General Chemistry I

Symbolic Logic

Experiment 3: Development of an Equation Samantha Perez Chemistry 1210 09/24/2013 T.A. – Marnie Janson Purpose: Identify patterns of chemical reactivity and learn to write chemical equations and particle diagrams. Procedure: Refer to pages 32-35 of General Chemistry 1210 Laboratory Manual by Department of Chemistry at the Ohio State University, Hayden-McNeil Publishing, 2013- 2014. Data and Results: See attached report sheet pages 37-40. Sample Calculations: Formula of Precipitate: PbI 2 Chemical equation: 2 KI (aq) + Pb(NO 3 ) 2 (aq)  PbI 2 (s) + 2 KNO 3 (aq) Reasoning: Lead nitrate and Potassium iodide are both soluble in water. When lead and iodide mix together in the solution they react and form the precipitate Lead Iodide which is not soluble in water which is why it forms a solid precipitate. Questions: See attached report sheet page 41. Conclusion: In this experiment potassium iodide and lead II nitrate were used to show a chemical reaction and how to test for different cations and anions. These tests can be used to find balanced equations for chemical reactions. First the lead nitrate and potassium iodide are mixed and the chemical reaction is observed. Next the anions are observed when reacting with hydrogen peroxide. Potassium nitrate and potassium iodide are mixed with hydrogen peroxide and observed. Then the cations, lead and potassium, are tested when mixed with hydrogen sulfide and observed. Using these tests, the precipitate is tested for specific cations and anions. Then different ratios of the precipitate are tested to find the balanced equation of the reaction. When potassium iodide and lead nitrate react a yellow precipitate is formed which was observed in the first part of the experiment. In part b of the experiment tests were done to find the different ions. Potassium iodide and potassium nitrate were mixed with hydrogen peroxide to show how the iodide and nitrate anions reacted. The iodide reacted and formed a dark red/brown precipitate while the nitrate formed a clear precipitate. Next cations were tested by mixing potassium nitrate and lead nitrate with hydrogen sulfide. The lead reacted and formed a black precipitate while the potassium showed no changes. Next the original precipitate that was formed when potassium iodide and lead nitrate reacted was tested for anions and cations. First tests for anions were done by adding hydrogen peroxide and resulted in finding iodide anions. Then tests for cations were done by using hydrogen sulfide and lead cations were found. The last part of the experiment was to find the balanced equation. Here different ratios were tested for supernatant ions. The ratios of one lead nitrate to two potassium iodide was found to have an extremely small amount of supernatant ions which led to finding the balanced equation 2 KI (aq) + Pb(NO 3 ) 2 (aq)  PbI 2 (s) + 2 KNO 3 (aq). One error that could have occurred during the experiment is the amount that was being put in the test tubes to find the exact ratio. Atoms are extremely small and it is nearly impossible to get an exact ratio using medicine droppers. This could have been avoided by having different, more precise, lab equipment. This lab showed that the ratio 1:2 lead nitrate to potassium iodide was balanced. Tests can be done to find different cations and anions and how they react with one another. Using lab techniques you can find and balance chemical equations.

