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"for studies of G-protein-coupled receptors" 萊夫科維茲( Robert Lefkowitz )和克比爾卡 ( Brian Kobilka )因發現 G 蛋白耦合性受體的 內部運作方式而獲獎,他們兩人研究讓身體細 胞對外界訊號有所反應的蛋白質,對 G 蛋白耦.

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Presentation on theme: ""for studies of G-protein-coupled receptors" 萊夫科維茲( Robert Lefkowitz )和克比爾卡 ( Brian Kobilka )因發現 G 蛋白耦合性受體的 內部運作方式而獲獎,他們兩人研究讓身體細 胞對外界訊號有所反應的蛋白質,對 G 蛋白耦."— Presentation transcript:

1 "for studies of G-protein-coupled receptors" 萊夫科維茲( Robert Lefkowitz )和克比爾卡 ( Brian Kobilka )因發現 G 蛋白耦合性受體的 內部運作方式而獲獎,他們兩人研究讓身體細 胞對外界訊號有所反應的蛋白質,對 G 蛋白耦 合性受體有突破性的發現。 G 蛋白耦合性受體為重要的受體家族,大約一 半的藥物都作用於這些受體,因此對這些受體 的認識將有助科學家研發效果更好的藥物。 As yet, how many Nobel Prize in Chemistry You know? Additional HW: (one page A4 ) 由學號選擇年份 : 如 1 號 選擇 2011; 2 號選擇 2010…

2 Reactions in Aqueous Solution Chapter 4

3 Why we talk about “solution” Most of chemical reaction and bio- processing occurs in “solution” phase 3 DNA in buffer solution Li-ion battery

4 4 A solution is a homogenous mixture of 2 or more substances The solute is(are) the substance(s) present in the smaller amount(s) The solvent is the substance present in the larger amount SolutionSolventSolute Soft drink(l) Air(g) Soft Solder(s) H2OH2O N2N2 Pb Sugar, CO 2 O 2, Ar, CH 4 Sn aqueous solutions of KMnO 4 焊錫 Definition of Solution

5 5 An electrolyte is a substance that, when dissolved in water, results in a solution that can conduct electricity. A nonelectrolyte is a substance that, when dissolved, results in a solution that does not conduct electricity. nonelectrolyte weak electrolyte strong electrolyte Electrolyte

6 6 Strong Electrolyte – 100% dissociation NaCl(s) Na + (aq) + Cl - (aq) H2OH2O Weak Electrolyte – not completely dissociated CH 3 COOH CH 3 COO - (aq) + H + (aq) Conduct electricity in solution? Cations (+) and Anions (-) The degree of dissociation

7 7 Ionization of acetic acid CH 3 COOH CH 3 COO - (aq) + H + (aq) A reversible reaction. The reaction can occur in both directions.(Until it reach the chemical equilibrium) Acetic acid is a weak electrolyte because its ionization in water is incomplete. 醋酸

8 8 Nonelectrolyte does not conduct electricity? No cations (+) and anions (-) in solution C 6 H 12 O 6 (s) C 6 H 12 O 6 (aq) H2OH2O

9 9 Electrolyte in Electrochemical Reaction: applications 1.Solar cell (Dye-sensitized solar cell) 2.Electroplating 3.Battery(Li-ion battery etc) 4.Chemical sensor or biomedical sensor 5.Etching …… and so on

10 10 Graphene Ink Electrochemical Exfoliation of graphite to produce thin graphene films Vedio clip C.Y. Su et al., ACS Nano (2011)

11 11 Hydration is the process in which an ion is surrounded by water molecules arranged in a specific manner.   H2OH2O Hydration( 水合 ) 水分子為一種良好的極性溶劑

12 12 Precipitation Reactions Precipitate – insoluble solid that separates from solution molecular equation ionic equation net ionic equation Pb NO Na + + 2I - PbI 2 (s) + 2Na + + 2NO 3 - Na + and NO 3 - are spectator ions PbI 2 Pb(NO 3 ) 2 (aq) + 2NaI (aq) PbI 2 (s) + 2NaNO 3 (aq) precipitate Pb I - PbI 2 (s) 旁觀離子

13 13 Precipitation of Lead Iodide PbI 2 Pb I - PbI 2 (s)

14 14 Solubility is the maximum amount of solute that will dissolve in a given quantity of solvent at a specific temperature. AgCl

