CHEM 103 Exam 1-6 Final - Complete Solutions, Graded A+ ( 2021/2022 latest update )

CHEM 103 Exam 1-6 Final - Complete Solutions, Graded A Exam 1-6 Question 1 1. Convert 1005.3 to exponential form and explain your answer. 2. Convert 4.87 x 10-6 to ordinary form and explain your answer. Answer: 1. Convert 1005.3 = larger than 1 = positive exponent, move decimal 3 places = 1.0053 x 103 2. Convert 4.87 x 10-6 =negative exponent =smaller than 1, move decimal 6 places = 0. Question 2 Using the following information, do the conversions shown below, showing all work: 1 ft = 12 inches 1 pound = 16 oz 1 gallon = 4 quarts 1 mile = 5280 feet 1 ton = 2000 pounds 1 quart = 2 pints kilo (= 1000) milli (= 1/1000) centi (= 1/100) deci (= 1/10) 1. 2.73 liters = ? ml 2. 8.6 pts = ? qts Answer: 1. 2.73 liters x 1000 ml / 1 liter = 2730 ml 2. 8.6 pts x 1 qt / 2 pts = 4.3 qts Question 3 Do the conversions shown below, showing all work: 1. 248oC = ? oK 2. 25oF = ? oC 3. 175oK = ? oF Answer: 1. 248oC + 273 = 521 oK oC → oK (make larger) +273 2. 25oF - 32 ÷ 1.8 = -3.88 oC oF → oC (make smaller) -32 ÷1.8 3. 175oK - 273 = -98 oC x 1.8 + 32 = -144.4 oF oK → oC → oF Question 4 Be sure to show the correct number of significant figures in each calculation. 1. Show the calculation of the mass of a 17.9 ml sample of chloroform with density of 1.49 g/ml 2. Show the calculation of the volume of 19.4 grams of cresol with density of 1.02 g/ml Answer: 1. M = D x V = 1.49 x 17.9 = 26.7 g 2. V = M / D = 19.4 / 1.02 = 19.0 ml Question 5 1. 1.35601 contains ? significant figures. 2. 0.151 contains ? significant figures. 3. 1.35601 + 0.151 = ? (give answer to correct number of significant figures) Answer: 1. 1.35601 contains 6 significant figures. 2. 0.151 contains 3 significant figures. 3. 1.35601 + 0.151 = 1.50701 = 1.507 (to the thousands place for 0.151) Question 6 Classify each of the following as an element, compound, solution or heterogeneous mixture and explain your answer. 1. Honey 2. Hydrogen iodide 3. Tea Answer: 1. Honey - is not on periodic table (not element) - no element names (not compound) appears to be one substance = Solution 2. Hydrogen iodide - is not on periodic table (not element) - has two element names = Compound 3. Tea - is not on periodic table (not element) - no element names (not compound) appears to be one substance = Solution Question 7 Classify each of the following as a chemical change or a physical change 1. A silver spoon forms a black tarnish coating 2. Food is digested 3. Rain freezes on a road on a very cold day Answer: 1. Silver spoon forms black coating - this is Ag → Ag2S (color change) = chemical change 2. Food is digested - breakdown of carbs, proteins, fats to new materials = chemical change 3. Rain freezes on a road on a very cold day - freezing = physical change Question 8 Show the full Nuclear symbol including any + or - charge (n), the atomic number (y), the mass number (x) and the correct element symbol (Z) for each element for which the protons, neutrons and electrons are shown - symbol should appear as follows: xZy+/- n 53 protons, 74 neutrons, 54 electrons Answer:53 protons = I53, 74 neutrons = 127I53, 54 electrons = (+53 - 54 = -1) = 127I53- 1 Question 9 Name each of the following chemical compounds. Be sure to name all acids as acids (NOT for instance as binary compounds) 1. SF6 2. (NH4)3PO4 3. H2S Answer: 1. SF6 - binary molecular = sulfur hexafluoride 2. (NH4)3PO4 - nonbinary ionic = ammonium phosphate 3. H2S - binary acid = hydrosulfuric acid Question 10 Write the formula for each of the following chemical compounds explaining the answer with appropriate charges and/or prefixes and/or suffixes. 