- WELCOME TO THE VIDEO TUTORIAL ON INTERMOLECULAR FORCES
IMF'S AND THEIR RELATIONSHIP TO BOILING POINTS,
ANOTHER GREAT EXAMPLE OF AN APPLICATION
OF INTERMOLECULAR FORCES.
SO LET'S LOOK AT WHAT IT MEANS ABOUT BOILING POINTS.
WE WANT TO UNDERSTAND THE CONCEPT OF VAPOR PRESSURE.
VAPOR PRESSURE IS THE PRESSURE EXERTED BY MOLECULES
IN GAS PHASES IN CONTACT WITH MOLECULES IN THE LIQUID PHASE.
SO HERE'S AN EXAMPLE WITH PERFUME.
SO WE CAN SEE THE CIRCLES REPRESENTING THE MOLECULES
IN THE LIQUID PHASE AND THEN IN THE GAS PHASE.
AND SO THIS IS THE VAPOR PRESSURE
HERE REPRESENTED AS THE GAS PHASE MOLECULES ARE
HITTING THE WALLS OF THE CONTAINER,
CREATING A PRESSURE FORCE.
NOW, FOR A SUBSTANCE TO BOIL,
THE TEMPERATURE HAS TO INCREASE,
SO REMEMBER THAT WHEN WE'RE TALKING ABOUT PHASE CHANGES,
IT'S THAT DYNAMICS BETWEEN KINETIC ENERGY
AND POTENTIAL ENERGY.
SO IF YOU NEED TO REVIEW ON THOSE CONCEPTS,
PLEASE LOOK AT THE ENERGY TUTORIAL.
BUT WHAT HAPPENS IS, RIGHT,
THAT THE POTENTIAL ENERGY OF THE MOLECULES CREATED
BY THE INTERMOLECULAR FORCES IS HOLDING THE MOLECULES
TOGETHER IN THE LIQUID PHASE.
SO FOR A LIQUID MOLECULE
TO JUMP OR EVAPORATE OR BOIL INTO THE GAS PHASE,
IT HAS TO OVERCOME THIS ATMOSPHERIC PRESSURE.
AND SO WE NEED KINETIC ENERGY.
SO THAT'S HOW TEMPERATURE COMES IN.
SO HERE, WE HAVE A GRAPH WHERE WE HAVE THE PRESSURE
AND WE COULD RELATE THIS.
SO THIS WOULD REALLY BE THE VAPOR PRESSURE OF THE SUBSTANCE,
AND THEN ACROSS THE X-AXIS, WE HAVE THE TEMPERATURE.
AND SO WE SEE THEN THAT AS WE INCREASE THE TEMPERATURE,
WE INCREASE THE VAPOR PRESSURE OF THE SUBSTANCE,
AND THEN WHEN THE VAPOR PRESSURE EQUALS THE ATMOSPHERIC PRESSURE,
THIS IS WHERE WE GET THE BOILING POINT.
AND SO WHAT IS SO MAGICAL ABOUT 760 MILLIMETERS OF MERCURY,
THAT'S THE EQUIVALENT OF ONE ATMOSPHERIC PRESSURE.
SO IT'S IMPORTANT TO NOTE
THAT ATMOSPHERIC PRESSURE DOES CHANGE WITH ALTITUDE.
AND SO AT SEA LEVEL,
THAT'S WHAT WE CONSIDER THE NORMAL CONDITION
OF ONE ATMOSPHERE.
AS WE RISE IN ELEVATION,
BY THE TIME WE GET TO 14,000 FEET,
ATMOSPHERIC PRESSURE HAS DROPPED TO 0.59 ATMOSPHERE.
VERY SIGNIFICANT.
AND THEN WHEN WE DROP BELOW SEA LEVEL,
THE OPPOSITE EFFECT OCCURS,
AND IF WE ARE 14,000 FEET BELOW SEA LEVEL,
THE PRESSURE HAS INCREASED ALL THE WAY TO 470 ATMOSPHERES.
SO ANOTHER WAY TO THINK ABOUT VAPOR PRESSURE IS TO JUST--
TO CONSIDER IT AS THE TENDENCY TO EVAPORATE.
SO AS WE INCREASE THE VAPOR PRESSURE,
WE HAVE AN INCREASED TENDENCY TO EVAPORATE.
ALL RIGHT.
