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1. Isotopes Atoms of the same element always have the same number of protons, but they may have different numbers of neutrons. 2.

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Presentation on theme: "1. Isotopes Atoms of the same element always have the same number of protons, but they may have different numbers of neutrons. 2."— Presentation transcript:

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2 Isotopes Atoms of the same element always have the same number of protons, but they may have different numbers of neutrons. 2

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4 The average atomic mass of an element is the weighted average mass of the mixture of an element’s isotopes. 4

5 Energy from Nuclear Reactions E=mc 2 Energy produced from the Nuclear Reaction Difference in mass between products and reactants Very Large Constant (Speed of Light) 2 Because the constant is so large, tiny amounts of mass produce tons of energy Where does the energy come from ? In chemical reactions the energy comes from the bonds, In nuclear reactions the energy comes from the conversion of mass to energy 5

6 Detection of Radioactivity and the Concept of Half-life Half-life – time required for half of the original sample of radioactive nuclides to decay 6

7 Half-Life Practice Calculate how long it would take for 120g of Radioactive Carbon-14 to decay to 15.0g if the half-life of Carbon-14 is 1 week (7days). 7

8 Decay Practice 8

9 Alpha Decay Pra 9

10 Alpha Decay 10

11 Beta Decay 11

12 Beta Decay 12

13 Which has the highest energy? – Alpha, Beta, Gamma 13

14 Draw a Gamma particle. 14

15 Beta Fusion 0/-1 e + 232/91 Pa 15

16 Alpha Fission 16

17 Le Chatelier’s Principle 25kj + N2O4 (g) 2NO2 (g) (Ea) CHANGE: SHIFT PREDICTION : 1. Addition of N2O4 Right 2. Addition of NO2Left 3. Removal of N2O4 Left 4. Removal of NO2Right 5. Decrease of container volumeLeft 6. Increase of container volume Right 7. Increase of temperatureRight 8. Decrease of temperature Left 17

18 Barometer – device that measures atmospheric pressure – Invented by Evangelista Torricelli in 1643 Measuring Pressure UNITS for Pressure 1 standard atmosphere = 1.000 atm (Atmospheres) = 760.0 mm Hg (millimeters Mercury) = 760.0 torr (Names after Torricelli, Same as mm Hg) = 101,325 Pa (Pascal) 18

19 Pressure and Volume: Boyle’s Law This graph has the shape of half of a hyperbola with an equation PV = k Volume and pressure are inversely proportional. –If one increases the other decreases. 19

20 Pressure and Volume: Boyle’s Law Another way of stating Boyle’s Law is P 1 V 1 = P 2 V 2 (constant temperature and amount of gas) 20

21 These graphs are lines with an equation V = bT (where T is in kelvins) Volume and Temperature: Charles’s Law Volume and temperature are directly proportional. –If one increases the other increases. Another way of stating Charles’s Law is V 1 = V 2 T 1 T 2 (constant pressure and amount of gas) 21

22 Volume and Moles: Avogadro’s Law 22

23 Volume and Moles: Avogadro’s Law Volume and moles are directly proportional. –If one increases the other increases. –V = an –constant temperature and pressure Another way of stating Avogadro’s Law is V 1 = V 2 n 1 n 2 (constant temperature and pressure) 23

24 The left side represents the gas before The right side represents the gas after Eliminate the Variables that are not changing!!! The Super Combined Law 24

25 Boyle’s Law Charles’s Law 25

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29 P V = nRT  P V = nR T PV = nR T  P V = nR T or P V = nR T or 29

30 Dalton’s Law of Partial Pressures The pressure is independent of the nature of the particles. 30

31 The Kinetic Molecular Theory of Gases - describes motion of ideal gases 31

32 Real Gases 32

33 Practice 33

34 Temperature7a Temperature is a measure of the average kinetic energy of molecular motion in a sample. ◦ In a hot sample, the molecules are moving much faster than in a cold sample. What is temperature? 34

35 Enthalpy Enthalpy (H): The amount of heat that a system can potentially give to other systems. If ∆H rxn is positive the reaction is endothermic (it has absorbed energy) and if ∆H rxn is negative, it is exothermic (and has given off energy). The enthalpy change that accompanies the heating/cooling of a pure substance is determined by the equation: ∆H = mC p ∆T 35

36 ΔH – Heat of Reaction(ENTHALPY) Endothermic ◦ The products have more energy than the reactants. ◦ Heat must be put into the reaction, so ΔH is positive. 36

37 Exothermic ◦ The products have less energy than the reactants. ◦ Heat must be released from the reaction, so ΔH is negative. 37

