|Ana_carolina Lopez 2/13/01 3:33:46
Hi....2 questions: 1)How many elemnts are known,
110 or 112? The book and the lessons desagree on
this. 2)About significant figures: When adding or
substracting two numbers in scientific notation
with different exponents, how do we know which
common exponent to convert the expressions?
|Jimmy_g Park 2/7/01 1:50:50 AM:
Umm, it looks like someone posted the same site as
earlier...doh! Anyhow, here's another dope site,
and for your convenience, I made an account for
public usage: username: jimgpark password:
hereto go to http://chemviz.ncsa.uiuc.edu/
There's a program that will produce 3-d images of
|Monluedee Luecha 2/5/01 5:20:10 PM:
Dr. Ferede: Finally the second project was
linked:) The subject is pretty much compatible to
what chem 11 provides, but the site got much more
in depth into the subject that it is a useful page
to research more on the subject if one has more
time. I found that it would be useful in the
future classes of chemistry, ie. higher level
chemistry classes. I hope it is useful for those
who come upon it in this VOH.
|Deepak Malhotra 2/5/01 2:41:34 PM:
This site involves basic organic chemistry for
students who are eager to practice their naming
procedure skills. This site is creative,
interesting, and motivational. click
to go to www.chemhelper.com.
|Michel_young Choi 2/5/01 1:55:39 PM:
This website provides a visual interpretation of
the table of elements, as well as detailed
properties for each element. This site is a good
way to familiarize oneself with the basic
fundamentals of Chemistry. click
to go to www.chemicool.com.
|Brian_c Caslin 2/5/01 11:09:19 AM:
here is a nice site to help with step by step
naming of organic molecules
|Marcey_y Sato 2/5/01 12:22:37 AM:
When you enter a chemical name, molecular formula,
or molecular weight at this site, you'll get some
info on it (physical properties, density,
solubility, etc). click
to go to www.chemfinder.com
|Jimmy_g Park 2/4/01 10:12:33 PM:
Ok, here's a very web-oriented Chemistry resource
site with a plethora of information on General
Chemistry, notably a section including interactive
3-D VSEPR models that you can click and drag with
your mouse, giving you 3-D rotational positioning.
Pretty interesting http://www.wbaileynet.com/wldchem/home/
|Kathie_j Nichols 2/4/01 3:02:01 PM:
This website gives details about the dispersion of
different elements in the ocean. The placement of
elements in the ocean is determined by ocean
currents. This site lets you click on the element
in the provided periodic table, which then brings
up an explanation of how and where that particular
element is located most commonly in the ocean. click
to go to www.mbari.org/chemsensor/pteo.htm
|Kathie_j Nichols 2/4/01 2:57:19 PM:
This website gives details about the dispersion of
different elements in the ocean. The placement of
elements in the ocean is determined by ocean
currents. This site lets you click on the element
in the provided periodic table, which then brings
up an explanation of how and where that particular
element is located most commonly in the ocean. .
|Marcey_y Sato 2/4/01 2:46:16 PM:
I don't know what happened, but somehow my web
address and explanation merged together with
Duraiya's. Anyway, here it is again. Columbia
University (NY) created this VSEPR
visual tutorial. It takes 15 min to download but
the Player comes with it. Features simple
explanation of hybridizations, rotations, and
geometry of VSEPR models.
|Duraiya Zirapury 2/4/01 2:23:36 PM:
This web-site has Variety of chemistry related
abstacts. One interesting topic is ozone depletion
caused by CFC molecules. Click here to view : VSEPR
visual tutorial. It takes 15 min to download but
the Player comes with it. Features simple
explanation of hybridizations, rotations, and
geometry of VSEPR models.
|Nadia_a Chowdhury 2/3/01 1:19:43 PM:
This is a basic organic chemistry site and gives a
brief intro to all the functional groups, their
formulas and what isomers are.
click to go to "http://www.chem.vt.edu/RVGS/ACT/notes/Organic_Intro.html"
|Randall_y Ishimaru 2/2/01 4:14:40 PM:
At this website any chemistry question can be
answered by an expert for free! All you have to do
is register, post a question, and wait a short a
time for a great answer. Do you want explanations
for the reactions in Chapter 22? Go to this
website. click to
go to www.askme.com.
|Ashley_ann Tucker 12/8/00 5:33:44 AM:
Extra credit website posting that deals with
simple organic chemistry:click here to go to
||this is the same web site that
is posted down below by Kevin Ha, "Nomeclature of
|Kevin_kwanghoon Ha 12/8/00 4:08:54
Dr. Ferede, you said that 40 questions on the
final will be exactly from self tests on modules
1-32, 34, 35, 37, 38 1*, 2*. But there is no self
tests on modules 1* and 2*. Does that mean they
will be excluded from the final? If you are
concerned that we should know the materials on
those two modules, we are going to take a quiz on
those modules tomorrow and we have to study for it
anyways. It would be nice if you tell us those are
not going to on the final.
||Sorry, since the final is
comprehensive it has to include all the modules you have
studied. That would mean that 1* and 2* will also be
included. Like you point it out modules 1* and 2* do not
have any self tests therefore mostly I am going to depend
on your homework questions from you text book and the
additional examples of module 1*. I would be sending you
the exact homework questions from the text book that you
have to know for the final.
|Kevin_kwanghoon Ha 12/7/00 5:54:41
Dr. Ferede, I am asking for the same thing as
Could you curve our final grades after the final
||Please read the answer I gave
Mai down below.
|Quynh-mai_tu Nguyen 12/7/00 4:32:41
3Dr. Ferede, 1. Would you curve our final grades
after the final? 2. Could you send me a copy of
Module 1 and 2 study guide questions, it would
help me studying for those modules. Thanks. 3.
