Wednesday, December 29, 2010
Wednesday, December 22, 2010
Sunday, December 19, 2010
sample exam ep 6 old questions
Match each item with the correct statement below.
a. acid dissociation constant h. Lewis acid
b. diprotic acid i. pH
c salt hydrolysis j. equivalence point
d. end point k. buffer capacity
e. oxidation number l. oxidizing agent
f. half-reaction m reducing agent
g. hydrogen-ion donor n titration
____ 1. can accept an electron pair
____ 2. acid with two ionizable protons
____ 3. Brønsted-Lowry acid
____ 4. negative logarithm of the hydrogen ion concentration
____ 5. ratio of the concentration of the dissociated to the undissociated form
____ 6. process of adding a known amount of solution of known concentration to determine the concentration of another solution
____ 7. The number of moles of hydrogen ions equals the number of moles of hydroxide ions.
____ 8. Indicator changes color.
____ 9. Cations or anions of a dissociated salt remove hydrogen ions from or donate hydrogen ions to water.
____ 10. the amount of acid or base that can be added to a buffer solution before a significant change in pH occurs
____ 11. substance that accepts electrons
____ 12. substance that donates electrons
____ 13. integer related to the number of electrons under an atom's control
____ 14. reaction showing either the reduction or the oxidation reaction
Match each item with the correct statement below.
a. substituent j asymmetric carbon
b. structural isomers k. trans configuration
c. geometric isomers l. cis configuration
d. aromatic compound m. lignite
e. aliphatic hydrocarbon n bituminous coal
f. condensed structural formula o. saturated compound
g. homologous series p. complete structural formula
h. unsaturated compound q Stereoisomers
i. anthracite coal
____ 15. atom or group of atoms that can take the place of a hydrogen in a parent hydrocarbon molecule
____ 16. compounds that have the same molecular formula, but the atoms are joined in a different order
____ 17. arrangement in which substituted groups are on the same side of a double bond
____ 18. molecules in which atoms are joined in the same order but differ in the arrangements of their atoms in space
____ 19. arrangement in which substituted groups are on opposite sides of a double bond
____ 20. compounds that differ in the orientation of groups around a double bond
____ 21. carbon atom to which four different atoms or groups are attached
____ 22. group of compounds in which there is a constant increment of change in molecular structure from one compound in the series to the next
____ 23. formula showing all the atoms and bonds in a molecule
____ 24. structural formula in which some bonds and/or atoms are left out
____ 25. organic compound that contains the maximum number of hydrogens per carbon atom
____ 26. organic compound that contains at least one double or triple carbon-carbon bond
____ 27. any straight-chain or branched-chain alkane, alkene, or alkyne
____ 28. any hydrocarbon compound in which a ring has bonding similar to benzene
____ 29. hard coal, having a carbon content of over 80%
____ 30. brown coal, having a carbon content of approximately 50%
____ 31. soft coal, having a carbon content of 70–80%
a. acid dissociation constant h. Lewis acid
b. diprotic acid i. pH
c salt hydrolysis j. equivalence point
d. end point k. buffer capacity
e. oxidation number l. oxidizing agent
f. half-reaction m reducing agent
g. hydrogen-ion donor n titration
____ 1. can accept an electron pair
____ 2. acid with two ionizable protons
____ 3. Brønsted-Lowry acid
____ 4. negative logarithm of the hydrogen ion concentration
____ 5. ratio of the concentration of the dissociated to the undissociated form
____ 6. process of adding a known amount of solution of known concentration to determine the concentration of another solution
____ 7. The number of moles of hydrogen ions equals the number of moles of hydroxide ions.
____ 8. Indicator changes color.
____ 9. Cations or anions of a dissociated salt remove hydrogen ions from or donate hydrogen ions to water.
____ 10. the amount of acid or base that can be added to a buffer solution before a significant change in pH occurs
____ 11. substance that accepts electrons
____ 12. substance that donates electrons
____ 13. integer related to the number of electrons under an atom's control
____ 14. reaction showing either the reduction or the oxidation reaction
Match each item with the correct statement below.
a. substituent j asymmetric carbon
b. structural isomers k. trans configuration
c. geometric isomers l. cis configuration
d. aromatic compound m. lignite
e. aliphatic hydrocarbon n bituminous coal
f. condensed structural formula o. saturated compound
g. homologous series p. complete structural formula
h. unsaturated compound q Stereoisomers
i. anthracite coal
____ 15. atom or group of atoms that can take the place of a hydrogen in a parent hydrocarbon molecule
____ 16. compounds that have the same molecular formula, but the atoms are joined in a different order
____ 17. arrangement in which substituted groups are on the same side of a double bond
____ 18. molecules in which atoms are joined in the same order but differ in the arrangements of their atoms in space
____ 19. arrangement in which substituted groups are on opposite sides of a double bond
____ 20. compounds that differ in the orientation of groups around a double bond
____ 21. carbon atom to which four different atoms or groups are attached
____ 22. group of compounds in which there is a constant increment of change in molecular structure from one compound in the series to the next
____ 23. formula showing all the atoms and bonds in a molecule
____ 24. structural formula in which some bonds and/or atoms are left out
____ 25. organic compound that contains the maximum number of hydrogens per carbon atom
____ 26. organic compound that contains at least one double or triple carbon-carbon bond
____ 27. any straight-chain or branched-chain alkane, alkene, or alkyne
____ 28. any hydrocarbon compound in which a ring has bonding similar to benzene
____ 29. hard coal, having a carbon content of over 80%
____ 30. brown coal, having a carbon content of approximately 50%
____ 31. soft coal, having a carbon content of 70–80%
Friday, December 17, 2010
EP 6 exam questions
1. what is the formula for sulfuric acid, nitrous acid, and phosphorous acid?
2. what is the difference between chemical and physical properties?
3. Compare an ester, ketone, and aldehyde structures.
4. using the scientific method, outline how the gas collection experiment combining ammonium sulfate and sodium nitrate works or doesn't work? what are sources of error?
5. You have an unknown gas that burns blue in a flame test what is it?
6. in the neutralization of HCl and NaOH, what is your product?
2. what is the difference between chemical and physical properties?
3. Compare an ester, ketone, and aldehyde structures.
4. using the scientific method, outline how the gas collection experiment combining ammonium sulfate and sodium nitrate works or doesn't work? what are sources of error?
5. You have an unknown gas that burns blue in a flame test what is it?
6. in the neutralization of HCl and NaOH, what is your product?
Thursday, December 16, 2010
EP 6 Exams topics
Acid naming on exam
Chemical/physical properties
Dilutions
Vocab ch 19-23
Hess’s law
Labs—Mg+HCl
Zn+H2SO4
Ammonium sulfate + sodium nitrate=
NaOH+ HCl=
Know your neutralization reactions
Acid + base= water + salt
Flame test
Hydrogen
Carbon dioxide
Oxygen
Methane
Flame test
Calcium
Sodium
copper
Chemical/physical properties
Dilutions
Vocab ch 19-23
Hess’s law
Labs—Mg+HCl
Zn+H2SO4
Ammonium sulfate + sodium nitrate=
NaOH+ HCl=
Know your neutralization reactions
Acid + base= water + salt
Flame test
Hydrogen
Carbon dioxide
Oxygen
Methane
Flame test
Calcium
Sodium
copper
EP 5 Review Questions for exams
Ep 5 review exam
Vocab
Ch 16, 17
Page 498 p. 534
Solubility page 477 # 3-7
Molarity page 481 # 8, 9
Find the moles of solute in solution page 482 # 10-11
Dilutions page 484 # 12-13
Page 486 #16-17, 19-21
Molality page 492 # 29-30
thermochemistry
Page 510 # 3-11
Page 513 Enthalpy #12-13
Page 517 # 16-17, 20
Page 521 #21, 22
Page 523 Heating curve
Page 524 heat of vaporization #23-24
Page 526 # 27-31
Page 535 #38-55
labs
Dilution m1v1=m2v2
Hess’s law
Flame test—hydrogen pops, carbon dioxide puts the fllame out, oxygen, reignites the flame, methane burns blue
Lab equipment
Gas collecting apparatus
Vocab
Ch 16, 17
Page 498 p. 534
Solubility page 477 # 3-7
Molarity page 481 # 8, 9
Find the moles of solute in solution page 482 # 10-11
Dilutions page 484 # 12-13
Page 486 #16-17, 19-21
Molality page 492 # 29-30
thermochemistry
Page 510 # 3-11
Page 513 Enthalpy #12-13
Page 517 # 16-17, 20
Page 521 #21, 22
Page 523 Heating curve
Page 524 heat of vaporization #23-24
Page 526 # 27-31
Page 535 #38-55
labs
Dilution m1v1=m2v2
Hess’s law
Flame test—hydrogen pops, carbon dioxide puts the fllame out, oxygen, reignites the flame, methane burns blue
Lab equipment
Gas collecting apparatus
Tuesday, December 14, 2010
Ep 4 exam things
All vocab ch 9, 10
All lab equipment
Video will be posted to acchemistry.blogspot site
Page 291—number of atoms to moles
Page 292 5, 6 moles to atoms
Page 296 molar mass of a compound 5, 6 -15
Page 298 moles to mass 16, 17
Page 299 mass to moles 18, 19
Page 301 volume of gas stp 20-21
Page 302 molar mass of a gas 22-23
Page 303 24-31
Page 306 percent composition 32-33
Page 307 34-35 percent composition from forula
Page 310 36-37 empirical formula
312 molecular formula 38-46
Page 315-317 more practice
All lab equipment
Video will be posted to acchemistry.blogspot site
Page 291—number of atoms to moles
Page 292 5, 6 moles to atoms
Page 296 molar mass of a compound 5, 6 -15
Page 298 moles to mass 16, 17
Page 299 mass to moles 18, 19
Page 301 volume of gas stp 20-21
Page 302 molar mass of a gas 22-23
Page 303 24-31
Page 306 percent composition 32-33
Page 307 34-35 percent composition from forula
Page 310 36-37 empirical formula
312 molecular formula 38-46
Page 315-317 more practice
Sunday, December 12, 2010
Friday, December 10, 2010
Tuesday, November 30, 2010
Thursday, November 25, 2010
Monday, November 22, 2010
Assignments Nov 22, 2010
EP 4--empirical formulae and molecular formulae calculations
EP 5--gas detection and flame tests along with heat of formation
ep 6--flame tests, liquid tests for cations, and mixing HCl and Zinc
EP 5--gas detection and flame tests along with heat of formation
ep 6--flame tests, liquid tests for cations, and mixing HCl and Zinc
Friday, November 12, 2010
experiments this week
ep 4 anions lab day 1 and cations lab day 2
ep 5 thermochemistry
ep 6 identifying unknowns-gases
:D have fun boys
ep 5 thermochemistry
ep 6 identifying unknowns-gases
:D have fun boys
Monday, November 8, 2010
textbooks and notebook checks this week
i'm checking to see you brought your textbook and did your notes and hw this week
Debate Results
AC 2 Teechayut, Tanachai, and Kasidey finished 5th overall at the 5th Thailand High School Debate Championships!
Thursday, November 4, 2010
Wednesday, November 3, 2010
Thermochemistry Lab: EP 5 EP 6
Thermochemistry and Hess's Law
In this experiment students determined the enthalpy change that occurs when sodium hydroxide and hydrochloric acid solutions are mixed. Next, the enthalpy change for the reaction between sodium hydroxide and ammonium chloride was measured. Lastly, they determined the enthalpy change for the reaction between ammonia and hydrochloric acid. An algebraic combination of the first two equations can lead to the third equation. Therefore, according to Hess's law, an algebraic combination of the enthalpy changes of the first two should lead to the enthalpy of the third reaction.
The molecular equations for the reactions are as follows:
(1) NaOH(aq) + HCl(aq) --> NaCl(aq) + H2O(l)
(2) NH4CI(aq) + NaOH(aq) --> 4NH3 (aq) + NaCl(aq) + H2O(l)
(3) NH3 (aq) + HCl(aq) --> NH4CI(aq)
There is no single instrument that can directly measure heat in the way a balance measures mass or a thermometer measures temperature. However, it is possible to determine the heat change when a chemical reaction occurs. The change in heat is calculated from the mass, temperature change, and specific heat of the substance which gains or loses heat.
The equation that is used to calculate heat gain or loss is:
q = (grams of substance) x (specific heat) x D T
where q = the heat energy gained or lost and DT is the change in temperature. Since DT = (final temperature minus initial temperature), an increase in temperature will result in a positive value for both DT and q, and a loss of heat will give a negative value. A positive value for q means a heat gain, while a negative value means a heat loss.
Acid-base neutralization is an exothermic process. Combining solutions containing an acid and a base results in a rise of solution temperature. The heat given off by the reaction (which will cause the solution temperature to rise) can be calculated from the specific heat of the solution, the mass of solution and the temperature change. This heat quantity can then be converted to the enthalpy change for the reaction in terms of kJ/mole by using the concentrations of the reactants.
According to Hess, if a reaction can be carried out in a series of steps, the sum of the enthalpies for each step should equal the enthalpy change for the total reaction. Another way of stating "Hess's Law" is: If two chemical equations can algebraically be combined to give a third equation, the values of DH for the two equations can be combined in the same manner to give DH for the third equation. An examination of the acid-base equations above shows that if equation (2) is subtracted from equation (1), equation (3) will result. Therefore, if the value of DH for equation (2) is subtracted from that of equation (1), the enthalpy change for equation (3) should result.
In this experiment students determined the enthalpy change that occurs when sodium hydroxide and hydrochloric acid solutions are mixed. Next, the enthalpy change for the reaction between sodium hydroxide and ammonium chloride was measured. Lastly, they determined the enthalpy change for the reaction between ammonia and hydrochloric acid. An algebraic combination of the first two equations can lead to the third equation. Therefore, according to Hess's law, an algebraic combination of the enthalpy changes of the first two should lead to the enthalpy of the third reaction.
The molecular equations for the reactions are as follows:
(1) NaOH(aq) + HCl(aq) --> NaCl(aq) + H2O(l)
(2) NH4CI(aq) + NaOH(aq) --> 4NH3 (aq) + NaCl(aq) + H2O(l)
(3) NH3 (aq) + HCl(aq) --> NH4CI(aq)
There is no single instrument that can directly measure heat in the way a balance measures mass or a thermometer measures temperature. However, it is possible to determine the heat change when a chemical reaction occurs. The change in heat is calculated from the mass, temperature change, and specific heat of the substance which gains or loses heat.
The equation that is used to calculate heat gain or loss is:
q = (grams of substance) x (specific heat) x D T
where q = the heat energy gained or lost and DT is the change in temperature. Since DT = (final temperature minus initial temperature), an increase in temperature will result in a positive value for both DT and q, and a loss of heat will give a negative value. A positive value for q means a heat gain, while a negative value means a heat loss.