1. Iron (III) cyanide 2. Diiodine pentoxide 3. Potassium phosphide 1. Iron (III) cyanide - Fe+3, CN-1 = Fe(CN)3 2. Diiodine pentoxide - ide = binary, two I, 5 O = I2O5 3. Potassium phosphide - ide = binary K+1, P-3 = K3P Question 1 Show the calculation of the molecular weight for the following compounds, reporting your answer to 2 places after the decimal. 1. (NH4)2CrO4 2. C8H8NOI Answer: 1. 2N + 8H + Cr + 4O = 152.08 2. 8C + 8H + N + O + I = 261.05 Question 2 Show the calculation of the number of moles in the given amount of the following substances. Report your answerto 3 significant figures. 1. 12.0 grams of (NH4)2CrO4 2. 15.0 grams of C8H8NOI Answer: 1. Moles = grams / molecular weight = 12.0 / 152.08 = 0.0789 mole 2. Moles = grams / molecular weight = 15.0 / 261.05 = 0.0575 mole Question 3 Show the calculation of the number of grams in the given amount of the following substances. Report your answer to 1 place after the decimal. 1. 1.05 moles of (NH4)2CrO4 2. 1.18 moles of C8H8NOI Answer: 1. Grams = Moles x molecular weight = 1.05 x 152.08 = 159.7 grams 2. Grams = Moles x molecular weight = 1.18 x 261.05 = 308.0 grams Question 4 Show the calculation of the percent of each element present in the following compounds. Report your answer to 2 places after the decimal. 1. (NH4)2SO4 2. C9H9NO3 Answer: 1. %N = 2 x 14.01/132.15 x 100 = 21.20% %H = 8 x 1.008/132.15 x 100 = 6.10% %S = 1 x 32.07/132.15 x 100 = 24.27% %O = 4 x 16/132.15 x 100 = 48.43% 2. %C = 9 x 12.01/179.17 x 100 = 60.33% %H = 9 x 1.008/179.17 x 100 = 5.06% %N = 1 x 14.01/179.17 x 100 = 7.82% %O = 3 x 16/179.17 x 100 = 26.79% Question 5 Show the calculation of the empirical formula for each compound whose elemental composition is shown below. 38.76% Ca, 19.87% P, 41.27% O Answer: 38.76% Ca / 40.08 = 0.9671 / 0.6416 = 1.5 x 2 = 3 19.87% P / 30.97 = 0.6416 / 0.6416 = 1 x 2 = 2 41.27% O / 16.00 = 2.579 / 0.6416 = 4 x 2 = 8 → Ca3P2O8 Question 6 Balance each of the following equations by placing coefficients in front of each substance. 1. C4H10 + O2 → CO2 + H2O 2. P + O2 → P2O5 3. Al + H2SO4 → Al2(SO4)3 + H2 Answer: 1. 2 C4H10 + 13 O2 → 8 CO2 + 10 H2O 2. 4 P + 5 O2 → 2 P2O5 3. 2 Al + 3 H2SO4 → Al2(SO4)3 + 3 H2 Question 7 Classify each of the following reactions as either: Combination Decomposition Combustion Double Replacement Single Replacement 1. C5H12 + 8 O2 → 5 CO2 + 6 H2O 2. Zn + CuSO4 → Cu + ZnSO4 3. 2 Fe + 3 Cl2 → 2 FeCl3 Answer: 1. C5H12 + 8 O2 → 5 CO2 + 6 H2O = Combustion, Hydrocarbon + O2 → CO2 + H2O 2. Zn + CuSO4 → Cu + ZnSO4 = Single Replacement, Metal displaces metal ion 3. 