Principles of Chemistry I - Solved Exam 1 - Fall 2016 | CH 301 | DOWNLOAD FOR QUALITY GRADES

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Explanation: A balanced equation has 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 stoich... iometry. For example, we find there are 12 H atoms on the reactant side: ? H atoms = 6 H2O × 2 H 1 H2O = 12 H The balanced equation is Mg3N2 + 6 H2O → 3 Mg(OH)2 + 2 NH3 , and has 3 Mg, 2 N, 12 H and 6 O atoms on each side. The equation coefficients are 1, 6, 3, 2. 007 4.0 points A sample of a gas occupies 460 mL at 70.0◦C and 1.00 atmosphere. At what temperature would the gas occupy 650 mL at the same pressure? 1. 99◦C 2. 188◦C 3. 111◦C 4. 154◦C 5. 435◦C 6. −30.0◦C 7. 212◦C correct Explanation: V1 = 460 mL V2 = 650 mL T1 = 70.0◦C + 273 = 343 K Charles’s law relates the volume and absolute (Kelvin) temperature of a sample of gas: V1 T1 = V2 T2 T2 = V2 T1 V1 = (650 mL) (343 K) 460 mL = 485 K ◦C = 485 K − 273 = 212◦C 008 4.0 points Consider the initial set up in the apparatus below at a temperature of 300 K. Each gas is at the same pressure of 200 torr while the valve is closed. 30 L 10 L H2 N2 valve The gases will not react at all at 300 K. After the valve is opened, the gases completely mix. What is the partial pressure of the hydrogen gas in this (opened valve) state? 1. 150 torr correct 2. 400 torr 3. 75 torr 4. 125 torr 5. 200 torr 6. 175 torr 7. 225 torr 8. 100 torr Explanation: When the valve is opened the total available volume for both gases is now 40 L. This means (Boyle’s Law) that the hydrogen gas pressure will drop by a factor of 30/40 or 3/4 or 0.75. 0.75(200) = 150 torr for the hydrogen. The nitrogen will drop to 10/40 or 1/4 or 0.25. 0.25(200) = 50 torr for the nitrogen. 009 4.0 points What is the density of xenon gas at 750 kPa and 25◦C? [Show More]

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