Test of Facts and Concepts 5 7

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1 Test of Facts and Concepts A strong electrolyte dissociates completely in solution while a weak electrolyte does not, but has a low percentage ionization or dissociation in solution. HCHO 2 (aq) + H 2 O H O + (aq) + CHO 2 (aq) 2. CH NH 2 (aq) + H 2 O CH NH + (aq) + OH (aq). Molecular equation: CH NH 2 + HCl CH NH + + Cl Ionic equation: CH NH 2 + H + + Cl CH NH + + H + + Cl Net ionic equation: CH NH 2 + H + + CH NH + + H + +. (a) Ca (PO ) 2 insoluble (b) Ni(OH) 2 insoluble (c) (NH ) 2 HPO soluble (d) SnCl 2 soluble (e) Sr(NO ) 2 soluble (f) Au(ClO ) soluble (g) Cu(C 2 H O 2 ) 2 soluble (h) AgBg insoluble (i) KOH soluble (j) Hg 2 Cl 2 insoluble (k) ZnSO soluble (l) Na 2 S soluble (m) CoCO insoluble (n) BaSO insoluble (o) MnS insoluble 5. (a) CuCl 2 (aq) + (NH ) 2 CO (aq) CuCO (s) + 2NH Cl(aq) Cu 2+ (aq) + 2Cl (aq) + 2NH + (aq) + CO (aq) CuCO (s) + 2NH + (aq) + 2Cl (aq) Cu 2+ (aq) + CO (aq) CuCO (s) (b) (c) (d) (e) (f) 2HCl(aq) + MgCO (s) MgCl 2 (aq) + CO 2 (g) + H 2 O(l) 2H + (aq) + 2Cl (aq) + MgCO (s) Mg 2+ (aq) + 2Cl (aq) + CO 2 (g) + H 2 O(l) 2H + (aq) + MgCO (s) Mg 2+ (aq) + CO 2 (g) + H 2 O(l) ZnCl 2 (aq) + 2AgC 2 H O 2 (aq) 2AgCl(s) + Zn(C 2 H O 2 ) 2 (aq) Zn 2+ (aq) + 2Cl (aq) + 2Ag + (aq) + 2C 2 H O 2 (aq) 2AgCl(s) + Zn 2+ (aq) + 2C 2 H O 2 (aq) 2Cl (aq) + 2Ag + (aq) 2AgCl(s) HClO (aq) + NaCHO 2 (aq) HCHO 2 (aq) + NaClO (aq) H + (aq) + ClO (aq) + Na + (aq) + CHO 2 (aq) HCHO 2 (aq) + Na + (aq) + ClO (aq) H + (aq) + CHO 2 (aq) HCHO 2 (aq) MnO(s) + H 2 SO (aq) MnSO (aq) + H 2 O(l) MnO(s) + 2H + (aq) + SO (aq) Mn 2+ (aq) + SO (aq) + H 2 O(l) MnO(s) + 2H + (aq) Mn 2+ (aq) + H 2 O(l) FeS(s) + 2HCl(aq) H 2 S(g) + FeCl 2 (aq) FeS(s) + 2H + (aq) + 2Cl (aq) H 2 S(g) + Fe 2+ (aq) + 2Cl (aq) FeS(s) + 2H + (aq) H 2 S(g) + Fe 2+ (aq) 6. H PO + NaOH Na PO + H 2 O 7. Acidic oxides: P O 6, SeO, SO 2 Basic oxides: Na 2 O, CaO, PbO 8. (a) KHSO (b) No acid salt (c) no acid salt (d) NaH 2 PO, Na 2 HPO (e) NaHCO 58

