First exam Exercises

1- Prefixes giga and deci represent, respectively: First Exam/ Exercises 1- Prefixes giga and deci represent, respectively: a) 10-9 and 10-1. b) 106 and 10-3. c) 103 and 10-3  d) 109 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

1- Prefixes giga and deci represent, respectively: First Exam/ Exercises 1- Prefixes giga and deci represent, respectively: a) 10-9 and 10-1. b) 106 and 10-3. c) 103 and 10-3  d) 109 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 10-9m Then from m to um 3.5 x 10-9 x 10+6 = 3.5 x 10-3

1- Prefixes giga and deci represent, respectively: First Exam/ Exercises 1- Prefixes giga and deci represent, respectively: a) 10-9 and 10-1. b) 106 and 10-3. c) 103 and 10-3  d) 109 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) Ga b) Be c) Al d) Mg

First Exam/ Exercises 5- Predict the formula of a compound formed from Sr and Cl? a) Sr2Cl b) Sr2Cl2 c) SrCl2 d) SrCl Sr+2 Cl- SrCl2

5- Predict the formula of a compound formed from Sr and Cl? First Exam/ Exercises 5- Predict the formula of a compound formed from Sr and Cl? a) Sr2Cl b) Sr2Cl2 c) SrCl2 d) SrCl 6- Which compound has the same empirical formula as C6H12O6? a) C12H20O4 b) C2H8O4 c) C6H3O6 d) C12H24O12 C6H12O6 divided by 6 = CH2O a- C12H20O4 divided by 4 = C3H5O b- C2H8O4 divided by 2 = CH4O2 c- C6H3O6 divided by 3 = C2HO2 d- C12H24O12 divided by 12 = CH2O

5- Predict the formula of a compound formed from Sr and Cl? First Exam/ Exercises 5- Predict the formula of a compound formed from Sr and Cl? a) Sr2Cl b) Sr2Cl2 c) SrCl2 d) SrCl 6- Which compound has the same empirical formula as C6H12O6? a) C12H20O4 b) C2H8O4 c) C6H3O6 d) C12H24O12 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

a)Compounds are: N2, NH3 , H2O and CH4 First Exam/ Exercises 8-The right answer is: a)Compounds are: N2, NH3 , H2O and CH4 b)Elements are: N2, H2O, H2 and F2 c) Molecules are: N2, H2, F2 and Cl2 and Fe2SO4 d) Compounds are: NH3, O2, H2O and CH4 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

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

a)Compounds are: N2, NH3 , H2O and CH4 First Exam/ Exercises 8-The right answer is: a)Compounds are: N2, NH3 , H2O and CH4 b)Elements are: N2, H2O, H2 and F2 c) Molecules are: N2, H2, F2 and Cl2 and Fe2SO4 d) Compounds are: NH3, O2, H2O and CH4 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- The correct systematic name for Cr2O3 is First Exam/ Exercises 11- The correct systematic name for Cr2O3 is a)Chromium (III) oxide. b) Dichromium trioxide. c) Chromium trioxide d) Chromium oxide.

First Exam/ Exercises Summery of naming compound Ionic Molecular Cation: metal or NH4+ Anion: monotomic or polytomic Nonmetal + nonmetal Nonmetal + metalloid Cation has only one charge Cation has more than one charge 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 Alkali metal Alkaline earth metal Ag+, Al+3, Cd+2, Zn+2 Other metal cations Name first element add ide to the name of second element Add the prefix (prefix mono usually omitted for the first element 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 Name metal first If monoatomic anion, add ide to the anion If polyatomic anion use name of anion from previous table

11- The correct systematic name for Cr2O3 is First Exam/ Exercises 11- The correct systematic name for Cr2O3 is a)Chromium (III) oxide. b) Dichromium trioxide. c) Chromium trioxide d) Chromium oxide. 12- The correct formula for calcium sulfate is a) CaHSO4 b) CaSO4 c) CaS d) Ca(HSO4)2 13- Gallium consists of 60.108% 69Ga with a mass of 68.9256 amu, and 39.892% 71Ga 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

11- The correct systematic name for Cr2O3 is First Exam/ Exercises 11- The correct systematic name for Cr2O3 is a)Chromium (III) oxide. b) Dichromium trioxide. c) Chromium trioxide d) Chromium oxide. 12- The correct formula for calcium sulfate is a) CaHSO4 b) CaSO4 c) CaS d) Ca(HSO4)2 13- Gallium consists of 60.108% 69Ga with a mass of 68.9256 amu, and 39.892% 71Ga 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- Two isotopes of the same element differ only in their 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  1023 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 1023 = 9.3  10-23 g

15- Two isotopes of the same element differ only in their 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  1023 g b) 1.66  10-24 g c) 9.3  10-23 g d) 55.85 g 17- A 4.691 g sample of MgCl2 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 10-2M

15- Two isotopes of the same element differ only in their 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  1023 g b) 1.66  10-24 g c) 9.3  10-23 g d) 55.85 g 17- A 4.691 g sample of MgCl2 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 (C6H3N3O7)? a) 1.3 % b) 18.3 % c) 31.4 % d) 48.9 %