Acid-base neutralization is an exothermic process. Combining solutions containing an acid and a base results in a rise of solution temperature. The heat given off by the reaction (which will cause the solution temperature to rise) can be calculated from the specific heat of the solution, the mass of solution and the temperature change. This heat quantity can then be converted to the enthalpy change for the reaction in terms of kJ/mole by using the concentrations of the reactants.
According to Hess, if a reaction can be carried out in a series of steps, the sum of the enthalpies for each step should equal the enthalpy change for the total reaction. Another way of stating "Hess's Law" is: If two chemical equations can algebraically be combined to give a third equation, the values of DH for the two equations can be combined in the same manner to give DH for the third equation. An examination of the acid-base equations above shows that if equation (2) is subtracted from equation (1), equation (3) will result. Therefore, if the value of DH for equation (2) is subtracted from that of equation (1), the enthalpy change for equation (3) should result.
Sunday, October 31, 2010
Debate practice This week
GSPR MOnday
Duel to death Tuesday
Wed?
Friday?
on Sat, Vichitpol will be the final checker for cases
Duel to death Tuesday
Wed?
Friday?
on Sat, Vichitpol will be the final checker for cases
Thursday, October 28, 2010
debate Practice
next monday, tuesday, wednesday, and friday.
the plan is to leave school at 7:30, i want to go through all the cases as we go there. we wont debate until 3
the plan is to leave school at 7:30, i want to go through all the cases as we go there. we wont debate until 3
Class Schedule of events
all classes you have a textbook check and quiz on the vocabulary
ep 4--quiz ch 10
ep 5 quiz ch 17.1 and 17.2
ep 6 ch 23 and basic naming of ketones and aldehydes
welcome back to school boys
ep 4--quiz ch 10
ep 5 quiz ch 17.1 and 17.2
ep 6 ch 23 and basic naming of ketones and aldehydes
welcome back to school boys
Wednesday, October 13, 2010
Friday, October 8, 2010
Wednesday, October 6, 2010
Ep 4/2 Failures and High Score
High Score; Chanawut Pornsuksawang and Tossapon Rattanajangwang 288/290
failures: 51124
page 281 #42-45, 47-50, 58-60,
page 247 #39-41, 42-45, and 57-60
page 207 #30-33, 41-43, and 48-50
failures: 51124
page 281 #42-45, 47-50, 58-60,
page 247 #39-41, 42-45, and 57-60
page 207 #30-33, 41-43, and 48-50
Ep 4/1 Exam Grades
High Score: pichet Neerapitak and Morris Horng 290/290
Failures...
NONE!! AWESOME JOB BOYS!
Failures...
NONE!! AWESOME JOB BOYS!
Monday, October 4, 2010
EP5/1 Failures and High score
The High Score: Chanayut Montrisuksirikul and Ronapart Srisupavanich 260/260
failures: 48011, 48021, 50342, 50380, 50399,
retest:
p 465 #22-25, 31-33, 48-51
p. 499 #42-45, 50-52, 56-58, and 61-63
failures: 48011, 48021, 50342, 50380, 50399,
retest:
p 465 #22-25, 31-33, 48-51
p. 499 #42-45, 50-52, 56-58, and 61-63
Sunday, October 3, 2010
Ep 5/2 failures and high score
high score is thanakorn panichaporn 260/260 Congratulations!
failures are: NONE! excellent job Ep 5/2
failures are: NONE! excellent job Ep 5/2
Ep 6/2 Failures
I'm proud to announce the failures for ep 6/2. none.
high score Vasin 257.5/260 Excellent job!
high score Vasin 257.5/260 Excellent job!
Sigh. Ep 6/1 failures plus high score
Thanatat Pasupa high score 250/260
fails--43179, 43183
make up exams:
page 625 #44-48, 51-53, 58-61, and 66-69
fails--43179, 43183
make up exams:
page 625 #44-48, 51-53, 58-61, and 66-69
Friday, October 1, 2010
Happy Holidays
no grades can be given out at this time, but failures will be posted by Tuesday. retests will be put up asap.
Have a GREAT holiday
Have a GREAT holiday
Wednesday, September 29, 2010
Debate Practice
i leave on Oct 9 for Khon Kaen for a family vacation. I want to practice the Tuesday through Thursday before I go
Wednesday, September 22, 2010
Monday, September 20, 2010
Ch 16 Review Questions
molarity
Calculate the molarity of each of the following solutions.
a. 0.40 mol of NaCl dissolved in 1.6 L of solution
20.2 g of potassium nitrate, KNO3, in enough water to make 250.0 mL of
solution
dilution
You must prepare 300.0 mL of 0.750M NaBr solution using 2.00M NaBr stock solution. How many milliliters of stock solution should you use?
5. In order to dilute 1.0 L of a 6.00M solution of NaOH to 0.500M solution
how
much water must you add?
v/v
6. What is the concentration in percent by volume, %(v/v), of the following
solutions?
a. 60.0 mL of methanol in a total volume of 500.0 mL
b. 25.0 mL of rubbing alcohol (C3H7OH) diluted to a volume of 200.0 mL
with water
7. How many grams of solute are needed to prepare each of the following
solutions?
a. 1.00 L of a 3.00% (m/m) NaCl solution?
b. 2.00 L of 5.00% (m/m) KNO3 solution?
How many particles in solution are produced by each formula unit of potassium carbonate, K2CO3?
3. How may moles of particles would 3 mol Na2SO4 give in solution?
Calculate the mole fraction of solute in each of the following solutions.
a. 3.0 moles of lithium bromide, LiBr, dissolved in 6.0 moles of water
Find the molality of each of the following solutions.
a. 2.3 moles of glucose dissolved in 500.0 g of water
b. 131 g of Ba(NO3)2 dissolved in 750.0 g of water
Calculate the molarity of each of the following solutions.
a. 0.40 mol of NaCl dissolved in 1.6 L of solution
20.2 g of potassium nitrate, KNO3, in enough water to make 250.0 mL of
solution
dilution
You must prepare 300.0 mL of 0.750M NaBr solution using 2.00M NaBr stock solution. How many milliliters of stock solution should you use?
5. In order to dilute 1.0 L of a 6.00M solution of NaOH to 0.500M solution
how
much water must you add?
v/v
6. What is the concentration in percent by volume, %(v/v), of the following
solutions?
a. 60.0 mL of methanol in a total volume of 500.0 mL
b. 25.0 mL of rubbing alcohol (C3H7OH) diluted to a volume of 200.0 mL
with water
7. How many grams of solute are needed to prepare each of the following
solutions?
a. 1.00 L of a 3.00% (m/m) NaCl solution?
b. 2.00 L of 5.00% (m/m) KNO3 solution?
How many particles in solution are produced by each formula unit of potassium carbonate, K2CO3?
3. How may moles of particles would 3 mol Na2SO4 give in solution?
Calculate the mole fraction of solute in each of the following solutions.
a. 3.0 moles of lithium bromide, LiBr, dissolved in 6.0 moles of water
Find the molality of each of the following solutions.
a. 2.3 moles of glucose dissolved in 500.0 g of water
b. 131 g of Ba(NO3)2 dissolved in 750.0 g of water
Chem review EP 5
Find the percent by mass of water in
NiCl2 * 6H2O.
In your own words, explain hydrogen bonds.
Draw a diagram of the hydrogen bonding between three water molecules.
Explain why the density of ice at 0°C is less than the density of liquid water at 0°C.
Draw a diagram of the hydrogen bonding between three water molecules.
Explain why the density of ice at 0°C is less than the density of liquid water at 0°C.
Identify the solute and solvent in a dilute aqueous solution of potassium chloride.
Which of the following compounds are soluble in water? Which are insoluble?
a. CaCl2
b. N2
c. HBr
d. NH4C2H3O2
Write the formulas for the following hydrates.
a. Calcium sulfate dihydrate
b. Cobalt(II) chloride hexahydrate
NiCl2 * 6H2O.
In your own words, explain hydrogen bonds.
Draw a diagram of the hydrogen bonding between three water molecules.
Explain why the density of ice at 0°C is less than the density of liquid water at 0°C.
Draw a diagram of the hydrogen bonding between three water molecules.
Explain why the density of ice at 0°C is less than the density of liquid water at 0°C.
Identify the solute and solvent in a dilute aqueous solution of potassium chloride.
Which of the following compounds are soluble in water? Which are insoluble?
a. CaCl2
b. N2
c. HBr
d. NH4C2H3O2
Write the formulas for the following hydrates.
a. Calcium sulfate dihydrate
b. Cobalt(II) chloride hexahydrate
Sunday, September 19, 2010
Boys Win SHC Debate Challenge
AC 2 comprised of EP 6/1 Chanwit, Thanakorn, and Ep 4/2 Patipol tied for first along with AC 4--comprised of EP 4/2 Kasidej, EP 6/1 Chaya, and Sirin of TU with a 3-0 record at the Sacred Hearts Convent High School Challenge. There were 7 schools with 20 teams competing.
Monday, September 13, 2010
ep 6 Powerpoint review text only
Identify the following acids as monoprotic, di-protic, or triprotic. Explain your reasoning. H2CO3
What is true about the relative concentrations of hydrogen ions and hydroxide ions in each kind of solution?
basic
How many moles of HCl are required to neutralize aqueous solutions of these bases?
2 mol NH3
Write complete balanced equations for the following acid-base reactions?
H2SO4(aq) + KOH(aq) →
What substances are combined to make a buffer?
What is the molarity of sodium hydroxide if 10.0 mL of the solution is neutralized by each of the following 2.00M solutions?
56.0 mL of HCl
predict
Acidic solution?
Naming acids
Hydrochloric
HBr
Acid Nomenclature Flowchart
What is true about the relative concentrations of hydrogen ions and hydroxide ions in each kind of solution?
basic
How many moles of HCl are required to neutralize aqueous solutions of these bases?
2 mol NH3
Write complete balanced equations for the following acid-base reactions?
H2SO4(aq) + KOH(aq) →
What substances are combined to make a buffer?
What is the molarity of sodium hydroxide if 10.0 mL of the solution is neutralized by each of the following 2.00M solutions?
56.0 mL of HCl
predict
Acidic solution?
Naming acids
Hydrochloric
HBr
Acid Nomenclature Flowchart
EP 4 Review Powerpoint text only
How many valence electrons are in each atom?
Carbon
Know your Bonds
Ionic
Covalent
Polar covalent
Coordinate covalent
Network solid
Write the correct chemical formula for the compounds formed from each pair of ions.
K+,S2−
Write formulas for each compound.
barium chloride
VSEPR
Two bonding atoms, one lone pair
Properties of metals
Sea of electrons
formula
Write the formula for these binary compounds.beryllium chloride
Naming acids
Give the names of these acids. HNO2
Carbon
Know your Bonds
Ionic
Covalent
Polar covalent
Coordinate covalent
Network solid
Write the correct chemical formula for the compounds formed from each pair of ions.
K+,S2−
Write formulas for each compound.
barium chloride
VSEPR
Two bonding atoms, one lone pair
Properties of metals
Sea of electrons
formula
Write the formula for these binary compounds.beryllium chloride
Naming acids
Give the names of these acids. HNO2
Test Review ep 5 text only
Test Review EP 5 Sept 2010
Gary Sakuma
Like dissolves like
What is a polar molecule?
Water
Dissolves ionics, sugars, not lipids, methane, hydrocarbons
Formulae and percent
calcium chloride dihydrate
Percent by water
Solubility is affected by
Temperature
Pressure
size
Mole fraction problem
Calculate the mole fraction of each component in a solution of 1.50 mol ethanoic acid (CH3COOH) in 12.00 mol of water.
Colligative properties
What is the freezing point of a solution of 20.0 g of CCl4 dissolved in 500.0 g of benzene? The freezing point of benzene is 5.48°C; Kf is 5.12°C/m.
m/m problem
Calculate the grams of solute required to make the following solutions.
1100 g of saline solution (0.90% NaCl (m/m))
molarity
How many grams of NaCl would you need to make 500 ml of 3.0M
Concentration of Solute
The amount of solute in a solution is given by its concentration.
Molality
What is the molality of 5 moles of KCl in 500 ml of water
Two Other Concentration Units
V/V problems
What is the concentration (in % (v/v)) of the following solutions?
55 mL of ethanol (C2H5OH) is diluted to a volume of 250 mL with water.
Calculating Concentrations
Calculate molality
Boiling Point Elevation and Freezing Point Depression
∆T = K•m•i
i = van’t Hoff factor = number of particles produced per molecule/formula unit. For covalent compounds, i = 1. For ionic compounds, i = the number of ions present (both + and -)
Compound Theoretical Value of i
glycol 1
NaCl 2
CaCl2 3
Ca3(PO4)2 5
Boiling Point Elevation and Freezing Point Depression
∆T = K•m•i
Change in Boiling Point
Dissolve 62.1 g of glycol (1.00 mol) in 250. g of water. What is the boiling point of the solution?
Kb = 0.52 oC/molal for water (see Kb table).
Solution ∆TBP = Kb • m • i
1. Calculate solution molality = 4.00 m
2. ∆TBP = Kb • m • i
∆TBP = 0.52 oC/molal (4.00 molal) (1)
∆TBP = 2.08 oC
BP = 100 + 2.08 = 102.08 oC (water normally boils at 100)
Freezing Point Depression
Calculate the Freezing Point of a 4.00 molal glycol/water solution.
Kf = 1.86 oC/molal (See Kf table)
Solution
∆TFP = Kf • m • i
= (1.86 oC/molal)(4.00 m)(1)
∆TFP = 7.44
FP = 0 – 7.44 = -7.44 oC (because water normally freezes at 0)
Freezing Point Depression
At what temperature will a 5.4 molal solution of NaCl freeze?
Solution
∆TFP = Kf • m • i
∆TFP = (1.86 oC/molal) • 5.4 m • 2
∆TFP = 20.1 oC
FP = 0 – 20.1 = -20.1 oC
Solution Molality
How many grams of potassium iodide must be dissolved in 500.0 grams of water to produce a 0.060 molal KI solution
.5 kg H20 * .6 mole KI/1 kg H2O *
166.0 gKI/1 mol KI
= answer
Types of problems on exam
Ch 15 and 16 vocabulary and reading
Mass/Mass m/m
Volume/volume v/v
Hydrates percent water and writing formulae
Molarity
Molality
Freezing point depression/boiling point elevation T=k m I
Gary Sakuma
Like dissolves like
What is a polar molecule?
Water
Dissolves ionics, sugars, not lipids, methane, hydrocarbons
Formulae and percent
calcium chloride dihydrate
Percent by water
Solubility is affected by
Temperature
Pressure
size
Mole fraction problem
Calculate the mole fraction of each component in a solution of 1.50 mol ethanoic acid (CH3COOH) in 12.00 mol of water.
Colligative properties
What is the freezing point of a solution of 20.0 g of CCl4 dissolved in 500.0 g of benzene? The freezing point of benzene is 5.48°C; Kf is 5.12°C/m.
m/m problem
Calculate the grams of solute required to make the following solutions.