2 Test of Facts and Concepts (a) iodic acid (b) hypobromous acid (c) nitrous acid (d) calcium hydrogen phosphate (e) iron(iii) hydrogen sulfate 10. (a) HBrO 2 (b) HIO (c) LiHSO (d) HBrO 11. BaCl 2 + Na 2 SO BaSO + 2NaCl L SO # ml BaCl 2 = 27.0 ml SO mol SO 1 mol Ba ml SO 1 L SO 1 mol SO 1 L Ba ml Ba mol Ba 2+ 1 L Ba 2+ = 81 ml Ba2+ solution 12. MgCl 2 + 2NaOH Mg(OH) 2 + 2NaCl This is a limiting reagent problem. Start with the MgCl 2 solution: # ml 0.20 M NaOH required for 0.0 ml M MgCl 2 1 L Mg mol Mg 2+ 2 mol OH # ml NaOH = 0.0 ml 1000 ml Mg 2+ 1 L Mg 2+ 1 mol Mg 2+ 1 L OH 1000 ml OH 0.20 mol OH 1 L OH = 28.6 ml of 0.20 M NaOH solution Only 25.0 ml of the 0.20 M NaOH solution are supplied; therefore the NaOH is the limiting reagent. The amount of Mg(OH) 2 formed is # g Mg(OH) 2 = 25.0 ml OH 1 L OH 0.20 mol OH 1000 ml OH 1 L OH 1 mol Mg(OH) 2 2 mol OH Ñ g Mg(OH) 2 = 0.06 g Mg(OH) 2 1 mol Mg(OH) 2 The total volume of the solution is 0.0 ml ml = 55.0 ml solution Mg 2+ : concentration is 0.00 M OH : concentration is 0.00 M Na + : concentration is: 25.0 ml Na + 1 L Na mol Na ml Na + 1 L Na + = mol Na mol Na ml solution 55.0 ml solution = M Na + 1 L solution Cl : concentration is: 1 L MgCl 0.0 ml MgCl mol MgCl 2 2 mol Cl Ñ 1000 ml MgCl 2 1 L MgCl 2 1 mol MgCl = mol Cl mol Cl Ñ 1000 ml solution 55.0 ml solution = M Cl 1 L solution 1. M 1 V 1 = M 2 V 2 (6.00 M HNO )(x ml) = (0.150 M HNO )(200 ml + x ml) 6.00x mol = 0 mol x mol 58

3 Test of Facts and Concepts x mol = 0 mol x = ml 6.00 M HNO need to be added 1. First determine the number of moles of HSO titrated. Then, find the mass of the HSO, and finally the percentage by wieght of NaHSO. # mole HSO 1 L solution mol NaOH = 2.60 ml NaOH 1 mol HSO 1000 ml solution 1 L solution 1 mol NaOH = mol HSO # g NaHSO = mol NaHSO mol HSO g NaHSO = g NaHSO 1 mol HSO 1 mol NaHSO percentage by mass of NaHSO in sample = g NaHSO g sample 100% = 62.0% 15. To find the molarity of the KOH solution, find the moles of HCl, then the moles of KOH and finally the molarity of the KOH solution. 1 L solution mol HCl 1 mol KOH # mole KOH = ml HCl 1000 ml solution 1 L solution 1 mol HCl = mol KOH mol KOH 1000 ml KOH molarity of KOH = ml KOH = M KOH 1 L KOH 16. M 1 V 1 = M 2 V 2 (125 ml)(0.1 M) = (x ml)(18.0 M) x = 1 ml 1.00 ml concentration H 2 SO is needed. 17. (a) 1 g H molar concentration = PO 1 mol H PO 1000 ml solution =1.7 M H PO 100 ml solution g H PO 1 L solution (b) 100 ml solution g solution # g solution = 50.0 g H PO = 58.6 g solution 1 g H PO 1 ml solution 18. Let x = g Li 2 CO x 1 mol Li 2 CO = mol Li 2 CO 7.89 g Li 2 CO Let y = g K 2 CO 1 mol K y 2 CO = mol K 2 CO g K 2 CO x + y =. g 2HCl + CO H 2 O + CO 2 + 2Cl # mole CO 1.8 mol HCl = L HCl solution 1 mol CO 2 = 0.09 mol CO 1 L solution 2 mol HCl x 1 mol Li 2 CO 1 mol K + y 2 CO = 0.09 mol CO 7.89 g Li 2 CO g K 2 CO x =. y 1 mol Li (. y) 2 CO 1 mol K + y 2 CO = 0.09 mol CO 7.89 g Li 2 CO g K 2 CO 585