2 Fe + 3 Cl2 → 2 FeCl3 = Combination. Two reactants→ One product Question 8 Show the calculation of the oxidation number (charge) of ONLY the atoms which are changing in the following redox equations. Na2HAsO3 + KBrO3 + HCl → NaCl + KBr + H3AsO4 Answer: Na2HAsO3: Na is metal in group I = +1 (total is +2), H = +1, each O is -2 (total is -6), so As is +3 H3AsO4: H is +1 (total is +3), each O is -2 (total is -8), so As is +5 KBrO3: K is metal in group I = +1, each O is -2 (total is -6), so Br is +5 KBr: K is metal in group I = +1, so Br is -1 Question 9 Show the balancing of the following redox equation, including the determination of the oxidation number (charge) of ONLY the atoms which are changing. Na2HAsO3 + KBrO3 + HCl → NaCl + KBr + H3AsO4 Answer: Na2HAsO3: Na is metal in group I = +1 (total is +2), H = +1, each O is -2 (total is -6), so As is +3 H3AsO4: H is +1 (total is +3), each O is -2 (total is -8), so As is +5 KBrO3: K is metal in group I = +1, each O is -2 (total is -6), so Br is +5 KBr: K is metal in group I = +1, so Br is -1 Since As (on left side) is +3 and As (on right side) is +5: As changes by 2 Since Br (on left side) is +5 and Br (on right side) is -1: Br changes by 6 Multiply As compounds by 3 and Br compounds by 1 and after balancing other atoms = 3 Na2HAsO3 + 1 KBrO3 + 6 HCl → 6 NaCl + 1 KBr + 3 H3AsO4 Question 10 Show the balanced equation and the calculation of the number of moles and grams of CO2 formed from 25.4 grams of C8H18. Show your answers to 3 significant figures. C8H18 + O2 → CO2 + H2O Answer: 2 C8H18 + 25 O2 → 16 CO2 + 18 H2O 25.4 g / (8 x 12.01 + 18 x 1.008) = 25.4 / 114.224 = 0.2223 mole x 16/2 = 1.78 mole CO2 1.78 mole CO2 x (12.01 + 2 x 16.00) = 78.3 g CO2 Exam 3 Question 1 A reaction between HCl and NaOH is being studied in a styrofoam coffee cup with NO lid and the heat given off is measured by means of a thermometer immersed in the reaction mixture. Enter the correct thermochemistry term to describe the item listed. 1. The type of thermochemical process 2. The amount of heat released in the reaction of HCl with NaOH Answer: 1. Heat given off = Exothermic process 2. The amount of heat released = Heat of reaction Question 2 1. Show the calculation of the final temperature for a 27.4 gram piece of aluminum heated to 100oC which has been added to a 32.5 gram sample of water at 25.6oC in a coffee cup calorimeter. c (water) = 4.184 J/g oC; c (Al) = 0.901 J/g oC 2. Show the calculation of the energy involved in condensation of 95.6 grams of steam at 100oC if the Heat of Vaporization for water is 2.26 kJ/g Answer: 1. - (mAl x cAl x ∆tAl) = (mH2O x cH2O x ∆tH2O) - [27.4 g x 0.901 J/g oC x (Tmix - 100oC)] = [32.5 g x 4.184 J/g oC x (Tmix - 25.6oC)] - [24.6874 J/oC x (Tmix - 100oC)] = [135.98 J/oC x (Tmix - 25.6oC)] - 24.6874 J/oC (Tmix) + 2468.74 J = 135.98 J/oC (Tmix) – 3481.088 J 5949.828 J = 160.6674 J/oC (Tmix) Tmix = 37.0oC 2. ql↔g = m x ∆Hvapor = 95.6 g x 2.26 kJ/g = 216.1 kJ (since heat is removed) = - 216.1 kJ Question 3 Show the calculation of the amount of heat involved if 18.3 g of S is reacted with excess O2 to yield sulfur trioxide by the following reaction equation. Report your answer to 4 significant figures. 2 S (s) + 3 O2 (g) → 2 SO3 (g) ΔH = - 792 kJ Answer: 2 S (s) + 3 O2 (g) → 2 SO3 (g) ΔH = - 792 kJ ΔHrx is for 2 mole of S reaction uses 18.3 g S = 18.3/32.07 = 0.5706 mole S q = ΔHrx x new moles / original moles q = -792 kJ x 0.5706 mole S / 2 mole S = 226.0 kJ given off Question 4 Show the calculation of the heat of reaction (ΔHrxn) for the reaction: 3 C (graphite) + 4 H2 (g) → C3H8 (g) by using the following thermochemical data: C (graphite) + O2 (g) → CO2 (g) ΔH = - 393.51 kJ 2 H2 (s) + O2 (g) → 2 H2O(l) ΔH = - 571.66 