After getting our final grades, will you post them
on your home page?
||1. No, I will not be curving
your final grades. I will not change the grading policy at
this stage of the semester, I will stick to what I put in
the syllabus. However I will try to help you in two ways.
a) if somebody misses a grade by 1 or 2 % I will give
them the higher grade.
b)if a student misses only 1 or 2 questions (out of the
60 questions) on the final then I will give them an
"A" no matter what they have been getting for
the rest of the semester as long as they have passed the
2) I will send you the guidelines
3) No, I am not allowed to post people's grade
publically. I will let you know the average, and how many
"A's, "B"s etc and also send email you your
||I can not connect to either of
the addresses please check the addresses and post them
again. Please try to hot link them properly by checking
everything and I will be emailing you the direction on how
to do it.
|Teresa_chanton Brown 12/5/00 4:18:30
Professor Ferede Are Hygrogen bonds stronger than
dispersion forces and Ion-dipole forces?
||The three main intermolecular
1) Hydrogen bonding
3) London dispersion force
if the size of the molecule (molecular weight) are very
close to each other then the one that has hydrogen bonding
has the stronger intermolecular force and then the one
that has the dipole -dipole force has next and the one
with only dispersion forces last.
But remember all molecules have London dispersion force
and therefore the one one with hydrogen bonding or
dipole-dipole force also have in addition London
For larger molecules London dispersion force becomes
more important and therefore as the size of the molecule
becomes larger London dispersion force becomes more
important than the hydrogen bonding if especially there is
only one or few sites of hydrogen bonding.
eg H2O and a compound called pentadecane(an organic
compound which is an alkane that has 15 carbons). If we
look at the main intermolecular force in H2O it is
hydrogen bonding and if we look at the main intermolecular
force in pentadecane it is London dispersion force. If the
molecular weights of these two molecules were relatively
close to each other then H2O should have the stronger
intermolecular force. However since pentadecane is a much
bigger molecule it has the higher intermolecular force
eventhough it only has London dispersion force.
the strength of ion-dipole force is dependent on the
polarity of the dipole but if the molecule is very polar
then ion-dipole force can be stronger than hydrogen bonds.
|Eric_m Yoshida 12/1/00 10:26:26 AM:
Extra Credit Website for MO Theory Duran: Chapter
Twelve Molecular Orbital Theory ,Eric
||The extra credit assignment is
to find a website not a chapter in a text book. I really
do not understand what you wrote here.
|Quynh-mai_tu Nguyen 12/1/00 1:36:53
1. How can you tell if it's a sigma or pi bond by
just knowing the sp or sp2 or sp3 hybridization?
2. Why is acetylene's bond is weaker than ethylene
when acetylene has double bond and ethylene only
have single bond? 3. When is our extra credit of
posting website on voh due?
||1. overlap of sp, sp2 or sp3
with any other orbital makes a sigma bond not a pi bond.
However a pi bond is formed when two p orbitals that
are parallel to each other overlap. There for an atom that
has sp or sp2 (not sp3) has a potential of forming an
added pi bond since they have unhybridized "p"
orbitals left over. In the case of sp orbital there are
two "p" orbitals left over (eg in acetylene HCCH)
or the case of sp2 orbital there is one "p"
orbital left over (eg in ethylene H2CCH2).
2. No, the bond in acetylene is stronger than the bond
in ethylene since there is a triple bond in acethylene and
there is a double bond in ethylene.
the one that has only single bond is called ethane
(H3CCH3), not ethylnene.
so the strenth of bonds go
3. I give you a chance to do it until the end of next
week i.e. Dec 7.
|Shahar Kalev 11/30/00 2:18:42 PM:
Can a molecule change its shape because of
polarity? For example, if one side is more
negative it`s pulls the atoms toward one side and
can change an angle? Shahar
||Yes the electron density of an
atom, and more importantly lone pairs of electron can
distort an angle and can give a shape that has a different
angle than expected.
For example if we look at the geometry of CH4 molecule
it has a tetrahedral anlge of 109.5o. We get
this angle since there are four pairs of electrons around
the central atom.
Also there are four pairs of electrons around the central
atom N in the molecule NH3. and we would expect a
tetrahedral geometry around the N with 109.5o.
However since lone pairs have diffused electron density
(occupies more space) than the bonding pairs, the lone
pairs on the nitrogen push away the bonding pairs towards
each other distorting the angle and making it 107 rather
You can see this distortion even more when there are
two lone pairs involved such as in H2O where the angle is
even more distorted and becomes 104.5 rather than 109.
strenght of repuslion is in this order.
lone pair, lone pair>lone pair, bonding pair>bonding
pair, bonding pair.
Distortion of angle from bonded atom can happen but it
|Shahar Kalev 11/30/00 2:15:54 PM:
Is it possible for an atom to have 0.5 hybridized
||No there is no such thing as
half a hybridized orbital. However an atom can use one
hybridized orbital and another unhybrid orbital(s) to form
bonds. For example if we describe the Carbon-Carbon bond
in acetylene, H2C2 we will discribe it as one
"sp" orbital of one carbon overlap with anther
"sp" orbital of the other carbon to form the
sigma bond and a "p" orbital (unhybridized) of
one carbon overlaps with a "p" orbital of the
other carbon to form one pi bond and another pi bond is
also formed by overlap of the left over "p"
orbitals. Therefore the carbon-carbon bond of acetylene
contains a bond that is formed from hybridized orbitals as
well as unhybridized orbitals
|Shahar Kalev 11/30/00 2:12:23 PM:
What is the shape of BrF3? When I tried, it turned
out that is T shape. However, in the additional
study for module 30 example # 1 it says trigonal.