4 Test of Facts and Concepts 5 7 Solve for y. y =.26 g K 2 CO x =. g.26 = 1.17 g Li 2 CO 1 mol NaIO 19. # mol I 2 = 16. g NaIO mol I 2 = 0.29 mole I g NaIO 1 mol NaIO 25.8 g I # g I 2 = 0.29 mol I 2 2 = 6.2 g I 2 1 mol I (a) As 0 (b) H +1 Cl + O 2 (c) Mn +2 Cl 1 (d) V + S + O (a) K 2 Cr 2 O 7 + HCl KCl + Cl 2 + H 2 O + CrCl Reactants: K + Cr 2 O 7 H + Cl Products K + Cl Cl 2 H 2 O Cr + Cl Cr 2 O 7 Cr + Cl Cl 2 Cr 2 O 7 2Cr + 2Cl Cl 2 Cr 2 O 7 2Cr + + 7H 2 O 2Cl Cl 2 1H + + Cr 2 O 7 2Cr + + 7H 2 O 2Cl Cl 2 6e + 1H + + Cr 2 O 7 2Cr + + 7H 2 O 2Cl Cl 2 + 2e 6e + 1H + + Cr 2 O 7 2Cr + + 7H 2 O (2Cl Cl 2 + 2e ) 6e + 1H + + Cr 2 O 7 2Cr + + 7H 2 O 6Cl Cl 2 + 6e 6e + 1H + + Cr 2 O 7 + 6Cl 2Cr + + 7H 2 O + Cl 2 + 6e 1H + + Cr 2 O 7 + 6Cl 2Cr + + 7H 2 O + Cl 2 (b) KOH + SO 2 + KMnO K 2 SO + MnO 2 + H 2 O Reactants: K + OH SO 2 K + MnO Products: K + SO MnO 2 H 2 O SO 2 SO MnO MnO 2 2H 2 O + SO 2 SO MnO MnO 2 + 2H 2 O 586

5 2H 2 O + SO 2 SO + H + H + + MnO MnO 2 + 2H 2 O 2H 2 O + SO 2 SO + H + + 2e e + H + + MnO MnO 2 + 2H 2 O Test of Facts and Concepts 5 7 (2H 2 O + SO 2 SO + H + + 2e ) 2(e + H + + MnO MnO 2 + 2H 2 O) 6H 2 O + SO 2 SO + 12H + + 6e 6e + 8H + + 2MnO 2MnO 2 + H 2 O 6H 2 O + SO 2 + 6e + 8H + + 2MnO 2MnO 2 + H 2 O + SO + 12H + + 6e 2H 2 O + SO 2 + 2MnO 2MnO 2 + SO + H + OH + 2H 2 O + SO 2 + 2MnO 2MnO 2 + SO + H + + OH OH + 2H 2 O + SO 2 + 2MnO 2MnO 2 + SO + H 2 O OH + SO 2 + 2MnO 2MnO 2 + SO + 2H 2 O 22. (a) Cr 2 O 7 + Br Br 2 + Cr + Cr 2 O 7 Cr + Br Br 2 Cr 2 O 7 2Cr + 2Br Br 2 Cr 2 O 7 2Cr + + 7H 2 O 2Br Br 2 1H + + Cr 2 O 7 2Cr + + 7H 2 O 2Br Br 2 6e + 1H + + Cr 2 O 7 2Cr + + 7H 2 O 2Br Br 2 + 2e 6e + 1H + + Cr 2 O 7 2Cr + + 7H 2 O (2Br Br 2 + 2e ) 6e + 1H + + Cr 2 O 7 2Cr + + 7H 2 O 6Br Br 2 + 6e 6e + 1H + + Cr 2 O 7 + 6Br Br 2 + 6e + 2Cr + + 7H 2 O 1H + + Cr 2 O 7 + 6Br Br 2 + 2Cr + + 7H 2 O (b) H AsO + MnO H 2 AsO + Mn 2+ H AsO H 2 AsO MnO Mn

