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Transcript of 200 300 500 400 ThermochemSolutions and suchAcids and BasesEquilibrium 100 200 100 500 400 300 200...
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Thermochem
Solutions and such
Acids and Bases
Equilibrium
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Thermochem 100
The total energy of the world is constant. Energy can neither be created
nor destroyed.
What is the law of conservation of energy?
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Thermochem 200
What is the equation for a change in temperature?
ΔT = Tf - Ti
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Thermochem 300
In the equation for calculation of heat transfer: q= mcΔT what does each letter
represent?Q = heat m= mass c = specific heat capacity ΔT = change in
temperature
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Thermochem 400
What is the equivalence of 0ºC in Kelvins?
273.15K
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Thermochem 500
How do you spell …
LOLCalculate the specific heat capacity of canola
oil. If the mass of the oil is 60g, the initial temperature is 35ºC and the final temperature
is 5.2ºC with a heat change of 4.0x103?
C= q/m∆T = -4.0x103J
(60.0g)(5.2ºC – 35.0ºC) = 2.24 J/gºC (pg 635)
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Solutions and Such 100
Is ethanol (CH3CH20H) soluble in water?
Yes, infinitely soluble!
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Solutions and Such 200
True or False, Polar compounds can dissolve in non-polar solvents. Explain.
False, polar compounds can only dissolve polar solvents and non-polar compounds can only dissolve in non-polar solvents!!
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Solutions and Such 300
Define Miscibility and give an example of a misicble and immisicble solution.
Miscible – compounds that dissolve readily in each other in any proportion. (example: water
and alcohol)
Immiscible – liquids that do not readily dissolve in each other (example oil and water)
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Solutions and Such 400
What is a dipole – dipole attraction?
The attraction between the opposite charge on two different
polar molecules is called a dipole – dipole attraction.
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Solutions and Such 500
Calculate the volume of isoproply alcohol used to make a 500mL solution.
The volume/volume % is 70%.
Volume/volume percent = volume of solute x 100% volume of solution
Volume of solute = 70% x 500mL100%
= 350mL (pg 262)
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Equilibrium 100
What is Le Châtelier’s Principle?
If a stress is applied to a system at equilibrium, the equilibrium will shift to
relieve the stress.
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Equilibrium 200
Given the balanced formula:
2SO2(g) + O2 2SO3(g) + heat Increase SO2 concentration
Increase temperatureWhich way will the reaction shift?
Increase SO2 concentration = shift right
Increase temperature = shift left
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Equilibrium 300
What is a catalyst, and how does it affect a reaction?
A catalyst is a substance that increases the rate of a chemical reaction without being consumed by the reaction. The catalyst is always left over.
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Equilibrium 400
For the Reaction: CO(g)+NO2(g) CO2(g)+NO(g)
Given the ΔE= -226kJ and the Ea(fwd)= 134kJ
What is the Ea(rev)?
ΔE=Ea(rev)-Ea(fwd)
Ea(rev)=ΔE+Ea(fwd)
Ea(rev)=360Kj
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Equilibrium 500
Step one: NO2(g) + NO2(g) NO3(g)+NO(g)
Step two: NO3(g)+CO(g) NO2(g)+CO2(g)
NO2(g)+NO2(g)+NO3(g)+CO(g) NO3(g)+NO(g)
+NO2(g)+CO2(g)
The Balanced Equation is: NO2(g)+CO(g) NO(g)+CO2(g)
The reaction Intermediate is NO3.
Given the following two step reaction:
Step one: NO2(g) + NO2(g) NO3(g)+NO(g)
Step two: NO3(g)+CO(g) NO2(g)+CO2(g)
What is the overall balanced equation for the reaction and what is the reaction intermediate.
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Acids and Bases 100
What are the properties of an acid?
•pH between -1 and 7
•Sour taste
•Strong acids dissociate completely
•Conduct electricity
•Turns blue litmus paper red
•Have no characteristic feel
•Neutralize basic solutions, forming salt and water
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Acids and Bases 200
Explain the Arrhenius, and the Brǿnsted Lowry Theories.
Arrhenius theory states: An acid is a substance that dissociates in water to produce one or more hydrogen ions, H+. A Base is a substance that dissociates in water to produce one or more Hydroxide ions, OH ֿ.
Brǿnsted Lowry theory states: An acid is a substance from which one proton (H+ ion) can be removed. A base is a substance that can
remove a proton (H+ ion) from an acid.
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Acids and Bases 300
Identify the conjugate acid-base pairs of the following reaction:
HBr(g)+H2O(l) H3O+(aq)+Brˉ
HBr(g)+H2O(l) H3O+(aq)
+Brˉ
Conjugate acid-base pair
Conjugate acid-base pair
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Acids and Bases 400
Calculate the pH of a solution with [H3O+]=3.8x10-3.
pH= -log(H3O+)
pH= -log(3.8x10-3)
pH= 2.42
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Acids and Bases 500Given:
CH3CH2COOH(aq)+H2O(l) CH3CH2COOֿ(aq)+H3O(aq)
The initial [CH3CH2COOH]= 0.10 mol/L and pH= 2.96
Find Ka.
Concentration (mol/L) CH3CH2COOH(aq)+ H2O(l) CH3CH2COOֿ(aq)+ H3O(aq)
Initial 0.10 0 0
Change -x +x +x
Equilibrium 0.10 – x +x +x
Ka= (x)(x) (0.10-x)
[H3O+]= 10-2.96
[H3O+]= 1.1 x 10-
3
Ka= (1.1x10-3)(1.1x10-3) (0.10-1.1x10-3)
Ka= 1.2x10-5