kJ C3H8 (g) + 5 O2 (g) → 3 CO2 (g) + 4 H2O(l) ΔH = - 2220.0 kJ Answer: 3 (C (graphite) + O2 (g) → CO2 (g) ΔH = - 393.51 kJ)2 (2 H2 (s) + O2 (g) → 2 H2O(l) ΔH = - 571.66 kJ) 3 CO2 (g) + 4 H2O(l) → C3H8 (g) + 5 O2 (g) ΔH = + 2220.0 kJ3 C (graphite) + 4 H2 (g) → C3H8 (g) ΔHrxn = - 103.85 kJ ΔHrxn = 3(- 393.51) + 2(- 571.66) + 2220.0 = - 103.85kJ Question 5 Show the calculation of the heat of reaction (ΔHrxn) for the reaction: 2 C6H6 (g) + 9 O2 (g) → 12 CO (g) + 6 H2O (l) by using the following thermochemical data: ΔHf0 C6H6 (g) = +49.1 kJ/mole, ΔHf0 CO (g) = -110.5 kJ/mole, ΔHf0 H2O (l) = -285.8 kJ/mole Answer: 2 C6H6 (g) + 9 O2 (g) → 12 CO (g) + 6 H2O (l) ΔHf0 C6H6 (g) = +49.1 kJ/mole, ΔHf0 CO (g) = -110.5 kJ/mole, ΔHf0 H2O (l) = -285.8 kJ/mole ΔHrxn = 2(-49.1) + 9(0) + 12(-110.5) + 6(-285.8) = - 3139.0 kJ/mole Question 6 Show the calculation of the new pressure of a gas sample which has an original volume of 560 ml when collected at 1.05 atm and 32oC when the volume becomes 1.35 liters at 50oC. Your Answer: (Pi x Vi ) / Ti = (Pf x Vf ) / Tf 560 ml/1000 = 0.560 liters = Vi 1.05 atm = Pi 1.35 liters = Pf 32oC + 273 = 305oK = Ti 50oC + 273 = 323oK = Tf (1.05) x (0.560) / 305 = Pf x (1.35) / 323 Pf = 0.461 atm Question 7 Show the calculation of the volume occupied by a gas sample containing 0.632 mole collected at 710 mm and 35oC. Answer:P x V = n x R x T 0.632 mole = n R = 0.0821 710 mm/760 = 0.934 atm = P 35oC + 273 = 308oK = T (0.934) x V = (0.632) x (0.0821) x (308) V = 17.1 liters Question 8 Show the calculation of the volume of H2O gas produced by the combustion of 33.5 grams of ethanol by excess O2 gas at 35oC and 1.05 atm. The combustion of ethanol (C2H5OH) takes place by the following reaction equation. (g) C2H5OH (l) + 3 O2 (g) → 2 CO2 (g) + 3 H2O Your Answer: MW = 46) C2H5OH (l) + (MW = 32) 3 O2 (g) (MW = 44) → 2 CO2 (g) + (MW = 18) 3 H2O (g) 33.5 grams 52.62 liters ↓ by V = nRT / P = (2.1849 mole) (0.0821)(308)/1.05 ↑ 0.7283 mol → 3/1 x 0.7283 mol Question 9 Show the calculation of the mole fraction of each gas in a 1.00 liter container holding a mixture of 3.62 g of He and 5.45 g of Ne at 25oC. Answer: nHe = gHe / (MWHe) = 3.62 g / 4.002 = 0.9045 mol nNe = gNe / (MWNe) = 5.45 g / 20.18 = 0.2701 mol XNe = 0.9045 / (0.9045 + 0.2701) = 0.7700 XNe = 0.2701 / (0.9045 + 0.2701) = 0.2300 Question 10 Show the calculation of the molecular weight of an unknown gas if the rate of effusion of Neon gas (Ne) is 1.86 times faster than that of an unknown gas. Your Answer: (rN2 /runknown)2 = MWunknown / MWN2 (1.86/1)2 = MWunknown / 20.18 MWunknown = (1.86)2 x 20.18 = 69.81 Exam 4 Question 1 Write the subshell electron configuration (i.e.1s2 2s2, etc.) for the Ni28 atom. Answer: Ni28 = 28 electrons = 1s2 2s2 2p6 3s2 3p6 4s2 3d8 Question 2 Write the subshell electron configuration (i.e.1s2 2s2, etc.) for the K19 atom. Answer: K19 = 19 electrons = 1s2 2s2 2p6 3s2 3p6 4s1 Question 3 Write the subshell electron configuration (i.e.1s2 2s2, etc.) for the K19 atom and identify which are valence (outer shell) electrons and determine how many valence electrons there are. Answer: K19 = 19 electrons = 1s2 2s2 2p6 3s2 3p6 4s1 = 1 valence electron Question 4 Using up and down arrows, write the orbital diagram for the Ni28 atom. Your Answer: Ni28 = 1s2 2s2 2p6 3s2 3p6 4s2 3d8 ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑ ↑ Question 5 Using up and down arrows, write the orbital diagram for the Ni28 atom and identify which are unpaired electrons and determine how many unpaired electrons there the Periodic Table.This may be helpful throughout the exam. Answer: Ni28 = 1s2 2s2 2p6 3s2 3p6 4s2 3d8 ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑ ↑ = 2 unpaired electrons Question 6 Write the subshell electron configuration (i.e.1s2 2s2, etc.) for the V23 atom and then identify the last electron to fill and write the 4 quantum numbers (n, l, ml and ms) for this electron. Answer: V23: 23 electrons: 1s2 2s2 2p6 3s2 3p6 4s2 3d3 3d3 is the last electron to fill n=3, l=2, ml=0, ms=+½ Question 6 Write the subshell electron configuration (i.e.1s2 2s2, etc.) for the V23 atom and then identify the last electron to fill and write the 4 quantum numbers (n, l, ml and ms) for this electron. Answer: V23 = 1s2 2s2 2p6 3s2 3p6 4s2 3d3 : n=3, l=2, ml = 0, ms = +½ Question 7 1. Arrange the following elements in a vertical list from largest (top) to smallest (bottom) atomic size: Cl, Br, I 2. Arrange the following elements in a vertical list from lowest (top) to highest (bottom) electronegativity: S, P, Cl 3. Arrange the following elements in a vertical list from highest (top) to lowest (bottom) ionization energy: S, O, Se Answer: 1. I Br Cl 2. P S Cl 3. O S Se Question 8 1. List and explain which of the following atoms holds its valence electrons less tightly. Si or Cl 2. List and explain which of the following atoms forms a positive ion with more difficulty. B or F Answer: 1. Si holds its valence electrons less tightly than Cl since electronegativity increases as you go to the right in a period which means that Si which is further to the left in the period has the lower electronegativity and therefore the lower attraction for its valence electrons. 2. F forms a positive ion less easily than B since ionization potential increases as you go to the right in a period which means that F with the higher ionization potential requires more energy to lose an electron and form a positive ion so it does so less easily. Question 9 On a piece of scratch paper, draw the orbital configuration of the C6 atom and use it to draw the Lewis structure for the C6 atom. Then choose the correct Lewis structure for C6 from the options listed below. Answer: C Question 10 On a piece of scratch paper, draw the orbital configuration of the P15 atom and use it to draw the Lewis structure for the P15 atom. Then choose the correct Lewis structure for P15 from the options listed below. Answer: B Exam 5 Question 1 Show the determination of the charge on the ion formed by the Ga31 atom. Answer: Ga31 (metal = lose electrons) 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p1 lose 3e → Ga+3 Question 2 H = 2.1 Li = 1.0 Be = 1.5 B = 2.0 C = 2.5 N = 3.0 O= 3.5 F = 4.0 Na = 1.0 Mg = 1.2 Al = 1.5 Si = 1.8 P = 2.1 S = 2.5 Cl = 3.0 K = 0.8 Ca = 1.0 Ga = 1.6 Ge = 1.8 As = 2.0 Se = 2.4 Br = 2.8 Using the electronegativities from the table above, show the determination of the polarity of each different type of bond in the following molecule Answer: H-O bond electronegativity difference = 3.5 - 2.1 = 1.4 1.6 - 0.5 bond is Polar Br-O bond electronegativity difference = 3.5 - 2.8 = 0.7 1.6 - 0.5 bond is Polar Question 