6 Test of Facts and Concepts 5 7 H 2 O + H AsO H 2 AsO MnO Mn 2+ + H 2 O H 2 O + H AsO H 2 AsO + H + 8H + + MnO Mn 2+ + H 2 O H 2 O + H AsO H 2 AsO + H + + 2e 5e + 8H + + MnO Mn 2+ + H 2 O 5(H 2 O + H AsO H 2 AsO + H + + 2e ) 2(5e + 8H + + MnO Mn 2+ + H 2 O) 5H 2 O + 5 H AsO 5H 2 AsO + 15H e 10e + 16H + + 2MnO 2Mn H 2 O 5H 2 O + 5 H AsO + 10e + 16H + + 2MnO 5H 2 AsO + 15H e + 2Mn H 2 O 5 H AsO + 1H + + 2MnO 5H 2 AsO + 2Mn 2+ +H 2 O 2. (a) I + CrO CrO 2 + IO I IO CrO CrO 2 H 2 O + I IO CrO CrO 2 + 2H 2 O H 2 O + I IO + 6H + H + + CrO CrO 2 + 2H 2 O H 2 O + I IO + 6H + + 6e e + H + + CrO CrO 2 + 2H 2 O H 2 O + I IO + 6H + + 6e 2(e + H + + CrO CrO 2 + 2H 2 O) H 2 O + I IO + 6H + + 6e 6e + 8H + + 2CrO 2CrO 2 + H 2 O H 2 O + I + 6e + 8H + + 2CrO 2CrO 2 + H 2 O + IO + 6H + + 6e H 2 O + I + 2H + + 2CrO 2CrO 2 + H 2 O + IO 2OH + H 2 O + I + 2H + + 2CrO 2CrO 2 + H 2 O + IO + 2OH 5H 2 O + I + 2CrO 2CrO 2 + H 2 O + IO + 2OH H 2 O + I + 2CrO 2CrO 2 + IO + 2OH (b) SO 2 + MnO MnO 2 + SO SO 2 SO 588

7 MnO MnO 2 2H 2 O + SO 2 SO MnO MnO 2 + 2H 2 O 2H 2 O + SO 2 SO + H + H + + MnO MnO 2 + 2H 2 O 2H 2 O + SO 2 SO + H + + 2e e + H + + MnO MnO 2 + 2H 2 O 2(2H 2 O + SO 2 SO + H + + 2e ) e + H + + MnO MnO 2 + 2H 2 O H 2 O + 2SO 2 2SO + 8H + + e e + H + + MnO MnO 2 + 2H 2 O Test of Facts and Concepts 5 7 H 2 O + 2SO 2 + e + H + + MnO MnO 2 + 2H 2 O + 2SO + 8H + + e 2H 2 O + 2SO 2 + MnO MnO 2 + 2SO + H + OH + 2H 2 O + 2SO 2 + MnO MnO 2 + 2SO + H + + OH OH + 2H 2 O + 2SO 2 + MnO MnO 2 + 2SO + H 2 O OH + 2SO 2 + MnO MnO 2 + 2SO + 2H 2 O 2. Question 22. (a) Oxidizing agent: Cr 2 O 7 Reducing agent: Br (b) Oxidizing agent: MnO Reducing agent: H AsO Question 2. (a) Oxidizing agent: CrO Reducing agent: I (b) Oxidizing agent: MnO Reducing agent: SO (a) Sn(s) + 2HCl(aq) Sn 2+ (aq) + H 2 (g) + 2Cl (aq) (b) Cu(s) +2HNO (concd) Cu 2+ (aq) + 2NO (aq) + H 2 (g) (c) Zn(s) + Cu 2+ (aq) Zn 2+ (aq) + Cu(s) (d) Ag(s) + Cu 2+ (aq) No Reaction 26. (a) 2C 16 H + 9O 2 2CO 2 + H 2 O (b) 2C 16 H + O 2 2CO + H 2 O (c) 2C 16 H + 17O 2 2C + H 2 O 27. C 17 H 5 CO 2 H + 26O 2 18CO H 2 O 28. 2CH SH + 6O 2 2CO 2 + H 2 O + 2SO (a) 2Mg + O 2 2MgO (b) Al + O 2 2Al 2 O (c) P + 5O 2 P O 10 (d) S + O 2 SO 2 589

