1. Final Exam/ Exercises 1- The element in group 3A and period 3 is: a) Gab) Bec) Ald) Mg 2- Which compound has the same empirical formula as C 3 H 12 O 6 ? a) C 12 H 20 O 4 b) C 2 H 8 O 4 c) C 6 H 3 O 6 d) C 12 H 24 O 12 We need to determine the empirical formula for all compounds C 3 H 12 O 6 ===  CH 4 O 2 C 12 H 20 O 4 ===  C 3 H 5 O C 2 H 8 O 4 ===-  CH 4 O 2 C 6 H 3 O 6 ====  C 2 HO 2 C 12 H 24 O 12 ===  CH 2 O

2. Final Exam/ Exercises 1- The element in group 3A and period 3 is: a) Gab) Bec) Ald) Mg 2- Which compound has the same empirical formula as C 3 H 12 O 6 ? a) C 12 H 20 O 4 b) C 2 H 8 O 4 c) C 6 H 3 O 6 d) C 12 H 24 O 12 3- What is the number of protons, electrons and neutrons in the atom of ? a) 31 protons, 34 electrons, 15 neutrons b) 16 protons, 16 electrons, 16 neutrons c) 15 protons, 18 electrons, 16 neutrons d) 18 protons, 15 electrons, 31 neutrons P = 15, e = 15+3 = 18, n = 31-15 = 16

3. Final Exam/ Exercises 1- The element in group 3A and period 3 is: a) Gab) Bec) Ald) Mg 2- Which compound has the same empirical formula as C 3 H 12 O 6 ? a) C 12 H 20 O 4 b) C 2 H 8 O 4 c) C 6 H 3 O 6 d) C 12 H 24 O 12 3- What is the number of protons, electrons and neutrons in the atom of ? a) 31 protons, 34 electrons, 15 neutrons b) 16 protons, 16 electrons, 16 neutrons c) 15 protons, 18 electrons, 16 neutrons d) 18 protons, 15 electrons, 31 neutrons 4- The correct systematic name for Fe 2 O 3 is a) Iron (II) oxide. b) Iron (III) oxide. c) Diiron trioxide d) Iron oxide

4. First Exam/ Exercises Summery of naming compound IonicMolecular Cation: metal or NH 4 + Anion: monotomic or polytomic Cation has only one charge Cation has more than one charge Alkali metal Alkaline earth metal Ag +, Al +3, Cd +2, Zn +2 Other metal cations Name metal first If monoatomic anion, add ide to the anion If polyatomic anion use name of anion from previous table Name metal first Specify charge of metal cation with roman numeral (STOCK SYSTEM) If monoatomic anion, add ide to the anion If polyatomic anion use name of anion from previous table Nonmetal + nonmetal Nonmetal + metalloid Pair Form one type of compound Pair Form more than one type of compound Name first element add ide to the name of second element Name first element add ide to the name of second element Add the prefix (prefix mono usually omitted for the first element

5. Final Exam/ Exercises 1- The element in group 3A and period 3 is: a) Gab) Bec) Ald) Mg 2- Which compound has the same empirical formula as C 3 H 12 O 6 ? a) C 12 H 20 O 4 b) C 2 H 8 O 4 c) C 6 H 3 O 6 d) C 12 H 24 O 12 3- What is the number of protons, electrons and neutrons in the atom of ? a) 31 protons, 34 electrons, 15 neutrons b) 16 protons, 16 electrons, 16 neutrons c) 15 protons, 18 electrons, 16 neutrons d) 18 protons, 15 electrons, 31 neutrons 4- The correct systematic name for Fe 2 O 3 is a) Iron (II) oxide. b) Iron (III) oxide. c) Diiron trioxide d) Iron oxide

6. Final Exam/ Exercises 5- A 5.691 g sample of MgCl 2 is dissolved in enough water to give 750 mL of solution. What is the molarity of this solution? a) 3.70  10 -2 M b) 1.05  10 -2 M c) 2.58  10 -2 M d) 7.99  10 -2 M First we calculate the number of mole n = 5.691 / 95 = 0.06 mole M=n/V = 0.06 / (750/1000) = 0.08 M = 8 X 10 -2 M