1100 g of saline solution (0.90% NaCl (m/m))
molarity
How many grams of NaCl would you need to make 500 ml of 3.0M
Concentration of Solute
The amount of solute in a solution is given by its concentration.
Molality
What is the molality of 5 moles of KCl in 500 ml of water
Two Other Concentration Units
V/V problems
What is the concentration (in % (v/v)) of the following solutions?
55 mL of ethanol (C2H5OH) is diluted to a volume of 250 mL with water.
Calculating Concentrations
Calculate molality
Boiling Point Elevation and Freezing Point Depression
∆T = K•m•i
i = van’t Hoff factor = number of particles produced per molecule/formula unit. For covalent compounds, i = 1. For ionic compounds, i = the number of ions present (both + and -)
Compound Theoretical Value of i
glycol 1
NaCl 2
CaCl2 3
Ca3(PO4)2 5
Boiling Point Elevation and Freezing Point Depression
∆T = K•m•i
Change in Boiling Point
Dissolve 62.1 g of glycol (1.00 mol) in 250. g of water. What is the boiling point of the solution?
Kb = 0.52 oC/molal for water (see Kb table).
Solution ∆TBP = Kb • m • i
1. Calculate solution molality = 4.00 m
2. ∆TBP = Kb • m • i
∆TBP = 0.52 oC/molal (4.00 molal) (1)
∆TBP = 2.08 oC
BP = 100 + 2.08 = 102.08 oC (water normally boils at 100)
Freezing Point Depression
Calculate the Freezing Point of a 4.00 molal glycol/water solution.
Kf = 1.86 oC/molal (See Kf table)
Solution
∆TFP = Kf • m • i
= (1.86 oC/molal)(4.00 m)(1)
∆TFP = 7.44
FP = 0 – 7.44 = -7.44 oC (because water normally freezes at 0)
Freezing Point Depression
At what temperature will a 5.4 molal solution of NaCl freeze?
Solution
∆TFP = Kf • m • i
∆TFP = (1.86 oC/molal) • 5.4 m • 2
∆TFP = 20.1 oC
FP = 0 – 20.1 = -20.1 oC
Solution Molality
How many grams of potassium iodide must be dissolved in 500.0 grams of water to produce a 0.060 molal KI solution
.5 kg H20 * .6 mole KI/1 kg H2O *
166.0 gKI/1 mol KI
= answer
Types of problems on exam
Ch 15 and 16 vocabulary and reading
Mass/Mass m/m
Volume/volume v/v
Hydrates percent water and writing formulae
Molarity
Molality
Freezing point depression/boiling point elevation T=k m I
Friday, September 10, 2010
photo highlights of ASDC2010
The ASDC2010 was a great journey for our students. Our first International competition, our students got to battle the various other schools from other countries.
The three boys were Tanachai ep 6/2, Chaya EP 6/1, and Kasidej EP4/2
ASDC 2010
Highlights of the tournament: The Boys beat TU in the final round. We finished 3-4, an outstanding first time at International Competition. Mr. Gary made the play-offs as an adjudicator judging up to the Semis in the tournament.
The three boys were Tanachai ep 6/2, Chaya EP 6/1, and Kasidej EP4/2
ASDC 2010
Highlights of the tournament: The Boys beat TU in the final round. We finished 3-4, an outstanding first time at International Competition. Mr. Gary made the play-offs as an adjudicator judging up to the Semis in the tournament.
Tuesday, September 7, 2010
Boys of AC Battle On in Asian Schools Debate Championships 2010
Young EP 4/2 Kasidej along with veterans EP 6/1 Chaya and EP 6/2 Tanachai face a large obstacle in the field of 11 countries with past Champions from Korea, World School players from Srilanka, last year's Champions from the Philippines, India, Malaysia, Indonesia, and others. It's been a great tournament so far with the boys standing at 2-3 going into the Final day.
Tuesday, August 31, 2010
make up Quiz ep 4
find me video links on Van Der Waals Forces, pi sigma bonding, hybridization, or metallic bonding
Sunday, August 29, 2010
Week Aug 30, 2010 Topics
Ep 4 Exam topics--chapter 6-9--periodic law from ch 6, covalent, ionic bonds, polyatomics ions, and naming--more details forthcoming.
Ep 5 Exam Topics: Ch 15 and 16--aqueous solutions, moles revisited, molarity, molality, dilutions, percent problems, and colligative properties.
Ep 6 Exam topics: Ch 19-- Buffers, neutralization, titrations, dilutions, and pH.
This week will be about the problems from this topic. Next week, we will start the review sheets for the exam.
Exam format--vocab--pick 10 out of 20 to define, and then, short answers that will comprise of problems, questions about application of chemical theories, and explaining certain theories to demonstrate understanding.
Mr. Gary leaves Sept 4 until Sept 10 for the Debate Tournament in the Phillipines.
Ep 5 Exam Topics: Ch 15 and 16--aqueous solutions, moles revisited, molarity, molality, dilutions, percent problems, and colligative properties.
Ep 6 Exam topics: Ch 19-- Buffers, neutralization, titrations, dilutions, and pH.
This week will be about the problems from this topic. Next week, we will start the review sheets for the exam.
Exam format--vocab--pick 10 out of 20 to define, and then, short answers that will comprise of problems, questions about application of chemical theories, and explaining certain theories to demonstrate understanding.
Mr. Gary leaves Sept 4 until Sept 10 for the Debate Tournament in the Phillipines.
Saturday, August 28, 2010
Wednesday, August 25, 2010
Make up quiz 1 ep 4, 5, and 6
I want a video link for an interesting 3-7 minute video on the following:
ep 4--stoichemistry, limiting reagents, moles
ep 5 solutions, colloids, tyndall effect, brownian motion
ep 6 acid bases, oxidation/reduction, or electrochemistry.
if i look at the video, and i'm bored in one minute, you get half bonus for your quiz.
also include a summation of the top three points in the quiz--what's going on, what's interesting, and a description proving you watched the video--example:
solutions video--link www.blahblah.com
The video starts out with a definition of solution formation. It has an interesting discussion of solute and solvents and shows a supersaturation of acetate. Students will like how the acetate solution freezes over after the temperature drops.
ep 4--stoichemistry, limiting reagents, moles
ep 5 solutions, colloids, tyndall effect, brownian motion
ep 6 acid bases, oxidation/reduction, or electrochemistry.
if i look at the video, and i'm bored in one minute, you get half bonus for your quiz.
also include a summation of the top three points in the quiz--what's going on, what's interesting, and a description proving you watched the video--example:
solutions video--link www.blahblah.com
The video starts out with a definition of solution formation. It has an interesting discussion of solute and solvents and shows a supersaturation of acetate. Students will like how the acetate solution freezes over after the temperature drops.
Tuesday, August 24, 2010
Ch 9.3 Polyatomics
MEMORIZE table 9.3 on page 257 for quiz
Sometimes the same two or three elements combine in different ratios to form different polyatomic ions. You can see examples in Table 9.3. Look for pairs of ions for which there is both an -ite and an -ate ending, for example, sulfite and sulfate. In the list below, examine the charge on each ion in the pair. Note the number of oxygen atoms and the endings on each name. You should be able to discern a pattern in the naming convention.
The charge on each polyatomic ion in a given pair is the same. The -ite ending indicates one less oxygen atom than the -ate ending. However, the ending does not tell you the actual number of oxygen atoms in the ion. For example, the nitrite ion has two oxygen atoms and the sulfite ion has three oxygen atoms. All anions with names ending in -ite or -ate contain oxygen.
Sometimes the same two or three elements combine in different ratios to form different polyatomic ions. You can see examples in Table 9.3. Look for pairs of ions for which there is both an -ite and an -ate ending, for example, sulfite and sulfate. In the list below, examine the charge on each ion in the pair. Note the number of oxygen atoms and the endings on each name. You should be able to discern a pattern in the naming convention.
The charge on each polyatomic ion in a given pair is the same. The -ite ending indicates one less oxygen atom than the -ate ending. However, the ending does not tell you the actual number of oxygen atoms in the ion. For example, the nitrite ion has two oxygen atoms and the sulfite ion has three oxygen atoms. All anions with names ending in -ite or -ate contain oxygen.
EP 4 ASSIGNMENTS: LAST THREE
assignment 1: p. 256 #1, 2 page 258 # 3, 4 page 263 # 10-11 skip b
assignment 2: p. 258 # 8, 9 p. 265 #12, 13 p. 266 #17-19
Vocab section 8.3 and 7.3
assignment 2: p. 258 # 8, 9 p. 265 #12, 13 p. 266 #17-19
Vocab section 8.3 and 7.3
Monday, August 23, 2010
Ch 19: Problems
(40) Key Concept What type of salt produces an acidic solution? A basic solution?Hint
(41) Key Concept What substances are combined to make a buffer? Hint
(42)Which of these salts would form an acidic aqueous solution?
a. KC2H3O2
b. LiCl
c. NaHCO3
d. (NH4)2SO4
(43)Using equations, show what happens when acid is added to an ammonium ion–ammonia buffer. What happens when base is added?
(41) Key Concept What substances are combined to make a buffer? Hint
(42)Which of these salts would form an acidic aqueous solution?
a. KC2H3O2
b. LiCl
c. NaHCO3
d. (NH4)2SO4
(43)Using equations, show what happens when acid is added to an ammonium ion–ammonia buffer. What happens when base is added?
Word Origins Buffer
Word Origins
Buffer comes from the Old English word buff, meaning “firmly or sturdily.” A buffer solution resists changes in its pH even when acidic or basic solutions are added to it. What does it mean if authorities set up a buffer zone around a fire investigation site?
Buffer comes from the Old English word buff, meaning “firmly or sturdily.” A buffer solution resists changes in its pH even when acidic or basic solutions are added to it. What does it mean if authorities set up a buffer zone around a fire investigation site?
Ch 19: Intro to Buffers 2
Buffers
The addition of 10 mL of 0.10M sodium hydroxide to 1 L of pure water increases the pH by 4.0 pH units (from 7.0 to 11.0). A solution containing 0.20 mol/L each of ethanoic acid and sodium ethanoate has a pH of 4.76. When moderate amounts of either acid or base are added to this solution, however, the pH changes little. The addition of 10 mL of 0.10M sodium hydroxide to 1 L of this solution, for example, increases the pH by only 0.01 pH unit, from 4.76 to 4.77. Figure 19.27 shows what happens when 1.0 mL of 0.01M HCl solution is added to an unbuffered solution.
Figure 19.27 A buffer is a solution in which the pH remains relatively constant. a. The indicator shows that the buffered solution on the left and the unbuffered solution on the right are basic—pH about 8. b. After the addition of 1.0 mL of 0.01M HCl solution, the pH of the buffered solution shows no visible change. The pH of the unbuffered solution, however, is now about 3—the solution is acidic.
The solution of ethanoic acid and sodium ethanoate is an example of a typical buffer. A buffer is a solution in which the pH remains relatively constant when small amounts of acid or base are added. A buffer is a solution of a weak acid and one of its salts, or a solution of a weak base and one of its salts.
A buffer solution is better able to resist drastic changes in pH than is pure water. Figure 19.28 illustrates how a buffer works. Ethanoic acid (CH3COOH) and its anion (CH3COO−) act as reservoirs of neutralizing power. They react with any hydroxide ions or hydrogen ions added to the solution. For example, consider the buffer solution in which the sodium ethanoate (CH3COONa) is completely ionized.
The addition of 10 mL of 0.10M sodium hydroxide to 1 L of pure water increases the pH by 4.0 pH units (from 7.0 to 11.0). A solution containing 0.20 mol/L each of ethanoic acid and sodium ethanoate has a pH of 4.76. When moderate amounts of either acid or base are added to this solution, however, the pH changes little. The addition of 10 mL of 0.10M sodium hydroxide to 1 L of this solution, for example, increases the pH by only 0.01 pH unit, from 4.76 to 4.77. Figure 19.27 shows what happens when 1.0 mL of 0.01M HCl solution is added to an unbuffered solution.
Figure 19.27 A buffer is a solution in which the pH remains relatively constant. a. The indicator shows that the buffered solution on the left and the unbuffered solution on the right are basic—pH about 8. b. After the addition of 1.0 mL of 0.01M HCl solution, the pH of the buffered solution shows no visible change. The pH of the unbuffered solution, however, is now about 3—the solution is acidic.
The solution of ethanoic acid and sodium ethanoate is an example of a typical buffer. A buffer is a solution in which the pH remains relatively constant when small amounts of acid or base are added. A buffer is a solution of a weak acid and one of its salts, or a solution of a weak base and one of its salts.
A buffer solution is better able to resist drastic changes in pH than is pure water. Figure 19.28 illustrates how a buffer works. Ethanoic acid (CH3COOH) and its anion (CH3COO−) act as reservoirs of neutralizing power. They react with any hydroxide ions or hydrogen ions added to the solution. For example, consider the buffer solution in which the sodium ethanoate (CH3COONa) is completely ionized.
Ch 19: Salt Hydrolysis
The ethanoate ion is a Brønsted-Lowry base, which means it is a hydrogen-ion acceptor. It establishes an equilibrium with water, forming electrically neutral ethanoic acid and negative hydroxide ions.
This process is called hydrolysis because it splits a hydrogen ion off a water molecule. The resulting solution contains a hydroxide-ion concentration greater than the hydrogen-ion concentration. Thus the solution is basic.
Ammonium chloride (NH4Cl) is the salt of a strong acid (hydrochloric acid, HCl) and a weak base (ammonia, NH3). It is completely ionized in solution.
NH4Cl(aq) → NH4+(aq) + Cl−(aq)
Figure 19.25 Vapors of the strong acid HCl(aq) and the weak base NH3(aq) combine to form the acidic white salt ammonium chloride (NH4Cl).
The ammonium ion (NH4+) is a strong enough acid to donate a hydrogen ion to a water molecule, although the equilibrium is strongly to the left.
This process is also called hydrolysis. It results in the formation of unionized ammonia and hydronium (hydrogen) ions. The [H3O+] is greater than the [OH−]. Thus, a solution of ammonium chloride is acidic. To determine if a salt solution is acidic or basic, remember the following rules:
Strong acid + Strong base → Neutral solution
Strong acid + Weak base → Acidic solution
Weak acid + Strong base → Basic solution
This process is called hydrolysis because it splits a hydrogen ion off a water molecule. The resulting solution contains a hydroxide-ion concentration greater than the hydrogen-ion concentration. Thus the solution is basic.
Ammonium chloride (NH4Cl) is the salt of a strong acid (hydrochloric acid, HCl) and a weak base (ammonia, NH3). It is completely ionized in solution.
NH4Cl(aq) → NH4+(aq) + Cl−(aq)
Figure 19.25 Vapors of the strong acid HCl(aq) and the weak base NH3(aq) combine to form the acidic white salt ammonium chloride (NH4Cl).
