. ii. Catalyst or enzyme
Question 8 At elevated temperature (997 C) limestone dissociates according to the equation - 9 - CaCO 3 (s) CaO(s) + CO 2 (g) ΔH +42.5 kj a. If 50.0 g CaCO 3 (00. g/mol) is placed in an evacuated 4.00 L container and heated up to 997 C, how many grams of CaCO 3 will decompose if the pressure at equilibrium is 392 kpa? Mole of CO 2 produced PV nrt PV n RT 392 kpa x 4.00 L 0.49 mole L.kPa 8.3 x (977 + 273)K K. mol Since it is a : ratio between CO 2 and CaCO 3 then: Mass of CaCO 3 consumed 0.49 mol CO 2 x CaCO CO 2 3 x 00. g.mol - 4.9 g b. If the volume of the container is expanded to 0.0 L at 997 C, what will be the CO 2 pressure at equilibrium? K [CO 2 ]. The concentration of CO 2 remains the same whatever the volume For a gas, concentration pressure. Therefore, pressure does not change 392 kpa or 3.87 atm. c. Calculate K c for this reaction at 997 C ( mark) ( mark) K [CO 2 ] since the other reactants are all solids. In part A, the number of mole of CO 2 for a container of 4.00 L was 0.49 mole. K 0.49 mol 4.00 L 0.0373 also K RT K p RT P where P is also obtained from PV nrt d. Predict the effect of each of the following changes will have on the equilibrium position. equilibrium position shift change to the left no change to the right i. CO 2 is added ii. iii. iv. CaCO 3 is added Pressure is increased (adding N 2 gas, volume unchanged) The temperature is increased
Question 9 Consider the following set of data: Formula K a (at 25 C) [Al(H 2 O) 6 ] 3+.4 0-5 HNO 2 4.0 0-4 HF 7.2 0-4 - 0 - a. What is the strongest acid in the table? HF ( mark) b. With the help of the table, arrange the following in order of most basic to least basic: H 2 O, NO 2, [Al(H 2 O) 5 OH] 2+ Most basic [Al(H 2 O) 5 OH] 2+ > NO 2 > H 2 O Least basic c. What is the value of K b for F at 25 C? K b 0-4 7.2x0-4.4x0 - Ans. K b : d. Write the chemical reaction represented by the K b for F in water and place the species involved in the appropriate place H 2 O + F - HF + OH - Acid Base Conjugate Conjugate acid base e. At 40 C, K w 2.9x0-4. What is the neutral ph of water at this temperature? H 2 O OH - + H + then K w [H + ][OH - ] at neutral ph: [H + ] [OH - ] then k w [H + ] 2.9x0-4.7x0-7 ph -log(.7x0-7 ) 6.77 Ans.
Question 0 a. A solution of the basic oxide CaO is prepared by adding water to 0.28 g CaO to make 0.50 L of solution. i. Write the equations for the reactions that occur when CaO is dissolved in water - - ( mark) ) CaO(s) + H 2 O(l) à Ca(OH) 2 (aq) 2) Ca(OH) 2 (aq) à Ca 2+ (aq) + 2OH (aq) or ) CaO(s) à Ca 2+ (aq) + O 2 (aq) 2) O 2 (aq) + H 2 O(l) à 2OH (aq) ii. Assuming that ion-pairing is non-existent, what is the expected ph of this solution? Mole of OH in solution: 0.28 g CaO x mol 56.08 g x Ca(OH) 2 CaO x 2 OH - Ca(OH) 2 0.00 mol OH Concentration of OH in solution: 0.00 mol OH- 0.50 L 0.020 M Finally, the ph will be: poh -log(0.020 M).70 ph 4 - poh 2.30 ans. ph: b. For which of the following salts will the solubility depend on ph? ph sensitive i. KClO 4 ii. Pb(OH) 2 ph independent iii. AgF iv. Ba(NO 3 ) 2 c. For each of the following salts dissolved in water, predict whether the aqueous solution will be acidic, neutral or basic. i. RbOH ii. NaIO acid neutral basic iii. NH 4 OH iv. LiClO 3
Question a. Consider 0.500 L of a buffer that consists of.50 M KClO (K a HClO 3.5x0-8 ) and 0.50 M HClO. What will be the ph of this buffer after the addition of 250 ml of.0 M HNO 3? (4 marks) - 2 - The system is the following: HClO(aq) H + (aq) + ClO (aq) (note H + H 3 O + ) Mol of HClO 0.500 L x 0.50 mol L Mol of ClO 0.500 L x.50 mol Mol of H + added 0.250 L x.0 mol L 0.25 mol 0.750 mol L 0.25 mol First we have a reaction between a strong acid and a week base: H + + ClO HClO Initial 0.25 0.75 0.25 Reaction -0.25-0.25 +0.25 After reaction 0 0.50 0.50 After, the system reach equilibrium: HClO H + + ClO I 0.50 0 0.50 C -x +x +x E 0.50-x +x 0.50+x Since we have a buffer, x << 0.50 AND the number of mole for the acid and the conjugated base are the same, then ph pka ph -log(3.5x0-8 ) 7.46 ans. ph: b. Which of the following mixtures would result in a buffer solution when 00 ml of each of the two solutions are mixed together? buffer not a buffer i. 0. M KOH and 0.2 M NH 3 ii. 0.2 M HCl and 0.2 M NH 3 iii. 0.2 M HNO 3 and 0.4 M NaNO 3 iv. 0. M HNO 3 and 0.2 M NaF