7. Final Exam/ Exercises 5- A 5.691 g sample of MgCl 2 is dissolved in enough water to give 750 mL of solution. What is the molarity of this solution? a) 3.70  10 -2 M b) 1.05  10 -2 M c) 2.58  10 -2 M d) 7.99  10 -2 M 6- Calculate the percent composition by mass of O in picric acid (C 6 H 3 N 3 O 7 )? a) 1.3 % b) 18.3 % c) 31.4 % d) 48.9 % Molar mass of C 6 H 3 N 3 O 7 =229.114g/mol n x molar mass of element molar mass of compound x 100% 7 x 16 229.11 x 100% = 48.9 %

8. Final Exam/ Exercises 5- A 5.691 g sample of MgCl 2 is dissolved in enough water to give 750 mL of solution. What is the molarity of this solution? a) 3.70  10 -2 M b) 1.05  10 -2 M c) 2.58  10 -2 M d) 7.99  10 -2 M 6- Calculate the percent composition by mass of O in picric acid (C 6 H 3 N 3 O 7 )? a) 1.3 % b) 18.3 % c) 31.4 % d) 48.9 % 7- After balancing the following equation the coefficients are: a Ba(OH) 2 + b HBr → c BaBr 2 + d H 2 O a) a=1, b=2, c=1, d=2 b) a=2, b=2, c=1, d=2 c) a=1, b=3, c=1, d=2 d) a=2, b=1, c=3, d=1

9. Final Exam/ Exercises 7- After balancing the following equation the coefficients are: a Ba(OH) 2 + b HBr → c BaBr 2 + d H 2 O a) a=1, b=2, c=1, d=2 b) a=2, b=2, c=1, d=2 c) a=1, b=3, c=1, d=2 d) a=2, b=1, c=3, d=1 Ba(OH) 2 + HBr → BaBr 2 + H 2 O Ba 1 Ba 1 Ba(OH) 2 + HBr → BaBr 2 + H 2 O O 2 O 1 X2 Ba(OH) 2 + HBr → BaBr 2 + 2H 2 O Br 1 Br 2 X 2 Ba(OH) 2 + 2HBr → BaBr 2 + 2H 2 O H 2 + 2 2X2 Ba(OH) 2 + 2HBr → BaBr 2 + 2H 2 O

10. Final Exam/ Exercises 5- A 5.691 g sample of MgCl 2 is dissolved in enough water to give 750 mL of solution. What is the molarity of this solution? a) 3.70  10 -2 M b) 1.05  10 -2 M c) 2.58  10 -2 M d) 7.99  10 -2 M 6- Calculate the percent composition by mass of O in picric acid (C 6 H 3 N 3 O 7 )? a) 1.3 % b) 18.3 % c) 31.4 % d) 48.9 % 7- After balancing the following equation the coefficients are: a Ba(OH) 2 + b HBr → c BaBr 2 + d H 2 O a) a=1, b=2, c=1, d=2 b) a=2, b=2, c=1, d=2 c) a=1, b=3, c=1, d=2 d) a=2, b=1, c=3, d=1

11. Final Exam/ Exercises 8- What is the theoretical yield of chromium (Cr) that can be produced by the reaction of 50.0 g of Cr 2 O 3 with 9.00 g of aluminum (Al) according to the chemical reaction? 2 Al + Cr 2 O 3  Al 2 O 3 + 2 Cr a) 15.4 g b) 17.33 g c) 19.4 g d) 13.48 g First we have to determine the limiting reagent: First start with Al 1-Convert to mole : n = 9 / 27 = 0.33 mol 2- from equation 2mole Al ========= 2 mole Cr 0.33 mole Al =====? Mole Cr 2x 0.33 = 2 x ? Mole of Cr = 0.33 mol Mass = n x molar mass = 0.33 x 52 =17.2g second start with Cr 2 O 3 1-Convert to mole : n = 50 / 152 = 0.33mol 2- from equation 1mole Cr 2 O 3 ========= 2 mole Cr 0.33mole Cr 2 O 3 =====? Mole Cr 2 x 0.33 = 1 x ? Mole of Cr = 0.66 mol