38 Heat added here doesn’t change the temperature of the substance, it is causing the phase change. 38

39 Heat Flow Problems To calculate heat, use the following formula: ◦ Energy = mass * specific heat * temperature change Always given to you! 39

40 Practice Problems If 200.0 grams of water is heated from 70.0° C to 100.0° C to make a cup of tea, how much heat must be added? Q = (m)(c)( ΔT) Joules Calculations: 4.184J/gC 40

41 Energy as a Driving Force Energy as a Driving Force Entropy, S – function which keeps track of the tendency for the components of the universe to become disordered 41

42 Entropy is a thermodynamic measure of the randomness in the universe.  Phase changes that result in greater molecular freedom have a positive ∆S, and those that result in less molecular freedom have a negative ∆S. Which one favors more Randomness?  Example: For melting, ∆S = +, for freezing ∆S = -. 42

43 Energy as a Driving Force Second law of thermodynamics –The entropy of the universe is always increasing. 43

44 Entropy Practice Which has a +∆S? ◦ Water Vapor Condensing or water vaporizing  Water Vaporizing – Forming a gas increases randomness Which has a -∆S? ◦ Ice melting or water freezing  Water Freezing – Solids do not move therefore there is a decrease in randomness of the particles 44

45 Br Ø nsted-Lowry Model of Acids and Bases5b Acids ◦ substances that are hydrogen-ion donors.  HCl + H 2 O → Cl - + H + Base ◦ substances that are hydrogen-ion acceptors.  NH 3 + H 2 O  NH 4 + + OH - Note: ◦ acid/base are always added to water acid hydrogen ion basehydroxide ion According to the Bronsted-Lowry Model, what is the definition of… An Acid? A Base? 45

46 The Bronsted-Lowry concept + ClH H H O + H H HO + acidbase conjugate acidconjugate base conjugate acid-base pairs 46

47 Practice problems Identify the acid, base, conjugate acid, conjugate base, and conjugate acid-base pairs: acidbase conjugate acidconjugate base HC 2 H 3 O 2 (aq) + H 2 O (l)  C 2 H 3 O 2 – (aq) + H 3 O + (aq) conjugate acid-base pairs acidbase conjugate acidconjugate base OH – (aq) + HCO 3 – (aq)  CO 3 2– (aq) + H 2 O (l) conjugate acid-base pairs 47

48 acidbase conjugate acidconjugate base HF (aq) + SO 3 2– (aq)  F – (aq) + HSO 3 – (aq) conjugate acid-base pairs acidbase conjugate acidconjugate base CO 3 2– (aq) + HC 2 H 3 O 2 (aq)  C 2 H 3 O 2 – (aq) + HCO 3 – (aq) conjugate acid-base pairs acidbase conjugate acidconjugate base H 3 PO 4 (aq) + OCl – (aq)  H 2 PO 4 – (aq) + HOCl (aq) conjugate acid-base pairs More Practice (a) (b) (c) 48

49 The pH Scale The “p scale” is used to express small numbers. pH =  log [H + ] 49

50 The pH Scale pOH scale pOH =  log [OH  ] pH + pOH = 14.00 50

51 Practice Problem Calculate pH for a solution that has [H + ] = 1.0 x 10 -9 M Step 1) [H + ] = 1.0 x 10 -9 M Step 2) -log (1.0 x 10 -9 )= 9.00 Step 3) pH = 9.00 51

52 Practice Problem #2 Calculate pH for a solution that has [OH - ] = 1.0 x 10 -6 M. 1) Kw = [H + ] [OH - ] = 1.0 x 10 -14 2) [H + ] = 1.0 x 10 -14 / [OH - ] 3) [H + ] = 1.0 x 10 -14 / 1.0 x 10 -6 = 1.0 x 10 -8 M 4) -log 1.0 x 10 -8 = 8.00 5) pH = 8.00 52

53 Practice Problem #3 Calculate [OH-] for a solution that has a pH = 6.20 [OH-] = 1.6 x 10 -8 53

54 Each set of equilibrium concentrations is called an equilibrium position. aA + bB  cC + dD –Equilibrium expression 54

55 Large values for K signify the reaction is “product favored” When equilibrium is achieved, most reactant has been converted to product 55

56 Small values for K signify the reaction is “reactant favored” When equilibrium is achieved, very little reactant has been converted to product 56

57 For the reaction For the reaction: Where K is the equilibrium constant, and is unitless 2A + 3B  4C + 2D 57

58 –K > 1  the equilibrium position is far to the right –K < 1  the equilibrium position is far to the left The Meaning of K 58


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