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Chemistry 11 Final Examination Review - Answers Part A - True or False. Indicate whether each of the following statements is true or false. Correct the false statements. F 1. The mass of an electron is equal to the mass of a proton. The mass of an electron is less than the mass of a proton. T 2. The mass of a proton is approximately equal to the mass of a neutron. T 3. The atomic number represents the number of protons in a nucleus. T 4. The proton has a mass of approximately 1 u. F 5. The difference in mass of isotopes of the same element is due to the different number of protons in the nucleus. The difference in mass of isotopes of the same element is due to the different number of neutrons in the nucleus. T 6. The isotope carbon-12 is used as the relative mass standard for the atomic mass scale. F 7. The mass of the most common isotope of each element is listed on the periodic table. The average mass of the naturally occurring isotopes of each element is listed on the periodic table. Part B - Multiple Choice D A 1. A(n) __ is used to represent a compound. a) symbol b) equation c) subscript d) formula 2. ___ atoms or groups of atoms are called ions. a) charged b) diatomic c) neutral d) monatomic A 3. For the formula of a compound to be correct, the algebraic addition of the charges on the atoms or ions in the compound must add up to __. a) zero b) one c) two d) four D 4. Potassium bromide is an example of a(n) __ compound. a) molecular b) organic c) polyatomic B d) ionic 5. The only common polyatomic ion that has a positive charge is the __ ion. a) phosphate b) ammonium c) sulfate d) nitrate Chemistry 11 Exam Review – Answers - Page 1 A B D 6. In the formula H2SO4, the number 4 would be called a(n) __. a) subscript b) oxidation number c) coefficient d) charge 7. Which subatomic particle contributes the least to the mass of an atom? a) nucleon b) electron c) proton d) neutron 8. In the free, or uncombined, state the number of protons in the nucleus of an element must equal the __. a) mass number c) mass number - atomic number b) number of neutrons in the nucleus d) number of electrons present C 9. Which of the following ideas of the Bohr model is not retained in the modern theory of atomic structure? a) Electrons can absorb or emit energy only in whole numbers of photons. b) Atoms have a central positively charged nucleus. c) Electrons move around the nucleus as planets orbit the sun. d) Most of the volume of an atom is empty space. B 10. Which of the following orbitals is spherical in shape? a) 3p b) 2s c) 4d d) 5f 11. The third energy level of an atom may have __ electrons. a) 2 b) 18 c) 8 d) 32 B C 12. How many sublevels are possible at the fourth energy level? a) 2 b) 3 c) 4 d) 18 A 13. Lustrous, malleable, ductile elements that are good conductors of electricity and heat are classified as __. a) metals b) nonmetals c) metalloids d) noble gases C 14. The electron configuration of a certain element ends with 3p5. Which of the following describes its position in the periodic table? a) period 5, group 13 b) period 3, group 15 c) period 3, group 17 d) period 5, group 15 B 15. Which of the following is an example of a metalloid? a) iodine b) boron c) bromine d) indium 16. The periodicity of the elements is basically a function of their __. a) nuclear stability b) atomic numbers c) mass numbers d) none of these 17. An element with seven electrons in the outer level would be a __. a) metal b) metalloid c) noble gas d) nonmetal B D A 18. As the atomic number in a period increases, the degree of nonmetallic character __. a) increases c) increases then decreases b) decreases d) remains the same Chemistry 11 Exam Review – Answers - Page 2 B 19. Elements in a group have similar chemical properties because of their similar __. a) nuclear configurations c) mass numbers b) outer electron configurations d) names D 20. The period number in the periodic table designates the __ for the row. a) total nuclear charge c) maximum number of outer electrons b) maximum number of nucleons d) highest energy level D 21. Compared to the stability of the original atom, the stability of its ion that resembles a noble gas configuration would be __. a) identical b) sometimes less c) less d) greater C 22. The formation of bonds between atoms depends on __. a) the electron configurations of the atoms involved b) the attraction the atoms have for electrons c) both of the preceding factors d) neither of the preceding factors D 23. The particle that results when two or more atoms form covalent bonds is a __. a) single charged atom b) molecule c) polyatomic ion d) b or c b) metallic d) compounds