12. Final Exam/ Exercises 8- What is the theoretical yield of chromium (Cr) that can be produced by the reaction of 50.0 g of Cr 2 O 3 with 9.00 g of aluminum (Al) according to the chemical reaction? 2 Al + Cr 2 O 3  Al 2 O 3 + 2 Cr a) 15.4 g b) 17.33 g c) 19.4 g d) 13.48 g 9- If the actual yield for the experiment in the above question (8) produced 13.0g, what is the percentage yield? a) 84.4% b) 75.0% c) 67% d) 96.4%

13. Final Exam/ Exercises 8- What is the theoretical yield of chromium (Cr) that can be produced by the reaction of 50.0 g of Cr 2 O 3 with 9.00 g of aluminum (Al) according to the chemical reaction? 2 Al + Cr 2 O 3  Al 2 O 3 + 2 Cr a) 15.4 g b) 17.33 g c) 19.4 g d) 13.48 g 9- If the actual yield for the experiment in the above question (8) produced 13.0g, what is the percentage yield? a) 84.4% b) 75.0% c) 67% d) 96.4% 10- What is the mass of carbon in 15 g carbon dioxide (CO 2 )? a) 4.1 g b) 2.73 g c) 5.45 g d) 6.82 g

14. Final Exam/ Exercises 10- What is the mass of carbon in 15 g carbon dioxide (CO 2 )? a) 4.1 g b) 2.73 g c) 5.45 g d) 6.82 g Mole of CO 2 = mass / molar mass = 15 / 44 = 0.34 mol From the formula 1mole C ====== 1 mol CO 2 ? Mole C ====== 0.34 mole CO 2 Mole of C = 0.34 mol Mass of C = n x molar mass = 0.34 x 12 = 4.09 g.

15. Final Exam/ Exercises 8- What is the theoretical yield of chromium (Cr) that can be produced by the reaction of 50.0 g of Cr 2 O 3 with 9.00 g of aluminum (Al) according to the chemical reaction? 2 Al + Cr 2 O 3  Al 2 O 3 + 2 Cr a) 15.4 g b) 17.33 g c) 19.4 g d) 13.48 g 9- If the actual yield for the experiment in the above question (8) produced 13.0g, what is the percentage yield? a) 84.4% b) 75.0% c) 67% d) 96.4% 10- What is the mass of carbon in 15 g carbon dioxide (CO 2 )? a) 4.1 g b) 2.73 g c) 5.45 g d) 6.82 g

16. Final Exam/ Exercises 11- Which of the following electron transitions would absorb the lowest energy by the hydrogen atom? a) from n = 1 to n = 4 b) from n = 1 to n = 5 c) from n = 3 to n = 4 d) from n = 1 to n = 6 12- What mass of helium is required to fill a 2.50 L balloon at STP? a) 0.134 g b) 0.267 g c) 0.447 g d) 267 g V= 2.5L, T= 273K, P= 1 atm, Mwt= 4 g/mol PV= nRT n= PV/RT = 1 x 2.5 /( 0.082 x 273) = 0. 11 mol Mass = n x molar mass = 0.11 x 4 = 0.44 g

17. Final Exam/ Exercises 11- Which of the following electron transitions would absorb the lowest energy by the hydrogen atom? a) from n = 1 to n = 4 b) from n = 1 to n = 5 c) from n = 3 to n = 4 d) from n = 1 to n = 6 12- What mass of helium is required to fill a 2.50 L balloon at STP? a) 0.134 g b) 0.267 g c) 0.447 g d) 267 g 13- The electronic configuration of Fe +3 is: a) [Ar] 4s 2 3d 6 b) [Ar] 4s 1 3d 5 c) [Ar] 3d 5 d) [Ar] 3d 6 14- If the mass of 0.210 L of gaseous compound was 3.6 g at 35 °C with 1.7atm pressure, the molar mass of the compound is equal to: a) 168.37 g/mol b) 254.68 g/mol c) 150.6 g/mol d) 100 g/mol

18. Final Exam/ Exercises 14- If the mass of 0.210 L of gaseous compound was 3.6 g at 35 °C with 1.7atm pressure, the molar mass of the compound is equal to: a) 168.37 g/mol b) 254.68 g/mol c) 150.6 g/mol d) 100 g/mol V = 0.21 L, m= 3.6 g, T=35 °C = 35+273= 308 K, P= 1.7 atm, d= mass/volume = 3.6 / 0.21 = 17.14 g/L