Question 2 Consider the following titration curve of trimethylamine (C 3 H 9 N) a weak base with 0.00 M HCl at 23 C. Initial solution: smaller K 50.0 ml of C 3 H 9 N, 4.00x0-2 b value M - 3 - ans. K b : a. Draw on the graph the shape of the titration curve if this base had a smaller K b value. b. Which letter (A to F) on the graph corresponds to each of the following? The equivalence point The point of half-neutralization The point corresponding to the pk a of C 3 H 9 NH + letter D B B c. When 5.0 ml of 0.00 M HCl is added, the ph of the solution is 9.255. Calculate K b of trimethylamine. Let's call trimethylamine TMA an its conjugated acid TMAH + Mole of TMA 4.00x0-2 M x 0.0500 L 2.00x0-3 mol Mole of acid added.00x0 - M x 0.050 L.50x0-3 mol The HCl reacts to completion (strong acid) with TMA according to: TMA + H + TMAH + Initial 2.00x0-3.50x0-3 0 Reaction -.50x0-3 -.50x0-3 +.50x0-3 After reaction 0.50x0-3 0.50x0-3 Then, the system will reach equilibrium from those initial conditions: TMA + H 2 O TMAH + + OH 0.50x0-3 -x mole.50x0-3 + x mole x mole In a buffer regime, x is small compared to [base] or [conjugated acid] and can be neglected. In this case ONLY the ratio of mole of TMAH + / TMA is the same as the one of concentrations. Therefore: K b [TMAH+ ][OH ] [TMA] (.50x0-3 )(0 (4 9.255) ) 0.50x0-3 5.4x0-5
- 4 - ( mark) (3 marks) (3 marks)
Question 3 a. A saturated aqueous solution of Mg(OH) 2 has a ph of 0.08, what is the K sp of Mg(OH) 2? - 5 - Mg(OH) 2 (s) Mg 2+ (aq) + 2 OH (aq) K sp [Mg 2+ ][OH ] 2 I Solid 0 0 C -x +x +2x E solid x +2x K sp (x)(2x) 2 4x 3 We can use the ph to know [OH ] that is equal to 2x poh 4-0.08 3.92, [OH ] 0-3.92.2x0-4 M. since 2x.2x0-4 M. then x 6.0x0-5. finally: K sp 4 (6.0x0-5 ) 3 K sp 8.6x0-3 ans. K sp: b. The K sp of cobalt(iii) hydroxide is 2.5x0-43. Calculate the solubility of Co(OH) 3 in water in mol/l According to the K sp value, the concentration of OH generated by the salt will be small compared to the one of natural ionization of water [OH ] 0-7 M therefore, common ion effect. Co(OH) 3 (s) Co 3+ (aq) + 3OH (aq) K sp [Co 3+ ][OH ] 3 I Solid 0 0-7 C -x +x +3x E solid x 3x+0-7 if x<< 0-7 then K sp (x)( 0-7 ) 3 And the solubility (or x) is x K sp -43 2.5x0-2 0 0-2 2.5x0-22 -22-7 M (check : x 2.5x0 << 0 ) ans. solubility (mol/l): c. Does a precipitate form when 25 ml of 0.0 M lithium nitrate LiNO 3, is mixed with 35 ml of 0.75 M sodium carbonate Na 2 CO 3? (K sp Li 2 CO 3 8.5x0-4 ) Show your work. (3 marks) A precipitation will occur if Q > Ksp Li 2 CO 3 (s) 2Li + (aq) + CO 3 (aq) Ksp [Li + ] 2 [NO 3 ] Li + concentration 0.025 L x 0.0 M (0.025 + 0.035)L 0.042 M CO 3 2- concentration 0.035 L x 0.75 M (0.025 + 0.035)L 0.44 M Q [Li + ] 2 [NO 3 ] (0.042) 2 (0.44) 7.8x0-4. Since Q < K sp then: No precipitation. ans: yes no
Question 4 a. A system is made of a cylinder of gas with a piston. When 4.0 kj of heat is transferred from the surroundings to the system, the gas in the piston expands from 2 L to 27 L and performs work on the surroundings. If the system gains 20 J of internal energy from this process, against what constant external pressure, in atmospheres, is the piston working? - 6 - (3 marks) H E + P V (first law: conservation of the energy) H - E P V H - E [+4000 J - (+ 20 J) ] x L.atm 0.3 J 37.50 L.atm 37.5 L.atm P V and V V final - V initial P 37.5 L.atm 27L - 2L 2.5 atm Ans. pressure (atm): b. Bromine is a liquid at room temperature. Calculate the freezing point of bromine if its heat of fusion is + 5.79 kj mol - and its entropy of fusion is 2.8 J K - mol -. (3 marks) G H - T S (second law: spontaneity of a system) The freezing point occurs when all the heat added to a system is converted to entropy. At this point, the system is highly reversible (equilibrium condition, therefore G 0) The equation becomes: 0 H - T S and ΔH ΔS T T ΔH ΔS 5.79x03 J.mol 2.8 J.K -. mol 266K or -7ºC Ans. T f bromine:
Question 5 a. Circle the substance in each of the following pairs that would have the greater entropy. - 7 - i. H 2 O (l, mol, 75 C, atm) or H 2 O (g, mol, 75 C, atm) ii. Fe (s, 50.0 g, 5 C, atm) or Fe (s, 0.80 mol, 5 C, atm) iii. Br 2 (l, mol, 8 C, atm) or Br 2 (s, mol, 8 C, atm) iv. SO 2 (g, 0.32 mol, 32.5 C, 0.0 atm) or SO 2 (g, 0.284 mol, 22.3 C, 5 atm) b. Methyl isothiocyanate, CH 3 N C S, is a highly irritating pesticide. It can be prepared by reacting carbon disulfide with methylamine. Given the thermodynamic data at 25 C below, calculate the standard molar entropy of methyl isothiocyanate. (4 marks) CS 2 (g) + CH 3 NH 2 (g) CH 3 N C S (g) + H 2 S (g) ΔG (kj mol - ) 67.5 32.09 44.35 33.56 ΔH (kj mol - ) 7.36 22.98 30.96 20.63 S (J mol - K - ) 237.73 243.30? 205.69 Gº (-33.56) +(44.35) - [(67.5) + (32.09)].55 kj Hº (-20.63) +(30.96) - [(7.36) + (-22.98)] 5.95 kj S ΔH - ΔG T (5950-550)J (25 + 273)K 4.8 J/K finally: S S product - S reactant 4.8 J/K (205.69 + x) J/K- (237.73 + 243.30)J/K x Sº CH3 -NSC 290. J/K Ans: :
Question 6 In the laboratory experiment 4, you want to determine the activation energy of the following reaction: - 8 - (5 marks) 2I (aq) + S 2 O 82 (aq) I2 (aq) + 2S 2 O 42 Where the reaction rate is: Rate - [ΔI ]/2Δt and the rate law for this reaction is: Rate k [I ][S 2 O 82 ] By recording the reaction rate of several experiments at different temperatures, the following graph based on the linear form of the Arrhenius equation is obtained. Arrhenius plot for the determination of the activation energy for the reaction of iodide with peroxydisulfate From this graph, calculate the activation energy (with units) for this reaction. We will be using the linear form of the Arrhenius equation: Ln k - E a R T + constant The graph of Ln k vs. T gives a slope E a R Slope Y 2 Y X 2 X (-4.80) (-5.60) (3.2x0 3-3.35x0 3 )K - 0.80 -.5x0 4 K - 5.3x0 3 K Since: slope E a R then E a (8.3 J K.mol ) x 5.3x03 K 44x0 3 J or 44 kj mol mol. Ans. E a :
- 9 - A Periodic Table of the Elements 8A 2 H He.008 2A 3A 4A 5A 6A 7A 4.003 3 4 5 6 7 8 9 0 2 Li Be B C N O F Ne 6.94 9.02 0.8 2.0 4.0 6.00 9.00 20.8 2 3 4 5 6 7 8 3 Na Mg Al Si P S Cl Ar 22.99 24.3 3B 4B 5B 6B 7B 8B 9B 0B B 2B 26.98 28.09 30.97 32.07 35.45 39.95 9 20 2 22 23 24 25 26 27 28 29 30 3 32 33 34 35 36 4 K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr 39.0 40.08 44.96 47.87 50.94 52.00 54.94 55.85 58.93 58.69 63.55 65.39 69.72 72.6 74.92 78.96 79.90 83.80 37 38 39 40 4 42 43 44 45 46 47 48 49 50 5 52 53 54 5 Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe 85.47 87.62 88.9 9.22 92.9 95.94 98.00 0. 02.9 06.4 07.9 2.4 4.8 8.7 2.8 27.6 26.9 3.3 55 56 57 72 73 74 75 76 77 78 79 80 8 82 83 84 85 86 6 Cs Ba La* Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn 32.9 37.3 38.9 78.5 8.0 83.8 86.2 90.2 92.2 95. 97.0 200.6 204.4 207.2 209.0 209.0 20.0 222.0 87 88 89 04 05 06 07 08 09 0 2 7 Fr Ra Acª Rf Db Sg Bh Hs Mt Uun Uuu Uub metalloid 223.0 226.0 227.0 26.0 262.0 263.0 262.0 265.0 266.0 269.0 272.0 277.0 58 59 60 6 62 63 64 65 66 67 68 69 70 7 *Lanthanides Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu 40 4 44 45 50 52 57 59 63 65 67 69 73 75 90 9 92 93 94 95 96 97 98 99 00 0 02 03 ªActinides Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr 232 23 238 237. 244 243 247 247 25 252 257 258 259 260