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Final Exam Review, Part I Get ready for Battleship! You need… 1)A calculator 2)A periodic table (with a back) 3)A formula chart 4)Your final review 5)Any notes you may have 6)Scratch paper to do the math on – lots of math. 7)Your game face! TAKE THOSE OTHER TEAMS DOWN! 8)To sit in groups of 3-4.
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Final Exam Review, Part I Get ready for Battleship!
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Set-Up: Number 1-10 on your piece of paper. Each team member must print their name next to ONE number only. Don’t let any other team see what your numbers are! These are your battleships.
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The Rules You gain the ability to take shots at other teams’ battleships by getting questions correct. Questions are worth 1, 2, or 3 shots at the team of your choice. When one of your battleships is sunk, the person who’s battleship was shot will be rescued by their teammates. THE TEAM ONLY DIES WHEN EVERY MEMBER HAS BEEN SHOT. Once a team is dead, if they get the next question right, they will come back to life as angry ghosts! Ghosts get double shots for every question they get right.
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The Rules: Part II Winners will be determined by which team has the most LIVING battleships at the end of the class. All students MUST participate. You show this by making sure everyone’s color appears in the answer. Sharing pens results in immediate disqualification for that question. Answers are only accepted if written on the group’s white board!
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Battleship: Last Questions? Let the games begin!
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Question 1: for 2 shots How many formula units are in 3 moles of AlF 3 ?
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Question 1: for 2 shots How many formula units are in 3 moles of AlF 3 ? 3 x (6.02 x 10 23 ) = 1.806 x 10 24 formula units
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Question 2: for 2 shots N 2 + 3 H 2 2NH 3 At STP, How many liters of Nitrogen gas, reacting with hydrogen gas, are needed to produce 25 L of ammonia (NH 3 )?
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Question 2: for 2 shots N 2 + 3 H 2 2NH 3 At STP, How many liters of Nitrogen gas, reacting with hydrogen gas, are needed to produce 25 L of ammonia (NH 3 )? 25 L NH 3 x 1 L N 2 = (25/2) = 12.5 L N 2 2 L NH 3
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Question 3: for 1 shot If you have 205 g of NaOH in 350 mL of water: How many moles of NaOH do you have?
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Question 3: for 1 shot If you have 205 g of NaOH in 350 mL of water: How many moles of NaOH do you have? 205 g x _1 mol NaOH_ = 5.13 mol NaOH 39.997 g NaOH
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Question 4: for 1 shot If you have 5.13 mol of NaOH in 350 mL of water: What is the concentration (M) of your solution?
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Question 4: for 1 shot If you have 5.13 mol of NaOH in 350 mL of water: What is the concentration (M) of your solution? 350 mL / 1000 = 0.35 L M = (5.13 mol/ 0.35 L) = 14.7 M
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Question 5: for 2 shots A sample that contains 4.38 mol of a gas at 15 o C has a pressure of 0.857 atm. What is its volume?
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Question 5: for 2 shots A sample that contains 4.38 mol of a gas at 15 o C has a pressure of 0.857 atm. What is its volume? 15 o C + 273 = 288 K PV = nRT (0.857 atm) V = (4.38 mol) (0.0821) (288 k) V = 103.564 / 0.857 = 120.8 L
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Question 6: for 1 shot If Ms. Park needs to make 2.5 L of 0.6 M hydrobromic acid, how much of 12 M acid would she need?
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Question 6: for 1 shot If Ms. Park needs to make 2.5 L of 0.6 M hydrobromic acid, how much of 12 M acid would she need? (0.6 M) (2.5 L) = (12 M) V 2 V 2 = (0.6 x 2.5) / 12 = 0.125 L
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Question 7: for 3 shots 2 KOH + H 2 CO 3 2 K 2 CO 3 + 2 H 2 O How many grams of water will be produced from the neutralization of KOH with 24.8 g of carbonic acid?
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Question 7: for 3 shots 2 KOH + H 2 CO 3 2 K 2 CO 3 + 2 H 2 O How many grams of water will be produced from the neutralization of KOH with 24.8 g of carbonic acid? 24.8 g H 2 CO 3 x 1 mol x 2 mol H 2 O x 18.015 g = 62.025 g 1 mol H 2 CO 3 1 mol H 2 O = (24.8 x 2 x 18.015) / 62.025 = 14.4 g H 2 O
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Question 8: for 1 shot Balance the following reaction! ___Cu 2 O 3 ___Cu + ___O 2
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Question 8: for 1 shot Balance the following reaction! 2 Cu 2 O 3 4 Cu + 3 O 2
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Question 9: for 2 shots What is the mass (in grams) of 3.7 moles of copper atoms?
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Question 9: for 2 shots What is the mass (in grams) of 3.7 moles of copper atoms? 3.7 mol x 63.546 g/mol = 235.1 grams
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Question 10: for 2 shots Identify ALL of the following compounds that would be insoluble (form precipitates) in water: (NH 4 ) 2 S PbCl 2 Cs 2 CO 3 Mg(OH) 2 HClO CaCrO 4
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Question 11: for 3 shots List any 3 (out of the 5) postulates of Kinetic Molecular Theory (KMT).
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Question 11: for 3 shots ANY 3: 1.Gas particles are very small (no volume) 2.Gas particles have elastic collisions (energy conserved) 3.Gas particles do not attract or repel each other 4.Gas particles are always in rapid motion 5.The temperature of a gas is directly related to the amount of kinetic energy of the particles
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Question 12: for 2 shots What volume is occupied by 15 moles of CO 2 gas at STP?
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Question 12: for 2 shots What volume is occupied by 15 moles of CO 2 gas at STP? 15 mol x 22.4 L/mol = 336 L CO 2
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Question 13: for 2 shots Write and balance the following reaction: Phosphorus pentachloride decomposes to yield phosphorus trichloride and chlorine gas
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Question 13: for 2 shots Write and balance the following reaction: Phosphorus pentachloride decomposes to yield phosphorus trichloride and chlorine gas PCl 5 PCl 3 + Cl 2
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