19. Final Exam/ Exercises 11- Which of the following electron transitions would absorb the lowest energy by the hydrogen atom? a) from n = 1 to n = 4 b) from n = 1 to n = 5 c) from n = 3 to n = 4 d) from n = 1 to n = 6 12- What mass of helium is required to fill a 2.50 L balloon at STP? a) 0.134 g b) 0.267 g c) 0.447 g d) 267 g 13- The electronic configuration of Fe +3 is: a) [Ar] 4s 2 3d 6 b) [Ar] 4s 1 3d 5 c) [Ar] 3d 5 d) [Ar] 3d 6 14- If the mass of 0.210 L of gaseous compound was 3.6 g at 35 °C with 1.7atm pressure, the molar mass of the compound is equal to: a) 168.37 g/mol b) 254.68 g/mol c) 150.6 g/mol d) 100 g/mol

20. Final Exam/ Exercises 15- What is the molecular formula of the compound in the above question if you know that it contained 33.0 percent Si and 67.0 percent F by mass? a) Si 2 F 6 b) SiF 3 c) Si 3 F 9 d) Si 3 F 1- we change from % to g 33 g of Si, 67 g of F 2- change from g to mole using Divided by the smallest number of mole which is 1.18 Thus the empirical formula is SiF 3 33 28 = 1.18 mol of Si n Si = 67 19 = 3.53 mol of F n F = 1.18 = 1 Si: 3.53 1.18 = 3 F: 254.68 85 = 3 Ratio = molecular formula = ratio x empirical formula = 3 x SiF 3 = Si 3 F 9

21. Final Exam/ Exercises 15- What is the molecular formula of the compound in the above question if you know that it contained 33.0 percent Si and 67.0 percent F by mass? a) Si 2 F 6 b) SiF 3 c) Si 3 F 9 d) Si 3 F 16-The element 'X' does not usually form compounds with atoms of other elements. Which one of the following could be the electron configuration of 'X'? a) 1s 2 2s 1 b) 1s 2 2s 2 2p 3 c) 1s 2 2s 2 2p 6 d)1s 2 2s 2 2p 1 17-''The energy required to remove an electron from a gaseous atom in its ground state" is known as: a) Electron affinity. b) Electronegativity. c) Ionization energy d) Ionic radius.

22. Final Exam/ Exercises 18-If the energy (E) of radiation is 6.63 x 10 -24 J, what is the frequency (υ) of this radiation? a) 1X10 10 Hz b) 1X10 8 Hz c) 1X10 -34 Hz d) 1X10 -8 Hz

23. Final Exam/ Exercises 18-If the energy (E) of radiation is 6.63 x 10 -24 J, what is the frequency (υ) of this radiation? a) 1X10 10 Hz b) 1X10 8 Hz c) 1X10 -34 Hz d) 1X10 -8 Hz 19-How many total valance electrons are available in SO 2 ? a) 10 b) 18 c) 16 d) 24 S=6, O=6 TOTAL = 6 + (6X2)=18

24. Final Exam/ Exercises 18-If the energy (E) of radiation is 6.63 x 10 -24 J, what is the frequency (υ) of this radiation? a) 1X10 10 Hz b) 1X10 8 Hz c) 1X10 -34 Hz d) 1X10 -8 Hz 19-How many total valance electrons are available in SO 2 ? a) 10 b) 18 c) 16 d) 24 20-What is the volume of 4.35×10 -3 mol gas at 21.2 °C and 0.83 atm? a) 0.13 L b) 0.2 L c) 0.1L d) 0.3 L n= 4.35×10 -3 mole, T= 21.2 °C= 21.2+273=294.2 K, P= 0.83 atm, V= ? PV=nRT V= nRT/P = 4.35×10 -3 x 0.082 x 294.2 / 0.83 = 0.13L