CHEM | 1210 | General Chemistry I | lab | Experiment 3: Development of a

CHAPTER 8: STATEMENT LOGIC: PROOFS 109 8. 1. ~C → (N ∨ I) 2. C → M 3. ~M 4. T → ~I 5. T ∴ N 6. ~I 4, 5, MP 7. ~C 2, 3, MT 8. N ∨ I 1, 7, MP 9. N 6, 8, DS 9. 1. L ∨ F 2. L → C 3. C → T 4. F → P 5. P → S 6. I → ~T 7. I ∴ S 8. ~T 6, 7, MP 9. ~C 3, 8, MT 10. ~L 2, 9, MT 11. F 1, 10, DS 12. P 4, 11, MP 13. S 5, 12, MP 10. 1. M ∨ ~M 2. M → R 3. ~M → K 4. (R ∨ K) → E 5. E → D ∴ D 6. R ∨ K 1, 2, 3, CD 7. E 6, 4, MP 8. D 6, 7, MP Exercise 8.2 Part A: Annotating *1. 1. ~~A → B ∴ A → B 2. A → B 1, DN 2. 1. ~C → ~D ∴ D → C 2. D → C 1, Cont 3. 1. ~(E • ~D) ∴ D ∨ ~E 2. ~E ∨ ~~D 1, DeM 3. ~E ∨ D 2, DN 4. D ∨ ~E 3, Com *4. 1. ~(E ∨ D) ∴ ~D 2. ~E • ~D 1, DeM 3. ~D 2, Simp 110 SOLUTIONS MANUAL 5. 1. ~A • [(A ∨ B) ∨ C] ∴ B ∨ C 2. ~A 1, Simp 3. (A ∨ B) ∨ C 2, Simp 4. A ∨ (B ∨ C) 3, As 5. B ∨ C 4, 2, DS 6. 1. F • (G • R) ∴ G • F 2. (F • G) • R 1, As 3. F • G 2, Simp 4. G • F 3, Com *7. 1. [(P → Q) → R] • (~Q → ~P) ∴ ~~R 2. ~Q → ~P 1, Simp 3. P → Q 2, Cont 4. (P → Q) → R 1, Simp 5. R 4, 3, MP 6. ~~R 5, DN 8. 1. [~(S • T) ∨ ~~U] • (T • S) ∴ U 2. T • S 1, Simp 3. S • T 2, Com 4. ~(S • T) ∨ ~~U 1, Simp 5. ~~(S • T) 3, DN 6. ~~U 4, 5, DS 7. U 6, DN 9. 1. ~W ∨ (~X ∨ ~Y) ∴ ~Y ∨ ~(W • X) 2. (~W ∨ ~X) ∨ ~Y 1, As 3. ~(W • X) ∨ ~Y 2, DeM 4. ~Y ∨ ~(W • X) 3, Com *10. 1. [~O → (~M → ~N)] • ~(N → M) ∴ O 2. ~O → (~M → ~N) 1, Simp 3. ~(N → M) 1, Simp 4. ~(~M → ~N) 3, Cont 5. ~~O 2, 4, MT 6. O 5, DN 11. 1. ~~P • ~P ∴ W 2. ~~P 1, Simp 3. ~~P ∨ W 2, Add 4. P ∨ W 3, DN 5. ~P 1, Simp 6. W 4, 5, DS 12. 1. ~T ∴ ~(T • ~~S) 2. ~T ∨ ~S 1, Add 3. ~(T • S) 2, DeM 4. ~(T • ~~S) 3, DN *13. 1. ~A ∴ ~[(B • C) • A] 2. ~(B • C) ∨ ~A 1, Add 3. ~[(B • C) • A] 2, DeM CHAPTER 8: STATEMENT LOGIC: PROOFS 111 14. 1. (S → G) • (G → T) ∴ ~T → ~S 2. S → G 1, Simp 3. G → T 1, Simp 4. S → T 2, 3, HS 5. ~T → ~S 4, Cont 15. 1. ~R → ~S ∴ ~P ∨ (S → ~~R) 2. S → R 1, Cont 3. S → ~~R 2, DN 4. ~P ∨ (S → ~~R) 3, Add Part B: Correct or Incorrect? *1. DeM 2. Cont 3. Incorrect. If we apply DeM to ~(F ∨ G), we get ~F • ~G. *4. DeM 5. Com 6. Cont *7. Incorrect use of DeM. Correct sequence: from ~S ∨ T to ~S ∨ ~~T by DN; from ~S ∨ ~~T to ~(S • ~T) by DeM. 8. Cont 9. Incorrect use of Cont; requires a step of DN as well as contraposition. *10. Cont 11. Incorrect; requires a step of association followed by a step of commutation. 12. DeM *13. Com 14. DeM 15. DeM *16. As 17. DeM 18. DN *19. DeM 20. Incorrect. The given conclusion follows validly, but only by a sequence of steps. Part C: Proofs *1. 1. ~(C • D) 2. ~ C → S 3. ~D → T ∴ S ∨ T 4. ~C ∨ ~D 1, DeM 5. S ∨ T 2, 3, 4, CD 2. 1. (W → U) • ~X ∴ ~U → ~W 2. W → U 1, Simp 3. ~U → ~W 2, Cont 3. 1. F → ~G 2. G ∴ ~F 3. ~~G 2, DN 4. ~F 1, 3, MT *4. 1. ~(~A ∨ B) ∴ A 2. ~~A • ~B 1, DeM 3. A • ~B 2, DN 4. A 3, Simp 112 SOLUTIONS MANUAL 5. 1. (~P → Q) • ~Q ∴ P 2. ~P → Q 1, Simp 3. ~Q 1, Simp 4. ~~P 2, 3, MT 5. P 4, DN 6. 1. ~(N ∨ M) 2. ~L → (M ∨ N) ∴ L 3. ~(M ∨ N) 1, Com 4. ~~L 2, 3, MT 5. L 4, DN *7. 1. (A ∨ B) ∨ C 2. ~A ∴ C ∨ B 3. A ∨ (B ∨ C) 1, As 4. B ∨ C 2, 3, DS 5. C ∨ B 4, Com 8. 1. (W • ~X) ∨ (Y • Z) 2. (~X • W) → U 3. (Y • Z) → T ∴ U ∨ T 4. (~X • W) ∨ (Y • Z) 1, Com 5. U ∨ T 4, 2, 3, CD 9. 1. ~(S ∨ R) 2. P → R ∴ ~P 3. ~S • ~ R 1, DeM 4. ~R 3, Simp 5. ~P 2, 4, MT *10. 1. F → (G • H) 2. (H • G) → J ∴ F → J 3. (G • H) → J 2, Com 4. F → J 1, 3, HS 11. 1. K ∨ (L ∨ S) 2. ~(K ∨ L) ∴ S 3. (K ∨ L) ∨ S 1, As 4. S 2, 3, DS 12. 1. ~P 2. ~(P ∨ Q) → ~R 3. ~Q ∴ ~R 4. ~P • ~Q 1, 3, Conj 5. ~(P ∨ Q) 4, DeM 6. ~R 2, 5, MP *13. 1. ~S → (T • U) 2. (~S → X) → ~Z 3. (U • T) → X ∴ ~Z 4. (T • U) → X 3, Com 5. ~S → X 1, 4, HS 6. ~Z 2, 5, MP

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