of polyatomic ions A 24. The most active __ have the highest electronegativities. a) nonmetals b) metalloids c) metals d) noble gases D 25. __ compounds have high melting points, conduct electricity in the molten phase, and tend to be soluble in water. a) hydrogen b) metallic c) covalent d) ionic A 26. The element in the following group that has the lowest electronegativity is __. a) potassium b) arsenic c) bromine d) chromium D 27. If there are only two electron pairs in the outer energy level of an atom in a molecule, they will be found __. a) at 90º to one another c) at 120º to one another b) on the same side of the nucleus d) on opposite side of the nucleus B 28. The __ molecule has two bonding pairs and two unshared pairs of electrons. a) CH4 b) H2O c) NH3 d) HF D 29. A certain atom contains 34 protons, 34 electrons, and 45 neutrons. This atom has a mass number of __. a) 34 b) 45 c) 68 d) 79 C 30. An example of a compound is a) oxygen b) mercury c) salt Chemistry 11 Exam Review – Answers - Page 3 d) diamond A 31. Carbon is classed as an element rather than as a compound because it a) cannot be chemically decomposed into two or more substances b) has been known for many centuries c) is formed when wood is heated out of contact with air d) combines with oxygen to form a gas A 32. The positively charged particles in the nucleus of an atom are called a) protons b) neutrons c) electrons d) ions C 33. The charged particles that are found outside the nucleus of an atom are called a) protons b) ions c) electrons d) mesons D 34. A nonmetallic atom generally becomes a negative ion by a) losing protons b) losing electrons c) gaining protons d) gaining electrons 35. Which of the following subatomic particles has the smallest mass? a) electron b) neutron c) proton d) nucleus A B 36. Which of the following symbols represents an atom that contains the largest number of neutrons? 235 a) 92 U 239 16 b) 92 U 239 239 c) 93 Np d) 94 Pu C 37. The nuclide symbol 8 O represents an oxygen atom with a) a mass of 8 u c) a mass of 16 u b) an atomic number of 16 d) 16 neutrons C 38. If Z represents the atomic number of an element and A represents the mass number, then the number of neutrons in one atom is a) A b) A + Z c) A - Z d) Z - A D B 24 25 39. Which of the following statements about the elemental species 11 X and 12 Z is correct? a) They are isotopes of the same element. b) They are nonmetals. c) They are members of the same chemical family. d) They have the same number of neutrons per atom. 40. An element X has a mass number of 32 and an atomic number of 16. The most common ion of element X is represented by a) X+ b) X2c) Xd) X2+ Chemistry 11 Exam Review – Answers - Page 4 Part C - Short Answer 1. What is the maximum number of electrons that may occupy one orbital? two 2. The Lewis electron dot diagram is used to represent only which electrons in the atom? Valence electrons 3. What is the diagonal rule used to predict? The order in which orbitals are filled 4. How many sublevels are possible at the third energy level? three 5. How many orbitals are there in the f sublevel? seven 6. What is the maximum number of electrons that can occupy a d sublevel? ten 7. Which sublevel may contain a maximum of three pairs of electrons? „p‟ sublevel 8. What must be true about the spins of two electrons occupying the same orbital? They must have opposite spin 9. Write the electron configuration for each of the following elements: a) lithium 1s22s1 b) radium 1s22s22p63s23p64s23d104p65s24d105p66s24f145d106p67s2 c) sodium 1s22s22p63s1 d) mercury 1s22s22p63s23p64s23d104p65s24d105p66s24f145d10 Chemistry 11 Exam Review – Answers - Page 5 10. Draw the energy level diagram for each of the following elements: a) tin 6s 5p 4d 5s 4p 3d 4s 3p 3s 2p 2s 1s b) krypton 5s 4p 3d 4s 3p 3s 2p 2s 1s Chemistry 11 Exam Review – Answers - Page 6 c) gold 6p 5d 4f 6s 5p 4d 5s 4p 3d 4s 3p 3s 2p 2s 1s d) potassium 3d 4s 3p 3s 2p 2s 1s Chemistry 11 Exam Review – Answers - Page 7 11. Write the energy level population for each of the following elements: a) calcium 2, 8, 8, 2 b) sulfur c) scandium d) tungsten 2. 8. 6 2, 8, 9, 2 2, 8, 18, 32, 12, 2 12. Element X has the following configuration: 1s22s22p63s22p64s23d104p65s1 a) What period is this element located in? five b) What group is this element in? one (alkali metals) c) Identify the element. Rubidium (Rb) d) Is the element a metal, nonmetal, or metalloid? Metal 13. List five properties of metals and five properties of nonmetals. Metals - good conductors of heat and electricity lustre malleable and ductile high densities high boiling points and melting points resist stretching and twisting solids at room temperature Nonmetals - no lustre poor conductors may be solid, liquid or gas low densities low melting and boiling points 14. Would an element with two outer electrons be a metal or a nonmetal? An element with two outer electrons would be a metal. 