1 / 18

1 st Issue of The Wall Street Journal (1889)

1 st Issue of The Wall Street Journal (1889). Reading: All of Chapter 13 HW 9: DUE 7/9/14 ( only assignment from this chapter) Chap. 9, #'s 13, 15, 21-29 odd, 37, 43-45, 47, 51, 57, 61, 62, 63, 65, 69, 71 HW 13.1: Due 7/10/14

draco
Télécharger la présentation

1 st Issue of The Wall Street Journal (1889)

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. 1st Issue of The Wall Street Journal (1889) • Reading: All of Chapter 13 • HW 9: DUE 7/9/14 (only assignment from this chapter) • Chap. 9, #'s 13, 15, 21-29 odd, 37, 43-45, 47, 51, 57, 61, 62, 63, 65, 69, 71 • HW 13.1: Due 7/10/14 • Chap. 13 #s 7, 9, 15, 17, 21, 23, 27, 31, 35, 37, 39, 43, 49, 55, 57, 59, 63 • HW 13.2: Due 7/14/14 • Chap. 13 #s 66, 67, 71, 73, 74, 78, 79, 85, 87, 89, 95, 112, 143 • Lab Tomorrow (WET LAB!) • Tomorrow – DA/Math Proficiency Quiz 2

  2. CrCl3 = Pb(NO3)2 = PbCl2 = 14.71 g of chromium (III) chloride is added to 23.41 g of lead (II) nitrate. How much of all reactants and products are present after the reaction? 2 CrCl3(aq) + 3 Pb(NO3)2(aq)  2 Cr(NO3)3(aq) + 3 PbCl2(s) Pb(NO3)2 is the L.R. (0 left) theoretical yield = 19.66g PbCl2(s)

  3. CrCl3 = Pb(NO3)2 = PbCl2 = 14.71 g of chromium (III) chloride is added to 23.41 g of lead (II) nitrate. How much of all reactants and products are present after the reaction? 2 CrCl3(aq) + 3 Pb(NO3)2(aq)  2 Cr(NO3)3(aq) + 3 PbCl2(s) Now calculate how much of the other product(s) will be formed. (Hint: start with the given amount of the limiting reactant.) Cr(NO3)3 =

  4. 14.71 g of chromium (III) chloride is added to 23.41 g of lead (II) nitrate. How much of all reactants and products are present after the reaction? 14.71 g of chromium (III) chloride is added to 23.41 g of lead (II) nitrate. How much of all reactants and products are present after the reaction? 2 CrCl3(aq) + 3 Pb(NO3)2(aq)  2 Cr(NO3)3(aq) + 3 PbCl2(s) reactants products CrCl3 = 7.24 g Cr(NO3)3 = 11.22 g 11.22 g PbCl2 = Pb(NO3)2 = 0 g (L.R.) 19.66 g 19.66 g Mass must be the same before and after the reaction!! mass before reaction  14.71 + 23.41 = 38.12 g mass after reaction  11.22 + 19.66 = 30.88 g difference will be the mass of excess reactant left over = 7.24 g

  5. 31.89 g of potassium sulfate is reacted with 25.00 g of lead (II) acetate. What are the masses of all reactants and products after the reaction? K2SO4(aq) + Pb(C2H3O2)2(aq)  2 KC2H3O2(aq) + PbSO4(s) 174.27 g/mol 325.3 g/mol 98.15 g/mol 303.3 g/mol limiting reactant theoretical yield of KC2H3O2 Pb(C2H3O2)2 is the L.R. (0 left) theoretical yield = 15.09g KC2H3O2

  6. 31.89 g of potassium sulfate is reacted with 25.00 g of lead (II) acetate. What are the masses of all reactants and products after the reaction? K2SO4(aq) + Pb(C2H3O2)2(aq)  2 KC2H3O2(aq) + PbSO4(s) 174.27 g/mol 325.3 g/mol 98.15 g/mol 303.3 g/mol Now calculate how much of the other product(s) will be formed. (Hint: start with the given amount of the limiting reactant.)

  7. 31.89 g of potassium sulfate is reacted with 25.00 g of lead (II) acetate. What are the masses of all reactants and products after the reaction? 31.89 g of potassium sulfate is reacted with 25.00 g of lead (II) acetate. What are the masses of all reactants and products after the reaction? K2SO4(aq) + Pb(C2H3O2)2(aq)  2 KC2H3O2(aq) + PbSO4(s) reactants products K2SO4 = 18.49 g KC2H3O3 = 15.09 g 15.09 g PbSO4= Pb(C2H3O2)2 = 0 g (L.R.) 23.31 g 23.31 g Mass must be the same before and after the reaction!! mass before reaction  31.89 + 25.00 = 56.89 g mass after reaction  15.09 + 23.31 = 38.40 g difference will be the mass of excess reactant left over = 18.49 g

  8. Pressure Pressure is: “Amount of force per unit area”

  9. Pressure Force = mass x acceleration how much how fast F = M x A F = M x A

  10. Pressure Pressure depends on two things: 1) Energy of collisions HEAT!!!

  11. LOWER pressure HIGHER pressure LOWER temperature HIGHER temperature

  12. Pressure Pressure depends on two things: 1) Energy of collisions HEAT!!! 2) Frequency of collisions Effected by how fast the particles are moving

  13. LOWER pressure HIGHER pressure LOWER temperature HIGHER temperature

  14. Pressure Pressure depends on two things: 1) Energy of collisions HEAT!!! 2) Frequency of collisions Effected by how fast the particles are moving Effected by how many particles there are

  15. LOWER pressure HIGHER pressure FEWER particles MORE particles

  16. way HIGHER pressure way lower pressure MORE particles and energy fewer particles and less energy

  17. Pressure Implications: Atmospheric pressure Vacuum How a straw works

  18. Pressure Units of standard pressure: atmosphere: 1 atm (exactly) millimeters of mercury: 760 mmHg (exactly) torr: 760 torr (exactly) kilopascal: 101 kPa (NOT exactly)

More Related