SO NOW LET'S TIE THIS ALL TOGETHER
TO CONNECT WHAT'S HAPPENING WITH INTERMOLECULAR FORCES
AND THE RESULTING BOILING POINTS
OF PURE SUBSTANCES. ALL RIGHT.
SO THE MOLECULES IN THE LIQUID ARE IN CONSTANT MOTION.
IF A MOLECULE NEAR THE SURFACE HAS ENOUGH ENERGY
IN THE FORM OF KINETIC ENERGY,
IT CAN ESCAPE INTO THE GAS STATE OR VAPOR.
SO IF WE LOOK AT THIS,
THEN AS WE INCREASE THE TEMPERATURE,
WE INCREASE THE KINETIC ENERGY, RIGHT,
WHICH CAN THEN OVERCOME THE POTENTIAL ENERGY.
AND REMEMBER THAT THE POTENTIAL ENERGY WAS CREATED
BY THE INTERMOLECULAR FORCES.
ALL RIGHTY.
SO LET'S LOOK AT AN EXAMPLE HERE, RIGHT,
SO THE BOILING POINT IN SUMMARY IS--
LET'S GET THIS CENTERED.
THE TEMPERATURE AT WHICH THE VAPOR PRESSURE
OF A LIQUID EQUALS THE EXTERNAL
ATMOSPHERIC PRESSURE EXERTED ON A LIQUID.
SO WE HAVE A SIMILAR GRAPH TO THE PREVIOUS PAGE
BUT NOW WE'VE ADDED MORE COMPOUNDS SO WE CAN START TO SEE
HOW THE INTERMOLECULAR FORCES AFFECT THE BOILING POINT.
SO WE HAVE THE SAME THING RIGHT ALONG THE X-AXIS,
WE HAVE THE TEMPERATURE.
AND THEN ALONG THE Y-AXIS,
WE HAVE THE VAPOR PRESSURE OF EACH SUBSTANCE.
SO HERE WE'RE STARTING WITH CHLOROFORM.
IF WE LOOK AT THE MOLECULE OF CHLOROFORM,
WE RECOGNIZE THAT IT HAS POLAR BONDS,
THREE POLAR BONDS, RADIATING OUT INTO THE CORNERS
OF THE TETRAHEDRON SO THIS MOLECULE IS POLAR.
REMEMBER THAT ONCE A MOLECULE IS POLAR,
WE ASK OURSELVES WHAT IS A DIPOLE-DIPOLE OR H-BONDING.
SINCE THERE'S NO NITROGEN, OXYGEN, OR FLUORINE,
THIS MOLECULE IS CLEARLY DIPOLE-DIPOLE. OKAY.
SO HERE AT ZERO DEGREES IS THE VAPOR PRESSURE OF CHLOROFORM,
AND THEN AS WE INCREASE THE TEMPERATURE,
WE INCREASE THE VAPOR PRESSURE
UNTIL IT REACHES ATMOSPHERIC PRESSURE
AND SO THE BOILING POINT OF CHLOROFORM IS ABOUT 62 DEGREES.
NOW, LET'S LOOK AT THE NEXT MOLECULE, ETHANOL.
HERE, WE SEE IT'S ALSO POLAR.
BUT HERE, WE RECOGNIZE THAT IT HAS THE ABILITY TO H-BOND
BECAUSE OF THAT OXYGEN-HYDROGEN BOND.
SO AS WE DISCUSSED WITH INTERMOLECULAR FORCES
IN THE INTRO TUTORIAL,
HYDROGEN BONDING IS A STRONGER INTERMOLECULAR FORCE
THAN DIPOLE-DIPOLE.
SO QUITE NATURALLY, THE VAPOR PRESSURE IS LOWER.
SO IT TAKES MORE ENERGY,
MORE HEAT APPLIED RAISING THE TEMPERATURE
TO REACH THE BOILING POINT.
SO NOW, WITH ETHANOL,
THE BOILING POINT IS JUST OVER 78 DEGREES.
IF WE LOOK AT WATER,
WATER IS DEFINITELY ANOTHER H-BONDER
AND IT'S ACTUALLY A DOUBLE H-BONDER
BECAUSE WE HAVE THE TWO POLAR HYDROGENS
WITH THE CORRESPONDING OXYGEN.