15. For the transition elements, as the atomic number increases, to which sublevel are the electrons being added? For the transition metals, the electrons are being added to the “d” sublevel. 16. According to the octet rule, how many pairs of outer electrons do the most stable atoms have? According to the octet rule, the most stable atoms have four pairs of outer electrons. 17. In the lanthanide series, as the atomic number increases, to which sublevel are electrons being added? In the lanthanide series, electrons are being added to the “f” sublevel. Chemistry 11 Exam Review – Answers - Page 8 18. Give examples of molecules with the following shapes: a) linear HCl, CO2, BeCl2 b) trigonal planer c) trigonal pyramidal d) tetrahedral e) bent BH3, BCl3, AlBr3 NH3, PH3 CH4, CCl4 H2O, H2S 19. Draw Lewis Structures, sketch and name the shape of the molecule, predict the bond polarity, and molecule polarity for the following molecules: a) PH3 Molecule b) CO2 c) H2O Lewis Structure d) HCN e) CF4 Shape f) BCl3 Bond Polarity Molecule Polarity a) PH3 nonpolar nonpolar b) CO2 polar nonpolar c) H2O polar polar d) HCN e) CH4 C-H = nonpolar C-N = polar polar nonpolar nonpolar polar nonpolar Cl f) BCl3 B Cl Cl Chemistry 11 Exam Review – Answers - Page 9 20. Use electron dot formulas to show the complete balanced bonding reactions between the following elements. Indicate the type of bond expected. a) sodium and oxygen b) nitrogen and hydrogen c) zirconium and sulfur d) magnesium and chlorine 21. Explain the process of fractional distillation and describe the variety of substances obtained from crude oil. Petroleum is a complex mixture of hundreds of thousands of compounds. Some of these compounds boil at temperatures as low as 20ºC. The least volatile compounds of crude oil, boil at temperatures above 400ºC. The differences in boiling points of the compounds making up petroleum enable the separation of these compounds in a process called fractional distillation, or fractionation. When crude oil is heated to 500ºC in the absence of air, most of its constituent compounds vaporize. The compounds with boiling points higher than 500ºC remain as mixtures called asphalts and tars. The vaporized components of the petroleum rise and gradually cool in a metal fractional distillation tower. A fractional distillation tower contains trays positioned at various levels. Heated crude oil enters near the bottom of the tower. The bottom of the tower is kept hot, and the temperature gradually decreases toward the top of the tower. Where the temperature in the higher parts of the tower is below the boiling point of the vaporized compounds, the substances in the vapour begin to condense. Those substances with high boiling points condense in the lower, Chemistry 11 Exam Review – Answers - Page 10 hotter parts of the tower, whereas those with lower boiling points condense near the cooler top of the tower. At various levels in the tower, trays collect mixtures of substances as they condense, each mixture containing compounds with similar boiling points. These mixtures are called petroleum fractions. The fractions with the lowest boiling points contain the smallest molecules. The low boiling points are due to the fact that small molecules have fewer electrons and weaker London dispersion forces, compared with large molecules. The fractions with higher boiling points contain much larger molecules. The physical process of fractionation is followed by chemical processes in which the fractions are converted into valuable products. Almost all of the products of petroleum refining are burned as fuel for heating and transportation. Only 5% of the original mass of petroleum is used as starting chemicals (feedstock) in the manufacture of solvents, greases, plastics, synthetic fibres, and pharmaceuticals. 