1. First exam Exercises

2. First Exam/ Exercises 1- Prefixes giga and deci represent, respectively: a) 10 -9 and 10 -1. b) 10 6 and 10 -3. c) 10 3 and 10 -3 d) 10 9 and 10 -1. 2- If 586 g of bromine occupies 188 mL, what is the density of bromine in g/mL? a) 3.12g/ml b) 0.321 g/ml c) 3.63g/ml d) 3.08g/ml d = m/V m = 586 g, V = 188 mL d = 586 / 188 = 3.12 g/ml

3. First Exam/ Exercises 1- Prefixes giga and deci represent, respectively: a) 10 -9 and 10 -1. b) 10 6 and 10 -3. c) 10 3 and 10 -3 d) 10 9 and 10 -1. 2- If 586 g of bromine occupies 188 mL, what is the density of bromine in g/mL? a) 3.12g/ml b) 0.321 g/ml c) 3.63g/ml d) 3.08g/ml 3- 3.5 nanometer contains how many micrometers? a) 3.5x 10 +3 b) 3.5x 10 -3 c) 3.5x 10 +8 d) 3.5x 10 -8 First convert from nm to m : 3.5 X 10 -9 m Then from m to um 3.5 x 10 -9 x 10 +6 = 3.5 x 10 -3

4. First Exam/ Exercises 1- Prefixes giga and deci represent, respectively: a) 10 -9 and 10 -1. b) 10 6 and 10 -3. c) 10 3 and 10 -3 d) 10 9 and 10 -1. 2- If 586 g of bromine occupies 188 mL, what is the density of bromine in g/mL? a) 3.12g/ml b) 0.321 g/ml c) 3.63g/ml d) 3.08g/ml 3- 3.5 nanometer contains how many micrometers? a) 3.5x 10 +3 b) 3.5x 10 -3 c) 3.5x 10 +8 d) 3.5x 10 -8 4- The element in group 2A and period 3 is: a) Gab) Bec) Ald) Mg

5. First Exam/ Exercises 5- Predict the formula of a compound formed from Sr and Cl? a) Sr 2 Cl b) Sr 2 Cl 2 c) SrCl 2 d) SrCl Sr +2 SrCl 2 Cl -

6. First Exam/ Exercises 5- Predict the formula of a compound formed from Sr and Cl? a) Sr 2 Cl b) Sr 2 Cl 2 c) SrCl 2 d) SrCl 6- Which compound has the same empirical formula as C 6 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 C 6 H 12 O 6 divided by 6 = CH 2 O a- C 12 H 20 O 4 divided by 4 = C 3 H 5 O b- C 2 H 8 O 4 divided by 2 = CH 4 O 2 c- C 6 H 3 O 6 divided by 3 = C 2 HO 2 d- C 12 H 24 O 12 divided by 12 = CH 2 O

7. First Exam/ Exercises 5- Predict the formula of a compound formed from Sr and Cl? a) Sr 2 Cl b) Sr 2 Cl 2 c) SrCl 2 d) SrCl 6- Which compound has the same empirical formula as C 6 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 7- An atom is a) Smallest unit of matter that maintains its chemical identity. b) The smallest unit of a compound. c) Always made of carbon d) Smaller than electron

8. First Exam/ Exercises 8-The right answer is: a)Compounds are: N 2, NH 3, H 2 O and CH 4 b)Elements are: N 2, H 2 O, H 2 and F 2 c) Molecules are: N 2, H 2, F 2 and Cl 2 and Fe 2 SO 4 d) Compounds are: NH 3, O 2, H 2 O and CH 4 9- Rubidium (Rb) and cesium (Cs) are members of which of the following categories? a) Halogens b) Alkali metals c) Alkaline earth metals d) Noble gases 10- What is the number of protons, electrons and neutrons in the atom of a) 32 protons, 34 electrons, 16 neutrons b) 16 protons, 16 electrons, 16 neutrons c) 16 protons, 18 electrons, 16 neutrons d) 18 protons, 16 electrons, 32 neutrons

9. First Exam/ Exercises 10- What is the number of protons, electrons and neutrons in the atom of a) 32 protons, 34 electrons, 16 neutrons b) 16 protons, 16 electrons, 16 neutrons c) 16 protons, 18 electrons, 16 neutrons d) 18 protons, 16 electrons, 32 neutrons P = 16, e = 16+2 = 18, n = 32-16 = 16