25. Final Exam/ Exercises 18-If the energy (E) of radiation is 6.63 x 10 -24 J, what is the frequency (υ) of this radiation? a) 1X10 10 Hz b) 1X10 8 Hz c) 1X10 -34 Hz d) 1X10 -8 Hz 19-How many total valance electrons are available in SO 2 ? a) 10 b) 18 c) 16 d) 24 20-What is the volume of 4.35×10 -3 mol gas at 21.2 °C and 0.83 atm? a) 0.13 L b) 0.2 L c) 0.1L d) 0.3 L 21-Calculate the pH of 0.05 M ammonium hydroxide (K b = 1.8x10 -5 at 25°C). NH 4 OH (aq) NH 4 + (aq) + OH - (aq) a) 3.02 b) 10.98 c) 11.55 d) 13

26. Final Exam/ Exercises 21-Calculate the pH of 0.05 M ammonium hydroxide (K b = 1.8x10 -5 at 25°C). NH 4 OH (aq) NH 4 + (aq) + OH - (aq) a) 3.02 b) 10.98 c) 11.55 d) 13 NH 3 NH 4 + OH - Initial (M) 0.05 0 0 Change -x x x Equilibrium 0.05 - x x x K b = x2x2 0.05 - x = 1.8 x 10 -5 0.05 – x  0.05 K b << 1 Kb  Kb  x2x2 0.05 = 1.8 x 10 -5 x 2 = 0.9 x 10 -6 x = 9.5 x 10 -4 M

27. Final Exam/ Exercises 21-Calculate the pH of 0.05 M ammonium hydroxide (K b = 1.8x10 -5 at 25°C). NH 4 OH (aq) NH 4 + (aq) + OH - (aq) a) 3.02 b) 10.98 c) 11.55 d) 13 When can I use the approximation? When x is less than 5% of the value from which it is subtracted. [OH - ] = 9.5 x 10 -4 M pOH = -log [OH - ] = 3.02 9.5 x 10 -4 M 0.05 M x 100% = 1.9% x = 9.5 x 10 -4 M

28. Final Exam/ Exercises 18-If the energy (E) of radiation is 6.63 x 10 -24 J, what is the frequency (υ) of this radiation? a) 1X10 10 Hz b) 1X10 8 Hz c) 1X10 -34 Hz d) 1X10 -8 Hz 19-How many total valance electrons are available in SO 2 ? a) 10 b) 18 c) 16 d) 24 20-What is the volume of 4.35×10 -3 mol gas at 21.2 °C and 0.83 atm? a) 0.13 L b) 0.2 L c) 0.1L d) 0.3 L 21-Calculate the pH of 0.05 M ammonium hydroxide (K b = 1.8x10 -5 at 25°C). NH 4 OH (aq) NH 4 + (aq) + OH - (aq) a) 3.02 b) 10.98 c) 11.55 d) 13

29. Final Exam/ Exercises 22-Calculate pH of final solution after diluting 10 mL of 0.1 M HCl with 90 mL water. a) 2.0 b) 2.5 c) 3.5 d) 1.0 First calculate the new concentration of HCl M 1 V 1 = M 2 V 2 0.1 X 10 =M 2 X (90+10) M 2 = 0.01 M

30. Final Exam/ Exercises 22-Calculate pH of final solution after diluting 10 mL of 0.1 M HCl with 90 mL water. a) 2.0 b) 2.5 c) 3.5 d) 1.0 23-Arrange the following salts according to increasing of solubility: AgCl (K sp =1.6x10 -10 ), AgI (K sp =8.3x10 -17 ), and AgBr (K sp =7.7x10 -13 ). a) AgCl >AgI> AgBr b) AgI<AgBr<AgCl c) AgI>AgBr>AgCl d) AgCl <AgI< AgBr 24-If the solubility of BaSO 4 is 1.05x10 -5 M at 25 °C, the K sp of this salt is: BaSO 4 (s) Ba +2 (aq) + SO 4 -2 (aq) a) 1.1x10 -10 b) 2.5x10 -12 c) 1.05x10 -5 d) 5x10 -20