||You are correct. It should be
T-shape. The geometry of BF3 is trigonal planar not BrF3
|Ashley_ann Tucker 11/30/00 4:38:37
Extra Credit Website Posting that deals with
Molecular Orbital Theory/Module 32: :Molecular
Orbital Theory http://www.wpi.edu/Academics/Depts/Chemistry//Courses/CH502/molecularorbitaltheory.html
Dr. Ferede I tried Hotlinking this site however,
I'm not sure how it works so, I apologize if it
doesn't go throught properly. If you don't mind
I'd like to get the instructions on how to do this
procedure onece more. I read your earlier steps
from the last assignment like this unfortunately I
still did not understand exactly how to create the
direct link. Thanks.
||I did the hot link for you. I
will send you an email to let you know how to do it.
this site a good site on MOs however it is more than
you need for chem 11 course. So choose and pick the topics
that are relevant for chem 11.
|Teresa_chanton Brown 11/29/00 4:27:19
Professor Ferede Will you help me with this
question from the practice exam you sent?
Question: Draw the molecular orbital energy
diagram and name the molecular orbitals. How do i
do that? And will we need to be able to do this on
||Yes, you need to do that for
the exam. Remember your exam #3 will be very similar (but
not exactly the same) as the practice exam.
When I say draw the molecular orbital energy diagram I
mean that you need to show the relative energies of the
atomic orbitals and the molecular orbitals to each other
and name the atomic and the molecular orbitals.
for example if we take two hydrogens atoms which
combine to form hydrogen molecule (a diatom)then we need
to show the relative energies of the atomic orbitals of
each hydrogen atoms and the relative energies of the
molecular orbitals they form when they combine to become
Each hydrogen atom will have "1s" atomic
orbital and each of this atomic orbitals will have one
electron. So now draw two straight lines one on the left
side and one on the right side and put "1s"
under each of them then put an arrow facing up for each of
them on the straight lines you draw.
Now you have shown the atomic orbitals and their
relative energies to each other (in this case both should
be same energies). and now you are ready to draw the
realtive energies of the molecular orbitals. You do that
in between the two "1s" atomic orbitals. You
know that when two atomic orbitals overlap they form two
molecular orbitals. one molecular orbital is lower energy
(it is bonding) than the atomic orbitals and therefore you
draw a line below the line you draw for the "1s"
orbitals and you call this molecular orbital "sigma
(use the Greek sign here) 1s". The second molecular
orbital will be higher in energy (antibonding) than the
individual "1s" atomic orbitals and therefore
you draw a line above the line you drew for the
"1s" orbital then you call this molecular
orbitals (sigma star (use the Greek letter) 1s"
then now you are ready to fill the molecular orbitals
with the appropriate number of electrons. You will notice
that there are two electrons involved (one coming from one
hydrogen and the other coming from the other hydrogen. As
you know molecular orbitals are filled according to the 'aufbau'
principle which means that the lowest energy is filled
first. In this case the lowest energy is "sigma
1s" and you know each orbital can hold a maximum of 2
electrons, so you put two arrows with one pointing up and
one pointing down on the line.
Ofcourse more is involved if we look at diatoms that
contain "p" orbitals. Look at what is shown in
your text book pages 479) or the CLICK THEN WEB SITE
ADDRESSES YOU AND KEVIN HAVE posted for diatoms of 1st and
second raw elements
|Teresa_chanton Brown 11/29/00
Professor Ferede Can you help to make this qestion
clear? Question 4 of 10 Self Test Mod 35. Consider
the three functional groups: alcohol, aldehyde or
ketone, and carboxylic acid. What is the possible
range of hydrogen atoms that can be on the carbon
atom to which the oxygen or oxygen atoms are
attached? alcohol aldehyde carboxylic acid
||Let us look at each of them.
this is a functional group that has a carbon bonded to -OH
i.e. C-OH. As you know each carbon must have four bonds to
each, so if it already has formed one bond with the -OH,
then it needs three more bonds. Therefore the possible
range of hydrogens it can have is 0 to 3.
this is a functional group that has a carbon double bonded
to oxygen, i.e. C=O (which is refered to as a carbonyl
group. Since carbon can bond to a maximum of four bonds
and it has already made two bonds with oxygen then it has
two more bonds left to bond. Therefore the possible range
of hydrogens it can have is 0 to 2 .
this is a functional group that has a carbonyl group (C=O)
and the carbon of this carbonyl group is also bonded to
two other carbons. and therefore this carbon has already
formed four bonds (2 with the oxygen and two with two
other carbon) and therefore this carbon can not have any
more hydrogens on it.
4. carboxylic acid:
this functional group has a carbonyl group (C=O) and on
this carbon is also bonded an -OH group so that means that
this carbon has three bonds to it already (2 with the
oxygen and one with the -OH) so that means one is left. so
the possible range of hydrogens that can be attached to it
is 0 to 1.
|Teresa_chanton Brown 11/28/00 8:41:57
YOu know I'd like to get the extra credit. but it
seems that I can't put the address in correctly to
make it clickable to the other students.
Anyway......I was wondering if we should know the
formulas for determinig the names and structures
of the organic molecules and functional groups?
||you get the extra points.
Yes you need to know the names of the organic compounds
of different functional groups up to 10 carbons. However
modules 34 and 35 will not be in exam #3
I sent you an outline for exanm#3 so that you would
know what things to know for each module.
|Teresa_chanton Brown 11/28/00 8:38:29
Here are two websites that deal with molecular
Orbital theory. Molecular
Orbital Theory http://www.discoverchemistry.com/dcv2-docroot/student/0200/0206a.html
Orbital Theory http://wilson-squier.ucsd.edu/education/gchem/molecorbs/moaccount.html
||I have hot-linked it for you.