10. First Exam/ Exercises 8-The right answer is: a)Compounds are: N 2, NH 3, H 2 O and CH 4 b)Elements are: N 2, H 2 O, H 2 and F 2 c) Molecules are: N 2, H 2, F 2 and Cl 2 and Fe 2 SO 4 d) Compounds are: NH 3, O 2, H 2 O and CH 4 9- Rubidium (Rb) and cesium (Cs) are members of which of the following categories? a) Halogens b) Alkali metals c) Alkaline earth metals d) Noble gases 10- What is the number of protons, electrons and neutrons in the atom of a) 32 protons, 34 electrons, 16 neutrons b) 16 protons, 16 electrons, 16 neutrons c) 16 protons, 18 electrons, 16 neutrons d) 18 protons, 16 electrons, 32 neutrons

11. First Exam/ Exercises 11- The correct systematic name for Cr 2 O 3 is a)Chromium (III) oxide. b) Dichromium trioxide. c) Chromium trioxide d) Chromium oxide.

12. 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

13. First Exam/ Exercises 11- The correct systematic name for Cr 2 O 3 is a)Chromium (III) oxide. b) Dichromium trioxide. c) Chromium trioxide d) Chromium oxide. 12- The correct formula for calcium sulfate is a) CaHSO 4 b) CaSO 4 c) CaS d) Ca(HSO 4 ) 2 13- Gallium consists of 60.108% 69 Ga with a mass of 68.9256 amu, and 39.892% 71 Ga with a mass of 70.9247 amu, the average atomic mass of gallium is: a) 70.93 amu b) 71.62 amu c) 68.93 amu d) 69.723 amu Average atomic mass = (atomic mass x abundenace) Ga-69 + (atomic mass x abundenace) Ga-71 Average atomic mass = (68.9256 X (60.108 /100)) Ga-69 + ( 70.9247 x (39.892 /100)) Ga-71 = 69.723

14. First Exam/ Exercises 11- The correct systematic name for Cr 2 O 3 is a)Chromium (III) oxide. b) Dichromium trioxide. c) Chromium trioxide d) Chromium oxide. 12- The correct formula for calcium sulfate is a) CaHSO 4 b) CaSO 4 c) CaS d) Ca(HSO 4 ) 2 13- Gallium consists of 60.108% 69 Ga with a mass of 68.9256 amu, and 39.892% 71 Ga with a mass of 70.9247 amu, the average atomic mass of gallium is: a) 70.93 amu b) 71.62 amu c) 68.93 amu d) 69.723 amu 14- Which pair of Atoms would be most likely to form an ionic compound? a) C and N b) K and Ca c) P and Ar d) Ba and S

15. First Exam/ Exercises 15- Two isotopes of the same element differ only in their a) Number of neutron b) Atomic number c) Number of protons d) Number of electron 16- What is the mass in grams of one atom of iron (Fe)? a) 6.02  10 23 g b) 1.66  10 -24 g c) 9.3  10 -23 g d) 55.85 g Mass of one atom = atomic mass/ avogadro number = 56 / 6.022 x 10 23 = 9.3  10 -23 g

16. First Exam/ Exercises 15- Two isotopes of the same element differ only in their a) Number of neutron b) Atomic number c) Number of protons d) Number of electron 16- What is the mass in grams of one atom of iron (Fe)? a) 6.02  10 23 g b) 1.66  10 -24 g c) 9.3  10 -23 g d) 55.85 g 17- A 4.691 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) 6.58  10 -2 M d) 4.93  10 -2 M M = n/V n = mass /molar mass = 4.691 / 95= 0.0493 mole M = 0.0493 / (750 / 1000)= 0.0658 M = 6.58 X 10 -2 M

17. First Exam/ Exercises 15- Two isotopes of the same element differ only in their a) Number of neutron b) Atomic number c) Number of protons d) Number of electron 16- What is the mass in grams of one atom of iron (Fe)? a) 6.02  10 23 g b) 1.66  10 -24 g c) 9.3  10 -23 g d) 55.85 g 17- A 4.691 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) 6.58  10 -2 M d) 4.93  10 -2 M 18- Calculate the percent composition by mass of C in picric acid (C 6 H 3 N 3 O 7 )? a) 1.3 % b) 18.3 % c) 31.4 % d) 48.9 %

18. First Exam/ Exercises 18- Calculate the percent composition by mass of C 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 g/mol % C= n x molar mass of element molar mass of compound x 100% 6x 12.01 229 x 100% = 31.4 %