SO WE HAVE ANOTHER H-BONDER,
AND BECAUSE OF THE SMALL SIZE OF WATER AND THE DOUBLE POLARITY,
WE GET A VERY STRONG INTERMOLECULAR FORCE,
AND SO WE SEE A VERY HIGH BOILING POINT FOR WATER.
AND THEN TO TAKE THIS TO AN EXTREME,
LET'S LOOK AT ACETIC ACID,
WHERE DEFINITELY WE HAVE H-BONDING
AND WE HAVE AN H-BOND ACCEPTOR WITH THE OXYGEN,
BUT THIS HYDROGEN HERE IS IONIZABLE,
WHICH IS WHY WE DESCRIBE IT AS AN ACID.
AND SO IT'S AN EQUILIBRIUM REACTION
WHERE IT CAN DISSOCIATE INTO AN ACETATE ION WITH A PROTON,
A HYDROGEN ION.
SO WE HAVE H-BONDING WITH FULL IONIC CHARACTER AND EQUILIBRIUM.
AND SO WE KNOW THAT HERE, WE'RE TALKING ABOUT H-BONDING,
AND HERE, WE'RE BACK TO IONIC.
SO THIS ACETIC ACID IS GOING TO HAVE THE STRONGEST
INTERMOLECULAR FORCES OF OUR SET AND CORRESPONDINGLY,
IT HAS THE HIGHEST BOILING POINT.
SO AS WE INCREASE THE INTERMOLECULAR FORCES,
WE INCREASE THE POTENTIAL ENERGY,
RIGHT? SO THEREFORE,
WE'RE GOING TO HAVE TO HAVE AN INCREASED TEMPERATURE
TO INCREASE THE KINETIC ENERGY TO REACH THE BOILING POINT.
ALL RIGHTY.
SO NOW THAT WE HAVE THE BASIC UNDERSTANDING,
LET'S PRACTICE APPLYING THIS KNOWLEDGE.
ALL RIGHT.
LET'S GET THIS CENTERED. OKAY.
SO BY NOW, THIS SHOULD BE REVIEW, RIGHT,
THAT THE RELATIVE STRENGTH OF THE BOILING POINTS,
H-BONDING IS STRONGER THAN DIPOLE-DIPOLE,
WHICH IS STRONGER THAN LONDON.
SO HERE, WE HAVE THE STRONGEST IMF,
WHICH MEANS WE'RE GOING TO HAVE THE HIGHEST BOILING POINT.
AND FOR LONDON, WE HAVE THE WEAKEST IMF.
SO THIS WILL BE THE LOWEST BOILING POINTS.
NOW, SO THIS MAKES IT VERY SIMPLE,
IF YOU ARE GIVEN THREE COMPOUNDS
WITH THREE DIFFERENT INTERMOLECULAR FORCES,
IT'S GOING TO BE VERY EASY
TO PREDICT THE RELATIVE BOILING POINTS.
HOWEVER, WHAT HAPPENS IF YOU'RE GIVEN TWO COMPOUNDS THAT H-BOND
OR TWO COMPOUNDS THAT ARE BOTH DIPOLE-DIPOLE OR LONDON?
WE NEED TO HAVE A WAY TO DO THE TIEBREAKER.
SO THE TIEBREAKERS ARE REALLY LINKED TO OUR UNDERSTANDING
OF LONDON FORCES, RIGHT?
EVERY COMPOUND DOES EXPERIENCE LONDON FORCES,
AND SO AS BASICALLY AS WE INCREASE THE SIZE
AND THE SHAPE OF A MOLECULE,
WE ARE GOING TO INCREASE THE LONDON FORCES,
AND THAT'S GOING TO INCREASE THE OVERALL INTERMOLECULAR FORCE.
ALL RIGHTY.
SO LET'S LOOK AT A COUPLE EXAMPLES
OF HOW WE CAN APPLY THE TIEBREAKER.
THIS FIRST EXAMPLE TAKES ADVANTAGE OF OUR UNDERSTANDING
OF THE PERIODIC TREND OF SIZE.
NOTICE HERE, WE HAVE FLUORINE,
CHLORINE, BROMINE, AND IODINE.
THESE MOLECULES ARE ALL NONPOLAR.
HOWEVER, WHEN WE LOOK AT THEIR BOILING POINTS,
WE SEE THAT FLUORINE AND CHLORINE ARE GASES
AT ROOM TEMPERATURE, RIGHT?
