slide1
Download
Skip this Video
Download Presentation
Chemistry Jeopardy Unit 6

Loading in 2 Seconds...

play fullscreen
1 / 26

Chemistry Jeopardy Unit 6 - PowerPoint PPT Presentation


  • 80 Views
  • Uploaded on

Chemistry Jeopardy Unit 6. The ingredient of a solution present in the greatest amount. Ions B. Liquid C. Solute D. Solvent. This factor has the largest effect of the solubility of solid compounds. Pressure Temperature Rate of Stirring Amount of solvent.

loader
I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
capcha
Download Presentation

PowerPoint Slideshow about 'Chemistry Jeopardy Unit 6' - meara


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.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript
slide3
This factor has the largest effect of the solubility of solid compounds.
  • Pressure
  • Temperature
  • Rate of Stirring
  • Amount of solvent
slide4
A solution that contains the maximum amount of solute possible.
  • Unsaturated
  • Saturated
  • Supersaturated
  • Megasaturated
slide5
To maximize the solubility of gases in solutions a student should.
  • Increase the temperature
  • Decrease the temperature
  • Decrease the pressure
  • Increase the amount of solvent
slide6
What saying allows us to predict if a solvent will dissolve a solute?
  • Likes dissolve likes
  • Polar dissolves nonpolar
  • Nonpolar dissolves polar
  • All solvents dissolve all solutes
slide7
Molarity has units of
  • grams/mole
  • moles/kilogram
  • liters/mole
  • moles/liter
slide8
What is the concentration of a solution made by dissolving 1.5 mol of NaCl to a volume of 500. mL
  • 1.5 M
  • 0.0030 M
  • 3.0 M
  • 2.0 M
slide9
The product of concentration times volume.
  • Moles
  • Molarity
  • Keq
  • Dilution
slide10
Proper solution preparation requires what piece of equipment.
  • Beaker
  • Erlenmeyer Flask
  • Graduated Cylinder
  • Volumetric Flask
slide11
What is the resulting concentration when 20. mL of a 2.0M solution is diluted to 40. mL?
  • 1.0 M
  • 0.50 M
  • 2.0 M
  • 4.0 M
slide12
Colligative Properties depend on the number of
  • Chemists Working
  • Particles in solution
  • Beakers used
  • Readings taken
slide13
A liquid reaches its boiling point when
  • It feels like it
  • When bubbles start form
  • The vapor pressure of the liquid equals the pressure of the atmosphere
  • Cannot be determined
slide14
Adding a solute to a solvent
  • Raises the freezing point
  • Lowers the boiling point
  • Lowers the freezing point
  • Does not change the boiling point
slide15
What is the advantage of using AlCl3 to melt ice on sidewalks as opposed to NaCl
  • A. The freezing point change will be 2x larger
  • The freezing point change will be 4x larger
  • The freezing point change will be 10x larger
  • There is no advantage
slide16
When ΔTf = 3.4 ºC for an aqueous solution of NaCl the new freezing point of the solution is
  • 3.4 ºC
  • -3.4 ºC
  • 103.4 ºC
  • 96.6 ºC
slide17
When a reaction reaches equilibrium the forward and reverse reactions
  • Occur at different rates
  • Occur at the same rate
  • Stop
  • Make the same amount of chemical
slide18
The only stress that can change the equilibrium constant is
  • Concentration
  • Temperature
  • Pressure
  • Final Exams
slide19
A large equilibrium constant mean the reaction is
  • Product favored
  • Reactant favored
  • Neutral
  • Happens very fast
slide20
Ksp expressions are written for
  • Chemical equilibrium
  • Force equilibrium
  • Solubility equilibrium
  • Balance Equilibrium
slide21
When a common ion is added to an equilibrium the equilibrium shifts
  • Toward the added ion
  • Away from the added ion
  • Toward the heat in the reaction
  • Toward the higher pressure
slide22
What is the equilibrium expression for the following equation?
  • 3A(g) + 2B(g) A3B2(g)
  • Keq = [A3B2] / ([A]3[B]2)
  • Keq = [A]3 [B]2 / [A3B2]
  • Keq = [A] [B] / [A3B2]
  • Keq = [A3] [B2] / [A3B2]
slide23
How can the following reaction be
  • made to favor the product side
  • of the reaction?
  • 3A(g) + 2B(g)  A3B2(g) + heat
  • Add more product
  • Increase the pressure
  • Increase the temperature
  • Add a catalyst
slide24
What is the Ksp expression for the following reaction
  • BaSO4(s)  Ba+2 + SO4-2
  • Ksp = [BaSO4] / ([Ba+2] [SO4-2])
  • Ksp = [Ba+2] [SO4-2] / [BaSO4]
  • Ksp = [BaSO4] / [Ba]2 [SO4] -2
  • Ksp = [Ba+2] [SO4-2]
slide25
Which of the following will have no effect on the reaction?
  • A (g) + B (g) AB (g) + heat
  • Increasing the amount of A
  • Decreasing the temperature
  • Increasing the amount of A2B2
  • Increasing the pressure
slide26
How could you lower the concentration of Ba+2 for the following reaction?
  • BaSO4(s) Ba+2 + SO4-2
  • Add solid Na2SO4
  • Increase the pressure
  • Add solid Ba(NO3)2
  • The concentration cannot be changed
ad