Both are good sites to use as a review, however the first
one is the same site as the one Kevin Ha has put down
||this is a good site to use as a
|Shahar Kalev 11/22/00 7:55:56 PM:
How can you predict if an element exists or not?
Module 32. Shahar
||I do not know what you are
referring to in this question. If you are referring to
predicting if a molecule exist or not using molecular
orbital theory then the answer is as follows:
1) count the number of electrons involved in the
2) do molecular orbital diagram for the molecule (this
can only be done with simple molecules, and it gets very
complicated for complicated molecules)
3) fill the molecular orbitals with the electrons for
4) subtract the antibonding electrons from the bonding
electrons. If this gives you 0 then the molecule can not
exist, it it gives you more than 0 then the molecule
Let me know if this answers your question or not.
|Shahar Kalev 11/22/00 7:55:00 PM:
Isn`t it true that when you increase the temp. the
pressure increase and so the volume? If so, how
come that when you increase the volume pressure
decrease? it sounds like a contradiction. Module
||When temperature increases,
pressure will increase if only the volume is kept
when we state T, P relationship we assume V and n are
When we state P, V relationship, i.e. Boyles law we
assume T and n(# of moles)are kept constant.
when we state T, V relationship, i.e. Charles' law we
assume P, and n are kept constant.
When we state n, V relationship i.e. Avogadros' law we
assume T, and P are kept constant
|Shahar Kalev 11/22/00 7:51:46 PM:
Can you send us outlines for Module 32? Thanks.
|Teresa_chanton Brown 11/19/00
Professor Ferede When module 32 says that the
bonding for the H2 molecule is delocalized. what
does the module mean by delocalized?
||when it says that the
electron(s) are delocalized, it means they are not only
located at only one atom but are being spread out. We
usually talk about delocalized pi electrons in which the
electrons are shared not by only two atoms but they are
being shared by more than two atoms. We rarely talk about
delocalization of sigma electrons as a pair of electrons
in hydrogen molecule. Here in hydrogen molecule
delocalization would refer to the electrons being spread
out over the two hydrogen atoms rather than one hydrogen
Like I mentioned the most common delocalization of
electrons is shown when a pi electron pairs are
delocalized. A good example to show the delocalization of
electrons is SO2. If we write a Lewis dot structure for
SO2 we can writetwo resonace structures. resonance
structure #1 has a double bond between sulfur and the
oxygen on the left and a single bond between the sulfur
and the oxygen on the right and resoncance structure #2
has a single bond between the sulfur and the oxygen on the
right and a double bond between the sulfur and the oxygen
on the left.
O=S-O <--> O-S=0
the sigma bonds are localized that is they are only
shown (being shared) between the O on the left and right
and S for both resonace structures. However the pi bond is
delocalized since it is being shared by the oxygen on the
right and sulfur in one resonace structure but also being
shred with the oxygen on the left and the S in the other
resonace structure. and therefore in this case we say that
the pi bond is being delocalized over all the three atoms
i.e. oxygen on the left, sulfur and oxygen on the right.
|Quynh-mai_tu Nguyen 11/18/00 11:52:40
If for some reason, you are not able to connect
when clicked on the website I recommended, just
retype the address into the address box of your
computer to get to those links. thanks.
||this is also a good web site
but you need to click at each hot linked items and each
title to get more information
|Quynh-mai_tu Nguyen 11/16/00 2:35:40
Dr. Ferede, 1) What does "cis" mean
literally? 2) If orbitals of 2 atoms overlapped,
what kind of bond would be formed? 3) Is there
such thing as "square bipyramidal" in
||1) "cis" means that
the two large groups are on the same face and
"trans" means that the two large groups are on
opposite face. This type of geometry occurs when a double
bond exist between the two large groups. It also happens
when there is a ring involved but you do not have to worry
about ring structures at this stage of chem 11.
The best way to explain "cis" and
"trans" structures is by giving examples.
eg: the molecule C2H2Cl2
before we can say if this molecule has a cis and trans
geometry we need to know the overall geometry of this
molecule. We find out the overall geometry by first
writting the correct Lewis dot structures using our
guideline for writing Lewis dot structures
The Lewis dot structure for this molecule has the two
carbons bonded to each other with a double bond and each
carbon has a hydrogen and a chlorine.
Now if we want to predict the geometry around each carbon
we see that each carbon has three sigma bonds and no lone
pair and therefore it will be trigonal planar and the over
all molecule will be trigonal planar where all the atoms
lay on the same plane and the angle around each carbon is
Now try to make a model of this molecule: At this stage
you relieze that you can make one model of this molecule
in which the two chlorines on the same side (cis) or you
can make another model in which the two chlorines are on
the opposite sides (trans). The two molecules i.e. the cis
and the trans are the same in every way except the way the
large groups (in this case the two chlorines) are facing
each other on one of the molecules and they are facing
opposite on the other molecule. these two molecules are
said to be Cis trans isomers.
As an anology you can think of two identical twins in
which on has one yellow glove on the right hand and one
yellow shoe on his right foot and not wearing anything on
his left hand or foot, and the other twin has one yellow
glove on his right hand and one yellow shoe on his left
foot and nothing on the other hand or foot
twin #1 has a cis geometry (both the yellow outfits are on
the same side) where as twin #2 has a trans geometry (the
yello outfits are on opposite side)
2) Your second question is "if orbitals of 2 atoms
overlapp, what kind of bond would be formed?" the
answer for this is it depends on the orientation of the
if the orbitals are overlapping along the same axis such
two "s" orbitals,
or an "s" orbital overlapping with
"sp3" hybrid orbital or
"Px" overlapping with "Px"
then they make sigma bonds.
but if the orbitals are overlapping and their axis are
parallel to each other such as "Py"
overlapping with a "Py" orbital or
"Pz" with "Pz"
then they form pi bonds.