15 15 Examples of Insoluble Compounds CdS PbSNi(OH) 2 Al(OH) 3

16 16 Writing Net Ionic Equations 1.Write the balanced molecular equation. 2.Write the ionic equation showing the strong electrolytes completely dissociated into cations and anions. 3.Cancel the spectator ions on both sides of the ionic equation 4.Check that charges and number of atoms are balanced in the net ionic equation AgNO 3 (aq) + NaCl(aq) AgCl(s) + NaNO 3 (aq) Ag + + NO Na + + Cl - AgCl(s) + Na + + NO 3 - Ag + + Cl - AgCl(s) Write the net ionic equation for the reaction of silver nitrate with sodium chloride.

17 17 Properties of Acids Have a sour taste. Vinegar owes its taste to acetic acid. Citrus fruits contain citric acid. React with certain metals to produce hydrogen gas. React with carbonates and bicarbonates to produce carbon dioxide gas Cause color changes in plant dyes. 2HCl(aq) + Mg(s) MgCl 2 (aq) + H 2 (g) 2HCl(aq) + CaCO 3 (s) CaCl 2 (aq) + CO 2 (g) + H 2 O(l) Aqueous acid solutions conduct electricity. (醋)(醋)

18 18 Have a bitter taste. Feel slippery. Many soaps contain bases. Properties of Bases Cause color changes in plant dyes. Aqueous base solutions conduct electricity. Examples: Ex: NaOH

19 19 Arrhenius acid is a substance that produces H + (H 3 O + ) in water Arrhenius base is a substance that produces OH - in water Arrhenius theory:

20 20 A Brønsted acid is a proton donor A Brønsted base is a proton acceptor acidbaseacidbase A Brønsted acid must contain at least one ionizable proton! Brønsted theory: Donate proton

21 21 Hydronium ion, hydrated proton, H 3 O + 水合質子   H2OH2O

22 22 Example: 鎖相萃取法 以蒸餾的純水帶走產物中的離子

23 23 Monoprotic acids HCl H + + Cl - HNO 3 H + + NO 3 - CH 3 COOH H + + CH 3 COO - Strong electrolyte, strong acid Weak electrolyte, weak acid Diprotic acids H 2 SO 4 H + + HSO 4 - HSO 4 - H + + SO 4 2- Strong electrolyte, strong acid Weak electrolyte, weak acid Triprotic acids H 3 PO 4 H + + H 2 PO 4 - H 2 PO 4 - H + + HPO 4 2- HPO 4 2- H + + PO 4 3- Weak electrolyte, weak acid Strong and Weak acid 單質子酸 雙質子酸 三質子酸

24 24 Identify each of the following species as a Brønsted acid, base, or both. (a) HI, (b) CH 3 COO -, (c) H 2 PO 4 - HI (aq) H + (aq) + I - (aq)Brønsted acid CH 3 COO - (aq) + H + (aq) CH 3 COOH (aq)Brønsted base H 2 PO 4 - (aq) H + (aq) + HPO 4 2- (aq) H 2 PO 4 - (aq) + H + (aq) H 3 PO 4 (aq) Brønsted acid Brønsted base Example:

25 25 Neutralization Reaction ( 中和反應 ) acid + base salt + water HCl(aq) + NaOH(aq) NaCl(aq) + H 2 O H + + Cl - + Na + + OH - Na + + Cl - + H 2 O H + + OH - H 2 O

26 26 Neutralization Reaction Involving a Weak Electrolyte weak acid + base salt + water HCN(aq) + NaOH(aq) NaCN(aq) + H 2 O HCN + Na + + OH - Na + + CN - + H 2 O HCN + OH - CN - + H 2 O

27 27 Neutralization Reaction Producing a Gas acid + base salt + water + CO 2 2HCl(aq) + Na 2 CO 3 (aq) 2NaCl(aq) + H 2 O +CO 2 2H + + 2Cl - + 2Na + + CO Na + + 2Cl - + H 2 O + CO 2 2H + + CO 3 2- H 2 O + CO 2

28 28 Oxidation-Reduction Reactions ( 氧化還原反應 ) (electron transfer reactions) 2Mg 2Mg e - O 2 + 4e - 2O 2- Oxidation half-reaction (lose e - ) Reduction half-reaction (gain e - ) 2Mg + O 2 + 4e - 2Mg O e - 2Mg + O 2 2MgO