The ammonium ion (NH4+) is a strong enough acid to donate a hydrogen ion to a water molecule, although the equilibrium is strongly to the left.
This process is also called hydrolysis. It results in the formation of unionized ammonia and hydronium (hydrogen) ions. The [H3O+] is greater than the [OH−]. Thus, a solution of ammonium chloride is acidic. To determine if a salt solution is acidic or basic, remember the following rules:
Strong acid + Strong base → Neutral solution
Strong acid + Weak base → Acidic solution
Weak acid + Strong base → Basic solution
Ch 19: Intro to Buffers
Salt Hydrolysis
A salt consists of an anion from an acid and a cation from a base. It forms as a result of a neutralization reaction. Although solutions of many salts are neutral, some are acidic and others are basic. Solutions of sodium chloride and of potassium sulfate are neutral. A solution of ammonium chloride is acidic. A solution of sodium ethanoate (sodium acetate) is basic. Figure 19.24 shows a titration curve obtained by adding a solution of sodium hydroxide, a strong base, to a solution of ethanoic (acetic) acid, a weak acid. An aqueous solution of sodium ethanoate exists at the equivalence point.
The pH at the equivalence point is 8.7—basic.
For a strong acid–strong base titration, the pH at the equivalence point is 7, or neutral. This difference exists because some salts promote hydrolysis. In salt hydrolysis, the cations or anions of a dissociated salt remove hydrogen ions from or donate hydrogen ions to water. Depending on the direction of the hydrogen-ion transfer, solutions containing hydrolyzing salts may be either acidic or basic. Hydrolyzing salts are usually derived from a strong acid and a weak base, or from a weak acid and a strong base. Sodium carbonate, washing soda, is the salt of the strong base sodium hydroxide and carbonic acid, a weak acid. Ammonium nitrate, used in fertilizers, is the salt of the weak base ammonia and nitric acid, a strong acid. Soap is the salt of a strong base, usually sodium hydroxide, and stearic acid, a weak acid present in fats. In general, salts that produce acidic solutions contain positive ions that release protons to water. Salts that produce basic solutions contain negative ions that attract protons from water.
Sodium ethanoate (CH3COONa) is the salt of a weak acid (ethanoic acid, CH3COOH) and a strong base (sodium hydroxide, NaOH). In solution, the salt is completely ionized.
A salt consists of an anion from an acid and a cation from a base. It forms as a result of a neutralization reaction. Although solutions of many salts are neutral, some are acidic and others are basic. Solutions of sodium chloride and of potassium sulfate are neutral. A solution of ammonium chloride is acidic. A solution of sodium ethanoate (sodium acetate) is basic. Figure 19.24 shows a titration curve obtained by adding a solution of sodium hydroxide, a strong base, to a solution of ethanoic (acetic) acid, a weak acid. An aqueous solution of sodium ethanoate exists at the equivalence point.
The pH at the equivalence point is 8.7—basic.
For a strong acid–strong base titration, the pH at the equivalence point is 7, or neutral. This difference exists because some salts promote hydrolysis. In salt hydrolysis, the cations or anions of a dissociated salt remove hydrogen ions from or donate hydrogen ions to water. Depending on the direction of the hydrogen-ion transfer, solutions containing hydrolyzing salts may be either acidic or basic. Hydrolyzing salts are usually derived from a strong acid and a weak base, or from a weak acid and a strong base. Sodium carbonate, washing soda, is the salt of the strong base sodium hydroxide and carbonic acid, a weak acid. Ammonium nitrate, used in fertilizers, is the salt of the weak base ammonia and nitric acid, a strong acid. Soap is the salt of a strong base, usually sodium hydroxide, and stearic acid, a weak acid present in fats. In general, salts that produce acidic solutions contain positive ions that release protons to water. Salts that produce basic solutions contain negative ions that attract protons from water.
Sodium ethanoate (CH3COONa) is the salt of a weak acid (ethanoic acid, CH3COOH) and a strong base (sodium hydroxide, NaOH). In solution, the salt is completely ionized.
Ch 19: Acids/Bases 19.5 Buffers
Ch 19
Key Concepts
• When is the solution of a salt acidic or basic?
• What are the components of a buffer?
Vocabulary
• salt hydrolysis
• buffer
• buffer capacity
Key Concepts
• When is the solution of a salt acidic or basic?
• What are the components of a buffer?
Vocabulary
• salt hydrolysis
• buffer
• buffer capacity
Ch 8: Problems
23) Key Concept How are atomic and molecular orbitals related?Hint
(24) Key Concept Explain how the VSEPR theory can be used to predict the shapes of molecules.Hint
(25) Key Concept How is orbital hybrization useful in describing molecules?Hint
(26)What shape would you expect a simple carbon-containing compound to have if the carbon atom has the following hybridizations?
a. sp2
b. sp3
c. sp
(27)What is a sigma bond? Describe, with the aid of a diagram, how the overlap of two half-filled 1s orbitals produces a sigma bond?
(28)How many sigma and how many pi bonds are in an ethyne molecule (C2H2)?
(29)The BF3 molecule is planar. The attachment of a fluoride ion to the boron in BF3, through a coordinate covalent bond, creates the BF4− ion. What is the geometric shape of this ion?
(24) Key Concept Explain how the VSEPR theory can be used to predict the shapes of molecules.Hint
(25) Key Concept How is orbital hybrization useful in describing molecules?Hint
(26)What shape would you expect a simple carbon-containing compound to have if the carbon atom has the following hybridizations?
a. sp2
b. sp3
c. sp
(27)What is a sigma bond? Describe, with the aid of a diagram, how the overlap of two half-filled 1s orbitals produces a sigma bond?
(28)How many sigma and how many pi bonds are in an ethyne molecule (C2H2)?
(29)The BF3 molecule is planar. The attachment of a fluoride ion to the boron in BF3, through a coordinate covalent bond, creates the BF4− ion. What is the geometric shape of this ion?
Ch 8: Hybridization Orbitals
Hybrid Orbitals
The VSEPR theory works well when accounting for molecular shapes, but it does not help much in describing the types of bonds formed. Orbital hybridization provides information about both molecular bonding and molecular shape. In hybridization, several atomic orbitals mix to form the same total number of equivalent hybrid orbitals.
Hybridization Involving Single Bonds
Recall that the carbon atom’s outer electron configuration is 2s2 2p2, but one of the 2s electrons is promoted to a 2p orbital to give one 2s electron and three 2p electrons, allowing it to bond to four hydrogen atoms in methane. You might suspect that one bond would be different from the other three. In fact, all the bonds are identical. This is explained by orbital hybridization.
The one 2s orbital and three 2p orbitals of a carbon atom mix to form four sp3 hybrid orbitals. These are at the tetrahedral angle of 109.5°. As you can see in Figure 8.19, the four sp3 orbitals of carbon overlap with the 1s orbitals of the four hydrogen atoms. The sp3 orbitals extend farther into space than either s or p orbitals, allowing a great deal of overlap with the hydrogen 1s orbitals. The eight available valence electrons fill the molecular orbitals to form four C —H sigma bonds. The extent of overlap results in unusually strong covalent bonds.
Figure 8.19
Hybridization Involving Double Bonds
Hybridization is also useful in describing double covalent bonds. Ethene is a relatively simple molecule that has one carbon–carbon double bond and four carbon–hydrogen single bonds.
Experimental evidence indicates that the H—C—H bond angles in ethene are about 120°. In ethene, sp2 hybrid orbitals form from the combination of one 2s and two 2p atomic orbitals of carbon. As you can see in Figure 8.20, each hybrid orbital is separated from the other two by 120°. Two sp2 hybrid orbitals of each carbon form sigma-bonding molecular orbitals with the four available hydrogen 1s orbitals. The third sp2 orbitals of each of the two carbons overlap to form a carbon–carbon sigma-bonding orbital. The nonhybridized 2p carbon orbitals overlap side-by-side to form a pi-bonding orbital. A total of twelve electrons fill six bonding molecular orbitals. Thus five sigma bonds and one pi bond hold the ethene molecule together. The sigma bonds and the pi bond are two-electron covalent bonds. Although they are drawn alike in structural formulas, pi bonds are weaker than sigma bonds. In chemical reactions that involve breaking one bond of a carbon–carbon double bond, the pi bond is more likely to break than the sigma bond.
Hybridization Involving Triple Bonds
A third type of covalent bond is a triple bond, such as is found in ethyne (C2H2), also called acetylene.
As with other molecules, the hybrid orbital description of ethyne is guided by an understanding of the properties of the molecule. Ethyne is a linear molecule. The best hybrid orbital description is obtained if a 2s atomic orbital of carbon mixes with only one of the three 2p atomic orbitals. The result is two sp hybrid orbitals for each carbon.
The carbon–carbon sigma-bonding molecular orbital of the ethyne molecule in Figure 8.21 forms from the overlap of one sp orbital from each carbon. The other sp orbital of each carbon overlaps with the 1s orbital of each hydrogen, also forming sigma-bonding molecular orbitals. The remaining pair of p atomic orbitals on each carbon overlap side-by-side. They form two pi-bonding molecular orbitals that surround the central carbons. The ten available electrons completely fill five bonding molecular orbitals. The bonding of ethyne consists of three sigma bonds and two pi bonds
The VSEPR theory works well when accounting for molecular shapes, but it does not help much in describing the types of bonds formed. Orbital hybridization provides information about both molecular bonding and molecular shape. In hybridization, several atomic orbitals mix to form the same total number of equivalent hybrid orbitals.
Hybridization Involving Single Bonds
Recall that the carbon atom’s outer electron configuration is 2s2 2p2, but one of the 2s electrons is promoted to a 2p orbital to give one 2s electron and three 2p electrons, allowing it to bond to four hydrogen atoms in methane. You might suspect that one bond would be different from the other three. In fact, all the bonds are identical. This is explained by orbital hybridization.
The one 2s orbital and three 2p orbitals of a carbon atom mix to form four sp3 hybrid orbitals. These are at the tetrahedral angle of 109.5°. As you can see in Figure 8.19, the four sp3 orbitals of carbon overlap with the 1s orbitals of the four hydrogen atoms. The sp3 orbitals extend farther into space than either s or p orbitals, allowing a great deal of overlap with the hydrogen 1s orbitals. The eight available valence electrons fill the molecular orbitals to form four C —H sigma bonds. The extent of overlap results in unusually strong covalent bonds.
Figure 8.19
Hybridization Involving Double Bonds
Hybridization is also useful in describing double covalent bonds. Ethene is a relatively simple molecule that has one carbon–carbon double bond and four carbon–hydrogen single bonds.
Experimental evidence indicates that the H—C—H bond angles in ethene are about 120°. In ethene, sp2 hybrid orbitals form from the combination of one 2s and two 2p atomic orbitals of carbon. As you can see in Figure 8.20, each hybrid orbital is separated from the other two by 120°. Two sp2 hybrid orbitals of each carbon form sigma-bonding molecular orbitals with the four available hydrogen 1s orbitals. The third sp2 orbitals of each of the two carbons overlap to form a carbon–carbon sigma-bonding orbital. The nonhybridized 2p carbon orbitals overlap side-by-side to form a pi-bonding orbital. A total of twelve electrons fill six bonding molecular orbitals. Thus five sigma bonds and one pi bond hold the ethene molecule together. The sigma bonds and the pi bond are two-electron covalent bonds. Although they are drawn alike in structural formulas, pi bonds are weaker than sigma bonds. In chemical reactions that involve breaking one bond of a carbon–carbon double bond, the pi bond is more likely to break than the sigma bond.
Hybridization Involving Triple Bonds
A third type of covalent bond is a triple bond, such as is found in ethyne (C2H2), also called acetylene.
As with other molecules, the hybrid orbital description of ethyne is guided by an understanding of the properties of the molecule. Ethyne is a linear molecule. The best hybrid orbital description is obtained if a 2s atomic orbital of carbon mixes with only one of the three 2p atomic orbitals. The result is two sp hybrid orbitals for each carbon.
The carbon–carbon sigma-bonding molecular orbital of the ethyne molecule in Figure 8.21 forms from the overlap of one sp orbital from each carbon. The other sp orbital of each carbon overlaps with the 1s orbital of each hydrogen, also forming sigma-bonding molecular orbitals. The remaining pair of p atomic orbitals on each carbon overlap side-by-side. They form two pi-bonding molecular orbitals that surround the central carbons. The ten available electrons completely fill five bonding molecular orbitals. The bonding of ethyne consists of three sigma bonds and two pi bonds
Ch 8: Molecular Orbitals
Molecular Orbitals
The model for covalent bonding you have been using assumes that the orbitals are those of the individual atoms. There is a quantum mechanical model of bonding, however, that describes the electrons in molecules using orbitals that exist only for groupings of atoms. When two atoms combine, this model assumes that their atomic orbitals overlap to produce molecular orbitals, or orbitals that apply to the entire molecule.
In some ways, atomic orbitals and molecular orbitals are similar. Just as an atomic orbital belongs to a particular atom, a molecular orbital belongs to a molecule as a whole. Each atomic orbital is filled if it contains two electrons. Similarly, two electrons are required to fill a molecular orbital. A molecular orbital that can be occupied by two electrons of a covalent bond is called a bonding orbital.
Sigma Bonds
When two atomic orbitals combine to form a molecular orbital that is symmetrical around the axis connecting two atomic nuclei, a sigma bond is formed, as illustrated in Figure 8.13. The symbol for this bond is the Greek letter sigma (σ).
In general, covalent bonding results from an imbalance between the attractions and repulsions of the nuclei and electrons involved. Because their charges have opposite signs, the nuclei and electrons attract each other. Because their charges have the same sign, nuclei repel other nuclei and electrons repel other electrons. In a hydrogen molecule, the nuclei repel each other, as do the electrons. In a bonding molecular orbital of hydrogen, however, the attractions between the hydrogen nuclei and the electrons are stronger than the repulsions. The balance of all the interactions between the hydrogen atoms is thus tipped in favor of holding the atoms together. The result is a stable diatomic molecule of H2.
Atomic p orbitals can also overlap to form molecular orbitals. A fluorine atom, for example, has a half-filled 2p orbital. When two fluorine atoms combine, as shown in Figure 8.14, the p orbitals overlap to produce a bonding molecular orbital. There is a high probability of finding a pair of electrons between the positively charged nuclei of the two fluorines. The fluorine nuclei are attracted to this region of high electron density. This attraction holds the atoms together in the fluorine molecule (F2). The overlap of the 2p orbitals produces a bonding molecular orbital that is symmetrical when viewed around the F —F bond axis connecting the nuclei. Therefore, the F —F bond is a sigma bond.