22. Explain the difference between saturated and unsaturated compounds and how this relates to fats and oil. Describe a laboratory technique that can be used to distinguish between saturated and unsaturated compounds. In saturated compounds, all carbon-carbon bonds are single bonds. In unsaturated compounds, there are double or triple carbon-carbon bonds. Fats and oils consist of saturated and unsaturated fatty acids (long chain carboxylic acids). Unsaturated fatty acids are a healthier choice. Many fats and oils contain a mixture of saturated and unsaturated fatty acids; a healthier choice would be one that had a low percentage of saturated fatty acids. A simple laboratory technique to distinguish between saturated and unsaturated compounds is to add a dilute solution of permanganate (usually potassium permanganate). Permanganate ions are purple. Permanganate ions will react with compounds that contain double or triple bonds and change to manganese dioxide, which is brown in colour. Therefore, if the solution remains purple, the substance is saturated, but if the solution turns brown, the substance is unsaturated. *23. Compare the boiling points of methane, ethane, propane, and butane. Use intermolecular bonding to explain why they are different. The boiling points, in order from least to greatest, are: methane, ethane, propane, and butane. All four compounds contain only carbon and hydrogen. All bonds are nonpolar, therefore all of the molecules are nonpolar. The only type of intermolecular attraction is dispersion for all four molecules. As the size of the atom increases (and the number of electrons) the amount of dispersion increases. Chemistry 11 Exam Review – Answers - Page 11 *24. Compare the boiling points of propanoic acid, 1-butanol, diethyl ether, butanal, and pentane. intermolecular bonding to explain why they are different. The boiling points, in order from least to greatest, are: diethyl ether, 1-butanol, and propanoic acid. Pentane CH3 CH2 CH2 CH2 Use pentane, butanal, CH3 42 electrons, all nonpolar bonds dispersion only lowest boiling point nonpolar molecule O Butanal CH3 CH2 CH2 CH 42 electrons approximately the same amount of dispersion nonpolar C-C and C-H bonds one polar C=O bond, dipole-dipole in addition to dispersion, causing a higher boiling point than pentane Diethyl ether CH3 CH2 O CH2 CH3 42 electrons approximately the same amount of dispersion nonpolar C-C and C-H bonds two polar C-O bonds, dipole-dipole in addition to dispersion, causing a higher boiling point than butanal because of the two polar bonds OH 1-Butanol CH2 CH2 CH2 CH3 42 electrons approximately the same amount of dispersion nonpolar C-C and C-H bonds one polar C-O bond and one polar O-H bond, dipole-dipole Hydrogen bonding (O-H) in addition to the dispersion and dipole-dipole, a higher boiling point than diethyl ether O Propanoic acid CH3 CH2 C OH 40 electrons approximately the same amount of dispersion nonpolar C-C and C-H bonds two polar C-O bonds and one polar O-H bond, dipole-dipole Hydrogen bonding (O-H) in addition to the dispersion and dipole-dipole a higher boiling point than 1-butanol because of the additional polar bonds Chemistry 11 Exam Review – Answers - Page 12 25. Write the correct names for the following chemical compounds. a) HCl(aq) hydrochloric acid m) SO3 sulfur trioxide b) KOH potassium hydroxide n) NaC2H3O2 sodium acetate c) HgOH mercury(I) hydroxide o) HFO2(aq) flourous acid d) FeCl3 iron(III) chloride p) Al(BrO4)3 aluminum perbromate e) Al2(SO4)3 aluminum sulfate q) SbF3 antimony(III) fluoride f) N2O5 dinitrogen pentoxide r) Pd(CN)2 palladium(II) cyanide g) HF hydrofluoric acid s) Ca(MnO3)2 calcium manganate h) Pb(OH)2 lead(II) hydroxide t) Be(NO4)2 beryllium pernitrate i) NH4NO3 ammonium nitrate u) NiSeO4 nickel(II) selenate j) NaHCO3 sodium bicarbonate v) H2SO3(aq) sulfurous acid k) Zn(NO2)2 zinc nitrite w) Ba(OH)2 barium hydroxide l) phosphoric acid x) PbS lead(II) sulfide H3PO4(aq) 26. Write the correct chemical formula for each compound and balance the equation. a) sodium carbonate + calcium hydroxide sodium hydroxide + calcium carbonate Na2CO3 + Ca(OH)2 2NaOH + CaCO3 b) carbon dioxide + water carbonic acid CO2 + H2O H2CO3 c) phosphorus + oxygen phosphorus pentoxide 2P + 5O2 2PO5 d) sodium + water sodium hydroxide + hydrogen 2Na + 2H2O 2NaOH + H2 e) zinc + sulfuric acid zinc sulfate + hydrogen Zn + H2SO4 ZnSO4 + H2 f) aluminum sulfate + calcium hydroxide aluminum hydroxide + calcium sulfate Al2(SO4)3 + 3Ca(OH)2 2Al(OH)3 + 3CaSO4 g) calcium oxide + water calcium hydroxide CaO + H2O Ca(OH)2 h) iron + copper(I) nitrate iron(II) nitrate + copper Fe + 2CuNO3 Fe(NO3)2 + 2Cu Chemistry 11 Exam Review – Answers - Page 13 i) iron(II) sulfide + hydrochloric acid hydrogen sulfide + iron(II) chloride j) potassium oxide + water potassium hydroxide k. carbon + ferric oxide iron + carbon dioxide FeS + 2HCl H2S + FeCl2 K2O + H2O 2KOH 3 C + 2 Fe2O3 4 Fe + 3 CO2 l. sulfur tetrafluoride + water sulfur dioxide + hydrofluoric acid SF4 + 2 H2O SO2 + 4 HF m. calcium hydroxide + phosphoric acid calcium phosphate + water 3 Ca(OH)2 + 2 H3PO4 Ca3(PO4)2 + 6 H2O n. ethane + oxygen carbon dioxide + water 2 C2H6 + o. 7 O2 4 CO2 + 6 H2O aluminum sulfate + ammonia + water aluminum hydroxide + ammonium sulfate Al2(SO4)3 + 6 NH3 + 6 H2O 2 Al(OH)3 + 3 (NH4)2SO4 27. Identify the type of reaction. Write the correct chemical formulas of the compounds, complete and balance the equations and name the products formed. calcium chloride + carbonic acid + H2CO3 a) calcium carbonate + hydrochloric acid CaCO3 + 2HCl CaCl2 double displacement b) iron + sodium bromide Fe + NaBr no reaction b/c iron is less active than sodium single displacement ammonium chloride + iron(II) acetate 2NH4Cl + Fe(C2H3O2)2 c) ammonium acetate + iron(II)chloride 2NH4C2H3O2 + FeCl2 double displacement silver sulfate + ammonium bromide 2AgBr + (NH4)2SO4 Ag2SO4 + 2NH4Br double displacement d) silver bromide + ammonium sulfate Chemistry 11 Exam Review – Answers - Page 14 e) zinc + sulfuric acid Zn + H2SO4 H2 single displacement hydrogen + zinc sulfate + ZnSO4 f) neon + potassium Ne + K no reaction b/c noble gases are unreactive combination lead(II) chloride + water + 2H2O g) lead(II )hydroxide + hydrochloric acid Pb(OH)2 + 2HCl PbCl2 double displacement iron(II) sulfide Fe + S FeS combination h) iron + sulfur i) potassium chloride + oxygen potassium chlorate (heated) 2KClO3 2KCl + 3O2 decomposition j) calcium hydroxide CaO + H2O Ca(OH)2 combination calcium oxide + water k) dinitrogen pentoxide + water nitric acid N2O5 + H2O 2HNO3 combination l) carbonic acid CO2 + H2O H2CO3 combination carbon dioxide + water m) chlorine + chromium(III) bromide bromine + chromium(III) chloride 3 Cl2 + 2 CrBr3 3 Br2 + 2 CrCl3 Single displacement n) sulfur + oxygen sulfur S + O2 SO2 dioxide or sulfur trioxide or S + O2 SO3 combination Chemistry 11 Exam Review – Answers - Page 15 o) zinc + hydrochloric acid zinc chloride + hydrogen Zn + 2 HCl ZnCl2 + H2 Single displacement p) sodium + water sodium hydroxide + hydrogen 2 Na + 2 H2O 2 NaOH + H2 single displacement q) magnesium + water Mg + H2O no reaction – Mg is not active enough to replace H in water Single displacement r) copper + stannic nitrate Cu + Sn(NO3)4 no reaction – Cu is less active than Sn Single displacement s) aluminum + cupric sulfate copper + aluminum sulfate 2 Al + 3 CuSO4 3 Cu + Al2(SO4)3 Single Displacement Part D – Calculations Show all work. Express your answers using significant digits and include units with your answers. 1. Calculate the average atomic mass of the following elements: a) Mg-24 Mg-25 Mg-26 mass = 23.985 u mass = 24.986 u mass = 25.983 u 78.70% 10.13% 11.17% (0.7870)(23.985) + (0.1013)(24.986) + (0.1117)(25.983) = 24.31 u b) Ir-191 Ir-193 mass = 191.0 u mass = 193.0 u 37.58% 62.42% (0.3758)(191.0) + (0.6242)(193.0) = 192.2 u Chemistry 11 Exam Review – Answers - Page 16 2. Calculate the average atomic mass for titanium based on the following data: 46 22Ti 47 22Ti = 8.0%, = 7.5%, 48 22Ti = 73.7%, 49 22Ti = 5.5%, and 50 22Ti = 5.3% Avg = (46)(0.080) + (47)(0.075) + (48)(0.737) + (49)(0.055) + (50)(0.053) Average Atomic Mass = 47.9 u or 47.9 amu 3. Calculate the percentage of each of the isotopes of silver if silver-107 has a mass of 106.905 u and silver-109 has a mass of 108.905 u and the average atomic mass of silver is 107.869 u 107 Ag = x, 109 Ag = y x+y=1 106.905x + 108.905y = 107.869 -106.905x - 106.905y = -106.905 106.905x + 108.905y = 107.869 2y = 0.964 107 109 Ag = 51.8%, use elimination or substitution y = 0.482 x = 0.518 Ag = 48.2% 4. Calculate the average atomic mass of selenium, which consists of the following naturally occurring 74 34Se isotopes: amu; 78 34Se 0.90%, mass = 73.9225 amu; 23.5%, mass = 77.9173 amu; 80 34Se 76 34Se 77 9.0%, mass = 75.9192 amu; 34Se 7.6%, mass = 76.9199 49.8%, mass = 79.9165 amu; and 82 34Se 9.2%, mass = 81.9167 amu. (73.9225)(0.0090) + (75.9192)(0.090) + (76.9199)(0.076) + (77.9173)(0.235) + (79.9165)(0.498) + (81.9167)(0.092) = avg Average Atomic Mass = 78.9892618 = 78.99 u or 78.99 amu *5. Naturally, occurring silicon consists of three isotopes, 28Si, 29Si, and 30Si, whose atomic masses are 27.9769, 28.9765, and 29.9738, respectively. The most abundant isotope is 28Si, which accounts for 92.23 percent of naturally occurring silicon. Given that the observed atomic mass of silicon is 28.0855, calculate the percentages of 29Si and 30Si in nature. Isotope Mass Percentage 28 Si 27.9769 g 92.23% 29 Si 28.9765 g x 30 Si 29.9738 g y Average 28.0855 (eqn 1) (eqn 1) (eqn 1) (0.9223)(27.9769) + (28.9765)(x) + (29.9738)(y) = 28.0855 25.8031 + 28.9765x + 29.9738y = 28.0855 28.9765x + 29.9738y = 2.2824 (eqn 2) (eqn 2) 0.9223 + x + y = 1 x + y = 0.0777 (eqn 1) (-28.9756)(eqn 2) y = 0.0310, use substitution or elimination 28.9765x + 29.9738y = 2.2824 -28.9765x – 28.9765y = -2.2515 0.9973y = 0.03093 x = 0.0467, 29 Si = 3.10% and Chemistry 11 Exam Review – Answers - Page 17 y = 0.03093 0.9973 30 Si = 4.67% 