CHEM 103 FINAL EXAM

CHEM 103 FINAL EXAM Question 1 Click this link to access the Periodic Table. This may be helpful throughout the exam. 1. Convert 845.3 to exponential form and explain your answer. 2. Convert 3.21 x 10 -5 to ordinary form and explain your answer. 1. Convert 845.3 = larger than 1 = positive exponent, move decimal 2 places = 8.453 x 10 2 2. Convert 3.21 x 10 -5 = negative exponent = smaller than 1, move decimal 5 places = 0.0000321 Question 2 Click this link to access the Periodic Table. This may be helpful throughout the exam. Do the conversions shown below, showing all work: 1. 246 oK = ? oC 2. 45 oC = ? oF 3. 18 oF = ? oK 1. 246 oK - 273 = -27 oC oK → oC (make smaller) -273 2. 45 oC x 1.8 + 32 = 113 oF oC → oF (make larger) x 1.8 + 32 3. 18 oF - 32 ÷ 1.8 = -7.8 + 273 = 265.2 oK oF → oC → oK Question 3 Click this link to access the Periodic Table. This may be helpful throughout the exam. Show the calculation of the number of moles in the given amount of the following substances. Report your answer to 3 significant figures. 1. Moles = grams / molecular weight = 12.0 / 132.15 = 0.0908 mole 2. Moles = grams / molecular weight = 15.0 / 179.17 = 0.0837 mole Question 4 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. f 2 6 f f 2 f 2 6 f f 2 1. Al2(SO4)3 2. C7H5NOBr 1. %Al = 2 x 26.98/342.17 x 100 = 15.77% %S = 3 x 32.07/342.17 x 100 = 28.12% %O = 12 x 16/342.17 = 56.11% 2. %C = 7 x 12.01/ 199.02 x 100 = 42.24% %H = 5 x 1.008/ 199.02 x 100 = 2.53% %N = 1 x 14.01/199.02 = 7.04% %O = 1 x 16.00/199.02 x 100 = 8.03% %Br = 79.90/199.02 x 100 = 40.15% Question 5 Click this link to access the Periodic Table. This may be helpful throughout the exam. Show the calculation of the heat of reaction (ΔHrxn) for the reaction: 2 C2H6 (g) + 5 O2 (g) → 4 CO (g) + 6 H2O (l) by using the following thermochemical data: ΔH0C H (g) = -84.0 kJ/mole, ΔH0 CO (g) = -110.5 kJ/mole, ΔH0H O (l) = -285.8 kJ/mole 2 C2H6 (g) + 5 O2 (g) → 4 CO (g) + 6 H2O (l) ΔH0C H (g) = -84.0 kJ/mole, ΔH0 CO (g) = -110.5 kJ/mole, ΔH0H O (l) = -285.8 kJ/mole ΔHrxn = 2(+84.0) + 5(0) + 4(-110.5) + 6(-285.8) = - 1988.8 kJ/mole Question 6 Click this link to access the Periodic Table. This may be helpful throughout the exam. Show the calculation of the number of moles of a 1.25 liter gas sample collected at 740 mm and 28 oC. P x V = n x R x T 740 mm/760 = 0.974 atm = P R = 0.0821 1.25 liters = V 28oC + 273 = 301oK = T (0.974) x (1.25) = n x (0.0821) x (301) n = 0.0493 mole Question 7 Click this link to access the Periodic Table.This may be helpful throughout the exam. Write the subshell electron configuration (i.e.1s 2 2s 2 , etc.) for the Fe26 atom and then identify the last electron to fill and write the 4 quantum numbers (n, l, ml and ms) for this electron. Fe26 = 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 6 : n=3, l=2, ml = -2, ms = -1/2 Question 8 Click this link to access the Periodic Table.This may be helpful throughout the exam. 1. List and explain which of the following is the smaller atom. C or N 2. List and explain which of the following atoms holds its valence electrons more tightly. Br or I 1. N is smaller than C since atomic size decreases as you go to the right in a period which means that N which is further to the right is smaller. 2. Br holds its valence electrons more tightly than I since electronegativity decreases as you go down a group which means that Br which is further up the group has the higher electronegativity and therefore the higher attraction for its valence electrons. Question 9 Click this link to access the Periodic Table. This may be helpful throughout the exam. Is CH4 Polar, Ionic or Nonpolar and List and Explain whether it is Soluble or Insoluble in Water? CH4 has all nonpolar bonds which makes it Nonpolar and since it is Nonpolar it is Insoluble in water. Question 10 Click this link to access the Periodic Table. This may be helpful throughout the exam. Show the determination of the charge on the ion formed by the Br35 atom. Br35 (nonmetal = gain electrons) 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 5 gain 1e → Br -1 Explain how and why the presence of a solute affects the boiling point of a solvent. The presence of a solute raises the boiling point of a solvent by lowering the vapor pressure of the solvent. With this lower vapor pressure, more heat (a higher boiling point) is required to raise the vapor pressure to atmospheric pressure. Question 12 Click this link to access the Periodic Table. This may be helpful throughout the exam. Question 11 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 -1.20 oC. Kf for water is 1.86 and the freezing point of pure water is 0 oC. Calculate your answer to 0.1 g/mole. Δtf= Kf × m ∆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 = 104.7