In summary for chem 11 you can say all orbitals overlap
to form sigma bonds except
"Py" orbital overlap with "Py"
or "Pz" with "Pz"
which form pi bonds.
3) Yes there is such a thing as "square
bipyramidal" except it is referred as
||These are all good web sites
(especially 1-4) to use for review.
You forgot to hot link the sites so I did them myself.
This is for everybody else, make sure you hot link the
sites you write here. I have given you an instructuion on
how to do that on your email.
|Teresa_chanton Brown 11/14/00 8:15:31
Professor ferede I do not understand this
question. How is the unhybrid orbital oriented
with respect to the two hybrid orbitals?
||Please read the anwser I gave
for the question you ask below this before you read this
if the central element has sp3 hybrid orbitals then that
means it has used all its p orbitals used for
hybridization. Remember for each pricniple energy levels
there are three p orbitals oriented 90 degrees from each
other, at an axis of Px, py, pz. If the central element
has sp3 orbitals then it forms four sigma bonds and no pi
bonds since pi bonds are formed by overlap of unhbride p
orbitals that are parallel to each other.
If however the central atom is sp2 hybridized it has
used only two of its p orbitals for hybridization and that
means it has one unhybride p orbital left. The geometry of
the three sp2 orbitals is trigonal planar (i.e. 120 degree
apart and all on the same plane). and the geometry of the
p orbital that is left unhybridized is perpendicular (90
degrees) to the three sp3 hybridized orbitals. You can
think of it as trigonal bipyramidal arrangement in which
the three sp2 orbitals are on the equatorial axis and the
p orbital is on the axial axis (one lope of the p up and
one lope of the p down)
If the central atom is sp hybridized it has used only
one of its p orbital for hybridization and that means that
it has two unhybridized p orbitals left. The geometry of
the two sp orbitals from each other is linear (i.e. one
180 degrees apart on a line) and the geometry of the two p
orbitals that are left unhybridized is that they are
perpendicular (90 degrees ) to each other and also
perpendicular to the hybridized sp orbitals. You can think
of it as an octahedral arrangement in which the two sp
orbitals are at linear geometry and the other two axis are
taken by the p orbitals.
|Teresa_chanton Brown 11/14/00 6:15:00
professor ferede just to confirm that I understnad.
when module 30 is referring to a sp3 orbital, as
in CH4, is it really saying that the s plus the 3
p orbitals make a sp3 orbital? Professor Ferede.
From what I understand anout hybridization when
the orbitals hybridize, they "in a
sense" move some of the electrons in the full
orbitals to the one orbitals that are not full? So
, for example in BF3 , how does it become a
trigonal planar molecule.
||The first part of your comment
is correct. one s orbital and three p orbitals mix to form
four (1 + 3) mixed (hybride) orbitals which contain one
part s and three parts p i.e. sp3.
. the same concept will apply if one s and two p
orbitals mix to form three (1 + 2) hybride orbitals whcih
contain one part s and two parts p i.e. sp2. In this case
one p orbital will be left unhybridized
the same concept will apply if one s and one p orbitals
mix to form two (1 + 1) hybridized orbitals which contin
one part s and one part p i.e. sp. In this case two p
orbitals will be left unhybridized.
the same concept will apply if one s, three p and one d
orbitals mix to form 5 (1 + 3 + 1) hybridized orbitals
which contain one part s and three parts p and one part d
the same concept will apply if one s, three p and two d
orbitals mix to form five (1 + 3 + 1) hybridized orbitals
which contain one part s and three parts p and two parts d
The second part of your comment which says that
"when orbitals hybridize, they move (the word here
used is excited) some of the electrons in the full
orbitals to the one orbital that are do not contain
any" is not always true. In some cases excitation of
an electron from an orbital that contains two electrons to
an orbital that contains no electrons is necessary. This
is for elements that have empty p orbitals such as Be, B,
and C however this is not necessary for elements that have
electrons (one or more) in each p orbital such as in N, O.
The simplest way to know what the hybridization of a
central atom is to first find out how many sigma bonds and
lone pairs it has :
If it has two sigma bonds and lone pairs, then it is sp
if it has three sigma bonds and lone pairs, then it is sp2
if it has four sigma bonds and lone pairs, then it is sp3
if it has five sigma bonds and lone pairs, then it is sp3d
if it has six sigma bonds and lone pairs, then it is sp3d2
Inorder to find out how many sigma bonds and lone pairs
a central atom has you have to write down the correct
Lewis dot structure.
Remember pi bonds are not considered in determining the
geometry of the molecule or the hybridization of the
This brings me to the question you ask on the top of
the reason pi bonds are not considered is because they are
formed by the overap of p orbitals that are parallel to
each other but are perpendicular to the hybrid orbitals.
Therefore they do not infuence the geometry of the central
|Kevin_kwanghoon Ha 11/14/00 2:03:17
I have included two websites that deals with Lewis
dot strucure, Geometry of molecules (VSEPR) and
here to go to the first website,
here to go to the second website.
