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Happy Mole Day!  Mole day is celebrated every October 23 rd between 6:02 in the morning and 6:02 in the evening.  Scientists celebrate mole day because.

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Presentation on theme: "Happy Mole Day!  Mole day is celebrated every October 23 rd between 6:02 in the morning and 6:02 in the evening.  Scientists celebrate mole day because."— Presentation transcript:

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3 Happy Mole Day!  Mole day is celebrated every October 23 rd between 6:02 in the morning and 6:02 in the evening.  Scientists celebrate mole day because the mole is the SI base unit for the amount of a substance.  The number of particles in 1 mole of any substance is 6.022 X 10 23, which is Avogadro’s number.

4 Just How Big is a Mole?  Avogadro's number of pennies placed side by side would stretch for more than 1 million light years!  This amount of pennies would also be good enough to repay the debt of United States of America 86 million times!

5  Why is Avogadro so rich? He's a multi-mole-ionaire!  What did Avogadro teach his students in math class? Moletiplication  Why did Avogadro stop going to a chiropractor on October 24th? He was only tense to the 23rd.

6 Wednesday, Oct. 23 rd : “A” Day Thursday, Oct. 24 th : “B” Day Agenda  Ch. 9 Stoichiometry Tests  Homework questions/collect  Sec.10.1 quiz: “Energy Transfer”  Section 10.2: “Using Enthalpy” Molar enthalpy changes, enthalpy changes of exothermic/endothermic reactions, enthalpy of a system, thermodynamics  Homework Sec 10.2 review, pg. 349: # 1-8 Concept Review: “Using Enthalpy” *Quiz next time over section 10.2*

7 Ch 9 Stoichiometry Tests ClassAverage Score (out of 60) Average Percentage 3A52.6887.80 % 4B52.8688.10 % In general, I was really pleased with these scores…well done!

8 Homework Questions/Problems  Pg. 344: 1-12, 14, 15

9 Ch 10: Causes of Change  What happened?  Why?

10 Quiz Section 10.1: “Energy Transfer”  You may use your notes and/or your book. (Question #10: M = molar mass) Good Luck!

11 Section 10.2: “Using Enthalpy”  Molar enthalpy change: the change in enthalpy for one mole of a pure substance.  Symbol: ΔH  Unit: J/mol (or kJ/mol)  When a pure substance is only heated or cooled, the amount of heat involved is the same as the change in enthalpy, ΔH.

12 Molar Enthalpy Change Fe (s, 300 K) Fe (s, 1,100 K) ΔH = 20.1 kJ/mol  This equation indicates that when 1 mol of solid iron is heated from 300 K to 1,100 K its molar enthalpy increases by 20,100 joules (20.1 kJ).

13 Molar Heat Capacity Governs the Changes  The iron that the blacksmith uses does not change state and is not involved in a chemical reaction.  When a pure substance is only heated or cooled, ΔH = q SO… ΔH = CΔT (Molar enthalpy change) = (molar heat capacity)(temp change)  This reaction does NOT apply to chemical reactions or changes of state.

14 Calculating Molar Enthalpy Change for Heating, Sample Problem B, pg. 346  How much does the molar enthalpy change when ice warms from -5.4˚C to -0.2˚C? ΔH = CΔT C = 37.4 J/K∙mol (heat capacity for solid water from Table 1 on pg. 343) ΔT = (-0.2) – (-5.4) = 5.2 K ΔH = (37.4 J/K∙mol) (5.2 K) ΔH = 190 J/mol (2 sig figs)

15 Calculating Molar Enthalpy Change for Heating, Practice #1, pg. 346  Calculate the molar enthalpy change of H 2 O(l) when liquid water is heated from 41.7˚C to 76.2˚C. ΔH = CΔT C =75.3 J/K∙mol (heat capacity for liquid water from Table 1 on pg. 343) ΔT = 76.2 – 41.7 = 34.5 K ΔH = (75.3 J/K∙mol) (34.5 K) ΔH = 2.60 X 10 3 J/mol (3 sig figs)

16 Calculating the Molar Enthalpy Change for Cooling, Sample Problem C, pg. 347  Calculate the molar enthalpy change when an aluminum can that has a temperature of 19.2˚C is cooled to a temperature of 4.00˚C. ΔH = CΔT C = 24.2 J/K∙mol (heat capacity for Al from Table 1 on pg. 343) ΔT = 4.00 – 19.2 = -15.2 K ΔH = (24.2 J/K∙mol) (-15.2 K) ΔH = -368 J/mol (3 sig figs)

17 Calculating the Molar Enthalpy Change for Cooling, Practice #1, pg. 347  The molar heat capacity of Al(s) is 24.2 J/K·mol. Calculate the molar enthalpy change when Al (s) is cooled from 128.5˚C to 22.6˚C. ΔH = CΔT C = 24.2 J/K∙mol ΔT = 22.6 – 128.5 = -105.9 K ΔH = (24.2 J/K∙mol) (-105.9 K) ΔH = -2,560 J/mol (3 sig figs)

18 Enthalpy Changes of Endothermic or Exothermic Processes  In example problem B, ice was warmed and the enthalpy change was positive, meaning that energy was added or absorbed.  In example problem C, an aluminum can was cooled and the enthalpy change was negative, meaning that energy was released. Positive ΔH = endothermic reaction Negative ΔH = exothermic reaction

19 Enthalpy of a System  Thermodynamics: the branch of science concerned with the energy changes that accompany chemical and physical changes.

20 In-Class/Homework  Section 10.2 Review, Pg 349: #1-8  Concept review: “Using Enthalpy” *Quiz next time over section 10.2*


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