6. Complete the chart below: Element Atomic # Mass # Protons Neutrons Electrons calcium-43 20 43 20 23 20 lead-211 82 211 82 129 82 plutonium-242 94 242 94 148 94 chromium-50 24 50 24 26 24 29 65 29 36 27 S2- 16 34 16 18 18 128 53 I 53 128 53 75 54 208 4+ 82 Pb 82 208 82 126 78 65 Cu2+ 34 7. Convert each of the following to moles. a. 8.8 g of potassium carbonate 8.8 g K2CO3 138.2052 g/mol = 0.0637 mol b. 0.257 g of arsenic pentachloride 252.1866 g/mol = 1.02 x 10-3 mol 0.257 g AsCl5 c. 12.5 L of carbon dioxide at STP 12.5 L CO2 22.4 L/mol = 0.558 mol 13 d. 5.00 x 10 atoms of iron 5.00 x 1013 atoms of Fe 6.022 x 1023 = 8.30 x 10-11 mol e. 8.63 x 1028 molecules of water 8.63 x 1028 molecules of H2O 6.022 x 1023 = 1.43 x 105 mol f. 450.0 mL helium at STP 0.4500 L He 22.4 L/mol = 0.0201 mol g. 236.0 g of ammonium phosphate 236.0 g (NH4)3PO4 149.086 74 g/mol = 1.58 mol h. 15.0 g of butanoic acid 15.0 g CH3CH2CH2COOH 88.106 32 g/mol = 0.170 mol Chemistry 11 Exam Review – Answers - Page 18 8. Calculate the mass of each of the following. a. 2.60 mol of sodium carbonate (2.60 mol)(105.988 74 g/mol) = 276 g b. five million atom of gold (5 000 000 atoms) (6.022 x 1023) = 8.30 x 10-18 mol (8.30 x 10-18 mol)(196.9665 g/mol) = 1.64 x 10-15 g c. 25.0 mL of carbon dioxide at STP (0.0250 L) (22.4 L/mol) = 0.001 12 mol (0.001 12 mol)(44.0098 g/mol) = 0.0491 g d. 4.50 x 1021 molecules of decanoic acid CH3CH2CH2CH2CH2CH2CH2CH2CH2COOH or C10H202 4.50 1021 6.022 1023 0.00747 mol g 0.00747mol 172.2676 mol 1.29 g 9. Determine the percentage composition of each element in gallium nitrate. % Ga %N %O 69.72 x 100 27.26% Ga 255.7347 3 14.0067 x 100 16.43% N 255.7347 9 15.9994 x 100 56.31% O 255.7347 10. Aspirin tablets contain Acetylsalicylic acid (ASA), which has the chemical formula C 9H8O4. Calculate the mass of carbon contained in a 250.0 mg sample of ASA. mass of carbon 9 12.011 250.0 mg 180.16012 150.0 mg of carbon 11. Calculate the molarity of 825 mL of solution, which contains 30.0 g of acetic acid. 30.0 g CH3 COOH 60.05256 g mol 0.4996 g 0.4996 g 0.825 L 0.606 Chemistry 11 Exam Review – Answers - Page 19 mol L 12. What volume of solution can be made from 80.0 g of sodium hydroxide if a 2.00 M solution is required? 80.0 g NaOH 38.887 11 n C V 2.00 mol g mol 2.00 mol 2.00 1.00 L mol L 13. What mass of calcium chloride is required to produce 750.0 mL of a 0.500 M solution? n C V mol L 0.500 0.375 mol 110.984 g mol 0.7500 L 0.375 mol 41.6 g 14. What volume 0f 14.0 M nitric acid is required to produce 750.0 mL of a 0.250 M solution? C1 V1 C2 V2 V1 0.250 C2 V2 C1 mol L 14 .0 0.7500 L 0.0134 L mol L or 13 .4 mL 15. What concentration results when 250.0 mL of a 0.125 M solution of hydrochloric acid is mixed with 125.0 mL of a 1.00 M solution? C ab C a Va Va 0.125 Cb Vb Vb 0.03125 mol mol L 0.2500 L 0.2500 L 0.125 mol 0.375 L 0.417 1.00 mol L 0.1250 L mol L Chemistry 11 Exam Review – Answers - Page 20 0.1250 L 16. What volume of 0.225 M sulfuric acid must be mixed with 500.0 mL of a 0.750 M solution in order to obtain a 0.500 M solution? C a Va Va C ab C b Vb Vb 0.225 0.500 mol L 0.500 mol L 0.500 mol Va L 0.275 mol L Va mol L Va Va mol L 0.5000 L 0.5000L 0.5000 L 0.225 mol L Va 0.375 mol 0.250 mol 0.225 mol L Va 0.375 mol Va Va 0.125 mol 0.275 0.750 mol L 0.125 mol 0.455 L 17. Determine the empirical formula for each compound listed below: a. 80.0% carbon; 20.0% hydrogen 80.0 g C 20.0 g H : 12.011 g mol 1.00794 g mol 6.66056 mol C 19.84245 mol H : 6.66056 6.66056 1 : 2.979 1:3 CH3 Chemistry 11 Exam Review – Answers - Page 21 b. 35.0% nitrogen; 5.0% hydrogen; 60.0% oxygen 35.0 g N 5.0 g H 60.0 g O : : 14.0067 g mol 1.00794 g mol 15.9994 g mol 2.498804 mol N 4.960613 mol H 3.750141mol H : : 2.498804 2.498804 2.498804 1 : 1.985 : 1.50 x 2 2 : 4 : 3 N2H4O3 c. 83.7% carbon; 16.3% hydrogen 83.7 g C 16.3 g H : 12.011 g mol 1.00794 g mol 6.969612 mol C 16.171598 mol H : 6.969612 6.969612 1 : 2.320634 x 3 3 : 6.962 3 : 7 C3H7 d. 26.6% potassium; 35.4% chromium; 38.0% oxygen e. *18. Chemical analysis of a 10.000 g sample of oil of wintergreen shows that it consists of 6.320 g of carbon, 0.530 g of hydrogen, and 3.16 g of oxygen. What is the simplest formula for oil of wintergreen? Calculate the molecular formula for the following compounds. a) Analysis of a compound shows that it consists of 24.3% carbon, 4.1% hydrogen, and 71.6% chlorine. The molecular mass of the compound is determined to be 99.8 g/mol. What molecular formula corresponds to these data? b) Chemical analysis of a gaseous compound show its composition to be 36.4% carbon, 57.5% fluorine, and 6.1% hydrogen. A sample of 1.00 L of this gas has a mass of 2,96 g. What molecular formula do these data suggest for this compound? c) Analysis of an organic compound indicates that it has a percentage composition as follows: 40.7% carbon; 5.0% hydrogen; 54.3% oxygen. When this compound is vapourized, 35.0 mL of the vapour has a mass of 0.184 g. Determine the molecular formula for this compound. d) A gaseous compound is found to have the following composition: 30,5% nitrogen and 69.5% oxygen. The molar mass of the gas if found to be 91.8 g/mol. What is the molecular formula of this gas? Chemistry 11 Exam Review – Answers - Page 22 E. STOICHIOMETRY - Begin each problem by writing a balanced chemical equation. 