Molar mass of C6H3N3O7 =229.114g/mol % C= n x molar mass of element First Exam/ Exercises 18- Calculate the percent composition by mass of C in picric acid (C6H3N3O7)? a) 1.3 % b) 18.3 % c) 31.4 % d) 48.9 % Molar mass of C6H3N3O7 =229.114g/mol % C= n x molar mass of element molar mass of compound x 100% 6x 12.01 229.11 x 100% = 31.4 %

15- Two isotopes of the same element differ only in their 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  1023 g b) 1.66  10-24 g c) 9.3  10-23 g d) 55.85 g 17- A 4.691 g sample of MgCl2 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 (C6H3N3O7)? a) 1.3 % b) 18.3 % c) 31.4 % d) 48.9 %

19- How many molecules of ethane (C2H6) are present in 0.30 g of C2H6? First Exam/ Exercises 19- How many molecules of ethane (C2H6) are present in 0.30 g of C2H6? a) 1.20 x 1022 b) 1.00 x 1022 c) 8.03 x 1021 d) 6.02 x 1021 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 1023 x 0.01 = 6.02 x 1021 molecules.

19- How many molecules of ethane (C2H6) are present in 0.30 g of C2H6? First Exam/ Exercises 19- How many molecules of ethane (C2H6) are present in 0.30 g of C2H6? a) 1.20 x 1022 b) 1.00 x 1022 c) 8.03 x 1021 d) 6.02 x 1021 20- The empirical formula of an organic compound with 85.7% C and 14.3% H is a) CH b) CH2 c) C2H d) CH4 1- we change from % to g 85.7 g of C, 14.3 g of H 2- change from g to mole using Divided by the smallest number of mole which is 7.14 7.14 = 1 7.14 14.3 = 2 H: C: 85.7 12 = 7.14 mol of C nc = Thus the empirical formula is CH2 14.3 1 = 14.3 mol of H nH =

19- How many molecules of ethane (C2H6) are present in 0.30 g of C2H6? First Exam/ Exercises 19- How many molecules of ethane (C2H6) are present in 0.30 g of C2H6? a) 1.20 x 1022 b) 1.00 x 1022 c) 8.03 x 1021 d) 6.02 x 1021 20- The empirical formula of an organic compound with 85.7% C and 14.3% H is CH b) CH2 c) C2H d) CH4 21- After balancing the following equation the coefficients are: a C3H8 + b O2 → c CO2 + d H2O 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

21- After balancing the following equation the coefficients are: First Exam/ Exercises 21- After balancing the following equation the coefficients are: a C3H8 + b O2 → c CO2 + d H2O 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 C3H8 + O2 → CO2 + H2O C 3 C 1 X3 C3H8 + O2 → 3 CO2 + H2O H 8 H2 X4 C3H8 + O2 → 3 CO2 + 4 H2O O 2 O 6 + 4 =10 X 5 C3H8 + 5O2 → 3 CO2 + 4 H2O

19- How many molecules of ethane (C2H6) are present in 0.30 g of C2H6? First Exam/ Exercises 19- How many molecules of ethane (C2H6) are present in 0.30 g of C2H6? a) 1.20 x 1022 b) 1.00 x 1022 c) 8.03 x 1021 d) 6.02 x 1021 20- The empirical formula of an organic compound with 85.7% C and 14.3% H is CH b) CH2 c) C2H d) CH4 21- After balancing the following equation the coefficients are: a C3H8 + b O2 → c CO2 + d H2O 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 (C12H22O11) a) 3.87 x 1022 b) 4.84 x 1022 c) 5.81 x 1022 d) 6.78 x 1022

22- Calculate the number of O atoms in 2.50 g of sucrose (C12H22O11) First Exam/ Exercises 22- Calculate the number of O atoms in 2.50 g of sucrose (C12H22O11) a) 3.87 x 1022 b) 4.84 x 1022 c) 5.81 x 1022 d) 6.78 x 1022 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 1023 x 0.007 = 4.4 x1022 molecules From the chemical formula of sucrose C12H22O11 1 molecules of sucrose = 11 atom of O 4.4 x 1021 molecules = ? Atom of O 11 x 4.4 x 1021 = 4.84 x1021 atoms

