Exam 3 T Th-solutions (1)

# Exam 3 T Th-solutions (1) - Version 186 – Exam 3 T Th –...

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Unformatted text preview: Version 186 – Exam 3 T Th – sutcliffe – (50985) 1 This print-out should have 28 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points What are the hybrid orbitals used by the underlined atoms in CH 3 C HCHC N, from left to right, respectively? 1. sp 3 and sp 2. sp 2 and sp 2 3. sp 2 and sp correct 4. sp and sp 3 5. sp 2 and sp 3 Explanation: C H 3 C H C H C N b b 002 10.0 points In the reaction CH 4 + 4 Cl 2 → CCl 4 + 4 HCl , how much methane is required to completely react with 85.6 grams of Cl 2 ? 1. 0.302 mol correct 2. 0.604 mol 3. 4.83 mol 4. 21.4 mol 5. 1.21 mol Explanation: m Cl 2 = 85.6 g The balanced equation for the reaction in- dicates that 1 mol CH 4 is needed for every 4 mol Cl 2 reacted. First we calculate the moles Cl 2 present: ? mol Cl 2 = 85 . 6 g Cl 2 × 1 mol Cl 2 70 . 9 g Cl 2 = 1 . 207 mol Cl 2 Using the mole ratio from the chemical equation, we find the moles CH 4 needed to completely react 1.207 mol Cl 2 : ? mol CH 4 = 1 . 207 mol Cl 2 × 1 mol CH 4 4 mol Cl 2 = 0 . 302 mol CH 4 003 10.0 points According to Boyle’s law, when the pressure on a gas in an enclosed container increases, the volume of that gas 1. remains the same. 2. decreases. correct 3. varies directly with the amount of pres- sure applied. 4. increases. Explanation: 004 10.0 points NOTE 1atm = 101.325 kPa The density of citronellal, a mosquito re- pellant, is 1 . 45 g · L − 1 at 365 ◦ C and 50.0 kPa. What is the molar mass of citronellal? 1. 88 . 0 g · mol − 1 2. 73 . 2 g · mol − 1 3. 95 . 7 g · mol − 1 4. 37 . 5 g · mol − 1 5. 154 g · mol − 1 correct Explanation: T = 365 ◦ C + 273.15 K = 638 . 15 K P = (50 kPa) 1 atm 101 . 325 kPa = 0 . 493462 atm Version 186 – Exam 3 T Th – sutcliffe – (50985) 2 ρ = 1 . 45 g / L The ideal gas law is P V = n R T n V = P R T with unit of measure mol/L on each side. Multiplying each by molar mass (MM) gives n V · MM = P R T · MM = ρ , with units of g/L. MM = ρ R T P = (1 . 45 g / L) ( . 08206 L · atm mol · K ) . 493462 atm × (638 . 15 K) = 153 . 875 g / mol 005 10.0 points NOTE: Balance using whole number coeffi- cients. The following reaction is important in the removal of sulfur dioxide, a major source of acid rain, from the smokestacks of coal burn- ing power plants. When the equation CaO + SO 2 + O 2 → CaSO 4 is balanced, the coefficient of calcium oxide (commonly called lime) is 1. 3. 2. 2. correct 3. 1. 4. 4. Explanation: A balanced equation must have the same number of each kind of atom on both sides of the equation. We find the number of each kind of atom using equation coefficients and composition stoichiometry. For example, we find there are 8 O atoms on the product side: ?O atoms = 2 CaSO 4 × 4 O 1 CaSO 4 = 8 O The balanced equation is 2 CaO + 2 SO 2 + O 2 → 2 CaSO 4 and the coefficient of calcuim oxide is 2....
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Exam 3 T Th-solutions (1) - Version 186 – Exam 3 T Th –...

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