31. Final Exam/ Exercises 24-If the solubility of BaSO 4 is 1.05x10 -5 M at 25 °C, the K sp of this salt is: BaSO 4 (s) Ba +2 (aq) + SO 4 -2 (aq) a) 1.1x10 -10 b) 2.5x10 -12 c) 1.05x10 -5 d) 5x10 -20 BaSO 4 Ba +2 SO 4 -2 Initial (M) 0 0 Change -s +s +s Equilibrium s s K sp = [Ba +2 ][SO 4 -2 ] K sp = s 2 s = 1.05x 10 -5 mol/L K sp = [Ba +2 ][SO 4 -2 ] K sp = (1.05x 10 -5 )(1.05x 10 -5 ) K sp = 1.0x10 -10

32. Final Exam/ Exercises 22-Calculate pH of final solution after diluting 10 mL of 0.1 M HCl with 90 mL water. a) 2.0 b) 2.5 c) 3.5 d) 1.0 23-Arrange the following salts according to increasing of solubility: AgCl (K sp =1.6x10 -10 ), AgI (K sp =8.3x10 -17 ), and AgBr (K sp =7.7x10 -13 ). a) AgCl >AgI> AgBr b) AgI<AgBr<AgCl c) AgI>AgBr>AgCl d) AgCl <AgI< AgBr 24-If the solubility of BaSO 4 is 1.05x10 -5 M at 25 °C, the K sp of this salt is: BaSO 4 (s) Ba +2 (aq) + SO 4 -2 (aq) a) 1.1x10 -10 b) 2.5x10 -12 c) 1.05x10 -5 d) 5x10 -20

33. Final Exam/ Exercises 25-For the reaction at equilibrium; N 2 (g) + 3 H 2 (g) 2 NH 3 (g) If [N 2 ] = 0.1 M, [H 2 ] = 0.2 M and [NH 3 ] = 0.2 M, the equilibrium constant (K c ) is: a) 10 b) 50 c) 112.5 d) 200

34. Final Exam/ Exercises 25-For the reaction at equilibrium; N 2 (g) + 3 H 2 (g) 2 NH 3 (g) If [N 2 ] = 0.1 M, [H 2 ] = 0.2 M and [NH 3 ] = 0.2 M, the equilibrium constant (K c ) is: a) 10 b) 50 c) 112.5 d) 200 26-The equilibrium constant for the following reaction at definite temperature is; H 2 (g) + I 2 (g) 2 HI (g) K 1 = 15 What is the value of the equilibrium constant for the following reaction? 2 HI (g) H 2 (g) + I 2 (g) a) 15 b) 6.6x10 -3 c) 6.6x10 -2 d) 5x10 -3 K2=1/K1 =1/15 =0.066

35. Final Exam/ Exercises 25-For the reaction at equilibrium; N 2 (g) + 3 H 2 (g) 2 NH 3 (g) If [N 2 ] = 0.1 M, [H 2 ] = 0.2 M and [NH 3 ] = 0.2 M, the equilibrium constant (K c ) is: a) 10 b) 50 c) 112.5 d) 200 26-The equilibrium constant for the following reaction at definite temperature is; H 2 (g) + I 2 (g) 2 HI (g) K 1 = 15 What is the value of the equilibrium constant for the following reaction? 2 HI (g) H 2 (g) + I 2 (g) a) 15 b) 6.6x10 -3 c) 6.6x10 -2 d) 5x10 -3 27-At equilibrium, the total gas pressure was found to be 0.033 atm. Calculate the equilibrium constant K P for the following decomposition; NH 4 CO 2 NH 2 (s) 2 NH 3 (g) + CO 2 (g) a) 5.32x10 -6 b) 5.55x10 -4 c) 2.22x10 -3 d) 5.0x10 -3

36. Final Exam/ Exercises 27-At equilibrium, the total gas pressure was found to be 0.033 atm. Calculate the equilibrium constant K P for the following decomposition; NH 4 CO 2 NH 2 (s) 2 NH 3 (g) + CO 2 (g) a) 5.32x10 -6 b) 5.55x10 -4 c) 2.22x10 -3 d) 5.0x10 -3 Pt = P(NH 3 )+ P(CO 2 ) = 2P + P = 3P 0.033 = 3P P= 0.033/3= 0.011 THEN P(CO 2 )= 0.011, P(NH 3 )= 2X0.011=0.022