ROOM TEMPERATURE IS ABOUT 25 DEGREES C,
DEPENDS ON WHO YOU ASK.
THEN BROMINE IS A LIQUID AT ROOM TEMPERATURE,
AND IODINE IS A SOLID.
SO WHY DO WE HAVE THESE DIFFERENT PHYSICAL STATES
WHEN THEY'RE ALL NONPOLAR?
SO LET'S THINK ABOUT WHAT WE KNOW ABOUT ATOMIC SIZE
AND HOW IT CHANGES ACROSS THE TABLE, ALL RIGHT?
SO WE REMEMBER THAT FLUORINE WILL BE THE SMALLEST,
AND THEN CHLORINE IS LARGER,
BROMINE IS LARGER,
AND THEN IODINE WOULD BE THE LARGEST.
SO HERE, WE HAVE AN EXAMPLE OF SIZE.
SO AS THE SIZE INCREASES,
THE INTERMOLECULAR FORCES INCREASE
CREATING HIGHER BOILING POINTS.
SO AS WE GO DOWN, WE'RE INCREASING SIZE,
WHICH INCREASES THE IMF,
WHICH INCREASES THE BOILING POINT.
NOW, WHAT ABOUT SHAPE?
IF WE LOOK AT THESE TWO MOLECULES HERE,
THEY BOTH ARE TECHNICALLY
THE SAME SIZE IN THE SENSE THAT THEY'RE BOTH C5H12.
THEY ARE--
BOTH MOLECULES ARE COMPRISED OF THE SAME NUMBER
AND TYPE OF ATOMS.
SO THIS IS AN EXAMPLE OF SHAPE.
WHEN WE BRANCH THE MOLECULE, SO HAVING THE CARBONS-- HERE,
THE CARBONS ARE IN A STRAIGHT CHAIN ONE AFTER ANOTHER.
HERE, WE HAVE CARBONS BRANCHING OFF OF EACH OTHER.
SO NOTICE THIS CREATES A VERY SPHERICAL MOLECULE.
SO IF WE HAVE TWO OF THESE BRANCHED MOLECULES,
WE SEE THAT THERE'S A VERY SMALL AREA
FOR THE LONDON FORCE TO OCCUR,
AND NOTICE THE VERY LOW BOILING POINT.
THIS COMPOUND WOULD BE A GAS AT ROOM TEMPERATURE.
WHEN WE HAVE THE STRAIGHT CHAIN, HOWEVER,
WE END UP WITH A MUCH LARGER SURFACE AREA.
AND SO NOW, WE HAVE ALL OF THE SURFACE AREA
THAT IS A POTENTIAL FOR THE LONDON FORCE, IMF.
SO WE CAN SEE--
HERE, IS AN EXAMPLE OF THE SHAPE,
STRAIGHT CHAIN ORGANIC MOLECULES WILL HAVE A HIGHER BOILING POINT
THAN BRANCHED ORGANIC MOLECULES. OKAY.
NOW, ONE LAST BIT HERE IS MORE FULLY--
LOOKING AT THE PRACTICE OF APPLYING THESE SKILLS,
WE HAVE TO LOOK AT THIS.
NOW, TALKING ABOUT BOILING POINTS,
THERE STARTS TO BE A QUESTION BEHIND THE QUESTION.
ANYTIME WE'RE ASKED ABOUT BOILING POINTS,
WE'RE REALLY BEING ASKED ABOUT INTERMOLECULAR FORCES.
AND WHEN WE'RE BEING ASKED ABOUT INTERMOLECULAR FORCES,
WE'RE REALLY BEING ASKED IF WE KNOW ABOUT THE POLARITY
OF A MOLECULE.
AND OF COURSE, THE POLARITY
OF A MOLECULE IS RELATED TO ITS SHAPE AND THE DIFFERENCE
IN ELECTRONEGATIVITY BETWEEN THE ATOMS.
SO THIS IS A TOPIC--
IT'S REALLY A MASTERY TOPIC
BECAUSE IT'S GIVING YOU THE OPPORTUNITY
TO INTEGRATE A LOT OF IMPORTANT CONCEPTS.
SO WHEN WE GO TO ANSWER A QUESTION LIKE THIS,
THERE IS A WHOLE, YOU KNOW,
LOGICAL APPROACH WHERE WE BUILD.