Pi Bonds
In the sigma bond of the fluorine molecule, the p atomic orbitals overlap end-to-end. In some molecules, however, orbitals can overlap side-by-side. As shown in Figure 8.15, the side-by-side overlap of atomic p orbitals produces what are called pi molecular orbitals. When a pi molecular orbital is filled with two electrons, a pi bond results. In a pi bond (symbolized by the Greek letter π), the bonding electrons are most likely to be found in sausage-shaped regions above and below the bond axis of the bonded atoms. It is not symmetrical around the F- —F bond axis. Atomic orbitals in pi bonding overlap less than in sigma bonding. Therefore, pi bonds tend to be weaker than sigma bonds.
VESPR—not very small elephants playing in the rain…
The valence-shell electron-pair repulsion theory, or VSEPR theory, explains the three-dimensional shape of methane. According to VSEPR theory, the repulsion between electron pairs causes molecular shapes to adjust so that the valence-electron pairs stay as far apart as possible. The methane molecule has four bonding electron pairs and no unshared pairs. The bonding pairs are farthest apart when the angle between the central carbon and its attached hydrogens is 109.5°. This is the H— C —H bond angle found experimentally.
Unshared pairs of electrons are also important in predicting the shapes of molecules. The nitrogen in ammonia (NH3) is surrounded by four pairs of valence electrons, so you might predict the tetrahedral angle of 109.5° for the H—N—H bond angle. However, one of the valence-electron pairs shown in Figure 8.16b is an unshared pair. No bonding atom is vying for these unshared electrons. Thus they are held closer to the nitrogen than are the bonding pairs. The unshared pair strongly repels the bonding pairs, pushing them together. The measured H—N—H bond angle is only 107°.
The model for covalent bonding you have been using assumes that the orbitals are those of the individual atoms. There is a quantum mechanical model of bonding, however, that describes the electrons in molecules using orbitals that exist only for groupings of atoms. When two atoms combine, this model assumes that their atomic orbitals overlap to produce molecular orbitals, or orbitals that apply to the entire molecule.
In some ways, atomic orbitals and molecular orbitals are similar. Just as an atomic orbital belongs to a particular atom, a molecular orbital belongs to a molecule as a whole. Each atomic orbital is filled if it contains two electrons. Similarly, two electrons are required to fill a molecular orbital. A molecular orbital that can be occupied by two electrons of a covalent bond is called a bonding orbital.
Sigma Bonds
When two atomic orbitals combine to form a molecular orbital that is symmetrical around the axis connecting two atomic nuclei, a sigma bond is formed, as illustrated in Figure 8.13. The symbol for this bond is the Greek letter sigma (σ).
In general, covalent bonding results from an imbalance between the attractions and repulsions of the nuclei and electrons involved. Because their charges have opposite signs, the nuclei and electrons attract each other. Because their charges have the same sign, nuclei repel other nuclei and electrons repel other electrons. In a hydrogen molecule, the nuclei repel each other, as do the electrons. In a bonding molecular orbital of hydrogen, however, the attractions between the hydrogen nuclei and the electrons are stronger than the repulsions. The balance of all the interactions between the hydrogen atoms is thus tipped in favor of holding the atoms together. The result is a stable diatomic molecule of H2.
Atomic p orbitals can also overlap to form molecular orbitals. A fluorine atom, for example, has a half-filled 2p orbital. When two fluorine atoms combine, as shown in Figure 8.14, the p orbitals overlap to produce a bonding molecular orbital. There is a high probability of finding a pair of electrons between the positively charged nuclei of the two fluorines. The fluorine nuclei are attracted to this region of high electron density. This attraction holds the atoms together in the fluorine molecule (F2). The overlap of the 2p orbitals produces a bonding molecular orbital that is symmetrical when viewed around the F —F bond axis connecting the nuclei. Therefore, the F —F bond is a sigma bond.
Pi Bonds
In the sigma bond of the fluorine molecule, the p atomic orbitals overlap end-to-end. In some molecules, however, orbitals can overlap side-by-side. As shown in Figure 8.15, the side-by-side overlap of atomic p orbitals produces what are called pi molecular orbitals. When a pi molecular orbital is filled with two electrons, a pi bond results. In a pi bond (symbolized by the Greek letter π), the bonding electrons are most likely to be found in sausage-shaped regions above and below the bond axis of the bonded atoms. It is not symmetrical around the F- —F bond axis. Atomic orbitals in pi bonding overlap less than in sigma bonding. Therefore, pi bonds tend to be weaker than sigma bonds.
VESPR—not very small elephants playing in the rain…
The valence-shell electron-pair repulsion theory, or VSEPR theory, explains the three-dimensional shape of methane. According to VSEPR theory, the repulsion between electron pairs causes molecular shapes to adjust so that the valence-electron pairs stay as far apart as possible. The methane molecule has four bonding electron pairs and no unshared pairs. The bonding pairs are farthest apart when the angle between the central carbon and its attached hydrogens is 109.5°. This is the H— C —H bond angle found experimentally.
Unshared pairs of electrons are also important in predicting the shapes of molecules. The nitrogen in ammonia (NH3) is surrounded by four pairs of valence electrons, so you might predict the tetrahedral angle of 109.5° for the H—N—H bond angle. However, one of the valence-electron pairs shown in Figure 8.16b is an unshared pair. No bonding atom is vying for these unshared electrons. Thus they are held closer to the nitrogen than are the bonding pairs. The unshared pair strongly repels the bonding pairs, pushing them together. The measured H—N—H bond angle is only 107°.
Ch 8: Molecular Orbitals
Molecular Orbitals
The model for covalent bonding you have been using assumes that the orbitals are those of the individual atoms. There is a quantum mechanical model of bonding, however, that describes the electrons in molecules using orbitals that exist only for groupings of atoms. When two atoms combine, this model assumes that their atomic orbitals overlap to produce molecular orbitals, or orbitals that apply to the entire molecule.
In some ways, atomic orbitals and molecular orbitals are similar. Just as an atomic orbital belongs to a particular atom, a molecular orbital belongs to a molecule as a whole. Each atomic orbital is filled if it contains two electrons. Similarly, two electrons are required to fill a molecular orbital. A molecular orbital that can be occupied by two electrons of a covalent bond is called a bonding orbital.
Sigma Bonds
When two atomic orbitals combine to form a molecular orbital that is symmetrical around the axis connecting two atomic nuclei, a sigma bond is formed, as illustrated in Figure 8.13. The symbol for this bond is the Greek letter sigma (σ).
In general, covalent bonding results from an imbalance between the attractions and repulsions of the nuclei and electrons involved. Because their charges have opposite signs, the nuclei and electrons attract each other. Because their charges have the same sign, nuclei repel other nuclei and electrons repel other electrons. In a hydrogen molecule, the nuclei repel each other, as do the electrons. In a bonding molecular orbital of hydrogen, however, the attractions between the hydrogen nuclei and the electrons are stronger than the repulsions. The balance of all the interactions between the hydrogen atoms is thus tipped in favor of holding the atoms together. The result is a stable diatomic molecule of H2.
Atomic p orbitals can also overlap to form molecular orbitals. A fluorine atom, for example, has a half-filled 2p orbital. When two fluorine atoms combine, as shown in Figure 8.14, the p orbitals overlap to produce a bonding molecular orbital. There is a high probability of finding a pair of electrons between the positively charged nuclei of the two fluorines. The fluorine nuclei are attracted to this region of high electron density. This attraction holds the atoms together in the fluorine molecule (F2). The overlap of the 2p orbitals produces a bonding molecular orbital that is symmetrical when viewed around the F —F bond axis connecting the nuclei. Therefore, the F —F bond is a sigma bond.
Pi Bonds
In the sigma bond of the fluorine molecule, the p atomic orbitals overlap end-to-end. In some molecules, however, orbitals can overlap side-by-side. As shown in Figure 8.15, the side-by-side overlap of atomic p orbitals produces what are called pi molecular orbitals. When a pi molecular orbital is filled with two electrons, a pi bond results. In a pi bond (symbolized by the Greek letter π), the bonding electrons are most likely to be found in sausage-shaped regions above and below the bond axis of the bonded atoms. It is not symmetrical around the F- —F bond axis. Atomic orbitals in pi bonding overlap less than in sigma bonding. Therefore, pi bonds tend to be weaker than sigma bonds.
VESPR—not very small elephants playing in the rain…
The valence-shell electron-pair repulsion theory, or VSEPR theory, explains the three-dimensional shape of methane. According to VSEPR theory, the repulsion between electron pairs causes molecular shapes to adjust so that the valence-electron pairs stay as far apart as possible. The methane molecule has four bonding electron pairs and no unshared pairs. The bonding pairs are farthest apart when the angle between the central carbon and its attached hydrogens is 109.5°. This is the H— C —H bond angle found experimentally.
Unshared pairs of electrons are also important in predicting the shapes of molecules. The nitrogen in ammonia (NH3) is surrounded by four pairs of valence electrons, so you might predict the tetrahedral angle of 109.5° for the H—N—H bond angle. However, one of the valence-electron pairs shown in Figure 8.16b is an unshared pair. No bonding atom is vying for these unshared electrons. Thus they are held closer to the nitrogen than are the bonding pairs. The unshared pair strongly repels the bonding pairs, pushing them together. The measured H—N—H bond angle is only 107°.
The model for covalent bonding you have been using assumes that the orbitals are those of the individual atoms. There is a quantum mechanical model of bonding, however, that describes the electrons in molecules using orbitals that exist only for groupings of atoms. When two atoms combine, this model assumes that their atomic orbitals overlap to produce molecular orbitals, or orbitals that apply to the entire molecule.
In some ways, atomic orbitals and molecular orbitals are similar. Just as an atomic orbital belongs to a particular atom, a molecular orbital belongs to a molecule as a whole. Each atomic orbital is filled if it contains two electrons. Similarly, two electrons are required to fill a molecular orbital. A molecular orbital that can be occupied by two electrons of a covalent bond is called a bonding orbital.
Sigma Bonds
When two atomic orbitals combine to form a molecular orbital that is symmetrical around the axis connecting two atomic nuclei, a sigma bond is formed, as illustrated in Figure 8.13. The symbol for this bond is the Greek letter sigma (σ).
In general, covalent bonding results from an imbalance between the attractions and repulsions of the nuclei and electrons involved. Because their charges have opposite signs, the nuclei and electrons attract each other. Because their charges have the same sign, nuclei repel other nuclei and electrons repel other electrons. In a hydrogen molecule, the nuclei repel each other, as do the electrons. In a bonding molecular orbital of hydrogen, however, the attractions between the hydrogen nuclei and the electrons are stronger than the repulsions. The balance of all the interactions between the hydrogen atoms is thus tipped in favor of holding the atoms together. The result is a stable diatomic molecule of H2.
Atomic p orbitals can also overlap to form molecular orbitals. A fluorine atom, for example, has a half-filled 2p orbital. When two fluorine atoms combine, as shown in Figure 8.14, the p orbitals overlap to produce a bonding molecular orbital. There is a high probability of finding a pair of electrons between the positively charged nuclei of the two fluorines. The fluorine nuclei are attracted to this region of high electron density. This attraction holds the atoms together in the fluorine molecule (F2). The overlap of the 2p orbitals produces a bonding molecular orbital that is symmetrical when viewed around the F —F bond axis connecting the nuclei. Therefore, the F —F bond is a sigma bond.
Pi Bonds
In the sigma bond of the fluorine molecule, the p atomic orbitals overlap end-to-end. In some molecules, however, orbitals can overlap side-by-side. As shown in Figure 8.15, the side-by-side overlap of atomic p orbitals produces what are called pi molecular orbitals. When a pi molecular orbital is filled with two electrons, a pi bond results. In a pi bond (symbolized by the Greek letter π), the bonding electrons are most likely to be found in sausage-shaped regions above and below the bond axis of the bonded atoms. It is not symmetrical around the F- —F bond axis. Atomic orbitals in pi bonding overlap less than in sigma bonding. Therefore, pi bonds tend to be weaker than sigma bonds.
VESPR—not very small elephants playing in the rain…
The valence-shell electron-pair repulsion theory, or VSEPR theory, explains the three-dimensional shape of methane. According to VSEPR theory, the repulsion between electron pairs causes molecular shapes to adjust so that the valence-electron pairs stay as far apart as possible. The methane molecule has four bonding electron pairs and no unshared pairs. The bonding pairs are farthest apart when the angle between the central carbon and its attached hydrogens is 109.5°. This is the H— C —H bond angle found experimentally.
Unshared pairs of electrons are also important in predicting the shapes of molecules. The nitrogen in ammonia (NH3) is surrounded by four pairs of valence electrons, so you might predict the tetrahedral angle of 109.5° for the H—N—H bond angle. However, one of the valence-electron pairs shown in Figure 8.16b is an unshared pair. No bonding atom is vying for these unshared electrons. Thus they are held closer to the nitrogen than are the bonding pairs. The unshared pair strongly repels the bonding pairs, pushing them together. The measured H—N—H bond angle is only 107°.
Ch 8 Orbitals
Ch 8
Key Concepts
• How are atomic and molecular orbitals related?
• How does VSEPR theory help predict the shapes of molecules?
• In what ways is orbital hybridization useful in describing molecules?
Vocabulary
• molecular orbitals
• bonding orbital
• sigma bond
• pi bond
• tetrahedral angle
• VSEPR theory
• hybridization
Key Concepts
• How are atomic and molecular orbitals related?
• How does VSEPR theory help predict the shapes of molecules?
• In what ways is orbital hybridization useful in describing molecules?
Vocabulary
• molecular orbitals
• bonding orbital
• sigma bond
• pi bond
• tetrahedral angle
• VSEPR theory
• hybridization
Problems Ch 16
(16) Key Concept How do you calculate the molarity of a solution? Hint
(17) Key Concept Compare the number of moles of solute before dilution with the number of moles of solute after dilution. Hint
(18) Key Concept What are two ways of expressing the concentration of a solution as a percent? Hint
(19)Calculate the molarity of a solution containing 400 g CuSO4 in 4.00 L of solution.
(20)How many moles of solute are present in 50.0 mL of 0.20M KNO3?
(21)How many milliliters of a stock solution of 2.00M KNO3 would you need to prepare 100.0 mL of 0.150M KNO3?
(22)What is the concentration, in percent (v/v), of a solution containing 50 mL of diethyl ether (C4H10O) in 2.5 L of solution?
(23)How many grams of K2SO4 would you need to prepare 1500 g of 5.0% K2SO4 (m/m) solution?
(17) Key Concept Compare the number of moles of solute before dilution with the number of moles of solute after dilution. Hint
(18) Key Concept What are two ways of expressing the concentration of a solution as a percent? Hint
(19)Calculate the molarity of a solution containing 400 g CuSO4 in 4.00 L of solution.
(20)How many moles of solute are present in 50.0 mL of 0.20M KNO3?
(21)How many milliliters of a stock solution of 2.00M KNO3 would you need to prepare 100.0 mL of 0.150M KNO3?
(22)What is the concentration, in percent (v/v), of a solution containing 50 mL of diethyl ether (C4H10O) in 2.5 L of solution?
(23)How many grams of K2SO4 would you need to prepare 1500 g of 5.0% K2SO4 (m/m) solution?