3 On a piece of scratch paper, draw the Lewis structure for the ClO3-1 ion. Then choose the correct Lewis structure for ClO3-1 from the options listed below. Answer: B Question 4 On a piece of scratch paper, draw the Lewis structure for HClO3. Then choose the correct Lewis structure for HClO3 from the options listed below. Your Answer: B Question 5 Determine the electron geometry and explain your answer for the S atom in H2S. Answer: The S atom in H2S has 4 groups of electrons around it in its Lewis structure, therefore, its electron geometry would be tetrahedral. Question 6 Determine the hybridization and explain your answer for the S atom in H2S. Your Answer:The S atom in H2S has 4 groups of electrons around it in its Lewis structure, therefore, its hybridization would be sp3. Question 7 Determine the shape and explain your answer for HCCH. Answer:The C atoms in HCCH has 2 groups of electrons around it in its Lewis structure, therefore, its electron geometry would be linear and since there are 2 atoms around each central C atom, the shape would be linear. Question 8 H = 2.1 Li = 1.0 Be = 1.5 B = 2.0 C = 2.5 N = 3.0 O= 3.5 F = 4.0 Na = 1.0 Mg = 1.2 Al = 1.5 Si = 1.8 P = 2.1 S = 2.5 Cl = 3.0 K = 0.8 Ca = 1.0 Ga = 1.6 Ge = 1.8 As = 2.0 Se = 2.4 Br = 2.8 Use the electronegativities above and your knowledge of the shape of PH3 to determine the molecular polarity of PH3 explaining your answer in detail. Answer: The shape of PH3 is triangular pyramid and since the P-H bonds are all nonpolar, PH3 would be nonpolar since all the bonds are nonpolar. Question 9 Is O2 Polar, Ionic or Nonpolar and List and Explain whether it is Soluble or Insoluble in Water? Answer:O2 has one nonpolar bond which makes it Nonpolar and since it is Nonpolar it is Insoluble in water. Question 10 Arrange the following compounds in a vertical list from highest boiling point (top) to lowest boiling point (bottom) and explain your answer on the basis of whether the substance is Polar, Nonpolar, Ionic, Metallic or Hydrogen bonding: Mg, H2O, Ne, HCl, LiCl LiCl (ionic) = Mg (metallic) H2O (Hydrogen Bonding) HCl (Polar) Ne (Nonpolar) Exam 6 Question 1 Explain the difference between amorphous and crystalline solids and give an example of each. Answer:Amorphous solids (like glass, plastic or rubber) have their particles arranged in a random fashion and crystalline solids (like salt, sugar, metals, quartz) have their particles arranged in orderly, repeating, geometric patterns. Question 2 Which is the least common state of matter among elements solids, liquids or gases? Explain your answer. Answer:Liquids are the least common state of matter among elements because liquids exist over a very narrow temperature range. Question 3 Rank and explain how the freezing point of 0.100 m solutions of the following ionic electrolytes compare, List from lowest freezing point to highest freezing point. GaCl3, Al2(SO4)3, NaI, MgCl2 3rd lowest FP 0.1 x 5 = lowest FP 0.1 x 2 = highest FP = 2nd lowest FP FP: Al2(SO4)3 < GaCl3 < MgCl2 < NaI Question 4 Answer: Mass % = (gsolute / gsolute + gsolvent) x 100% Mass % = (18.9 / 18.9 + 400) x 100 = 4.51% Question 5 Show the calculation of the molality of a solution made by dissolving 28.5 grams Answer: molality = (gsolute / MW) / (gsolvent / 1000) molality = (28.5 / 240.208) / (400 / 1000) = 