Dr. Ferede, please tell me if these websites are
good enough to get the 3 extra credit points, or I
will find more websites and post them.
||They are excellent web sites! I
advise all the other students to click there and view the
contents of the web sites. In the first web site is also
discussed "intermolecular forces" which we will
be discussing in the last week of class. Good Job!
||this is just an example for
extra crdit assignment #2 which is posting a web site
address in which you can go to it by just clicking it . I
have sent you the html format for this in your email
|Shahar Kalev 11/12/00 1:48:39 AM:
I had a little problem with ques. # 1 in Quiz #9.
which i didn't understand why the smallest
possible of formal charges is the best contributor
and if so, when I calculated it I got always # 4
is the smallest possible former charges but it`s
not the right answer? Shahar
||I am glad you asked this
question since there is a problem with the question
itself. First let me answer the first part of your
question before I point out what is wrong with the
the question: Why is the Lewis dot sturucture with the
smallest possible charges the best contributor?
answer: Since an species with an excess charge on it is
not stablilized i.e there is not enough positive charge or
negative charge to stabilize it. For example if an species
has a formal charge of +2 compared to +1 then the one with
+2 formal charge is less stable since it is in need of
more negative charge than the one that has +1 charge.
Now let me point out what is wrong with question #1 of
module 25, even though the answer given is correct.
Inorder to write valid Lewis dot sturtures you first
have to count the valence electrons involved. And only
these valence electrons must be shown on the Lewis dot
sturucture. Here for O3 the total valence electrons must
be (6 + 6 + 6) = 18. If you are showing different resonace
forms (i.e. different Lewis dot structures for the same
molecule or ion) each of them must have these total
valence electrons (18 electrons) to be valid. In this case
Lewis dot structure #1 and #2 have 18 valence electrons
however Lewis dot structure #3 and #4 do not and therefore
they should not even be considered as a resonance
However you could write third resonance stucture (not
shown in your module) in which all the valence electrons
are used but is not as good contributor as #1 and #2. The
third resonace strucutre would be like #4 (shown in your
module) but with one less lone pair of electrons on the
Now if we have these three resonace stucures i.e. #1
and #2 from your module and #3 the one you just draw and
if you want to find out which is the best contributor then
we should follow the rule of formal charges i.e. The one
that has the smallest formal charge will be the best. Let
us determine the formal charge on each atom on structure
#1, #2, #3.
resonace structure #1.
the left O has 0 formal charge
the middle O has +1 formal charge
the right O has -1 formal charge
resonace structure #2
the left O has -1 formal charge
the middle O has +1 formal charge
the right O has 0 formal charge
resonace structure #3
the left O has -1 formal charge
the middle O has +2 formal charge
the right O has -1 formal charge
As you can see resonance #3 has more excess formal
charges on each of its atoms compared to resonance
sturcutres #1 and #2 (which have the same amount of formal
charges). Therefore resonces structures #1 and #2 will be
better contributors than #3.
|Kevin_kwanghoon Ha 11/9/00 10:56:37
Do we have to memorize the electronegativity
values for even common atoms or are they going to
be given on quizes or exams?
||No you do not have to memorize
indivitual electronegativity values, however you need to
know the general trends and some general guidelines such
i) the right upper corner of the periodic table holds
elements that have high electronegativity values and is
vice versa for the left lower corner of the periodic
table. i) F is the highest electronegativity
ii) nonmetal as a whole have higher electrongegativity
iii) the halogens as a group have higher electronegativity
values than any other group.
|Kevin_kwanghoon Ha 11/9/00 10:54:12
Module 27 Self Test #9: From the value of H for
the atomization reaction of ammonia, calculate the
average value of H(N–H) in ammonia. 2 NH3(g)
N2(g) + 3 H2(g); H = 1173 kJ/mol
||You can answer this question in
two ways. the first way which is really a simpler one or
the second way which is given in your module. I will put
both of them down.
The atomization energy in this case 1173 kj/mole is the
energy needed to change 1 mole Of NH3 to its atoms N and
and there are three bonds of N-H that have to be broken
and it takes 1173 kj/mol to do all three. So if we just
want to calculate how much energy is needed to break only
one N-H bond all we need to do is divide the 1173/3 = 391
kj/mole to get the answer.
the second way (described in your module self test
Bonds broken: 6 mol NH bonds Energy input = 6(1173) kJ for
6 mol NH bonds = 7038 kJ Bonds formed: 1 mol NN bonds and
3 mol HH bonds Energy released = 1173 kJ for 1 mol NN
bonds + 3(1173 kJ) for 3 mol HH bonds = 4692 kJ Htotal =
7038 kJ – 4692 kJ = 2346 kJ H(N–H) = 2346 kJ/6 mol NH
bonds = 391 kJ/mol
|Kevin_kwanghoon Ha 11/9/00 10:52:37
This question is about the previous exam: #13 I
know that the radius size for atoms increases as
you go down the group and as you go left the
period, and the opposite for the ionization
energy. In #13a, Se>S, Se>Br, and Br
||Please look at the explanation
I gave to Shahar Kalev, down below on 10/27/00 at 12:22:23
|Quynh-mai_tu Nguyen 11/9/00 4:05:59
Dr. Ferede, 1. How much energy does the plant
needed to break ammonia from the fertilizer to get
nitrogen? 2. Why is energy of fluorine's valence
orbitals is lower than the energy of hydrogen's
valence orbitals? 3. Polarity apply for ionic or
covalent bond? 4. What is the "force of
||1. actually it does not use
ammonia. the ammonia is used to make the feritilizer. You
can see how much energy is obtained if ammonia is broken
to give H2 and N2, by looking up the the standard heat of
formation of NH3 and N2 and H2 and use the equation
delta H of a reaction = the summation of all the heats
of products - the summation of the heats of reactants.
However the reaction does not actually happen in plants.
2) because fluorine is a smaller atom since it has more
protons and the electrons have more attraction to the
3) Polarity is really for covalent compounds when we
describe the uneven sharing of electons. But if the
sharing is so uneven and one take the electron and the
other gives up the electron then they form ionic bonding.