Ch 16: Percent Mass
Concentration in Percent (Mass/Mass)
Another way to express the concentration of a solution is as a percent (mass/mass), which is the number of grams of solute in 100 grams of solution. Percent by mass is sometimes a convenient unit of concentration when the solute is a solid. For example, a solution containing 7 g of sodium chloride in 100 grams of solution is 7 percent (mass/mass), or 7% (m/m).
Suppose you want to make 2000 g of a solution of glucose in water that has a 2.8% (m/m) concentration of glucose. How much glucose should you use? In a 2.8 percent solution, each 100 g of solution contains 2.8 g of solute. Thus, to find the grams of solute, you can multiply the mass of the solution by the ratio of grams of solute to grams of solution.
How much solvent (water) should be used? The mass of the solvent equals the mass of the solution minus the mass of the solute or 1944 g (2000 g − 56 g). Thus a 2.8% (m/m) glucose solution contains 56 g of glucose dissolved in 1944 g of water.
Information is often expressed as percent composition on food labels. For example, the label on a fruit drink or on maple-flavored pancake syrup should show the percent of fruit juice or maple syrup contained in the product. Such information can be misleading unless the units are given. When you use percentages to express concentration, be sure to state the units (v/v) or (m/m).
Another way to express the concentration of a solution is as a percent (mass/mass), which is the number of grams of solute in 100 grams of solution. Percent by mass is sometimes a convenient unit of concentration when the solute is a solid. For example, a solution containing 7 g of sodium chloride in 100 grams of solution is 7 percent (mass/mass), or 7% (m/m).
Suppose you want to make 2000 g of a solution of glucose in water that has a 2.8% (m/m) concentration of glucose. How much glucose should you use? In a 2.8 percent solution, each 100 g of solution contains 2.8 g of solute. Thus, to find the grams of solute, you can multiply the mass of the solution by the ratio of grams of solute to grams of solution.
How much solvent (water) should be used? The mass of the solvent equals the mass of the solution minus the mass of the solute or 1944 g (2000 g − 56 g). Thus a 2.8% (m/m) glucose solution contains 56 g of glucose dissolved in 1944 g of water.
Information is often expressed as percent composition on food labels. For example, the label on a fruit drink or on maple-flavored pancake syrup should show the percent of fruit juice or maple syrup contained in the product. Such information can be misleading unless the units are given. When you use percentages to express concentration, be sure to state the units (v/v) or (m/m).
CH 16: Percent Solutions
Percent Solutions
Another way to describe the concentration of a solution is by the percent of a solute in the solvent. The concentration of a solution in percent can be expressed in two ways: as the ratio of the volume of the solute to the volume of the solution or as the ratio of the mass of the solute to the mass of the solution.
Concentration in Percent (Volume/Volume)
If both the solute and the solvent are liquids, a convenient way to make a solution is to measure the volumes of the solute and the solution. The concentration of the solute is then expressed as a percent of the solution by volume. For example, isopropyl alcohol (2-propanol) is sold as a 91% solution, as shown in Figure 16.12. This solution consists of 91 mL of isopropyl alcohol mixed with enough water to make 100 mL of solution. The concentration can be expressed as 91 percent (volume/volume), or 91% (v/v). The relationship between percent by volume and the volumes of solute and solution is
Figure 16.12 The label clearly distinguishes this solution of isopropyl alcohol from rubbing alcohol which is a 70% solution of isopropyl alcohol. Applying Concepts How many milliliters of isopropyl alcohol are in 100 mL of 91% alcohol?
Another way to describe the concentration of a solution is by the percent of a solute in the solvent. The concentration of a solution in percent can be expressed in two ways: as the ratio of the volume of the solute to the volume of the solution or as the ratio of the mass of the solute to the mass of the solution.
Concentration in Percent (Volume/Volume)
If both the solute and the solvent are liquids, a convenient way to make a solution is to measure the volumes of the solute and the solution. The concentration of the solute is then expressed as a percent of the solution by volume. For example, isopropyl alcohol (2-propanol) is sold as a 91% solution, as shown in Figure 16.12. This solution consists of 91 mL of isopropyl alcohol mixed with enough water to make 100 mL of solution. The concentration can be expressed as 91 percent (volume/volume), or 91% (v/v). The relationship between percent by volume and the volumes of solute and solution is
Figure 16.12 The label clearly distinguishes this solution of isopropyl alcohol from rubbing alcohol which is a 70% solution of isopropyl alcohol. Applying Concepts How many milliliters of isopropyl alcohol are in 100 mL of 91% alcohol?
Ch 16: Dilutions
Figure 16.10 The student is preparing 100 mL of 0.40M MgSO4 from a stock solution of 2.0M MgSO4. She measures 20 mL of the stock solution with a 20-mL pipet. She transfers the 20 mL to a 100-mL volumetric flask. She carefully adds water to the mark to make 100 mL of solution. Inferring How many significant figures does the new molarity have?
Moles of solute before dilution = moles of solute after dilution
Recall the definition of molarity.
Rearranging the equation gives an expression for moles of solute.
Moles of solute = molarity (M) × liters of solution (V)
The total number of moles of solute remains unchanged upon dilution, so you can write this equation.
Moles of solute = M1 × V1 = M2 × V2
M1 and V1 are the molarity and volume of the initial solution, and M2 and V2 are the molarity and volume of the diluted solution. Volumes can be in liters or milliliters, as long as the same units are used for both V1 and V2.
Moles of solute before dilution = moles of solute after dilution
Recall the definition of molarity.
Rearranging the equation gives an expression for moles of solute.
Moles of solute = molarity (M) × liters of solution (V)
The total number of moles of solute remains unchanged upon dilution, so you can write this equation.
Moles of solute = M1 × V1 = M2 × V2
M1 and V1 are the molarity and volume of the initial solution, and M2 and V2 are the molarity and volume of the diluted solution. Volumes can be in liters or milliliters, as long as the same units are used for both V1 and V2.
Ch 16: Molarity Calculations
Molarity revisited
Sometimes, you may need to determine the number of moles of solute dissolved in a given volume of solution. You can do this if the molarity of the solution is known. For example, how many moles are in 2.00 L of 2.5M lithium chloride (LiCl)? Rearrange the formula for molarity to solve for the number of moles.
Moles of solute = molarity (M) × liters of solution (V)
Thus 2.00 L of 2.5 M lithium chloride solution contains 5.0 mol LiCl.
Sometimes, you may need to determine the number of moles of solute dissolved in a given volume of solution. You can do this if the molarity of the solution is known. For example, how many moles are in 2.00 L of 2.5M lithium chloride (LiCl)? Rearrange the formula for molarity to solve for the number of moles.
Moles of solute = molarity (M) × liters of solution (V)
Thus 2.00 L of 2.5 M lithium chloride solution contains 5.0 mol LiCl.
Ch 16: Molarity Calculations
Key Concepts
• How do you calculate the molarity of a solution?
• What effect does dilution have on the total moles of solute in solution?
• What are two ways to express the percent concentration of a solution?
Vocabulary
• concentration
• dilute solution
• concentrated solution
• molarity (M)
Reading Strategy
Summarizing When you summarize, you restate the key ideas in your own words. As you read about molarity, making dilutions, and percent solutions, summarize the main ideas in the text. In your summary, be sure to include all the key terms and the sentences in boldfaced type.
• How do you calculate the molarity of a solution?
• What effect does dilution have on the total moles of solute in solution?
• What are two ways to express the percent concentration of a solution?
Vocabulary
• concentration
• dilute solution
• concentrated solution
• molarity (M)
Reading Strategy
Summarizing When you summarize, you restate the key ideas in your own words. As you read about molarity, making dilutions, and percent solutions, summarize the main ideas in the text. In your summary, be sure to include all the key terms and the sentences in boldfaced type.
Ch 16. Solubility factors: Pressure
Pressure
Changes in pressure have little affect on the solubility of solids and liquids, but pressure strongly influences the solubility of gases. Gas solubility increases as the partial pressure of the gas above the solution increases. Carbonated beverages are a good example. These drinks contain large amounts of carbon dioxide (CO2) dissolved in water. Dissolved CO2 makes the liquid fizz and your mouth tingle. The drinks are bottled under a high pressure of CO2 gas, which forces large amounts of the gas into solution. When a carbonated-beverage container is opened, the partial pressure of CO2 above the liquid decreases. Immediately, bubbles of CO2 form in the liquid and escape from the open bottle,
Changes in pressure have little affect on the solubility of solids and liquids, but pressure strongly influences the solubility of gases. Gas solubility increases as the partial pressure of the gas above the solution increases. Carbonated beverages are a good example. These drinks contain large amounts of carbon dioxide (CO2) dissolved in water. Dissolved CO2 makes the liquid fizz and your mouth tingle. The drinks are bottled under a high pressure of CO2 gas, which forces large amounts of the gas into solution. When a carbonated-beverage container is opened, the partial pressure of CO2 above the liquid decreases. Immediately, bubbles of CO2 form in the liquid and escape from the open bottle,
Ch 16. Solubility factors: Pressure
Pressure
Changes in pressure have little affect on the solubility of solids and liquids, but pressure strongly influences the solubility of gases. Gas solubility increases as the partial pressure of the gas above the solution increases. Carbonated beverages are a good example. These drinks contain large amounts of carbon dioxide (CO2) dissolved in water. Dissolved CO2 makes the liquid fizz and your mouth tingle. The drinks are bottled under a high pressure of CO2 gas, which forces large amounts of the gas into solution. When a carbonated-beverage container is opened, the partial pressure of CO2 above the liquid decreases. Immediately, bubbles of CO2 form in the liquid and escape from the open bottle,
Changes in pressure have little affect on the solubility of solids and liquids, but pressure strongly influences the solubility of gases. Gas solubility increases as the partial pressure of the gas above the solution increases. Carbonated beverages are a good example. These drinks contain large amounts of carbon dioxide (CO2) dissolved in water. Dissolved CO2 makes the liquid fizz and your mouth tingle. The drinks are bottled under a high pressure of CO2 gas, which forces large amounts of the gas into solution. When a carbonated-beverage container is opened, the partial pressure of CO2 above the liquid decreases. Immediately, bubbles of CO2 form in the liquid and escape from the open bottle,
Ch 16. Solubility factors: Temperature on Gases
The effect of temperature on the solubility of gases in liquid solvents is opposite that of solids. The solubilities of most gases are greater in cold water than in hot. For example, Table 16.1 shows that the most important component of air for living beings—oxygen—becomes less soluble in water as the temperature of the solution rises. This fact has some important consequences. When an industrial plant takes water from a lake to use for cooling and then dumps the resulting heated water back into the lake, the temperature of the entire lake increases. Such a change in temperature is known as thermal pollution. Aquatic animal and plant life can be severely affected because the increase in temperature lowers the concentration of dissolved oxygen in the lake water.
Ch 16: Super saturation
Suppose you make a saturated solution of sodium ethanoate (sodium acetate) at 30°C and let the solution stand undisturbed as it cools to 25°C. Because the solubility of this compound is greater at 30°C than at 25°C, you expect that solid sodium ethanoate will crystallize from the solution as the temperature drops. But no crystals form. You have made a supersaturated solution. A supersaturated solution contains more solute than it can theoretically hold at a given temperature. The crystallization of a supersaturated solution can be initiated if a very small crystal, called a seed crystal, of the solute is added. The rate at which excess solute deposits upon the surface of a seed crystal can be very rapid, as shown in Figure 16.6. Crystallization can also occur if the inside of the container is scratched.
Figure 16.6 A supersaturated solution crystallizes rapidly when disturbed. The solution is clear before a seed crystal is added. Crystals begin to form in the solution immediately after the addition of a seed crystal. Excess solute crystallizes rapidly. Applying Concepts When the crystallization has ceased, will the solution be saturated or unsaturated?
Another example of crystallization in a supersaturated solution is the production of rock candy. A solution is supersaturated with sugar. Seed crystals cause the sugar to crystallize out of solution onto a string for you to enjoy!
Figure 16.6 A supersaturated solution crystallizes rapidly when disturbed. The solution is clear before a seed crystal is added. Crystals begin to form in the solution immediately after the addition of a seed crystal. Excess solute crystallizes rapidly. Applying Concepts When the crystallization has ceased, will the solution be saturated or unsaturated?
Another example of crystallization in a supersaturated solution is the production of rock candy. A solution is supersaturated with sugar. Seed crystals cause the sugar to crystallize out of solution onto a string for you to enjoy!
Ch 16. Solubility factors
Factors Affecting Solubility
You have read that solubility is defined as the mass of solute that dissolves in a given mass of a solvent at a specified temperature. Temperature affects the solubility of solid, liquid, and gaseous solutes in a solvent; both temperature and pressure affect the solubility of gaseous solutes.
Temperature
The solubility of most solid substances increases as the temperature of the solvent increases. Figure 16.4 shows how the solubility of several substances changes as temperature increases. The mineral deposits around hot springs, such as the one shown in Figure 16.5, result from the cooling of the hot, saturated solution of minerals emerging from the spring. As the solution cools in air, it cannot contain the same concentration of minerals as it did at a higher temperature, so some of the minerals precipitate.
Figure 16.5 Mineral deposits form around the edges of this hot spring because the hot water is saturated with minerals. As the water cools, some of the minerals crystallize because they are less soluble at the lower temperature.
For a few substances, solubility decreases with temperature. For example, the solubility of ytterbium sulfate (Yb2(SO4)3) in water drops from 44.2 g per 100 g of water at 0°C to 5.8 g per 100 g of water at 90°C. Table 16.1 lists the solubilities of some common substances at various temperatures.
You have read that solubility is defined as the mass of solute that dissolves in a given mass of a solvent at a specified temperature. Temperature affects the solubility of solid, liquid, and gaseous solutes in a solvent; both temperature and pressure affect the solubility of gaseous solutes.
Temperature
The solubility of most solid substances increases as the temperature of the solvent increases. Figure 16.4 shows how the solubility of several substances changes as temperature increases. The mineral deposits around hot springs, such as the one shown in Figure 16.5, result from the cooling of the hot, saturated solution of minerals emerging from the spring. As the solution cools in air, it cannot contain the same concentration of minerals as it did at a higher temperature, so some of the minerals precipitate.
Figure 16.5 Mineral deposits form around the edges of this hot spring because the hot water is saturated with minerals. As the water cools, some of the minerals crystallize because they are less soluble at the lower temperature.
For a few substances, solubility decreases with temperature. For example, the solubility of ytterbium sulfate (Yb2(SO4)3) in water drops from 44.2 g per 100 g of water at 0°C to 5.8 g per 100 g of water at 90°C. Table 16.1 lists the solubilities of some common substances at various temperatures.
Word Origins Miscible
Word Origins
Miscible comes from the Latin word miscere, meaning “to mix.” Completely miscible liquids dissolve in each other in all proportions. If the prefix im- means “not,” what would you call two liquids that are insoluble in each other?