19. First Exam/ Exercises 15- Two isotopes of the same element differ only in their a) Number of neutron b) Atomic number c) Number of protons d) Number of electron 16- What is the mass in grams of one atom of iron (Fe)? a) 6.02  10 23 g b) 1.66  10 -24 g c) 9.3  10 -23 g d) 55.85 g 17- A 4.691 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) 6.58  10 -2 M d) 4.93  10 -2 M 18- Calculate the percent composition by mass of C in picric acid (C 6 H 3 N 3 O 7 )? a) 1.3 % b) 18.3 % c) 31.4 % d) 48.9 %

20. First Exam/ Exercises 19- How many molecules of ethane (C 2 H 6 ) are present in 0.30 g of C 2 H 6 ? a) 1.20 x 10 22 b) 1.00 x 10 22 c) 8.03 x 10 21 d) 6.02 x 10 21 Molar mass of ethane = 2 x 12 + 6 x 1 = 30 g/mole First we should calculate the number of mole Number of mole = mass / molar mass Number of mole = 0.30 / 30 = 0.01 mole We know that Number of molecules = avogadro’s number x number of mole = 6.022 x 10 23 x 0.01 = 6.02 x 10 21 molecules.

21. First Exam/ Exercises 19- How many molecules of ethane (C 2 H 6 ) are present in 0.30 g of C 2 H 6 ? a) 1.20 x 10 22 b) 1.00 x 10 22 c) 8.03 x 10 21 d) 6.02 x 10 21 20- The empirical formula of an organic compound with 85.7% C and 14.3% H is a) CH b) CH 2 c) C 2 H d) CH 4 1- change from % to g 85.7 g of C, 14.3 g of H 2- change from g to mole 85.7 12 = 7.14 mol of C n c = 14.3 1 = 14.3 mol of H n H = 7.14 = 1 C: 7.14 14.3 = 2 H: Divided by the smallest number of mole which is 7.14 Thus the empirical formula is CH 2

22. First Exam/ Exercises 19- How many molecules of ethane (C 2 H 6 ) are present in 0.30 g of C 2 H 6 ? a) 1.20 x 10 22 b) 1.00 x 10 22 c) 8.03 x 10 21 d) 6.02 x 10 21 20- The empirical formula of an organic compound with 85.7% C and 14.3% H is a)CH b) CH 2 c) C 2 H d) CH 4 21- After balancing the following equation the coefficients are: a C 3 H 8 + b O 2 → c CO 2 + d H 2 O a) a=1, b=5, c=2, d=4 b) a=2, b=5, c=3, d=4 c) a =1, b=3, c=3, d=4 d) a=1, b=5, c=3, d=4

23. First Exam/ Exercises 21- After balancing the following equation the coefficients are: a C 3 H 8 + b O 2 → c CO 2 + d H 2 O a) a=1, b=5, c=2, d=4 b) a=2, b=5, c=3, d=4 c) a =1, b=3, c=3, d=4 d) a=1, b=5, c=3, d=4 C 3 H 8 + O 2 → CO 2 + H 2 O C 3 C 1 X3 C 3 H 8 + O 2 → 3 CO 2 + H 2 O H 8 H2 X4 C 3 H 8 + O 2 → 3 CO 2 + 4 H 2 O O 2 O 6 + 4 =10 X 5 C 3 H 8 + 5O 2 → 3 CO 2 + 4 H 2 O

24. First Exam/ Exercises 19- How many molecules of ethane (C 2 H 6 ) are present in 0.30 g of C 2 H 6 ? a) 1.20 x 10 22 b) 1.00 x 10 22 c) 8.03 x 10 21 d) 6.02 x 10 21 20- The empirical formula of an organic compound with 85.7% C and 14.3% H is a)CH b) CH 2 c) C 2 H d) CH 4 21- After balancing the following equation the coefficients are: a C 3 H 8 + b O 2 → c CO 2 + d H 2 O a) a=1, b=5, c=2, d=4 b) a=2, b=5, c=3, d=4 c) a =1, b=3, c=3, d=4 d) a=1, b=5, c=3, d=4 22- Calculate the number of O atoms in 2.50 g of sucrose (C 12 H 22 O 11 ) a) 3.87 x 10 22 b) 4.84 x 10 22 c) 5.81 x 10 22 d) 6.78 x 10 22