19- How many molecules of ethane (C2H6) are present in 0.30 g of C2H6? First Exam/ Exercises 19- How many molecules of ethane (C2H6) are present in 0.30 g of C2H6? a) 1.20 x 1022 b) 1.00 x 1022 c) 8.03 x 1021 d) 6.02 x 1021 20- The empirical formula of an organic compound with 85.7% C and 14.3% H is CH b) CH2 c) C2H d) CH4 21- After balancing the following equation the coefficients are: a C3H8 + b O2 → c CO2 + d H2O 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 (C12H22O11) a) 3.87 x 1022 b) 4.84 x 1022 c) 5.81 x 1022 d) 6.78 x 1022

a) C6H8O4 b) C12H16O8 c) C9H12O6 d) C15H20O10 First Exam/ Exercises 23- An empirical formula of an organic compound is C3H4O2 , if the molecular weight of the compound is (360 g/mol), the molecular formula of the compound will be: a) C6H8O4 b) C12H16O8 c) C9H12O6 d) C15H20O10 FIRST we calculate the molar mass of emperical formula C3H4O2 = 72 g/mol Ratio = 360 / 72 = 5 molecular formula = ratio x empirical formula = 5 x C3H4O2 = C15H20O10

a) C6H8O4 b) C12H16O8 c) C9H12O6 d) C15H20O10 First Exam/ Exercises 23- An empirical formula of an organic compound is C3H4O2 , if the molecular weight of the compound is (360 g/mol), the molecular formula of the compound will be: a) C6H8O4 b) C12H16O8 c) C9H12O6 d) C15H20O10 24- How many moles of FeSO4 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

a) C6H8O4 b) C12H16O8 c) C9H12O6 d) C15H20O10 First Exam/ Exercises 23- An empirical formula of an organic compound is C3H4O2 , if the molecular weight of the compound is (360 g/mol), the molecular formula of the compound will be: a) C6H8O4 b) C12H16O8 c) C9H12O6 d) C15H20O10 24- How many moles of FeSO4 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 Na2CO3 to prepare 0.05 M Na2CO3? a) 200 mL b) 300 mL c) 400 mL d) 450 mL M1 V1 = M2 V2 0.15 X 200 = 0.05 X V2 V2 = 0.15 X 200 / 0.05 = 600 ml Add water = 600-200 = 400 ml

a) C6H8O4 b) C12H16O8 c) C9H12O6 d) C15H20O10 First Exam/ Exercises 23- An empirical formula of an organic compound is C3H4O2 , if the molecular weight of the compound is (360 g/mol), the molecular formula of the compound will be: a) C6H8O4 b) C12H16O8 c) C9H12O6 d) C15H20O10 24- How many moles of FeSO4 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 Na2CO3 to prepare 0.05 M Na2CO3? 200 mL b) 300 mL c) 400 mL d) 450 mL

First we have to determine the limiting reagent: First Exam/ Exercises 26- What is the theoretical yield of chromium (Cr) that can be produced by the reaction of 40.0 g of Cr2O3 with 8.00 g of aluminum (Al) according to the chemical reaction? 2 Al + Cr2O3  Al2O3 + 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 Cr2O3 1-Convert to mole : n = 40 / 152 = 0.263mol 2- from equation 1mole Cr2O3 ========= 2 mole Cr 0.263mole Cr2O3 =====? Mole Cr 2 x 0.263 = 1 x ? Mole of Cr = 0.526 mol

First Exam/ Exercises 26- What is the theoretical yield of chromium (Cr) that can be produced by the reaction of 40.0 g of Cr2O3 with 8.00 g of aluminum (Al) according to the chemical reaction? 2 Al + Cr2O3  Al2O3 + 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%

2Na(s) + 2H2O(l) → H2(g) + 2NaOH(aq) First Exam/ Exercises 26- What is the theoretical yield of chromium (Cr) that can be produced by the reaction of 40.0 g of Cr2O3 with 8.00 g of aluminum (Al) according to the chemical reaction? 2 Al + Cr2O3  Al2O3 + 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) + 2H2O(l) → H2(g) + 2NaOH(aq) a) 5.75 g b) 6.90 g c) 8.05 g d) 9.20 g

2Na(s) + 2H2O(l) → H2(g) + 2NaOH(aq) First Exam/ Exercises 28- The mass of Na that will react with excess of water to produce 16.0 g NaOH is 2Na(s) + 2H2O(l) → H2(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

2Na(s) + 2H2O(l) → H2(g) + 2NaOH(aq) First Exam/ Exercises 26- What is the theoretical yield of chromium (Cr) that can be produced by the reaction of 40.0 g of Cr2O3 with 8.00 g of aluminum (Al) according to the chemical reaction? 2 Al + Cr2O3  Al2O3 + 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) + 2H2O(l) → H2(g) + 2NaOH(aq) a) 5.75 g b) 6.90 g c) 8.05 g d) 9.20 g

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 M1 V1 = M2 V2 0.25 X 100 = M2 X 200 M2= 0.25 X 100 / 200 = 0.125 M

30- What is the mass of carbon in 10 g carbon dioxide (CO2)? 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 (CO2)? a) 4.1 g b) 2.73 g c) 6.82 g d) 5.45 g Mole of CO2 = mass / molar mass = 10 / 44 = 0.227 mol From the formula 1mole C ====== 1 mol CO2 ? Mole C ====== 0.227 mole CO2 Mole of C = 0.227 mol Mass of C = n x molar mass = 0.227 x 12 = 2.73 g.

30- What is the mass of carbon in 10 g carbon dioxide (CO2)? 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 (CO2)? a) 4.1 g b) 2.73 g c) 6.82 g d) 5.45 g

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