**VII-1 C NC I-3 C NC II-2 C NC *VII-2 C NC I-4 C NC. CHEMISTRY 131 Quiz 5 Fall 2010 Form B
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1 **VII-1 N I-3 N II-2 N *VII-2 N I-4 N EMISTRY 131 Quiz 5 Fall 2010 Form B NAME: Key hapter 11: States of Matter A. (2 pts) onsider the structures of the compounds shown below, and determine which intermolecular forces would be important in each of them. O l N N O O A. Water B. Vinyl chloride. Dinitrogen monoxide D. Methanol l O O O O E. ydrogen chloride F. Oxygen G. Methane. arbon dioxide 1. In which of the compounds is hydrogen bonding present? List the letter(s) before the names of the compounds you have selected. A, D 2. In which of the compounds are dispersion forces the only intermolecular forces present? List the letter(s) before the names of the compounds you have selected. F, G, 3. In which of the compounds are dipole-dipole interactions present, but not hydrogen bonds? List the letter(s) before the names of the compounds you have selected. B,, E
2 *B. (1 pt) onsider the following molecules, then answer the questions below. O S A. Methanol B. Methyl sulfide 1. Give the name of the compound that will have the higher boiling point. Methanol 2. What forces are present in that molecule, but not in the other one? ydrogen bonding. (2.5 pts) Mercury is the only metal that exists as a liquid at room temperature. Its specific heat is 1.25 J/g. Its f is 11.6 J/g and its v is 29.5 J/g. It freezes at 39.0 and boils at ow much heat would be required to raise the temperature of 25.0 g mercury from its freezing point to its boiling point? 1.24 x 10 4 J = 12.4 kj = 1.25 J x 25.0 g (357 ( 39.0)) = 1.24 x 10 4 J = 12.4 kj g 2. ow much heat would be released if 25.0 g of liquid mercury, at its freezing point, froze to solid metal? 290 J = kj = 11.6 J x 25.0 g = 290 J = kj g 3. ow much heat would be required to boil 25.0 g of liquid mercury, at its boiling point, to mercury vapor? 738 J = kj = 29.5 J x 25.0 g = 738 J = kj g 4. While the mercury is boiling, what does its temperature do? heck the correct answer. X A. Remain the same B. Increase. Decrease
3 hapter 12: Gases A. (3 pts) A tank of N 2 gas has a volume of 30 L and contains 125 moles of gas. The tank is in a cool store room at 15. T **1. What is the pressure of the gas in the tank? P = x V = 30 L n = 125 mol T = 15º = 288 K 98.5 atm PV = nrt P = nrt = 125 mol x 288 K x L atm= 98.5 atm V 30 L mol K 2. What is the density of N 2 in the tank? 117 g/l 125 mol N 2 x g N 2 = 117 g/l 30 L mol N 2 3. What volume would the gas occupy at STP? 2.80 x 10 3 L 22.4 L x 125 mol = 2.80 x 10 3 L mol B. (0.5 pt) A student performs a harles Law experiment similar to the one you did. She weighs her flask and stopper, finding a mass of g. She fills the flask with water and weighs it again. The final mass is g. What is the volume of water in the flask? 133 ml g g x 1 ml = 133 ml g 1.00 g g
4 B. (3 points) A student performed a harles' Law experiment very similar to the one you did in lab. Make the following calculations with her data. Show your equations. 1. er flask had an internal volume of 136 ml. She heated it in a boiling water bath and found its temperature to be 101.7º. She then closed the clamp and cooled it in an icewater bath that had a temperature of 0.3º. When she released the clamp, what final volume of air should she have obtained? 99.0 ml V 1 = 136 ml V 1 = V 2 V 1 x T 2 = V 2 T 1 = 101.7º = 375 K T 1 T 2 T 1 T 2 = 0.3º = 273 K 136 ml x 273 K = 99.0 ml 375 K 2. What was the pressure of the air in the flask after she cooled it but before she opened the stopcock and allowed water to fill it? The atmospheric pressure was mm g. 549 mm g = atm P 1 = mm g P 1 = P 2 P 1 x T 2 = P 2 T 1 = 101.7º = 375 K T 1 T 2 T 1 T 2 = 0.3º = 273 K mm g x 273 K = 549 mm g 375 K torr x 1 atm = atm atm x 273 K = atm 760 torr 375 K Note that volume is constant the flask can t shrink! Results are acceptable in any pressure unit. The conversion between torr or mm g and atmospheres is exact, so the result has the same number of significant figures as the measured pressure. *3. ow many moles of gas were present in the cooled, unopened flask described in Part 1 of this problem? mol P = 549 torr x 1.00 atm = atm PV = nrt PV = n 760 torr RT V = 136 ml x L = L 1000 ml T = 0.3º = 273 K atm x L x mol K = mol 273 K L atm
5 ompetencies from Previous Quizzes: I-3 Determine the empirical formula for ethylene glycol, which is used in anti-freeze. Its elemental composition is 38.69%, 9.76%, and 51.55% O. Show your work. 3 O g x 1 mol = mol mol O = mol O = 1 O g mol mol g x 1 mol = mol mol = mol = g mol mol g O x 1 mol O = mol O 3 O g O I-4. Glyceraldehyde is the simplest carbohydrate. Its empirical formula is 2 O. Its molar mass is g/mol What is its molecular formula? Show your work. 3 6 O 3 1 mol x g = g g x 1 formula = formula mol mol g mol 2 mol x 1.01 g = 2.02 g mol 3( 2 O) = 3 6 O 3 1 mol O x g O = g O mol O g/formula II-2 What mass of water is formed if 5.00 g of O 2 is treated with excess 2? The balanced equation is shown below. Show your work g 2 O O O 5.00 g O 2 x 1 mol O 2 x 2 mol 2 O x g 2 O = 5.63 g 2 O g O 2 1 mol O 2 mol 2 O
**VII-1 C NC *VII-2 C NC I-2 C NC I-3 C NC I-4 C NC II-1 C NC II-2 C NC II-3 C NC IV-1 C NC IV-2 C NC VIII-1 C NC
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