29 29

30 30 Zn(s) + CuSO 4 (aq) ZnSO 4 (aq) + Cu(s) Zn is oxidizedZn Zn e - Cu 2+ is reducedCu e - Cu Zn is the reducing agent Cu 2+ is the oxidizing agent Copper wire reacts with silver nitrate to form silver metal. What is the oxidizing agent in the reaction? Cu(s) + 2AgNO 3 (aq) Cu(NO 3 ) 2 (aq) + 2Ag(s) Cu Cu e - Ag + + 1e - AgAg + is reducedAg + is the oxidizing agent Electron transfer reactions

31 Oxidation number( 氧化數 ) The charge the atom would have in a molecule (or an ionic compound) if electrons were completely transferred. 4.4 分子化合物的生成過程,沒有電子轉移,因此定義氧化數 的概念來說明氧化還原態 EX: H 2 (g)+Cl 2 (g)  2HCl(g) S(s)+ O 2 (g)  SO 2 (g) 氧化數減少,該元素即被還原 (Cl 及 O) 氧化數增加,該元素即被氧化 (H 及 S)

32 Determine the oxidation number ( 如何決定氧化數 ) 32 1.Free elements (uncombined state) have an oxidation number of zero. Na, Be, K, Pb, H 2, O 2, P 4 = 0 2.In monatomic ions, the oxidation number is equal to the charge on the ion. Li +, Li = +1; Fe 3+, Fe = +3; O 2-, O = -2 3.The oxidation number of oxygen is usually –2. In H 2 O 2 and O 2 2- it is –1. 元素態 單原子離子

33 33 4.The oxidation number of hydrogen is +1 except when it is bonded to metals in binary compounds. In these cases, its oxidation number is –1.( 如 LiH, NaH, CaH 2 ) 6. The sum of the oxidation numbers of all the atoms in a molecule or ion is equal to the charge on the molecule or ion.( 如 NH 4 + ) 5.Group IA metals are +1, IIA metals are +2 and fluorine is always –1. HCO 3 - O = – 2H = +1 3x( – 2) ? = – 1 C = +4 What are the oxidation numbers of all the elements in HCO 3 - ? 7. Oxidation numbers do not have to be integers. Oxidation number of oxygen in the superoxide ion, O 2 -, is –½.

34 34 The Oxidation Numbers of Elements in their Compounds 1. 金屬離子氧化數皆正數,非金屬離子則有正有負 2. 紅色表示常見的氧化數

35 35 NaIO 3 Na = +1O = -2 3x(-2) ? = 0 I = +5 IF 7 F = -1 7x(-1) + ? = 0 I = +7 K 2 Cr 2 O 7 O = -2K = +1 7x(-2) + 2x(+1) + 2x(?) = 0 Cr = +6 What are the oxidation numbers of all the elements in each of these compounds? NaIO 3 IF 7 K 2 Cr 2 O 7 Example:

36 36 Types of Oxidation-Reduction Reactions Combination Reaction A + B C 3Mg + N 2 MgN 2 Decomposition Reaction 2KClO 3 2KCl + 3O 2 C A + B ( 結合反應 ) ( 分解反應 ) 1Al+Br 2  2AlBr 3

37 37 Types of Oxidation-Reduction Reactions Combustion Reaction( 燃燒 ) A + O 2 B S + O 2 SO Mg + O 2 2MgO

38 38 Displacement Reaction( 取代反應 ) A + BC AC + B Sr + 2H 2 O Sr(OH) 2 + H 2 TiCl 4 + 2Mg Ti + 2MgCl 2 Cl 2 + 2KBr 2KCl + Br 2 Hydrogen Displacement Metal Displacement Halogen Displacement Types of Oxidation-Reduction Reactions 鹵素

39 39 The Activity Series for Metals M + BC MC + B Hydrogen Displacement Reaction M is metal BC is acid or H 2 O B is H 2 Ca + 2H 2 O Ca(OH) 2 + H 2 Pb + 2H 2 O Pb(OH) 2 + H 2

40 40 The Activity Series for Halogens Halogen Displacement Reaction Cl 2 + 2KBr 2KCl + Br F 2 > Cl 2 > Br 2 > I 2 I 2 + 2KBr 2KI + Br 2