So you can think of ionic bonding as the highest polarity
4. It is the attraction that exist between opposite
|Teresa_chanton Brown 11/9/00 9:59:31
Which of the following molecules has a zero dipole
moment? CO2 C2H2 C2H4 BCl3 (1) (2) (3) (4)
Professor Ferede What I do not understand about
this question is the "zero dipole
moment", what does it mean exactly?
||A molecule that has "zero
dipole moment" is a molecule that is not polar.
Inorder to see if a molecule is polar or nonpolar you have
to consider the following.
1) a molecule that has no polar bonds (equal sharing of
electrons because there is no electronegativity difference
between the two atoms) is not polar eg all the homonuclear
diatoms such as H2, Cl2 etc
2) a molecule that has a polar bond (unequal sharing of
bonds because the atoms have different electronegativity)
can be polar if the molecule is assymetrical or nonpolar
if the molecule is symmetrical.
Inorder to know if a molecule is symmetrical or not you
have to determine the geometry of the molecule using the
Now let us look where each of the given molecules are
polar or nonpolar
i) CO2 is nonpolar eventhough it has polar bonds (C=O
bond) but the polarity cancels out since the molecule is
symmetrical. If you draw the Lewis dot strucute of CO2 (
the carbon in the middle with the two oxygens bonded to it
in a double bond) and determine its shape using VSEPR
theory you will find out that CO2 is a linear molecule
i) C2H2 is a nonpolar molecule with the same reason as
given for #1 i.e. the individaul bonds are polar but the
overall molecule is nonpolar because of symmetry.
The C-H bond is a polar bond (not so strong but still
polar) but the overall molecule is nonpolar. The molecule
is nonpolar since it symmetrical
If you draw the Lewis dot structure of C2H2 (the two
carbons are attached to each other with a triple bond and
each carbon has one hydrogen on it) and determine its
shape using VSEPR theory you will find out that C2H2 is a
linear around each carbon and therefore is linear molecule
overall and therefore symmetrical molecule molecule
iii) C2H4 is a nonpolar molecule with the same reason
as given for #1 i.e. the individaul bonds are polar but
the overall molecule is nonpolar because of symmetry.
The C-H bond is polar (not so strong but still polar) but
the overall molecule is nonpolar. The molecule is nonpolar
since it symmetrical
If you draw the Lewis dot structure of C2H4 (the two
carbons attached to each other with a double bond and each
carbon has 2 hydrogens on it) and determine its shape you
will find out that C2H4 is a trigonal planar around each
carbon and therefore trigonal planar overall and therefore
is a symmetrical molecule molecule
iv) BCl3 is a nonpolar molecule with the same reason as
given for #1 i.e. the individaul bonds are polar but the
overall molecule is nonpolar because of symmetry.
The B-Cl bond is polar strong but the overall molecule is
nonpolar. The molecule is nonpolar since it symmetrical
If you draw the Lewis dot structure of BCl3 (the B in
the middle with the three Cl attached to it) and determine
its shape you will find out that BCl3 using VSEPR model
you will find that BCL3 is a trigonal planar molecule and
So all the molecules mentioned are nonpolar i.e. they
have zero dipole moment.
|Teresa_chanton Brown 11/7/00 9:20:09
Professor Ferede. How do I answer this question
from the study guide. The question seems to be
repeating itself.. for 5 pairs around a Central
atom ? If all are bonding (eg PCl5) i. what is the
geometry of electron pairs? What is the name? ii.
and what is the molecular geometry? What is the
name of this molecular geometry
||the answer for i) is trigonal
bipyramidal and ii) is also trigonal bipyramidal. However
the question is not repeating itself because in some cases
the geometry of the electrons pairs could be different
than the geometry of the molecule (the molecular geometry)
eventhough the angle is determined by the total electron
pairs around the central atom. This happens when all the
pairs are not bonding. For example let us look at the case
for five pairs around a central atom.
in this case the geometry of the electron pairs is
trigonal bipyramidal, but the geometry of the molecule
will be different depending on the number of bonding and
nonbonding (lone) pair electrons.
Let us see what the molecular geometry would be if
there are 0, 1, 2, 3, lone pairs around a central atom.
1) if there are five pairs around a central atom and
all are bonding pairs (i.e. no lone pairs) then the
geometry of the molecule will be trigonal bipyramidal,
with three 120o angles (called equatorial
positions) and two 90o angles (called axial
2) If there are five pairs around a central atom and
one of them is a lone pair then the geometry of the
molecule will be see-saw in which the lone pair takes the
120o (the equatorial) position.
3)If there are five pairs around a central atom and two
of them are lone pairs then the geometry of the molecule
will be T-shape in which the lone pairs take the two 120o
(the equatorial) positions.
4) If there are five pairs around a central atom and
three of them are lone pairs then the geometry of the
molecule will be T-shape in which the lone pairs take the
three 120o (the equatorial) positions.
Notice for five pairs of electrons around a central
atom, the lone pair(s) occupy the 120o (the
equatorial) position in all cases and that is because that
is where they will have the least amount of repulsion.
To visualize all this it is better to work with models.
You will be doing this type of excercise in the lab next
week, however it is also advisable to get your own model
sets especially if you are going to continue with chem 12.