37. Final Exam/ Exercises 25-For the reaction at equilibrium; N 2 (g) + 3 H 2 (g) 2 NH 3 (g) If [N 2 ] = 0.1 M, [H 2 ] = 0.2 M and [NH 3 ] = 0.2 M, the equilibrium constant (K c ) is: a) 10 b) 50 c) 112.5 d) 200 26-The equilibrium constant for the following reaction at definite temperature is; H 2 (g) + I 2 (g) 2 HI (g) K 1 = 15 What is the value of the equilibrium constant for the following reaction? 2 HI (g) H 2 (g) + I 2 (g) a) 15 b) 6.6x10 -3 c) 6.6x10 -2 d) 5x10 -3 27-At equilibrium, the total gas pressure was found to be 0.033 atm. Calculate the equilibrium constant K P for the following decomposition; NH 4 CO 2 NH 2 (s) 2 NH 3 (g) + CO 2 (g) a) 5.32x10 -6 b) 5.55x10 -4 c) 2.22x10 -3 d) 5.0x10 -3

38. Final Exam/ Exercises 28-For the reaction at equilibrium; Cl 2 (g) + 3 F 2 (g) 2 ClF 3 (g) K C = 3.0x10 -2 at 25 °C, the K P of this reaction is: a) 4.86x10 -6 b) 2.28x10 -6 c) 5.0x10 -5 d) 1.33x10 -5 Δn= 2-(1+3)= -2

39. Final Exam/ Exercises 28-For the reaction at equilibrium; Cl 2 (g) + 3 F 2 (g) 2 ClF 3 (g) K C = 3.0x10 -2 at 25 °C, the K P of this reaction is: a) 4.86x10 -6 b) 2.28x10 -6 c) 5.0x10 -5 d) 1.33x10 -5 29-Which of the following is the correct form of the equilibrium constant expression for the reaction; Cu (s) + 2Ag + (aq) Cu 2+ (aq) + 2 Ag (s) a) K c = [Cu 2+ ] [2Ag] 2 b) K c = [Ag] 2 [Cu +2 ] / [Cu] [ Ag] 2 c) K c = [Cu] [2Ag + ] / [Cu +2 ] [2Ag] 2 d) K c = [Cu 2+ ] / [Ag + ] 2 30-For the following reaction, at equilibrium which choice gives a change that will shift the position of equilibrium to favor formation of NO? 2NOBr (g) 2NO (g) + Br 2 (g),  Hº = 30 kJ/mol a) Increase the total pressure. b) Increasing the temperature. c) Decreasing the temperature. d) Removing NOBr selectively.

40. Final Exam/ Exercises 31-The correct name for the following compound is : a) o-methyltoluene b) m-methyltoluene c) o-ethyltoluene d) m- ethyltoluene 32-The right systematic name for the following organic molecule is : a) 5-methyl-3-hexene b) 5-methylhexane c) 2-methylhexane d) 3-methylhexene 33-HF/NaF system is used as buffer solution. What is the pH if it has [HF] = 0.1M and [NaF] = 0.15M? (K a = 7.1x10 -4 at 25 °C). HF (aq) H + (aq) + F - (aq) NaF (s) Na + (aq) + F - (aq) a) 2.12 b) 3.33 c) 0.83 d) 1.4

41. Final Exam/ Exercises 33-HF/NaF system is used as buffer solution. What is the pH if it has [HF] = 0.1M and [NaF] = 0.15M? (K a = 7.1x10 -4 at 25 °C). HF (aq) H + (aq) + F - (aq) NaF (s) Na + (aq) + F - (aq) a) 2.12 b) 3.33 c) 0.83 d) 1.4 pH = pK a + log [A - ] [HA] pH = -log 7.1 x 10 -4 +log 0.15 0.1 pH = 3.33

42. Final Exam/ Exercises 31-The correct name for the following compound is : a) o-methyltoluene b) m-methyltoluene c) o-ethyltoluene d) m- ethyltoluene 32-The right systematic name for the following organic molecule is : a) 5-methyl-3-hexene b) 5-methylhexane c) 2-methylhexane d) 3-methylhexene 33-HF/NaF system is used as buffer solution. What is the pH if it has [HF] = 0.1M and [NaF] = 0.15M? (K a = 7.1x10 -4 at 25 °C). HF (aq) H + (aq) + F - (aq) NaF (s) Na + (aq) + F - (aq) a) 2.12 b) 3.33 c) 0.83 d) 1.4

