1. The Battle for Chemical Domination On the Open Seas
Group Slaughter
The Battle for Chemical Domination
On the Open Seas

2. 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.

3. 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.

4. 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!
NO FRIENDLY FIRE ;)

5. May the odds be ever in your favor…
Last Questions?
Let the games begin!
May the odds be ever in your favor…

6. C 2 H 5 OH 𝑎𝑞 +3 H 2 O 𝑙 → 2 CO 2 𝑔 +12 H + 𝑎𝑞 +12 𝑒 −
Question 1: for 2 shots
C 2 H 5 OH 𝑎𝑞 +3 O 2 𝑔 → 2 CO 2 𝑔 +3 H 2 O 𝑙 𝐸 𝑐𝑒𝑙𝑙 𝑜 = 𝑉
A student constructs a galvanic cell using the redox reaction above. Below is the oxidation half-reaction.
C 2 H 5 OH 𝑎𝑞 +3 H 2 O 𝑙 → 2 CO 2 𝑔 +12 H + 𝑎𝑞 +12 𝑒 −
𝐸 𝑜𝑥 𝑜 = 𝑉
Write the equation for the reduction half-reaction.
Calculate 𝐸 𝑟𝑒𝑑 𝑜 for this half-reaction.

7. ─( C 2 H 5 OH 𝑎𝑞 +3 H 2 O 𝑙 → 2 CO 2 𝑔 +12 H + 𝑎𝑞 +12 𝑒 − )
Question 1: for 2 shots
Overall cell rxn ─ oxidation half-rxn = reduction half-rxn!
C 2 H 5 OH 𝑎𝑞 +3 O 2 𝑔 → 2 CO 2 𝑔 +3 H 2 O 𝑙
─( C 2 H 5 OH 𝑎𝑞 +3 H 2 O 𝑙 → 2 CO 2 𝑔 +12 H + 𝑎𝑞 +12 𝑒 − )
3 O 2 𝑔 +12 H + 𝑎𝑞 +12 𝑒 − →6 H 2 O 𝑙
Simplify!
O 2 𝑔 +4 H + 𝑎𝑞 +4 𝑒 − →2 H 2 O 𝑙
𝐸 𝑟𝑒𝑑 𝑜 =1.314−0.085=𝟏.𝟐𝟐𝟗 𝐕

8. Question 2: for 2 shots
Given the above reaction, which has an experimentally determined rate law of
Rate = k[B]2
The following mechanism has been proposed. Is it an acceptable mechanism? Justify your answer!

9. Question 2: for 2 shots
One point for each valid answer (up to 2):

10. Question 3: for 1 shot
Dichloromethane has a greater solubility in water than carbon tetrachloride has. Account for this observation in terms of the intermolecular forces between each of the solutes and water.

11. Question 3: for 1 shot

12. Question 4: for 1 shot
An external direct-current power supply is connected to two platinum electrodes immersed in a beaker containing 1.0 M CuSO4(aq). As the cell operates, copper metal is deposited onto one electrode.
If a current of 1.50 amps passes through the cell for 40.0 minutes, what mass of copper will be plated out?

13. Question 4: for 1 shot
An external direct-current power supply blah blah beaker containing 1.0 M CuSO4(aq). As the cell operates, copper metal is deposited onto one electrode. If a current of 1.50 amps passes through the cell for 40.0 minutes, what mass of copper will be plated out?
40.0 min× 60 s 1 min × 1.50 C 1 s × 1 mol 𝑒 − 96,485 C × 1 mol Cu 2 mol 𝑒 − × g Cu 1 mol Cu
= 1.19 g Cu

14. Question 5: for 1 shot
The pH of a M solution of benzoic acid, C6H5COOH (a monoprotic acid), is What is the ionization constant, Ka, for benzoic acid?

15. Question 5: for 1 shot 𝐾 𝑎 =𝟔.𝟔× 𝟏𝟎 −𝟓
The pH of a M solution of benzoic acid, C6H5COOH (a monoprotic acid), is What is the ionization constant, Ka, for benzoic acid?
H 3 O + = 10 −pH = 10 −3.09 =8.1× 10 −4 𝑀
𝐾 𝑎 = H 3 O + [ C 6 H 5 COO − ] [ C 6 H 5 COOH] = 𝑥 −𝑥 ≈ 𝑥 = (8.1× 10 −4 )
(assume x << 0.010)
𝐾 𝑎 =𝟔.𝟔× 𝟏𝟎 −𝟓

16. Question 6: for 1 shot
Flask A contains helium gas and Flask B contains argon gas. If the contents of both flasks are combined into a previously evacuated 4.0 L flask (Flask C), what would be the total pressure in Flask C? (Assume T constant.)

17. Question 6: for 1 shot
1 atm + 3 atm = 4 atm, but…  volume increased to 4.0 L (from 1.0 L) 4 atm/4.0 = 1 atm

18. Question 7: for 2 shots 2 Al 𝑠 +6 HCl 𝑎𝑞 → 2 AlCl 3 𝑎𝑞 +3 H 2 𝑔
∆ H rxn o =−1,049 kJ/ mol rxn
A 0.65 g sample of aluminum, the limiting reactant, was added to 50.0 mL of HCl solution in a coffee cup calorimeter. Calculate the temperature change in the calorimeter if the specific heat of the solution is 4.18 J/goC, and the density of the solution = 1.0 g/mL.

19. Question 7: for 2 shots 2 Al 𝑠 +6 HCl 𝑎𝑞 → 2 AlCl 3 𝑎𝑞 +3 H 2 𝑔
∆ H rxn o =−1,049 kJ/ mol rxn
0.65 g Al× 1 mol Al g Al × 1 mol rxn 2 mol Al × −1,049 kJ 1 mol rxn =− kJ
12,636 J=(50.65 g)(4.18 𝐽 𝑔℃ )∆𝑇
∆𝑇= 12, ×4.18 =𝟔𝟎.℃

20. Question 8: for 1 shot
Draw two molecules of NH3 (with the correct Lewis dot structure) AND draw and label a hydrogen bond that can form between them.

21. Question 8: for 1 shot
Draw two molecules of NH3 (with the correct Lewis dot structure) AND draw and label a hydrogen bond that can form.
hydrogen bond

22. You must identify ALL correct elements to get the shot!
Question 9: for 1 shot
Which of the following elements can form covalent bonds in a way that violates the octet rule?
H, He, Li, Be, B, C, Ne, S, Xe
You must identify ALL correct elements to get the shot!

23. Question 9: for 1 shot H, Be, B, S, Xe
Which of the following elements can form covalent bonds in a way that violates the octet rule?
H, Be, B, S, Xe

24. Question 10: for 1 shot
Write the orbital notation and electron configuration for the cation in aluminum chloride.

25. Question 10: for 1 shot
Write the orbital notation and electron configuration for for the cation in aluminum chloride (Al3+).
Orbital notation:
Electron configuration: 1s22s22p6

26. Question 11: for 1 shot
1. Draw the correct Lewis Dot structure for the molecule SF4
2. Identify its electron geometry
3. Identify its VESPR shape

27. Question 11: for 1 shot 1. SF4 Lewis structure:
2. Identify its electron geometry: trigonal bipyramidal
3. Identify its VESPR shape: seesaw

28. Question 12: for 1 shot
What are two variables you can change to increase the attractive force between two particles,
AND: is each variable directly or inversely proportional to the attraction between the particles?

29. Question 12: for 1 shot Charge: directly proportional
Distance between particles: inversely proportional