1. Sulfur dioxide may be catalytically oxidized to sulfur trioxide. How many grams of sulfur dioxide could be converted by this process if 100.0 g of oxygen are available for the oxidation? 2SO2 + O2 2SO3 100.0 g O2 31.9988 g mol 2 mol SO2 1 mol 64.0648 O2 g mol 400.4 g SO2 2. Phosphoric acid is produced in the reaction between calcium phosphate and sulfuric acid. What mass of phosphoric acid would be produced from 55.0 g of the calcium phosphate? Ca3(PO4)2 + 3H2SO4 3CaSO4 + 2H3PO4 55.0 g Ca3 (PO 4 )2 310.183 g mol 2 mol 1 mol H3PO 4 97.995 Ca3 (PO 4 )2 g mol 34.8 g H3PO 4 3. How much magnesium sulfate is needed to completely react with 145 g of sodium chloride? How much sodium sulfate could be produced by this reaction? MgSO4 + 2NaCl Na2SO4 + MgCl2 145 g 58.443 145 g 58.443 NaCl 1 mol g mol MgSO4 2 mol NaCl 1 mol g mol 120.3686 NaCl Na2 SO4 2 mol NaCl 142.043 g mol g mol 149 g MgSO4 176 g Na2 SO4 4. Gold will dissolve in the acid mixture known as aqua regia according to the following reaction: Au + HNO3 + 3 HCl AuCl3 + NO + 2 H2O How much gold(III)chloride will be produced in this reaction when one starts with 5.0 mg of gold? 5.0 m g 196.9665 Au mg mmol 1 m ol 1 m ol AuCl3 Au 303.3255 mg mmol Chemistry 11 Exam Review – Answers - Page 23 7.7 m g AuCl3 How much hydrochloric acid must be added initially to dissolve the all this gold? 5.0 mg Au 196.9665 mg mmol 3 mol HCl 1 mol Au 36.461 mg mmol 2.8 mg HCl 5. How many grams of sulfuric acid will react with 400.0 g of aluminum metal? 400.0 g 26.98154 6. Al 3 m ol g mol H2 SO4 2 m ol 98.079 Al g mol 2181 g H2 SO4 An unknown amount of potassium chlorate was heated until no more oxygen was evolved. potassium chloride remained in the test tube. 15.824 g of 2KClO3 2KCl + 3O2 What mass of potassium chlorate had originally been placed in the tube? 15.824 g 74.5513 KCl 2 mol g mol KClO3 2 mol 122.5495 KCl g mol 26.0 g What volume of oxygen gas was evolved in the process? 15.824 g KCl 74.5513 7. g mol 3 mol O2 2 mol KCl L mol 22.4 7.13 L O2 How many grams of carbon can be completely burned in 15.0 L of oxygen? 15.0 L 22.4 O2 L mol 1 mol C 1 mol O2 12.011 g mol 8.04 g C How many litres of carbon dioxide gas are produced in this reaction? 15.0 L 22.4 O2 L mol 1 mol CO2 1 mol O 2 22.4 L mol Chemistry 11 Exam Review – Answers - Page 24 15.0 L CO2 KClO 3 8. How many grams of magnesium metal are required to liberate 250.0 mL of hydrogen gas from hydrochloric acid? 0.2500 L 22.4 H2 1 mol Mg 1 mol H2 L mol 24.305 g mol 0.271 g Mg Exactly how much acid would be used up in this reaction? 0.2500 L 22.4 9. L mol H2 2 mol 1 mol HC lg H2 36.46094 g mol 0.8141 g HCl Will 30.0 L of fluorine gas completely react with 50.0 L of hydrogen gas? Which gas is in excess and by how much? What volume of hydrogen fluoride is formed in this reaction? 30.0 L 22.4 50.0 L 22.4 F2 L mol H2 L mol 2 m ol HF 1 m ol F2 2 m ol HF 1 m ol H2 22.4 22.4 L mol L mol 60.0 L HF 100. L HF 60.0 L of hydrogen fluoride is produced. Fluorine is the limiting reactant and hydrogen is in excess. 30.0 L 22.4 F2 L mol 1 m ol H2 1 m ol O2 22.4 L mol 30.0 L HF needed 50.0 L of hydrogen available - 30.0 L needed 20.0 L in excess Chemistry 11 Exam Review – Answers - Page 25 10. A mixture containing 100.0 g of H2 and 100.0 g of O2 is sparked so that water is formed. How much water is formed? 100.0 g H2 2 m ol g mol 2.01588 100.0 g O2 2 m ol 2 m ol g mol 31.9988 H2 O H2 H2 O 1 m ol o2 18.01528 g mol 894 g H2 O 18.01528 g mol 113 g H2 O Oxygen is the limiting reactant and hydrogenis in excess. 113 g of water are formed. 11. When copper is heated with sulfur, Cu2S is formed. How many grams of Cu2S could be produced if 100.0 g of copper is heated with 50.0 g of sulfur? 100.0 g 63.546 Cu g mol 50.0 g S 32.066 g mol 1 m ol Cu2 S 2 m ol 1 m ol Cu Cu2 S 1 m ol S 159.158 159.158 g mol g mol 125.2 g 248 g Cu2 S Cu2 S Copper is the limiting reactant and sulfur is in excess. 125.2 g of copper(I)sulfide is formed. 