43. Final Exam/ Exercises 34-The percent ionization of hydrofluoric acid (HF) at the concentrations of 0.50 M ( K a = 7.1x10 -4 ) is: HF (aq) H + (aq) + F - (aq) a) 3.8% b) 0.6% c) 0.011% d) 2.83% K a = [H + ][F - ] [HF] HF H + F - Initial (M) 0.5 0 0 Change -x x x Equilibrium 0.5 - x x x K a = x2x2 0.50 - x = 7.1 x 10 -4 0.50 – x  0.50 K a << 1

44. Final Exam/ Exercises 34-The percent ionization of hydrofluoric acid (HF) at the concentrations of 0.50 M ( K a = 7.1x10 -4 ) is: HF (aq) H + (aq) + F - (aq) a) 3.8% b) 0.6% c) 0.011% d) 2.83% Ka  Ka  x2x2 0.50 = 7.1 x 10 -4 x 2 = 3.55 x 10 -4 x = 0.019 M x = 0.019 0.019 M 0.50 M x 100% = 3.8% Less than 5% Approximation ok. [H + ] = 0.019 M percent ionization = Ionized acid concentration at equilibrium Initial concentration of acid x 100% 0.19 0.5 x 100% percent ionization == 3.8%

45. Final Exam/ Exercises 34-The percent ionization of hydrofluoric acid (HF) at the concentrations of 0.50 M ( K a = 7.1x10 -4 ) is: HF (aq) H + (aq) + F - (aq) a) 3.8% b) 0.6% c) 0.011% d) 2.83% 35-The general molecular formula of alkynes is: a) C n H 2n b) C n H 2n-2 c) C n H 2n+2 d) C n H 2n+1 36-The concentration of hydroxide ion [OH - ] in an aqueous solution is 3.5x10 -9 M.What is the concentration of hydrogen ion [H + ]? a) 2.86 x 10 -6 M b) 2.86 x 10 6 M c) 3.5 x 10 -5 M d) 3.5 x 10 5 M

46. Final Exam/ Exercises 34-The percent ionization of hydrofluoric acid (HF) at the concentrations of 0.50 M ( K a = 7.1x10 -4 ) is: HF (aq) H + (aq) + F - (aq) a) 3.8% b) 0.6% c) 0.011% d) 2.83% 35-The general molecular formula of alkynes is: a) C n H 2n b) C n H 2n-2 c) C n H 2n+2 d) C n H 2n+1 36-The concentration of hydroxide ion [OH - ] in an aqueous solution is 3.5x10 -9 M.What is the concentration of hydrogen ion [H + ]? a) 2.86 x 10 -6 M b) 2.86 x 10 6 M c) 3.5 x 10 -5 M d) 3.5 x 10 5 M

47. Final Exam/ Exercises 37-Which of the following compounds has geometrical isomer? a) b) c) d) 38-The right name for the following molecule is: CH 3 -CH 2 -CH 2 -CO-CH 3 a) propyl methyl alchol b) 1-methyl-2-propanal c) 2-pentanone d) methyl propyl ester 39-Calculate the H + ion concentration in lemon juice having a pH= 2. a) 0.01 M b) 2.5x10 -4 M c) 10.0 M d) 3 M

48. Final Exam/ Exercises 39-Calculate the H + ion concentration in lemon juice having a pH= 2. a) 0.01 M b) 2.5x10 -4 M c) 10.0 M d) 3 M

49. Final Exam/ Exercises 37-Which of the following compounds has geometrical isomer? a) b) c) d) 38-The right name for the following molecule is: CH 3 -CH 2 -CH 2 -CO-CH 3 a) propyl methyl alchol b) 1-methyl-2-propanal c) 2-pentanone d) methyl propyl ester 39-Calculate the H + ion concentration in lemon juice having a pH= 2. a) 0.01 M b) 2.5x10 -4 M c) 10.0 M d) 3 M

50. Final Exam/ Exercises 40-Which expression correctly relates K p to K c for the reaction; 2A (g) + B (g) C (g) a) K p = K c (RT) -1 b) K p = K c (RT) -2 c) K p = K c (RT) d) K p = K c