Miscible comes from the Latin word miscere, meaning “to mix.” Completely miscible liquids dissolve in each other in all proportions. If the prefix im- means “not,” what would you call two liquids that are insoluble in each other?
Ch 16.1 Solution Formation
1. Stirring and Solution Formation
If a teaspoon of granulated sugar (sucrose) is placed in a glass of tea, the crystals dissolve slowly. If the contents of the glass are stirred, however, the crystals dissolve more quickly. The dissolving process occurs at the surface of the sugar crystals. Stirring speeds up the process because fresh solvent (the water in tea) is continually brought into contact with the surface of the solute (sugar). It's important to realize, however, that agitation (stirring or shaking) affects only the rate at which a solid solute dissolves. It does not influence the amount of solute that will dissolve. An insoluble substance remains undissolved regardless of how vigorously or for how long the solvent/solute system is agitated.
Temperature and Solution Formation
Temperature also influences the rate at which a solute dissolves. Sugar dissolves much more rapidly in hot tea than in iced tea. At higher temperatures, the kinetic energy of water molecules is greater than at lower temperatures so they move faster. The more rapid motion of the solvent molecules leads to an increase in the frequency and the force of the collisions between water molecules and the surfaces of the sugar crystals.
Particle Size and Solution Formation
The rate at which a solute dissolves also depends upon the size of the solute particles. A spoonful of granulated sugar dissolves more quickly than a sugar cube because the smaller particles in granulated sugar expose a much greater surface area to the colliding water molecules. Remember, the dissolving process is a surface phenomenon. The more surface of the solute that is exposed, the faster the rate of dissolving.
If a teaspoon of granulated sugar (sucrose) is placed in a glass of tea, the crystals dissolve slowly. If the contents of the glass are stirred, however, the crystals dissolve more quickly. The dissolving process occurs at the surface of the sugar crystals. Stirring speeds up the process because fresh solvent (the water in tea) is continually brought into contact with the surface of the solute (sugar). It's important to realize, however, that agitation (stirring or shaking) affects only the rate at which a solid solute dissolves. It does not influence the amount of solute that will dissolve. An insoluble substance remains undissolved regardless of how vigorously or for how long the solvent/solute system is agitated.
Temperature and Solution Formation
Temperature also influences the rate at which a solute dissolves. Sugar dissolves much more rapidly in hot tea than in iced tea. At higher temperatures, the kinetic energy of water molecules is greater than at lower temperatures so they move faster. The more rapid motion of the solvent molecules leads to an increase in the frequency and the force of the collisions between water molecules and the surfaces of the sugar crystals.
Particle Size and Solution Formation
The rate at which a solute dissolves also depends upon the size of the solute particles. A spoonful of granulated sugar dissolves more quickly than a sugar cube because the smaller particles in granulated sugar expose a much greater surface area to the colliding water molecules. Remember, the dissolving process is a surface phenomenon. The more surface of the solute that is exposed, the faster the rate of dissolving.
Ch 16: Solutions Key Terms/Questions
16.1
Key Concepts
1. What factors determine the rate at which a substance dissolves?
2. How is solubility usually expressed?
3. What conditions determine the amount of solute that will dissolve in a given solvent?
Vocabulary
• saturated solution
• solubility
• unsaturated solution
• miscible
• immiscible
• supersaturated solution
• Henry’s law
Key Concepts
1. What factors determine the rate at which a substance dissolves?
2. How is solubility usually expressed?
3. What conditions determine the amount of solute that will dissolve in a given solvent?
Vocabulary
• saturated solution
• solubility
• unsaturated solution
• miscible
• immiscible
• supersaturated solution
• Henry’s law
Changes to AC Chemistry
Mr. Gary is implementing a new program in lieu of our action research plan for the academic year 2010. The impact of multimedia and online assistance for class supplementation of work, quizzes, and laboratory assignments will be tested as well as an extensive assessment of student understanding, providing tiered lessons for both the advanced and remedial student. Pragmatically, what does this mean? More structure, more testing, and more accountability combined with more opportunities for bonus and enrichment activities.
Step 1. www.acchemistry.blogspot.com will be updated weekly to include all assignments for the week for each section—from reading, notes, questions to be answered, and quiz topics.
Step 2. Every double period will involve a quiz on the vocabulary of that week’s section, pertinent problems in the notes, and past problems.
Step 3. Notebooks will be checked every two weeks along with progress reports for any deficient students being sent home to parents.
Step 4. The website will have multimedia sections, further notes, and problem sets that can be done for improvement as well as an opportunity to raise quiz scores with a quiz make up—up to one grade.
Step 5. Grades will be tracked and students surveyed at the end of sessions comparing past performance with current performance to see if there are any improvements. Hits will be tracked on the website to see if activity increases over time. Finally, notebook samples will be checked to see any improvement.
Step 1. www.acchemistry.blogspot.com will be updated weekly to include all assignments for the week for each section—from reading, notes, questions to be answered, and quiz topics.
Step 2. Every double period will involve a quiz on the vocabulary of that week’s section, pertinent problems in the notes, and past problems.
Step 3. Notebooks will be checked every two weeks along with progress reports for any deficient students being sent home to parents.
Step 4. The website will have multimedia sections, further notes, and problem sets that can be done for improvement as well as an opportunity to raise quiz scores with a quiz make up—up to one grade.
Step 5. Grades will be tracked and students surveyed at the end of sessions comparing past performance with current performance to see if there are any improvements. Hits will be tracked on the website to see if activity increases over time. Finally, notebook samples will be checked to see any improvement.
Wednesday, August 18, 2010
Debate goes to PDS Tuesday Aug 24, 2010
The Debate Team will be going to PDS on Tuesday Aug 24, 2010 for a crosstraining from 4:30 p.m. to 7:30 p.m.
Retest EP 4
do this hand written in your notebook:
page 123 #58, 61, 62, 70, 73,
page 124 #85
page 125 #1-9
page 123 #58, 61, 62, 70, 73,
page 124 #85
page 125 #1-9
Monday, July 19, 2010
EP 5 and 6 Matching Sample
Match each item with the correct statement below.
a. anode d. half-cell
b. battery e. cathode
c. fuel cell
____ 1. the electrode at which oxidation occurs
____ 2. one part of a voltaic cell in which either oxidation or reduction occurs
____ 3. the electrode at which reduction occurs
____ 4. a group of cells that are connected together
____ 5. a voltaic cell in which a fuel substance undergoes oxidation and from which electrical energy is obtained continuously
Match each item with the correct statement below.
a. electrode d. voltaic cell
b. electrolysis e. dry cell
c. salt bridge
____ 6. a tube containing a conducting solution
____ 7. a conductor in a circuit that carries electrons to or from a substance other than a metal
____ 8. an electrochemical cell that is used to convert chemical energy to electrical energy
____ 9. a voltaic cell in which the electrolyte is a paste
____ 10. a process in which electrical energy is used to bring about a chemical change
Match each item with the correct statement below.
a. activity series of metals c. combustion reaction
b. single-replacement reaction d. decomposition reaction
____ 24. a reaction in which a single compound is broken down into simpler substances
____ 25. a reaction in which oxygen reacts with another substance, often producing heat or light
____ 26. a reaction in which the atoms of one element replace the atoms of a second element in a compound
____ 27. a list of metals in order of decreasing reactivity
matching answers
MATCHING
1. ANS: A DIF: L1 REF: p. 665 OBJ: 21.1.2
2. ANS: D DIF: L1 REF: p. 665 OBJ: 21.1.2
3. ANS: E DIF: L1 REF: p. 665 OBJ: 21.1.2
4. ANS: B DIF: L1 REF: p. 668 OBJ: 21.1.4
5. ANS: C DIF: L1 REF: p. 669 OBJ: 21.1.5
6. ANS: C DIF: L1 REF: p. 665 OBJ: 21.1.2
7. ANS: A DIF: L1 REF: p. 665 OBJ: 21.1.2
8. ANS: D DIF: L1 REF: p. 665 OBJ: 21.1.2
9. ANS: E DIF: L1 REF: p. 667 OBJ: 21.1.3
10. ANS: B DIF: L1 REF: p. 678 OBJ: 21.3.1
24. ANS: D DIF: L1 REF: p. 332 OBJ: 11.2.1
25. ANS: C DIF: L1 REF: p. 336, p. 337
OBJ: 11.2.1
26. ANS: B DIF: L1 REF: p. 333 OBJ: 11.2.1
27. ANS: A DIF: L1 REF: p. 333 OBJ: 11.2.2
28. ANS: C DIF: L1 REF: p. 294 OBJ: 10.1.3
29. ANS: B DIF: L1 REF: p. 294, p. 295
OBJ: 10.1.3, 10.1.4
30. ANS: A DIF: L1 REF: p. 300 OBJ: 10.2.2
p. 681 OBJ: 21.3.3
a. anode d. half-cell
b. battery e. cathode
c. fuel cell
____ 1. the electrode at which oxidation occurs
____ 2. one part of a voltaic cell in which either oxidation or reduction occurs
____ 3. the electrode at which reduction occurs
____ 4. a group of cells that are connected together
____ 5. a voltaic cell in which a fuel substance undergoes oxidation and from which electrical energy is obtained continuously
Match each item with the correct statement below.
a. electrode d. voltaic cell
b. electrolysis e. dry cell
c. salt bridge
____ 6. a tube containing a conducting solution
____ 7. a conductor in a circuit that carries electrons to or from a substance other than a metal
____ 8. an electrochemical cell that is used to convert chemical energy to electrical energy
____ 9. a voltaic cell in which the electrolyte is a paste
____ 10. a process in which electrical energy is used to bring about a chemical change
Match each item with the correct statement below.
a. activity series of metals c. combustion reaction
b. single-replacement reaction d. decomposition reaction
____ 24. a reaction in which a single compound is broken down into simpler substances
____ 25. a reaction in which oxygen reacts with another substance, often producing heat or light
____ 26. a reaction in which the atoms of one element replace the atoms of a second element in a compound
____ 27. a list of metals in order of decreasing reactivity
matching answers
MATCHING
1. ANS: A DIF: L1 REF: p. 665 OBJ: 21.1.2
2. ANS: D DIF: L1 REF: p. 665 OBJ: 21.1.2
3. ANS: E DIF: L1 REF: p. 665 OBJ: 21.1.2
4. ANS: B DIF: L1 REF: p. 668 OBJ: 21.1.4
5. ANS: C DIF: L1 REF: p. 669 OBJ: 21.1.5
6. ANS: C DIF: L1 REF: p. 665 OBJ: 21.1.2
7. ANS: A DIF: L1 REF: p. 665 OBJ: 21.1.2
8. ANS: D DIF: L1 REF: p. 665 OBJ: 21.1.2
9. ANS: E DIF: L1 REF: p. 667 OBJ: 21.1.3
10. ANS: B DIF: L1 REF: p. 678 OBJ: 21.3.1
24. ANS: D DIF: L1 REF: p. 332 OBJ: 11.2.1
25. ANS: C DIF: L1 REF: p. 336, p. 337
OBJ: 11.2.1
26. ANS: B DIF: L1 REF: p. 333 OBJ: 11.2.1
27. ANS: A DIF: L1 REF: p. 333 OBJ: 11.2.2
28. ANS: C DIF: L1 REF: p. 294 OBJ: 10.1.3
29. ANS: B DIF: L1 REF: p. 294, p. 295
OBJ: 10.1.3, 10.1.4
30. ANS: A DIF: L1 REF: p. 300 OBJ: 10.2.2
p. 681 OBJ: 21.3.3
Exam ep 5 and 6
KNOW fractionating column on page 713 and coal on 714
page 726 the table of functional groups
page 726 the table of functional groups
Thursday, July 8, 2010
Tuesday, July 6, 2010
midterm exam topics July 21
ep 4 ch 1-6--protons, neutrons, electrons, atomic theories rutherford thomson bohr, electron configuration
ep 5: ch 22-23, 15 functional groups and hydro carbons plus aqueous systems
ep 6 ch 22-23, ch 20 and 21 functional groups and hydro carbons plus electrochemistry/oxidation-reduction
ep 5: ch 22-23, 15 functional groups and hydro carbons plus aqueous systems
ep 6 ch 22-23, ch 20 and 21 functional groups and hydro carbons plus electrochemistry/oxidation-reduction
Monday, July 5, 2010
Exams July 21
The Exams are coming around the corner. please be aware of topics and chapters. see website from school for specific pages
Monday, June 28, 2010
Debate Practices this week
Monday, Wednesday afterschool. We are planning to practice at PDS on Thursdays. Next Saturday is a July 10, 2010 Sarass Ektra
Monday, June 21, 2010
Thursday, June 10, 2010
Mr. G has Debate this weekend and Monday
so on Monday--ep 5/2, 6/2, and 4/2 you're taking tests...enjoy
Friday, June 4, 2010
Thursday, June 3, 2010
Make up quiz ep 6 and ep 5
these names are wrong, draw the formula, and give the proper name.
a. 2-ethylbutane
b. 2-propylpentane
c. 1, 1 dimethylpentane
d. 2,2-dimethyl-4-ethylpentane
a. 2-ethylbutane
b. 2-propylpentane
c. 1, 1 dimethylpentane
d. 2,2-dimethyl-4-ethylpentane
Tuesday, June 1, 2010
Monday, May 31, 2010
Friday, May 28, 2010
Debate Activiy on Thursday period 4
it's work and difficult at times, but you learn to speak english extremely well, argue, and write papers.
Thursday period 4 ep 6/1 room
Thursday period 4 ep 6/1 room
Sunday, May 23, 2010
EP 4 Assignments for Week 1 May 24
This is Summer topics
Ch 4--Dalton's atomic theory--page 100-103
JJ Thompson p. 104-106
Rutherford Gold-foil Experiment and Model p. 107
Atomic Number, Mass number, atomic mass, and isotopes p. 110-115
Calculating Atomic Mass p. 117
CH 5
Light p. 138-141
This week's focus:
Ch 5 Atomic Models overview p. 127-129
Quantum Mechanical Model p. 130
Atomic orbitals: p. 131-133
Electron Configurations p. 133-135
exceptional electron configurations p. 136
Balmer Series--p. 143 --next week--
Quantum Mechanics--p. 144-146
Test--two weeks from now.
assignments: write the questions--back of notebook
#1: p. 132 #1-7,
#2: p. 135 #8-9, p. 136 #10-13
Ch 4--Dalton's atomic theory--page 100-103
JJ Thompson p. 104-106
Rutherford Gold-foil Experiment and Model p. 107
Atomic Number, Mass number, atomic mass, and isotopes p. 110-115
Calculating Atomic Mass p. 117
CH 5
Light p. 138-141
This week's focus:
Ch 5 Atomic Models overview p. 127-129
Quantum Mechanical Model p. 130
Atomic orbitals: p. 131-133
Electron Configurations p. 133-135
exceptional electron configurations p. 136
Balmer Series--p. 143 --next week--
Quantum Mechanics--p. 144-146
Test--two weeks from now.
assignments: write the questions--back of notebook
#1: p. 132 #1-7,
#2: p. 135 #8-9, p. 136 #10-13
Wednesday, May 19, 2010
Monday, May 10, 2010
Wednesday, May 5, 2010
bonus: Looking for cool websites--send me a chem link
I'm making an ebook for chem and looking for cool websites to spice it up
Saturday, May 1, 2010
Thursday, April 22, 2010
last notebook check
if i haven't checked your notebook this summer, your grade is BAD... last time to see notebooks... friday 10 a.m.