SO THE FIRST THING WE WANT TO DO
IS WRITE THE LEWIS STRUCTURE FOR EACH MOLECULE.
SO WE KNOW-- WE'RE AWARE OF ITS SHAPE AND CHARGE DISTRIBUTION.
OXYGEN, OF COURSE, IS GOING TO BE NONPOLAR
BECAUSE WE HAVE TWO IDENTICAL ATOMS.
HYDROCHLORIC ACID WILL BE POLAR
BECAUSE THE CHLORINE ATOM HAS GREATER ELECTRONEGATIVITY
THAN THE HYDROGEN.
CARBON DIOXIDE,
OF COURSE, HAS TWO POLAR BONDS,
BUT WE HAVE REVIEWED THAT THEY CANCEL
TO CREATE A NONPOLAR MOLECULE.
AND THEN LAST BUT NOT LEAST, WE'LL WRITE
THE THREE-DIMENSIONAL SHAPE OF AMMONIA,
AND SO WE CAN SEE THE DIPOLE MOMENTS
ALL POINTING DOWN INTO THE CORNERS
OF THE TETRAHEDRON CREATING A NET DIPOLE, RIGHT?
SO OUR FIRST THING IS WE HAVE NONPOLAR OXYGEN.
NONPOLAR COMPOUNDS, IT'S ALWAYS A LONDON FORCE.
WITH OUR SECOND EXAMPLE HERE, HYDROCHLORIC ACID,
ANYTIME YOU HAVE ONLY ONE DIPOLE MOMENT,
THEN WE KNOW IT'S A POLAR MOLECULE
AND THAT WE ARE MISSING OUR NOF,
NITROGEN, OXYGEN, AND FLUORINE.
SO THIS WILL BE DIPOLE-DIPOLE.
CO2, WE HAVE EQUAL AND OPPOSITE DIPOLE MOMENTS CANCELING OUT,
CREATING A NONPOLAR MOLECULE.
ANYTIME THE MOLECULES, NONPOLAR,
WE KNOW IT'S LONDON.
AND THEN LAST BUT NOT LEAST, WE HAVE A POLAR MOLECULE,
WHICH ALSO HAS THE NITROGEN-HYDROGEN BONDS,
SO THIS WILL BE A H-BONDER.
SO NOW, WE HAVE THE ABILITY.
SO SEE THE SYSTEMATIC APPROACH.
DRAW THE LEWIS STRUCTURE, IDENTIFY THE POLARITY,
AND THEN DETERMINE THE INTERMOLECULAR FORCES.
NOW, WE CAN ARRANGE THESE COMPOUNDS
IN DECREASING BOILING POINT.
SO THAT MEANS,
WE'RE GOING TO START WITH THE HIGHEST BOILING POINT
AND THEN DECREASE TO THE LOWEST BOILING POINT.
SO WE UNDERSTAND THAT THAT MEANS WE HAVE THE STRONGEST IMF HERE,
THE GREATEST AMOUNT OF POTENTIAL ENERGY.
AND HERE, WE HAVE THE WEAKEST IMF,
THE WEAKEST POTENTIAL ENERGY.
SO WE ONLY HAVE ONE H-BONDER.
AMMONIA WILL BE OUR HIGHEST BOILING POINT.
IT HAS THE STRONGEST INTERMOLECULAR FORCE.
THEN WOULD COME HYDROCHLORIC ACID,
BECAUSE DIPOLE-DIPOLE IS IN THE MIDDLE,
AND THEN THAT LEAVES US WITH TWO LONDON MOLECULES.
SO NOW, WE NEED TO APPLY THE TIEBREAKER.
SO REMEMBERING, RIGHT, THAT IT'S ABOUT SIZE AND SHAPE.
SO SIMPLY LOOKING AT CO2,
WE CAN RECOGNIZE THREE ATOMS ARE GOING TO BE BIGGER THAN TWO
AND SO WE WOULDN'T RECOGNIZE
THAT CO2 IS GOING TO HAVE MORE SURFACE AREA
AND STRONGER INTERMOLECULAR FORCES THAN OXYGEN.
SO THERE WE HAVE--
USE OUR UNDERSTANDING OF INTERMOLECULAR FORCES
TO PREDICT THE RELATIVE BOILING POINTS OF SUBSTANCES.
