CHEM 103 Exam 1 to 6 Questions With Correct Answers Latest Update 2024 (Portage Learning)

Show the calculation of the molecular weight for the following compounds, reporting your answer to 2 places after the decimal. 1. Al2(CO3)3 2. C8H6NO4Cl 1. 2Al + 3C + 9O = 233.99 2. 8C + 6H + N + 4O + Cl = 215.59 Question 2 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 answerto 3 significant figures. 1. 13.0 grams of (NH4)2CO3 2. 16.0 grams of C8H6NO4Br 1. Moles = grams / molecular weight = 13.0 / 96.09 = 0.135 mole 2. Moles = grams / molecular weight = 16.0 / 260.04 = 0.0615 mole Question 3 Click this link to access the Periodic Table. This may be helpful throughout the exam. 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. 1.20 moles of (NH4)2CO3 2. 1.04 moles of C8H6NO4Br 1. Grams = Moles x molecular weight = 1.20 x 96.09 = 115.3 grams 2. Grams = Moles x molecular weight = 1.04 x 260.04 = 270.4 grams 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. 1. (NH4)2CrO4 2. C8H8NOI 1. %N = 2 x 14.01/152.08 x 100 = 18.43% %H = 8 x 1.008/152.08 x 100 = 5.30% %Cr = 1 x 52.00/152.08 x 100 = 34.20% %O = 4 x 16.00/152.08 x 100 = 42.08% 2. %C = 8 x 12.01/261.05 x 100 = 36.80% %H = 8 x 1.008/261.05 x 100 = 3.09% %N = 1 x 14.01/261.05 x 100 = 5.37% %O = 1 x 16.00/261.05 x 100 = 6.13% %I = 1 x 126.9/261.05 x 100 = 48.61% Question 5 Click this link to access the Periodic Table. This may be helpful throughout the exam. 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 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 Click this link to access the Periodic Table. This may be helpful throughout the exam. Balance each of the following equations by placing coefficients in front of each substance. 1. C6H6 + O2 → CO2 + H2O 2. As + O2 → As2O5 3. Al2(SO4)3 + Ca(OH)2 → Al(OH)3 + CaSO4 1. 2 C6H6 + 15 O2 → 12 CO2 + 6 H2O 2. 4 As + 5 O2 → 2 As2O5 3. Al2(SO4)3 + 3 Ca(OH)2 → 2 Al(OH)3 + 3 CaSO4 Question 7 Click this link to access the Periodic Table. This may be helpful throughout the exam. Classify each of the following reactions as either: Combination Decomposition Combustion Double Replacement Single Replacement 1. H2SO4 → SO3 + H2O 2. S + 3 F2 → SF6 3. H2 + NiO → Ni + H2O 1. H2SO4 → SO3 + H2O = Decomposition, One reactant → Two Products 2. S + 3 F2 → SF6 = Combination. Two reactants→ One product 3. H2 + NiO → Ni + H2O = Single Replacement, Hydrogen displaces metal ion Question 8 Click this link to access the Periodic Table. This may be helpful throughout the exam. 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 Na2HAsO3 + KBrO3 + HCl → NaCl + KBr + H3AsO4 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 Click this link to access the Periodic Table. This may be helpful throughout the exam. Show the balancing of the following redox equation, including the determination of the oxidation number (charge) of ONLY the atoms which are changing. KMnO4 + KI + H2O → KIO3 + MnO2 + KOH Mn compounds x 2 ; I compounds x 1 = 2KMnO4 + 1 KI + 1 H2O → 1 KIO3 + 2MnO2 + 2KOH KMnO4 + KI + H2O → KIO3 + MnO2 + KOH KMnO4: K is metal in group I = +1, each O is -2 (total is -8), so Mn is +7 MnO2: Each O is -2 (total is -4), so Mn is +4 KI: K is metal in group I = +1, so I is -1 KIO3: K is metal in group I = +1, each O is -2 (total is -6), so I is +5 Since Mn (on left side) is +7 and Mn (on right side) is +4: Mn changes by 3 Since I (on left side) is -1 and I (on right side) is +5: I changes by 6 Multiply Mn compounds by 2 and I compounds by 1 and after balancing other atoms = 2 KMnO4 + 1 KI + 1 H2O → 1 KIO3 + 2 MnO2 + 2 KOH Question 10 Click this link to access the Periodic Table. This may be helpful throughout the exam. Show the balanced equation and the calculation of the number of moles and grams of CO2 formed from 20.6 grams of C6H6. Show your answers to 3 significant figures. C6H6 + O2 → CO2 + H2O Molar mass of CO2 = 44.01 g/mol mass of CO2 = 1.5822 ml x 44.01 g/mol =69.63 grams mass of CO2 = 69.63 grams moles of CO2 = 1.58 mole 2 C6H6 + 15 O2 → 12 CO2 + 6 H2O 20.6 g / (6 x 12.01 + 6 x 1.008) = 20.6 / 78.108 = 0.2637 mole x 12/2 = 1.58 mole CO2 1.582 mole CO2 x (12.01 + 2 x 16.00) = 69.6 g CO2 MODULE 3 EXAM Question 1 Click this link to access the Periodic Table. This may be helpful throughout the exam. 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 1. Heat given off = Exothermic process 2. The amount of heat released = Heat of reaction Question 2 Click this link to access the Periodic Table. This may be helpful throughout the exam. 1. Show the calculation of the final temperature of the mixture when a 22.8 gram sample of water at 74.6oC is added to a 14.3 gram sample of water at 24.3oC in a coffee cup calorimeter. c (water) = 4.184 J/g oC 2. Show the calculation of the energy involved in freezing 54.3 grams of water at 0oC if the Heat of Fusion for water is 0.334 kJ/g 1. - (mwarn H2O x cwarn H2O x ∆twarn H2O) = (mcool H2O x ccool H2O x ∆tcool H2O) - [22.8 g x 4.184 J/g oC x (Tmix - 74.6oC)] = [14.3 g x 4.184 J/g oC x (Tmix - 24.3oC)] - [95.3952 J/oC x (Tmix - 74.6oC)] = [59.8312 J/oC x (Tmix - 24.3oC)] Tmix = 55.2oC 2. ql↔ s = m x ∆Hfusion = 54.3 g x 0.334 kJ/g = 18.14 kJ (since heat is removed) = - 18.14 kJ Question 3 Click this link to access the