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Group Slaughter

# Group Slaughter

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## Group Slaughter

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1. 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. • If your team has less than 4 members, add sufficient names to reach 4 total people.

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

6. Question 1: for 1 shot Answer based on the Ka values provided. • Is this reaction reactant-favored or product-favored? • Will K be less than, equal to, or greater than 1? • Explain how you answered parts (a) and (b). HN3 + ClO2− ⇋ N3− + HClO2 Ka of HN3 = 2.5 x 10−5 Ka of HClO2 = 1.1 x 10−2

7. Question 1: for 1 shot • reactant-favored • K < 1 • HN3 is the weaker acid (smaller Ka), and the favored reaction direction is when the weaker acid/base is produced. HN3 + ClO2− ⇋ N3− + HClO2 Ka of HN3 = 2.5 x 10−5 Ka of HClO2 = 1.1 x 10−2

8. Question 2: for 1 shot Draw both resonance structures for the nitrite ion, NO2─. Label all atoms in each structure with their formal charge.

9. Question 2: for 1 shot Draw both resonance structures for the nitrite ion, NO2─. Label all atoms in each structure with their formal charge.

10. Question 3: for 2 shots 25.0 mL of 0.12 M HBr completely neutralized 15.0 mL of RbOH in a titration. • What was the initial concentration of the RbOH solution? • What was the initial pH of the HBr solution?

11. Question 3: for 2 shots 25.0 mL of 0.12 M HBr completely neutralized 15.0 mL of RbOH in a titration. • What was the initial [RbOH]? MaVa=MbVb Mb = 25.0 x 0.12 / 15.0 = 0.20 M RbOH • What was the initial pH of the HBr solution? pH = - log(0.12) = 0.92

12. Question 4: for 2 shots Calculate the energy, in kJ/mol, of light with a wavelength of 827 nm.

13. Question 4: for 2 shots

14. Question 5: for 1 shot Identify the molecular geometry and hybridization for the specified elements below:

15. Question 5: for 1 shot Identify the molecular geometry and hybridization for the specified elements below: trigonal planar, sp2 trigonal pyramidal, sp3 bent, sp3

16. Question 6: for 1 shot What is the net ionic equation for the neutralization reaction between LiOH(aq) and HBrO2(aq)?

17. Question 6: for 1 shot What is the net ionic equation for the neutralization reaction between LiOH(aq) and HBrO2(aq)? OH− (aq) + HBrO2(aq) → H2O(l) + BrO2−(aq)

18. Question 7: for 1 shot Which is going to be more soluble in water, CH3NH2 or CH3F? Justify your answer.

19. Question 7: for 1 shot Which is going to be more soluble in water, CH3NH2 or CH3F? Justify your answer. CH3NH2 is going to be more soluble in water than CH3F, because CH3NH2 can form energetically favorable hydrogen bonds with water molecules, but CH3F can only form less favorable dipole-dipole interactions which are weaker than the hydrogen bonds between water molecules.

20. Question 8: for 1 shot A 0.250 M solution of a weak acid is 0.75% ionized. Determine the pH and pKa of this solution.

21. Question 8: for 2 shots −5 Ka = − log(1.4 x 10−5) = 4.85

22. Question 9: for 1 shot The contents of beaker 1 (25.0 mL) are poured into beaker 2 (25.0 mL) and stirred. Assume the volumes are additive. Calculate the pH of the resulting solution. (Kb of NH3 = 1.8 x 10−5) Beaker 1 Beaker 2 0.100 M NH3 0.100 M NH4Cl

23. Question 9: for 1 shot

24. Question 10: for 1 shot Which of the following compounds should have the highest melting point and why? LiI, KI, LiF, NaF

25. Question 10: for 1 shot Which of the following compounds should have the highest melting point and why? LiI, KI, LiF, NaF NaF should have the highest melting point, because Na+ and F─ are the smallest cation and anion, respectively, and according to Coulomb’s Law, decreasing the distance between charged particles increases the attraction between them.

26. Question 11: for 1 shot Identify one of the compounds below that can be dissolved in water to form a basic solution. Write the net ionic equation for the reaction that occurs to cause the solution to be basic. LiI, KI, LiF, NaF

27. Question 11: for 1 shot Identify one of the compounds below that can be dissolved in water to form a basic solution. Write the net ionic equation for the reaction that occurs to cause the solution to be basic. LiI, KI, LiF, NaF Either LiF or NaF F− + H2O ↔ HF+ OH−

28. Question 12: for 1 shot The pH of a soft drink is 3.37 after the addition of KC6H7O2(aq). Which species, HC6H7O2 or C6H7O2 −, has a higher concentration in the soft drink? Justify your answer. (Ka of HC6H7O2 = 1.7 x 10−5)

29. Question 12: for 1 shot The pH of a soft drink is 3.37 after the addition of KC6H7O2(aq). Which species, HC6H7O2 or C6H7O2 −, has a higher concentration? (Ka of HC6H7O2 = 1.7 x 10−5) HC6H7O2 has a higher concentration. If [HC6H7O2] = [C6H7O2−], pH = pKa = −log(1.7 x 10−5) = 4.77 Since 3.37 < 4.77 (pH < pKa), the protonated form, HC6H7O2, will have a higher concentration.