41 41 Ca 2+ + CO 3 2- CaCO 3 NH 3 + H + NH 4 + Zn + 2HCl ZnCl 2 + H 2 Ca + F 2 CaF 2 Precipitation Acid-Base Redox (H 2 Displacement) Redox (Combination) Classify each of the following reactions. Examples:

42 42 Solution Stoichiometry( 溶液的化學計量 ) The concentration of a solution is the amount of solute present in a given quantity of solvent or solution. M = molarity = moles of solute liters of solution What mass of KI is required to make 500. mL of a 2.80 M KI solution? volume of KI solutionmoles KIgrams KI M KI 500. mL= 232 g KI 166 g KI 1 mol KI x 2.80 mol KI 1 L soln x 1 L 1000 mL x 莫耳濃度或體積莫耳濃度

43 43 Preparing a Solution of Known Concentration

44 44 Dilution is the procedure for preparing a less concentrated solution from a more concentrated solution. Dilution Add Solvent Moles of solute before dilution (i) Moles of solute after dilution (f) = MiViMiVi MfVfMfVf = 稀釋

45 45 How would you prepare 60.0 mL of M HNO 3 from a stock solution of 4.00 M HNO 3 ? M i V i = M f V f M i = 4.00 M M f = MV f = L V i = ? L V i = MfVfMfVf MiMi = M x L 4.00 M = L = 3.00 mL Dilute 3.00 mL of acid with water to a total volume of 60.0 mL.

46 46 Gravimetric Analysis( 重量分析 ) 1.Dissolve unknown substance in water 2.React unknown with known substance to form a precipitate 3.Filter and dry precipitate 4.Weigh precipitate 5.Use chemical formula and mass of precipitate to determine amount of unknown ion

47 47 Titrations( 滴定分析 ) In a titration a solution of accurately known concentration is added gradually added to another solution of unknown concentration until the chemical reaction between the two solutions is complete. Equivalence point – the point at which the reaction is complete Indicator – substance that changes color at (or near) the equivalence point Slowly add base to unknown acid UNTIL the indicator changes color 當量點 : 酸鹼完全中和反應的 點 指示劑

48 48 Titrations can be used in the analysis of Acid-base reactions Redox reactions H 2 SO 4 + 2NaOH 2H 2 O + Na 2 SO 4 5Fe 2+ + MnO H + Mn Fe H 2 O 酸鹼滴定 氧化還原滴定 當達到當量點, H 離子總莫耳數等於 OH 離子的總莫耳數 使用氧化還原指示劑來觀察當量點的發生

49 49 What volume of a M NaOH solution is required to titrate mL of a 4.50 M H 2 SO 4 solution? WRITE THE CHEMICAL EQUATION! volume acidmoles redmoles basevolume base H 2 SO 4 + 2NaOH 2H 2 O + Na 2 SO mol H 2 SO mL soln x 2 mol NaOH 1 mol H 2 SO 4 x 1000 ml soln mol NaOH x mL = 158 mL M acid rxn coef. M base Examples

50 50 WRITE THE CHEMICAL EQUATION! volume redmoles redmoles oxidM oxid mol KMnO 4 1 L x 5 mol Fe 2+ 1 mol KMnO 4 x L Fe 2+ x L = M M red rxn coef. V oxid 5Fe 2+ + MnO H + Mn Fe H 2 O mL of M KMnO 4 solution is needed to oxidize mL of an acidic FeSO 4 solution. What is the molarity of the iron solution? mL = L25.00 mL = L

51 51 1.Give the oxidation number of the underlined atoms in the following molecules and ions: (a) ClF, (b) IF 7, (c) CH 4, (d) C 2 H 2, (e) C 2 H 4, (f) K 2 CrO 4, (g) K 2 Cr 2 O 7, (h) KMnO 4 2.Describe how to prepare 1.00 L of M HCl solution, starting with a 2.00 M HCl solution. 3.Calculate the concentration (in molarity) of a NaOH solution if 25.0 mL of the solution are needed to neutralize 17.4 mL of a M HCl solution. HW ch4


Download ppt ""for studies of G-protein-coupled receptors" 萊夫科維茲( Robert Lefkowitz )和克比爾卡 ( Brian Kobilka )因發現 G 蛋白耦合性受體的 內部運作方式而獲獎,他們兩人研究讓身體細 胞對外界訊號有所反應的蛋白質,對 G 蛋白耦."

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