0.297 m Question 6 Show the calculation of the molarity of a solution made by dissolving 27.3 grams of Ca(NO3)2 to make 450 ml of solution. Report your answer to 3 significant figures. Answer: Molarity = (gsolute / MW) / (mlsolvent / 1000) Molarity = (27.3 / 164.10) / (450 / 1000) = 0.370 M Question 7 Show the calculation of the mass of Ba(MnO4)2 needed to make 250 ml of a 0.200 M solution. Report your answer to 3 significant figures. Answer: Molarity = (moles) / (mlsolvent / 1000) 0.200 = (moles) / (250 / 1000) Moles = 0.200 x 0.250 = 0.0500 Moles = (gsolute / MW) 0.0500 = (gsolute / 375.41) gsolute = 0.0500 x 375.41 = 18.8 g Question 8 Show the calculation of the volume of 0.667 M solution which can be prepared using 37.5 grams of Ba(NO3)2. Answer: molessolute = gsolute / MW molessolute = 37.5 g / 261.55 = 0.1434 mol Molarity = moles / (mL /1000) 0.667 = 0.1434 / (mL / 1000) mL / 1000 = 0.1434 / 0.667 = 0.2150 mL = 0.2150 x 1000 = 215 mL Question 9 Show the calculation of the boiling point of a solution made by dissolving 20.9 grams of the nonelectrolyte C4H8O4 in 250 grams of water. Kb for water is 0.51, BP of pure water is 100oC. Calculate your answer to 0.01oC. ∆tb = Kb x m Your Answer: molality = (gsolute / MW) / (gsolvent / 1000) molality = (20.9 / 120.104) / (250 / 1000) = 0.6961 m ∆tb = Kb x m = 0.51 x 0.6961 = 0.355oC BPsolution = BPsolvent - ∆tb = 100oC + 0.355 = 100.35oC Question 10 Show the calculation of the molar mass (molecular weight) of a solute if a solution of 13.5 grams of the solute in 200 grams of water has a freezing point of -1.20oC. Kf for water is 1.86 and the freezing point of pure water is 0oC. Calculate your answer to 0.1 g/mole. Answer: ∆tf = Kf x m molality = ∆tf / Kf = 1.20 / 1.86 = 0.645 m molality = (gsolute / MW) / (gsolvent / 1000) 0.645 = (moles) / (200 / 1000) Moles = 0.645 x 0.200 = 0.129 0.129 = (13.5 / MW) MW = 13.5 / 0.129 = 104.7g/mole Final: Question 1 1. Convert 0. to exponential form and explain your answer. 2. Convert 5.82 x 103 to ordinary form and explain your answer. Answer: 1. Convert 0. = smaller than 1 = negative exponent, move decimal 5 places = 7.26 x 10-5 2. Convert 5.82 x 103 = positive exponent = larger than 1, move decimal 3 places = 5820 Question 2 Do the conversions shown below, showing all work: 1. 28oC = ? oK 2. 158oF = ? oC 3. 343oK = ? oF Answer: 1. 28oC + 273 = 301 oK oC → oK (make larger) +273 2. 158oF - 32 ÷ 1.8 = 70 oC oF → oC (make smaller) -32 ÷1.8 3. 343oK - 273 = 70 oC x 1.8 + 32 = 158 oF oK → oC → oF Question 3 Show the calculation of the number of moles in the given amount of the following substances. Report your answerto 3 significant figures. 1. 12.0 grams of Ca3(PO4)2 2. 15.0 grams of C9H8NO4Cl Answer: 1. Moles = grams / molecular weight = 12.0 / 310.18 = 0.0387 mole 2. Moles = grams / molecular weight = 15.0 / 229.61 = 0.0653 mole Question 4 Not yet graded / 10 pts Click this link to access the Periodic Table. This may be helpful throughout the exam. Show the calculation of the percent of each element present in the following compounds. Report your answer to 2 places after the decimal. 