Chemistry Exam Solutions for CHEM 103: Molecular Weight, Empirical Formula, Redox, Exams of Chemistry

Download Chemistry Exam Solutions for CHEM 103: Molecular Weight, Empirical Formula, Redox and more Exams Chemistry in PDF only on Docsity! CHEM 103 EXAMS 1 TO 6 AND FINAL EXAM WITH ACCURATE ANSWERS BEST RATED A+ DOWNLOAD TO PASS 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 = 0.00000487 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) mille (= 1/1000) cent (= 1/100) Deci (= 1/10) 1. 2.73 liters =? Ml 2. 8.6 pts =? Qtrs. Answer: 1. 2.73 liters x 1000 ml / 1 liter = 2730 ml 2. 8.6 pts x 1 qtr. / 2 pts = 4.3 qtrs. Question 3 Do the conversions shown below, showing all work: 1. 248oC =? Oak 2. 25oF =? Co. 3. 175oK =? Of Answer: 1. 248oC + 273 = 521 oak co. → oak (make larger) +273 2. 25oF - 32 ÷ 1.8 = -3.88 co. of → co. (make smaller) - 32 ÷1.8 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: zee+/- 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 - no binary ionic = ammonium phosphate 3. H2S - binary acid = hydro sulfuric 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 EXAM 2 2. Iodine pent oxide 3. Potassium phosphide 1. Iron (III) cyanide - Fe+3, CN-1 = Fe(CN)3 2. Iodine pent oxide - 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 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 Decompositi on 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 + Hall → Nalco + Kerr + 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 Kerr: K is metal in group I = +1, so Br is -1 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 co.; c (Al) = 0.901 J/g co. 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. - (mall x call x ∆tall) = (mH2O x cH2O x ∆tH2O) - [27.4 g x 0.901 J/g co. x (Mix - 100oC)] = [32.5 g x 4.184 J/g co. x (Mix - 25.6oC)] - [24.6874 J/co. x (Mix - 100oC)] = [135.98 J/co. x (Mix - 25.6oC)] - 24.6874 J/co. (Mix) + 2468.74 J = 135.98 J/co. (Mix) – 3481.088 J 5949.828 J = 160.6674 J/co. (Mix) Mix = 37.0oC 2. Leg = m x ∆Vapor = 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 Hr is for 2 mole of S Reaction uses 18.3 g S = 18.3/32.07 = 0.5706 mole S q = Hr 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: 0.7283 moll → 3/1 x 0.7283 moll 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: He = ghee / (Maher) = 3.62 g / 4.002 = 0.9045 moll nine = gene / (Manes) = 5.45 g / 20.18 = 0.2701 moll NE = 0.9045 / (0.9045 + 0.2701) = 0.7700 NE = 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 /run known) 2 = Unknown / MWN2 (1.86/1)2 = Unknown / 20.18 Unknown = (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 recess the 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 election to fill and write the 4 quantum numbers (n, l, ml and mos) for this electron. Answer: V23: 23 electrons: 1s2 2s2 2p6 3s2 3p6 4s2 3d3 is the last election to fill n=3, l=2, ml=0, mos=+½ Question 6 Write the subshell electron configuration (i.e.1s2 2s2, etc.) for the V23 atom and then identify the last election to fill and write the 4 quantum numbers (n, l, ml and mos) for this electron. 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 Cal = 3.0 K = 0.8 CA = 1.0 GA = 1.6 Gee = 1.8 as = 2.0 Se = 2.4 Br = 2.8 Using the electronegativity’s 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 F = 4.0 Na = 1.0 Mg = 1.2 Al = 1.5 Si = 1.8 P = 2.1 S = 2.5 Cal = 3.0 K = 0.8 CA = 1.0 GA = 1.6 Gee = 1.8 as = 2.0 Se = 2.4 Br = 2.8 Use the electronegativity’s 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, Hall, and Local Local (ionic) = Mg (metallic) H2O (Hydrogen Bonding) Hall (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. Show the calculation of the mass percent solute in a solution of 18.9 grams of Ba(MnO4)2 in 400 grams of water. Report your answer to 3 significant figures. 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, Nail, 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 < Nail Question 4 GaCl3 → Ga+3 + 3 Cl- ∆tf = 1.86 x 0.1 x 4 = Al2(SO4)3 → 2 Al+3 + 3 SO4-3 ∆tf = 1.86 x NaI → Na+ + I- ∆tf = 1.86 x MgCl2 → Mg+2 + 2 Cl- ∆tf = 1.86 x 0.1 x 3 Moles solute = 37.5 g / 261.55 = 0.1434 moll 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. ∆tx = Kb x m Your Answer: Molality = (solute / MW) / (solvent / 1000) Molality = (20.9 / 120.104) / (250 / 1000) = 0.6961 m ∆tx = Kb x m = 0.51 x 0.6961 = 0.355oC Solution = Solvent - ∆tx = 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. Kef for water is 1.86 and the freezing point of pure water is 0oC. Calculate your answer to 0.1 g/mole. Answer: ∆tx = Kef x m Molality = ∆tx / Kef = 1.20 / 1.86 = 0.645 m Molality = (solute / MW) / (solvent / 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 0.0000726 to exponential form and explain your answer. 2. Convert 5.82 x 103 to ordinary form and explain your answer. Answer: 1. Convert 0.0000726 = 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 =? Oak 2. 158oF =? Co. 3. 343oK =? Of Answer: %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% %Cal = 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 election to fill and write the 4 quantum numbers (n, l, ml and mos) for this electron. Answer: V23 = 1s2 2s2 2p6 3s2 3p6 4s2 3d3: n=3, l=2, ml = 0, mos = +½ Question 8 1. List and explain which of the following atoms holds its valence electrons less tightly. Si or Cal 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 Cal 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 Are 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