Monday, April 19, 2010
make up quiz ep 6
draw the following structures:
1-methyl butane, 2, 3-dimethyl 3-pentyne, 1-ethyl, 2,3 dimethyl, 4-proplyl hexene.
what is hydrogenation? and what is the effect of adding chlorine to 4-octene.
1-methyl butane, 2, 3-dimethyl 3-pentyne, 1-ethyl, 2,3 dimethyl, 4-proplyl hexene.
what is hydrogenation? and what is the effect of adding chlorine to 4-octene.
Wednesday, April 14, 2010
Thursday, April 8, 2010
Tuesday, April 6, 2010
Random bonus: Driver for Samsung USB Modem 1.0
I need computer help from my wonderful computer experts out there, find me a link to download this or the driver. We have a Candy samsung and the cable: APCBS10UBE but it won't link up to my computer...
Thursday, April 1, 2010
Bonus for chem due April 7
define for me chemical and physical properties
give me a website address and print out the first page
give me a website address and print out the first page
Tuesday, March 30, 2010
Saturday, March 27, 2010
EU-Thailand Debate Workshop
This is what we do at debate, work in groups, cooperate, and present together.
The participants came from 11 schools and numbered over 70.
Mr. Gary along with Mr. LP, the Debate Director of PDS and organizer of this function.
Wednesday, March 24, 2010
Debate Activity on Thursday and Friday
any last minute additions to Debate must be cleared by the EP office and myself. ASAP
Sunday, March 21, 2010
Bonus Question for Summer School
EP 4--find me websites describing the following:
physical chemistry, analytical chemistry, organic chemistry, inorganic chemistry.
physical chemistry, analytical chemistry, organic chemistry, inorganic chemistry.
Thursday, March 4, 2010
no retest required for ep 4 and 5
Congratulations! Everyone has passed the semester so you don't have to retest for the exam, even if you failed.
Tuesday, March 2, 2010
High Scores EP 4-5
EP 4/1 Jiramate Chanaturakarnnon
EP 4/2 Chale Silpitaksakul
EP 5/1 Chanwit Iamudon
EP 5/2 Vasin Dumrongprechachan and Thanatham Julaphatachote PERFECT Score
EP 4/2 Chale Silpitaksakul
EP 5/1 Chanwit Iamudon
EP 5/2 Vasin Dumrongprechachan and Thanatham Julaphatachote PERFECT Score
EP 5/2 Failures
47255
High Score: Vasin and Thanatham PERFECT score... bummers...I mean CONGRATULATIONS!
High Score: Vasin and Thanatham PERFECT score... bummers...I mean CONGRATULATIONS!
Thursday, February 18, 2010
To balance redox reactions, assign oxidation numbers to the reactants and products to determine how many moles of each species are needed to conserve mass and charge. First, separate the equation into two half-reactions, the oxidation portion and the reduction portion. This is called the half-reaction method of balancing redox reactions or the ion-electron method. Each half-reaction is balanced separately and then the equations are added together to give a balanced overall reaction. We want the net charge and number of ions to be equal on both sides of the final balanced equation.
For this example, let's consider a redox reaction between KMnO4and HI in an acidic solution:
MnO4- + I- → I2 + Mn2+
from this link: http://chemistry.about.com/od/generalchemistry/ss/redoxbal.htm
http://www.science.uwaterloo.ca/~cchieh/cact/c123/balance.html
For this example, let's consider a redox reaction between KMnO4and HI in an acidic solution:
MnO4- + I- → I2 + Mn2+
from this link: http://chemistry.about.com/od/generalchemistry/ss/redoxbal.htm
http://www.science.uwaterloo.ca/~cchieh/cact/c123/balance.html
Tuesday, February 16, 2010
Monday, February 15, 2010
Debate: Capitalism is better than Communism
PM speech
LO speech
DPM
DLO i seem to be missing this speech
Govt whip
OPP whip
LO speech
DPM
DLO i seem to be missing this speech
Govt whip
OPP whip
Exam Content for Final
EP 5 Ch 19.3 plus an acid dissociation constant, ch 20 all--half reaction, oxidation change method, redox numbers, ch 21 electrochemical cells--half cells and potential.
EP 4 Ch 12.3 limiting reagent and percent yield, ch 13--gases, liquids, solids, change of state--triple point, ch 14 gas laws, graham's law
EP 4 Ch 12.3 limiting reagent and percent yield, ch 13--gases, liquids, solids, change of state--triple point, ch 14 gas laws, graham's law
Friday, February 12, 2010
Message to EP 6
1. Remember if you do well in Chemistry, mention my name. if you don't, say Dr. Dominic.
2. If you get caught doing something bad, you're old enough to be responsible now so don't get caught.
3. Debate
4. If you're in jail and have one phone call, and you call me to bail you out. I will after I leave you in jail for one night...
5. Challenge yourself to be a better person ALWAYS
moving on
2. If you get caught doing something bad, you're old enough to be responsible now so don't get caught.
3. Debate
4. If you're in jail and have one phone call, and you call me to bail you out. I will after I leave you in jail for one night...
5. Challenge yourself to be a better person ALWAYS
moving on
Tuesday, February 9, 2010
Mr. G is fine
But they're not letting me out of the hospital for one more day until Thursday due to a fever that comes and goes...SIGH
Monday, February 8, 2010
status of Mr. G
Thank you to all of the EP 5 students for their wonderful gifts and visit. I'm doing great and should be out by Wednesday as I'm under observation...
Tuesday, February 2, 2010
Sunday, January 31, 2010
Debate things I learned
case construction:
harms
model/principle
advantages.
leave one harm for the next speaker and one advantage.
ARGUE both ways.
example--they say access is required for farmers to reach market.
you say a.
there are roads in place that allow access.
and you say b
access is not required because the farmers do not produce commercial crops, but they produce crops only for sustenance.
that way if they don't buy a, you are covered with b.
government--your case has to be something that can work in the real world. that's how you win. it's something the current govt will not do at all.
poi's are not important whatsoever--taking them are unnecessary. i took none, and got best speaker of the round.
definitional challenge
a. you challenge the definition
b. you argue the case under the appropriate definition
c. you argue the opposition case against that case under the appropriate definition
d. you argue the case under their crazy definition.
therefore you are arguing three cases at the same time. this must be done by the LO
opposition case construction
present harms of the current system under the resolution--you're guessing where they are going.
you present two arguments in the first LO and give two arguments to the DLO
you refute the govt case--if you have to do the definitional challenge you do so.
if not, you rebut the govt case, and present your arguments against the resolution.
giving the split of what your second speaker will say.
BE SURE TO ALWAYS PRESENT YOUR OPTION--COUNTER MODEL COUNTER PRINCIPLE.
so you attack the govt case and build your status quo case
two separate situations in your speech at all times.
argue both ways as we suggested earlier.
govt proposed that voting in a flawed system is worthwhile
a. you state that voting in a flawed system such as thailand which has a weak democracy accomplishes nothing, when the ruling power are corrupt--buying votes, manipulating the system--voting accomplishes nothing at the end of the day, except costing money in setting up the vote, regulating/security for the fascade of the vote.
b. even if you vote in honesty, your vote is not heard as it is buried under the dominant power's heel.
c. you don't vote, and nothing happens anyway--you're not involved in the democracy and thus, eliminates your voice.
d. option is opposition solution, revolution because by erasing the flawed system, you accomplish a true democracy. yes, there are disadvantages, but then your vote will count for something, it's a true mechanism of change.
so in A--we accept the voting in a flawed system--means nothing. system never changes.
in B, we vote honestly, yet our vote is countered by bought votes, corruption. system never changes.
in C, we despair and don't vote, thus the system never changes.
in D, the system changes. there will be short term disadvantages, but there will be change. moving to a true democracy.
you cover all your bases in an opposition type argumentation and create massive linking.
harms
model/principle
advantages.
leave one harm for the next speaker and one advantage.
ARGUE both ways.
example--they say access is required for farmers to reach market.
you say a.
there are roads in place that allow access.
and you say b
access is not required because the farmers do not produce commercial crops, but they produce crops only for sustenance.
that way if they don't buy a, you are covered with b.
government--your case has to be something that can work in the real world. that's how you win. it's something the current govt will not do at all.
poi's are not important whatsoever--taking them are unnecessary. i took none, and got best speaker of the round.
definitional challenge
a. you challenge the definition
b. you argue the case under the appropriate definition
c. you argue the opposition case against that case under the appropriate definition
d. you argue the case under their crazy definition.
therefore you are arguing three cases at the same time. this must be done by the LO
opposition case construction
present harms of the current system under the resolution--you're guessing where they are going.
you present two arguments in the first LO and give two arguments to the DLO
you refute the govt case--if you have to do the definitional challenge you do so.
if not, you rebut the govt case, and present your arguments against the resolution.
giving the split of what your second speaker will say.
BE SURE TO ALWAYS PRESENT YOUR OPTION--COUNTER MODEL COUNTER PRINCIPLE.
so you attack the govt case and build your status quo case
two separate situations in your speech at all times.
argue both ways as we suggested earlier.
govt proposed that voting in a flawed system is worthwhile
a. you state that voting in a flawed system such as thailand which has a weak democracy accomplishes nothing, when the ruling power are corrupt--buying votes, manipulating the system--voting accomplishes nothing at the end of the day, except costing money in setting up the vote, regulating/security for the fascade of the vote.
b. even if you vote in honesty, your vote is not heard as it is buried under the dominant power's heel.
c. you don't vote, and nothing happens anyway--you're not involved in the democracy and thus, eliminates your voice.
d. option is opposition solution, revolution because by erasing the flawed system, you accomplish a true democracy. yes, there are disadvantages, but then your vote will count for something, it's a true mechanism of change.
so in A--we accept the voting in a flawed system--means nothing. system never changes.
in B, we vote honestly, yet our vote is countered by bought votes, corruption. system never changes.
in C, we despair and don't vote, thus the system never changes.
in D, the system changes. there will be short term disadvantages, but there will be change. moving to a true democracy.
you cover all your bases in an opposition type argumentation and create massive linking.
Saturday, January 30, 2010
Thursday, January 28, 2010
Assignments for January 28, 2010 week
ep 4: page 417 #1,2 pages 419-423 # 7-14
ep 5: page 652 #21 page 657 #40 b,c
ep 5: page 652 #21 page 657 #40 b,c
Sunday, January 24, 2010
RETEST
Make up exam must be handwritten and turned in by January 20
EP 4 page 347 # 40, 43, 53, 58. page 379 # 36, 39, 41-44
EP 5 page 581 # 44, 51, 53, 57, 60, 65 page 582 # 70-72
EP 6 p. 687 # 26-36
EP 4 page 347 # 40, 43, 53, 58. page 379 # 36, 39, 41-44
EP 5 page 581 # 44, 51, 53, 57, 60, 65 page 582 # 70-72
EP 6 p. 687 # 26-36
Wednesday, January 20, 2010
Assignments for EP 4
Assignment #1: page 374-75 #29-32 Page 380 #49-50
Assignment #2: page 395 #8, 9, 11, 12, 14
Assignment #2: page 395 #8, 9, 11, 12, 14
Assignments for Chem EP 5
EP 5 #1 assignment: page 634 #1, 2 page 638 #6, 7
#2 Assignment: page 641 #9, 10, page 643 #11, 12, 15, 16
#2 Assignment: page 641 #9, 10, page 643 #11, 12, 15, 16
Tuesday, January 19, 2010
Debate Tournament Jan 30-31 Triamudom Suksa
There will be the Legends Tournament there and our team is going. If you are interested in going, please contact me
Thursday, January 14, 2010
Monday, January 11, 2010
Make Up Exams for Chem Midterm
Make up exam must be handwritten and turned in by January 20
EP 4 page 347 # 40, 43, 53, 58. page 379 # 36, 39, 41-44
EP 5 page 581 # 44, 51, 53, 57, 60, 65 page 582 # 70-72
EP 6 p. 687 # 26-36
EP 4 page 347 # 40, 43, 53, 58. page 379 # 36, 39, 41-44
EP 5 page 581 # 44, 51, 53, 57, 60, 65 page 582 # 70-72
EP 6 p. 687 # 26-36
Tuesday, January 5, 2010
Military Camps for EP 5 and 6
EP 6... your exams are upon your return
EP5 we pick up where you left off.
EP5 we pick up where you left off.
Topics for Rest of the Semester
ep 6--stoichemistry ch 12
ep 5 acid/bases then oxidation/reduction
ep 4 limiting reactants, states of matter, and gas laws
ep 5 acid/bases then oxidation/reduction
ep 4 limiting reactants, states of matter, and gas laws
Monday, January 4, 2010
Sunday, January 3, 2010
EP 5 and 4 Grades for RETEST
EP 5 High Scores: EP 5/1 Thanakorn, Dhanachai, 445/450 EP 5/2 Vasin 450/450
ep 5/1 retest: none
ep 5/2 retest: none
Congratulations!
EP 4 High Scores: EP 4/1 Tirawit 385/420 EP 4/2 Chale 395/420
retests EP 4/1
43513, 47885, 48021, 50382, 50399, 50406
retests EP 4/2
50383
ep 5/1 retest: none
ep 5/2 retest: none
Congratulations!
EP 4 High Scores: EP 4/1 Tirawit 385/420 EP 4/2 Chale 395/420
retests EP 4/1
43513, 47885, 48021, 50382, 50399, 50406
retests EP 4/2
50383
EP 6 Exam Results for RETEST
High Scores: EP 6/1 Apinut 295/320 EP 6/2 Natee 315/320
retests due by January 14
ep 6/1 42277, 42300, 42354, 46524, 48846, 48855, 48861
ep 6/2
42346, 42357, 42416, 46547
retests due by January 14
ep 6/1 42277, 42300, 42354, 46524, 48846, 48855, 48861
ep 6/2
42346, 42357, 42416, 46547
Saturday, January 2, 2010
Posting of failures will be done when school starts
retests are due January 14. sorry...but hey, Happy New Year
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About Me
- Gary Hi10spro Sakuma
- I have played for 25 years and coached for the last 17 years--certified United States Professional Tennis Association Professional One--worked for Punahou Schools-voted the #1 Sports School in the United States, as a Program Supervisor, in charge of coaching the High Performance Players as well as coordinating programs for K-12 and Tennis Pro Education.