1 Test 2 Study Guide Thursday February 28 8:00 p.m. Dobo 134.
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Transcript of 1 Test 2 Study Guide Thursday February 28 8:00 p.m. Dobo 134.
1
Test 2 Study Guide
Thursday February 28
8:00 p.m. Dobo 134
2
Top 7 Topics
• Limiting reactants
• Ionic compounds in water
• Molarity
• Energy
• Specific heat
• Enthalpy
• Hess’ Law
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Limiting Reactants - 3
• Theoretical Yield
• How much is used, formed, left over
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Ionic Compounds 5
• In water
• Strong / weak electrolytes
• Ionic equations
• Metathesis reactions
• Ion Solubility
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Molarity 4
• Interconverting molarity, moles, volume
• Dilution
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Energy 2
E = q + w
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Specific heat 1
• Calculate energy given specific heat
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Enthalpy 3
• Endo , exo, sign of H
• Given heat for a certain mass, calculate reaction H
• Given H and mass, calculate heat
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Hess’ Law 2
• Addition of reactions to obtain H
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Top 7 TopicsLet’s do some examples!
• Limiting reactants
• Ionic compounds in water
• Molarity
• Energy
• Specific heat
• Enthalpy
• Hess’ Law
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Limiting Reactants - 3
• Theoretical Yield – how much you expect based upon the amount of reactants.
• If one reactant is is present in excess, then the other limits how much can be made and is used to determine the theoretical yield.
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2H2 + O2 2 H2O• If 10 moles of oxygen and 10 moles of hydrogen….• The hydrogen limits the amount of water that can be
formed
• Based upon 10 moles H2, we can make 10 moles water
• Based upon 10 moles O2, we can make 20 moles water.
• H2 limits. Theoretical Yield = 10 moles ( 180 g) H2O
• At end of reaction, will have 5 moles O2 left over
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• How much is used, formed, left over
• Q 1 on Test
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Ionic Compounds 5
• In water
• Strong / weak electrolytes
• Ionic equations
• Metathesis reactions
• Ion Solubility
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• Group 1A ( alkali metals) cations soluble
• NH4+ soluble
• NO3- soluble
• CH3CO2- = C2H3O2
- = acetate soluble
• Will be given a chart for other ions
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• Soluble ionic compounds in water exist as free ions surrounded by water molecules.
• Soluble ionic compounds are strong electrolytes
• Strong acids or bases (HCl, H2SO4), NaOH) are strong electrolytes
• Weak acids or bases (CH3CO2H, acetic acid) are weak electrolytes
• Molecules that do not disassociate (CO2, sugar) are non electrolytes.
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potassium sulfate with barium nitrate
• 2K+(aq) + SO4-2 (aq) + Ba+2 (aq) + 2 NO3
- (aq)
2K+(aq) + 2 NO3- (aq) +BaSO4(s)
• SO4-2 (aq) + Ba+2 (aq) BaSO4(s)
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Molarity 4
• Interconverting molarity, moles, volume
• M means moles per liter = moles/L
• M = moles/L
• If you know two things, can determine the third
• If you have moles and volume, can determine molarity
• 3 moles dissolved in 0.5 L = 3moles/0.5L = 6M
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Molarity - dilution
• (Vconc )(Mconc ) = (Vdil )(Mdil )
• If you know 3, can solve for the fourth• How many mL of 3M HCl is needed to make
100mL of 1.5 M HCl?
• (Vconc)(3M) = (100mL)(1.5M)
• Vconc = 50 mL
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E = q + w• q > 0 heat transferred from the surroundings to
the system (endothermic)• q < 0 heat transferred from the system to the
surroundings ( exothermic)• w > 0 work is done by the surroundings on the
system• w < 0 work is done by the system on the
surroundings• q > 0, w > 0 E > 0• q < 0, w < 0 E< 0
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Specific heat
• Calculate heat energy given specific heat
• q = (specific heat) ( mass in grams)(T)
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• 18. When 72 g of a metal at 97.0 C is added to 100.0 g of water at 25.0 C, the final temperature is 29.1 C. What is the heat capacity of the metal if cwater = 4.184
J/g.K?
• Heat lost by metal = heat gained by water
• qmetal = - qwater
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Enthalpy
• Endothermic H > 0
• Exothermic H < 0
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Given heat for a certain mass, calculate reaction H
• If it takes 60 kJ to melt 180 grams of ice, what is H for the following reaction?
• H2O(s) H2O(l)
• (60kJ / 180g) ( 18 g/mole) = 6 kJ/moleH = 6kJ
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Given H and mass, calculate heat
H2O(s) H2O(l) H = 6 kJ
How much heat is needed to melt 900 grams of ice?
(900g)(6 kJ/mole)(1mole/18 g) = 300 kJ
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Hess’ Law
• If a reaction is carried out in a series of steps, H for the reaction will be equal to the sum of the enthalpy changes for the individual steps.
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• 19. What is the value of H for
IF5(g) IF3(g) + F2(g) H = ?
given the following thermochemical equations?
IF(g) + F2(g) IF3(g) H = -390 kJ
IF(g) + 2 F2(g) IF5(g) H = -745 kJ