1. Al2(CO3)3 2. C8H6NO4Cl Answer: 1. %Al = 2 x 26.98/233.99 x 100 = 23.06% %C = 3 x 12.01/233.99 x 100 = 15.40% %O = 9 x 16.00/233.99 x 100 = 61.54% 2. %C = 8 x 12.01/215.59 x 100 = 44.57% %H = 6 x 1.008/215.59 x 100 = 2.81% %N = 1 x 14.01/215.59 x 100 = 6.50% %O = 4 x 16.00/215.59 x 100 = 29.69% %Cl = 1 x 35.45/215.59 x 100 = 16.44% Question 5 Show the calculation of the heat of reaction (ΔHrxn) for the reaction: CH4 (g) + 2 O2 (g) → CO2 (g) + 2 H2O (l) by using the following thermochemical data: ΔHf0 CH4 (g) = -74.6 kJ/mole, ΔHf0 CO2 (g) = -393.5 kJ/mole, ΔHf0 H2O (l) = -285.8 kJ/mole Answer: CH4 (g) + 2 O2 (g) → CO2 (g) + 2 H2O (l) ΔHf0 CH4 (g) = -74.6 kJ/mole, ΔHf0 CO2 (g) = -393.5 kJ/mole, ΔHf0 H2O (l) = -285.8 kJ/mole ΔHrxn = (+74.6) + 2(0) + (-393.5) + 2(-285.8) = - 890.5 kJ/mole Question 6 Show the calculation of the molecular weight of a gas sample with a mass of 0.456 grams which has a volume of 230 ml when collected at 29oC and 740 mm. Answer: P x V = (g/MW) x R x T 740 mm/760 = 0.974 atm = P R = 0.0821 V = 230 ml/1000 = 0.230 liters 29oC + 273 = 302oK = T 0.456 = g (0.974) x (0.230) = (0.456/MW) x (0.0821) x (302) MW = 50.5 Question 7 Write the subshell electron configuration (i.e.1s2 2s2, etc.) for the V23 atom and then identify the last electron to fill and write the 4 quantum numbers (n, l, ml and ms) for this electron. Answer: V23 = 1s2 2s2 2p6 3s2 3p6 4s2 3d3 : n=3, l=2, ml = 0, ms = +½ Question 8 1. List and explain which of the following atoms holds its valence electrons less tightly. Si or Cl 2. List and explain which of the following atoms forms a positive ion with more difficulty. B or F Answer: 1. Si holds its valence electrons less tightly than Cl since electronegativity increases as you go to the right in a period which means that Si which is further to the left in the period has the lower electronegativity and therefore the lower attraction for its valence electrons. 2. F forms a positive ion less easily than B since ionization potential increases as you go to the right in a period which means that F with the higher ionization potential requires more energy to lose an electron and form a positive ion so it does so less easily. Question 9 Is KNO3 Polar, Ionic or Nonpolar and List and Explain whether it is Soluble or Insoluble in Water? KNO3 is Ionic since it has Ionic bonds and since it is Ionic it is Soluble in water. Question 10 Show the determination of the charge on the ion formed by the Ga31 atom. Answer: Ga31 (metal = lose electrons) 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p1 lose 3e → Ga+3 Question 11 Water (H2O) with a molecular weight of 18 has a boiling point of 100oC and methane (CH4) with a somewhat similar molecular weight of 16 has a boiling point of -161oC. This difference is much higher than can be explained by the polarity of water. Explain what causes this difference. Answer: Hydrogen bonds which are strong intermolecular attractive forces form between water molecules requiring much more energy to separate the molecules than for other molecules. Question 12 Show the calculation of the molar mass (molecular weight) of a solute if a solution of 12.5 grams of the solute in 200 grams of water has a freezing point of -1.30oC. Kf for water is 1.86 and the freezing point of pure water is 0oC. Calculate your answer to 0.1 g/mole. ∆tf = Kf x m molality = ∆tf / Kf = 1.30 / 1.86 = 0.699 m molality = (gsolute / MW) / (gsolvent / 1000) 0.699 = (moles) / (200 / 1000) Moles = 0.699 x 0.200 = 0.1398 0.1398 = (12.5 / MW) MW = 12.5 / 0.1398 = 89.4