25. First Exam/ Exercises 22- Calculate the number of O atoms in 2.50 g of sucrose (C 12 H 22 O 11 ) a) 3.87 x 10 22 b) 4.84 x 10 22 c) 5.81 x 10 22 d) 6.78 x 10 22 First we calculate the number of mole n = 2.5/ 342 = 0.007 mole Number of molecules = Avogadro's number x number of mole = 6.022 x 10 23 x 0.007 = 4.4 x10 22 molecules From the chemical formula of sucrose C 12 H 22 O 11 1 molecules of sucrose = 11 atom of O 4.4 x 10 22 molecules = ? Atom of O 11 x 4.4 x 10 22 = 4.84 x10 21 atoms

26. First Exam/ Exercises 19- How many molecules of ethane (C 2 H 6 ) are present in 0.30 g of C 2 H 6 ? a) 1.20 x 10 22 b) 1.00 x 10 22 c) 8.03 x 10 21 d) 6.02 x 10 21 20- The empirical formula of an organic compound with 85.7% C and 14.3% H is a)CH b) CH 2 c) C 2 H d) CH 4 21- After balancing the following equation the coefficients are: a C 3 H 8 + b O 2 → c CO 2 + d H 2 O a) a=1, b=5, c=2, d=4 b) a=2, b=5, c=3, d=4 c) a =1, b=3, c=3, d=4 d) a=1, b=5, c=3, d=4 22- Calculate the number of O atoms in 2.50 g of sucrose (C 12 H 22 O 11 ) a) 3.87 x 10 22 b) 4.84 x 10 22 c) 5.81 x 10 22 d) 6.78 x 10 22

27. First Exam/ Exercises 23- An empirical formula of an organic compound is C 3 H 4 O 2, if the molecular weight of the compound is (360 g/mol), the molecular formula of the compound will be: a) C 6 H 8 O 4 b) C 12 H 16 O 8 c) C 9 H 12 O 6 d) C 15 H 20 O 10 FIRST we calculate the molar mass of emperical formula C 3 H 4 O 2 = 72 g/mol Ratio = 360 / 72 = 5 molecular formula = ratio x empirical formula = 5 x C 3 H 4 O 2 = C 15 H 20 O 10

28. First Exam/ Exercises 23- An empirical formula of an organic compound is C 3 H 4 O 2, if the molecular weight of the compound is (360 g/mol), the molecular formula of the compound will be: a) C 6 H 8 O 4 b) C 12 H 16 O 8 c) C 9 H 12 O 6 d) C 15 H 20 O 10 24- How many moles of FeSO 4 are present in a 500 mg of this salt? a) 1.64 x 10 -3 b) 2.30 x 10 -3 c) 3.29 x 10 -3 d) 4.93 x 10 -3 Mass = 500mg = 0.5 g n= mass / molar mass = 0.5 / 152 = 0.00329 = 3.29 x 10 -3 g

29. First Exam/ Exercises 23- An empirical formula of an organic compound is C 3 H 4 O 2, if the molecular weight of the compound is (360 g/mol), the molecular formula of the compound will be: a) C 6 H 8 O 4 b) C 12 H 16 O 8 c) C 9 H 12 O 6 d) C 15 H 20 O 10 24- How many moles of FeSO 4 are present in a 500 mg of this salt? a) 1.64 x 10 -3 b) 2.30 x 10 -3 c) 3.29 x 10 -3 d) 4.93 x 10 -3 25- How many milliliter of water must be added to 200 mL of 0.15M Na 2 CO 3 to prepare 0.05 M Na 2 CO 3 ? a) 200 mL b) 300 mL c) 400 mL d) 450 mL M 1 V 1 = M 2 V 2 0.15 X 200 = 0.05 X V 2 V 2 = 0.15 X 200 / 0.05 = 600 ml Add water = 600-200 = 400 ml

30. First Exam/ Exercises 23- An empirical formula of an organic compound is C 3 H 4 O 2, if the molecular weight of the compound is (360 g/mol), the molecular formula of the compound will be: a) C 6 H 8 O 4 b) C 12 H 16 O 8 c) C 9 H 12 O 6 d) C 15 H 20 O 10 24- How many moles of FeSO 4 are present in a 500 mg of this salt? a) 1.64 x 10 -3 b) 2.30 x 10 -3 c) 3.29 x 10 -3 d) 4.93 x 10 -3 25- How many milliliter of water must be added to 200 mL of 0.15M Na 2 CO 3 to prepare 0.05 M Na 2 CO 3 ? a)200 mL b) 300 mL c) 400 mL d) 450 mL