12. What volume (at STP) of carbon monoxide is required to produce 100.0 g of iron according to the equation: Fe2O3 + CO Fe + CO2 (not balanced) 100.0 g 55.847 Fe g mol 3 m ol CO 2 m ol F2 22.4 L mol Chemistry 11 Exam Review – Answers - Page 26 60.2 L CO 13. What is the molarity of a NaOH solution if 25.00 cm3 is required to completely neutralize 40.00 cm3 of a 1.50 M solution of H2SO4? H2SO4 + 2NaOH Na2SO4 + 2H2O Cb C aVaR b VbR a 1.50 mol L 0.04000 L 2 0.02500 L 1 4.80 mol L NaOH 14. Calculate the volume of a 0.600 M solution of HNO 3 necessary to neutralize 28.55 cm3 of a 0.450 M solution of KOH. HNO3 + KOH KNO3 + H2O Va Cb VbR a C aR b 0. 450 mol L 0.02855 L 1 0. 600 mol L 1 0.0214 L HNO 3 15. A titration of 15.00 cm3 of household ammonia, NH4OH(aq), required 38.57 cm3 of 0.780 M HCl. Calculate the molarity of the ammonia. HCl + NH4OH NH4Cl + H2O Cb C aVaR b VbR a 0.780 mol L 0.03857 L 1 0.01500 L 1 2.01 mol L NH 4OH 16. What volume of 0.250 M H3PO4 is required to neutralize 30.00 cm3 of a 0.0500 M Ba(OH)2 solution? 2H3PO4 + 3Ba(OH)2 Ba3(PO4)2 + 6H2O Va Cb VbR a C aR b 0. 0500 mol L 0.03000 L 2 0. 250 mol L 3 0.00400 L H3PO 4 17. What mass of Ca(OH)2 would be required to completely neutralize 50.0 cm3 of 0.125 M HCl? 2HCl + Ca(OH)2 CaCl2 + H2O n C V 0.125 mol L 0.0500 L 0.00625 mol HCl 0.00625 mol HCl 1 mol Ca(OH)2 74.09268 g mol 2 mol HCl Chemistry 11 Exam Review – Answers - Page 27 0.232 g Ca(OH)2 18. What mass of Mg(OH)2 would be required to completely neutralize 70.0 cm3 of 0.175 M HNO3? 2HNO3 + Mg(OH)2 Mg(NO3)2 + 2H2O n C V 0.175 mol L 0.0700 L 0.01225 mol HNO3 0.01225 mol HNO3 1 mol Mg(OH)2 58.31968 g mol 2 mol HNO3 0.357 g Mg(OH)2 *19. Hydrazine is a nitrogen-hydrogen compound having the formula N2H4. It is an oily, colourless liquid that freezes at 1.5°C and boils at 113.5°C. The principal use of hydrazine and certain compounds derived from it is a rocket fuels, but it is also used in fuel cells, in the treatment of water in boilers to removed dissolved oxygen gas, and in the plastics industry. One widely used method for the manufacture of hydrazine is the Raschig process. The Raschig process involves three reaction steps. In the first step, sodium hydroxide is reacted with chlorine to produce sodium hypochlorite, sodium chloride and water. The sodium hypochlorite produced in the first step is reacted with ammonia in the second step to produce chloramine (NH 2Cl) and sodium hydroxide. The chloramine and sodium hydroxide produced in the second step is reacted with ammonia in the third step to produce hydrazine, sodium chloride and water. A chemical plane using the Raschig process obtains 0.299 kg of 98.0% hydrazine for every 1.00 kg of chlorine. What are the theoretical, actual, and percent yields of pure hydrazine? 2NaOH + Cl2 NaOCl + NaCl + H2O NaOCl + NH3 NH2Cl + NaOH NH2Cl + NaOH + NH3 N2H4 + NaCl + H2O 1000 g Cl2 1 m ol 70.906 g mol 452 g N2 H 4 actual yield 1 m ol NaOCl 1 m ol NH 2 Cl 1 m ol N2 H 4 Cl2 1 m ol NaOCl 1 m ol NH 2 Cl theoretical yield 98.0% 299 g percentage yield 293 g actual yield theoretical yield actual yield 100 64.8% yield Chemistry 11 Exam Review – Answers - Page 28 32.04516 g mol *20. The characteristic odour of pineapple is due to ethyl butyrate, a compound containing carbon, hydrogen, and oxygen. Combustion of 2.78 mg of ethyl butyrate produces 6.32 mg of carbon dioxide and 2.58 mg of water. What is the empirical formula of the compound? __ CxHyOz + __ O2 2.78 mg mass of carbon __ CO2 + 6.32 mg 2.58 mg 12.011 g 0.00632g 44.0098g mass of hydrogen 2 1.00794 g 18.01528g __ H2O 0.001725g of carbon 0.00258g total mass mass of carbon mass of hydrogen 0.00278g 0.001725g 0.0002887g 0.0002887g of hydrogen mass of oxygen 0.0007665g of oxygen EmpiricalFormula 0.001725g of carbon 12.011 g mol 0.0001436mol C 0.00004791 2.998 : 5.979 : 1 0.0002887g of hydrogen 1.00794 g mol 0.0002864mol H 0.00004791 3: 6:1 0.0007665g of oxygen 15.9994g mol 0.00004791mol Ol 0.00004791 EmpiricalFormula C3H6O Chemistry 11 Exam Review – Answers - Page 29 *21. Nicotine, a component of tobacco, is composed of carbon, hydrogen, and nitrogen. A 5.250 mg sample of nicotine was combusted, producing 14.242 mg of carbon dioxide and 4.083 mg of water. What is the empirical formula for nicotine? mass of carbon 12.011 g 0.014242g 44.0098g mass of hydrogen 2 1.00794 g 18.01528 g 0.003887g of carbon 0.004083g total mass mass of carbon mass of hydrogen 0.005250g 0.003887g 0.00045688g 0.00045688g of hydrogen mass of nitrogen 0.0009062g of nitrogen EmpiricalFormula 0.003887g of carbon 12.011 g mol 0.0003236mol C 0.00006471 5.002 : 7.006 : 1 0.00045688g of hydrogen 1.00794 g mol 0.0004533mol H 0.00006471 5 : 7 :1 0.0009062g of nitrogen 14.0067 g mol 0.00006471mol N 0.00006471 EmpiricalFormula C5H7 N Chemistry 11 Exam Review – Answers - Page 30