AND NOW, FOR A LITTLE MORE PRACTICE
TO GIVE YOU CONFIDENCE IN DOING YOUR HOMEWORK,
LET'S LOOK AT--
HERE, WE'LL JUST COMPARE TWO COMPOUNDS AT A TIME.
WE WANT TO DETERMINE THE DOMINANT
INTERMOLECULAR FORCE OF EACH COMPOUND AND THEN PREDICT
WHICH COMPOUND HAS THE HIGHER BOILING POINT, RIGHT?
SO ONCE AGAIN, OUR FIRST STEP IS TO DRAW THE LEWIS STRUCTURE
SO THAT WE CAN PREDICT THE MOLECULAR POLARITY.
WE'VE ALREADY DRAWN CARBON DIOXIDE
BUT WE'LL PUT IT DOWN HERE FOR COMPLETION.
NOW, REMEMBER, CARBON IS A GROUP FOUR ELEMENT,
SO WE END UP WITH A LINEAR GEOMETRY.
HOWEVER, SULFUR IS A GROUP SIX ELEMENT,
SO WE HAVE THAT EXTRA LONE PAIR,
AND SO WE END UP WITH A BENT MOLECULE.
NOW, WE LOOK AT THE DIFFERENCES IN ELECTRONEGATIVITY
AND DRAW IN THE DIPOLE MOMENTS.
SO WITH CO2, OF COURSE,
IT'S A LINEAR ARRANGEMENT, NONPOLAR,
SO IT'S A LONDON FORCE.
WHEREAS HERE WITH THE SULFUR DIOXIDE,
WE HAVE TWO DIPOLE MOMENTS
BOTH POINTING DOWN IN THE SAME DIRECTION,
AND SO WE END UP WITH A POLAR MOLECULE WITH DIPOLE-DIPOLE.
NOTICE THERE ARE OXYGENS PRESENT BUT NO HYDROGEN.
NO-- NONE OF THE OXYGENS ARE BONDED TO HYDROGEN
SO THERE ARE NO H-BONDS HERE.
BECAUSE WE KNOW THAT DIPOLE-DIPOLE IS A STRONGER
INTERMOLECULAR FORCE THAN LONDON,
THE SULFUR DIOXIDE WOULD HAVE THE HIGHER BOILING POINT.
NOW, LET'S LOOK AT AN EXAMPLE
WITH METHANE AND CARBON TETRACHLORIDE.
SO WE'LL DRAW THESE MOLECULES
OUT AND THEY HAVE A THREE-DIMENSIONAL SHAPE,
SO WE'LL GO AHEAD AND DRAW
THEM IN THEIR THREE-DIMENSIONAL SHAPE.
SO THERE'S THE METHANE
AND THEN HERE WOULD BE THE CARBON TETRACHLORIDE,
AND I AM GOING TO LEAVE THE LONE PAIRS OFF OF THE CHLORINE ATOM.
FOR THIS EXAMPLE,
YOU CAN MENTALLY ADD THEM IF YOU'D LIKE OR WITH YOUR PENCIL.
SO IN THIS CASE, ANYTIME WE HAVE A HYDROCARBON,
WE KNOW THAT IT IS NONPOLAR AND A LONDON.
SO IT'LL HAVE THE LONDON IMFS.
HERE, THIS IS SIMILAR TO THE CARBON DIOXIDE.
WE HAVE FOUR POLAR BONDS.
EACH DIPOLE MOMENT IS THE SAME MAGNITUDE
AND THEY'RE RADIATING OUT INTO THE CORNERS OF THE TETRAHEDRON.
SO ONCE AGAIN, WE HAVE A NONPOLAR MOLECULE, SO LONDON.
TIME TO APPLY THE TIEBREAKER.
SO WE TALKED ABOUT MOLECULAR SIZE ON A PREVIOUS PAGE.
REMEMBER THAT CHLORINE HAS A LARGER SIZE THAN HYDROGEN.
SO THE CARBON TETRACHLORIDE WILL HAVE THE HIGHER BOILING POINT
BECAUSE OF THE LARGER SIZE OF THE CHLORINE ATOMS.
SO I HOPE THAT YOU FOUND THIS TUTORIAL HELPFUL
AND THAT YOU'LL TAKE THE TIME TO REINFORCE YOUR UNDERSTANDING
BY WORKING A FEW HOMEWORK PROBLEMS
WHILE IT'S FRESH IN YOUR MIND.