8 Test of Facts and Concepts g sample Reactions that occurred in the titrations: 2Cu I I + 2CuI I + 2S 2 O I + S O 6 # mol Cu = ml S 2 O 2 solution mol S2O 1mol I 1L solution 2 mols2o = mol Cu # g Cu = g Cu mol Cu = g Cu 1 mol Cu Percentage by weight = g Cu 100% = 12.98% 0.28 g sample 1. heat capacity = (specific heat)(mass) = (.18 J g 1 C 1 )(225 g) = 91. J C 1 specific heat =.18 J g 1 C 1 2. Ozone, O, is not the most stable form of oxygen at 25 C and 1atm.. He # J = (5.19 J g 1 C 1 )(.00 g)(1 C) = 20.8 J N 2 # J = (1.0 J g 1 C 1 )(28.02 g)(1 C) = 29.1 J 2 5 mol I 1mol I 2 mol Cu 5 mol I. # J released/0.686 g = ( J C 1 )(26.76 C C) = J/0.686 g J 80.9 g C Molar heat capacity = 2 H 6 O g C 2 H 6 O = J/mol C 2 H 6 O mol C 2 H 6 O 2 = 1981 kj/mol C 2 H 6 O 2 5. A state function does not depend on a system's history or future, and H is a state function. 6. Amount of heat needed = (.18 J g 1 C 1 )(250 g H 2 O)(50.0 C 25 C) = J 1 kj # g CH = J 1 mol CH 16.0 g CH = g CH 1000 J 802. kj 1 mol CH 7. (a) specific heat intensive (b) heat capacity extensive (c) H f intensive (d) H extensive (e) molar heat capacity intensive 8. Internal energy is the sum of all of the kinetic energies and potential energies of the particles within a system. The first law of thermodynamics is that the energy of the universe is constant; it can be neither created nor destroyed but only transformed and transferred. 9. Pressure-volume work is the energy transferred as work when a system expands or contracts against the pressure exerted by the surroundings. At constant pressure, work = P V. 0. The change in internal energy is equal to the "heat of reaction at constant volume" because no work is done. 1. H will be larger than E

9 Test of Facts and Concepts (a) 2CO(g) + O 2 (g) 2CO 2 (g) H = kj (b) CO 2 (g) CO(g) + 1/2O 2 (g) H = kj. C 20 H 2 (s) + 61/2O 2 (g) 20CO 2 (g) + 21H 2 O(l) H comb = kj mol 1 H comb = kj mol 1 = [(20 mol CO 2 )(9 kj mol 1 ) + (21 mol H 2 O)(285.9 kj mol 1 )] [(mol C 20 H 2 )( H f C 20 H 2 ) + (61/2 mol O 2 )(0 kj mnol 1 )] H f C 20 H 2 = kj mol 1 20C(s) + 16H 2 (g) C 20 H 2 (s) H f = kj mol 1. (a) H 2 SO (l) 2SO (l) + H 2 O(l) H reaction = [(2 mol SO (l))(96 kj mol 1 ) + (1 mol H 2 O(l))(285.9 kj mol 1 )] [(1 mol H 2 SO (l))(81.8 kj mol 1 )] = 26 kj (b) C 2 H 6 (g) C 2 H (g) + H 2 (g) H reaction =[(1 mol C 2 H (g))(51.9 kj mol 1 ) + 1 mol H 2 (g)(0.0 kj mol 1 )] [(1 mol C 2 H 6 (g))(8.5 kj mol 1 )] = 16. kj 5. Ca(s) + 2C(s) CaC 2 (s) H f Ca(s) + 2H 2 O(l) Ca(OH) 2 (s) + H 2 (g) H 2 (g) + 1/2O 2 (g) H 2 O(l) Ca(OH) 2 (s) CaO(s) + H 2 O(l) CaO(s) + C(s) CaC 2 (s) + CO(g) CO(g) C(s) + 1/2O 2 (g) Ca(s) + 2C(s) CaC 2 (s) H = 1.79 kj H = kj H = kj H = 62. kj H = kj H f = 62.7 kj 591