31. First Exam/ Exercises 26- What is the theoretical yield of chromium (Cr) that can be produced by the reaction of 40.0 g of Cr 2 O 3 with 8.00 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 = 8 / 27 = 0.296 mol 2- from equation 2mole Al ========= 2 mole Cr 0.296 mole Al =====? Mole Cr 2x 0.296 = 2 x ? Mole of Cr = 0.296 mol Mass = n x molar mass = 0.296 x 52 =15.4g second start with Cr 2 O 3 1-Convert to mole : n = 40 / 152 = 0.263mol 2- from equation 1mole Cr 2 O 3 ========= 2 mole Cr 0.263mole Cr 2 O 3 =====? Mole Cr 2 x 0.263 = 1 x ? Mole of Cr = 0.526 mol

32. First Exam/ Exercises 26- What is the theoretical yield of chromium (Cr) that can be produced by the reaction of 40.0 g of Cr 2 O 3 with 8.00 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 27- If the actual yield for the experiment in question (26) produced 13.0g, what is the percentage yield? a) 84.4% b) 75.0% c) 67% d) 96.4%

33. First Exam/ Exercises 26- What is the theoretical yield of chromium (Cr) that can be produced by the reaction of 40.0 g of Cr 2 O 3 with 8.00 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 27- If the actual yield for the experiment in question (26) produced 13.0g, what is the percentage yield? a) 84.4% b) 75.0% c) 67% d) 96.4% 28- The mass of Na that will react with excess of water to produce 16.0 g NaOH is 2Na(s) + 2H 2 O(l) → H 2 (g) + 2NaOH(aq) a) 5.75 g b) 6.90 g c) 8.05 g d) 9.20 g

34. First Exam/ Exercises 28- The mass of Na that will react with excess of water to produce 16.0 g NaOH is 2Na(s) + 2H 2 O(l) → H 2 (g) + 2NaOH(aq) a) 5.75 g b) 6.90 g c) 8.05 g d) 9.20 g 1-First make sure the equation is balanced 2- g to mole n = mass / molar mass = 16 / 40 = 0.4 mol From equation 2 mole Na =======2mole NaOH ?mole Na ======== 0.4 Mole NaOH 2 X 0.4 = 2 X ? Mole of Na = 0.4 mole Mass = n x molar mass = 0.4 x 23 = 9.2 g

35. First Exam/ Exercises 26- What is the theoretical yield of chromium (Cr) that can be produced by the reaction of 40.0 g of Cr 2 O 3 with 8.00 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 27- If the actual yield for the experiment in question (26) produced 13.0g, what is the percentage yield? a) 84.4% b) 75.0% c) 67% d) 96.4% 28- The mass of Na that will react with excess of water to produce 16.0 g NaOH is 2Na(s) + 2H 2 O(l) → H 2 (g) + 2NaOH(aq) a) 5.75 g b) 6.90 g c) 8.05 g d) 9.20 g

36. First Exam/ Exercises 29- A 100.0 mL of 0.25 M HCl is diluted with water to a total volume of 200.0 mL. What is the final concentration in the resulting solution? a) 0.125M b) 0.083 M c) 0.05 M d) 0.0625 M M 1 V 1 = M 2 V 2 0.25 X 100 = M 2 X 200 M 2 = 0.25 X 100 / 200 = 0.125 M

37. First Exam/ Exercises 29- A 100.0 mL of 0.25 M HCl is diluted with water to a total volume of 200.0 mL. What is the final concentration in the resulting solution? a) 0.125M b) 0.083 M c) 0.05 M d) 0.0625 M 30- What is the mass of carbon in 10 g carbon dioxide (CO 2 )? a) 4.1 g b) 2.73 g c) 6.82 g d) 5.45 g Mole of CO 2 = mass / molar mass = 10 / 44 = 0.227 mol From the formula 1mole C ====== 1 mol CO 2 ? Mole C ====== 0.227 mole CO 2 Mole of C = 0.227 mol Mass of C = n x molar mass = 0.227 x 12 = 2.73 g.

38. First Exam/ Exercises 29- A 100.0 mL of 0.25 M HCl is diluted with water to a total volume of 200.0 mL. What is the final concentration in the resulting solution? a) 0.125M b) 0.083 M c) 0.05 M d) 0.0625 M 30- What is the mass of carbon in 10 g carbon dioxide (CO 2 )? a) 4.1 g b) 2.73 g c) 6.82 g d) 5.45 g