You can buy the model sets at the SMC book store or any
other college book store.
|Kevin_kwanghoon Ha 11/2/00 11:51:28
Could you explain the difference between
monoprotic and diprotic acid?
||a monoprotic acid has one mole
of H+ for every mole of the acid. for example HCl is a
monomprotic acid since 1 mole of HCl has one mole of H+.
a diprotic acid has two moles of H+ for every mole of
the acid. for example H2SO4 is a diprotic acid
since for every mole of H2SO4 there are 2H+
a triprotic acid has three moles of H+ for every mole
of the acid an example of a triprotic acid is H3PO4
|Kevin_kwanghoon Ha 11/2/00 10:59:14
Module 16 Additional Study #3 --> one mole of
gas at STP occupies a volume of 22.414L, but how
do you determine the volume of one mole of gas
when it's at different temperature(eg. at 25
||We will discuss this in details
in module 37 when we talk about gas laws and volume of
gases at different temperatures and pressures. So do not
worry about it right now.
|Shahar Kalev 11/2/00 9:03:13 PM:
Could you explain about the exeption in writing
electron configuration ( i.e Cr)? Shahar
||Cr has 24 electrons so if it
has an electron configuration with out an exception it
1s2 2s2 2p6 3s2 3p6 ds2 3d4 (let us call this
"electron configuration A"). However It does not
have this electronic configuration. Its electron
1s2 2s2 2p6 3s2 3p6 4s1 3d5 (let us call this
"electron configuration B").
To understand the reason given for this exception you
would have to draw the box notation for the last two
important orbitals, the 4s and 3d orbitals. Draw one box
for the 4s orbital and 5 boxes for the d orbitals. If you
fill these boxes with the "electron configuration
A" (the wrong one) you will see that one of the boxes
in d orbitals is empty. However if you fill these boxes
with the "electron configuration B" then all the
boxes in d orbitals will have the same number of electrons
in them (i.e. one each). This state, where the degenerate
(equal energy) orbitals are half-filled (in the case of
Cr) or fully filled (in the case of Cu) is said to be at a
more stable state. This is given by "Hund's
. Following this rule if we write electron
configuration for Cu it will be
1s2 2s2 2p6 3s2 3p6 4s1 3d10
and not 1s2 2s2 2p6 3s2 3p6 4s1 3d9.
|Shahar Kalev 11/2/00 8:39:27 PM:
Which of the following sets of quantum # are
allowed for H atom? Shahar
||For the ground state of H then
the electronic configuration is 1s1 i.e. n =1,
l = 0 and ml = 0 and ms = 1/2 or
-1/2 . However if you are not only considering the ground
state but also the excited state then the possible sets of
quantum numbers depend on the n, l, ml, ms values
a) for all values of n the possible values of l are 0
--> n-1( i.e. 0 and all positive integers up to n-1)
for example if n = 4, the value for l will be 0, 1, 2, 3
b) and for all possible values of l, the possible
values of ml are -l <--> l (i.e. all positive and
negative integers ranging from -l to l including 0)
for example if l = 3 then the possible values of ml = -3,
-2, -1, 0, 1, 2, 3
d) the possible values of ms are always 1/2 or -1/2
|Shahar Kalev 11/2/00 8:37:06 PM:
How come that when K and H2O react we see a great
amount of energy released compared with Na and
H2O, and when K reacts with Cl there is less
energy released compared with Na with Cl? Shahar
||In the reaction of K and H2O
and Na and H2O, the products are a gas (H2) and aqueous
solution of the salt i.e KOH in the case of K and NaOH in
the case of Na. Since the valence electrons of K are less
tightly held than Na (further away from the nucleus) it
will react more vigarously to give more energy.
When Na reacts with Cl and K reacts with Cl, we would
expect the same trend in reactivity. However the trend is
reversed since there is more process involved because the
products for both reactions are ionic solids. the process
involved is the lattice energy in which the ions that are
formed (K+and Cl-, Na+
and Cl-) are attracted to each other to form
solids. the lattice energy is an exothermic process and is
directly proportional to the charge and inversely
proportional to radius of the ions involved. In this case
K+, and Na+, have the same charge
but they have different radius in which K+, has
a larger radius and therefore lower lattice energy.
|Shahar Kalev 11/2/00 8:33:57 PM:
When we reffer to IE, is it a (-) value or (+)
value of Energy? Module 23, Shahar
||Please refer to the answer I
gave below to find the answer for this.
|Shahar Kalev 11/2/00 8:32:21 PM:
If IE and AE have the same trend, how come that it
easy to form cations at the left side of the
periodic table and easy to form also anaions at
the right side? (AE is supposed to be high at the
right side) module 23, Shahar
||Ionization energy is an
endothermic process is positive for all elements. So IE is
reported as positive. so IE increases from left to right
i.e. IE is high on the right side
Electron affinity on the other hand is an exothermic
process for most elements . So EA is reported as negative
(this point is not stated very clearly in your module).
i.e. EA increases from left to right negatively and
therefore EA is high at the right side. for example an
element that has EA = -400 kj/mol has higher electron
affinity than an element that has EA = -300 kj/mol.
Eventhough the actual number -400 is less than -300. But
remember here the (-) and (+) signs are just used to
indicate exohtermic and endothermic processes.
|Teresa_chanton Brown 11/2/00 8:17:42
Professor Ferede Could you quickly explain what
heat of fusion is and what heat of vaporization
is. I think that the words "fusion" and
"vaporization", have me confused.
||heat of fusion is the same as
heat of melting, i.e. the amount of heat needed to take a
sample from a solid to a liquid state, usually given per
heat of vaporization is the amount of heat needed to take
a sample from a liquid state to a gaseous state
|Teresa_chanton Brown 11/2/00 8:13:16
Professor Ferede. If module 26 is not going to be
on the exam. can I post you some questions later?
I have questions yet I don't know how to ask them.
||Yes, you can ask me questions
on module 26 next week. Right now I think your time is
best spend on concetnrating on modules 13-24 and asking
questions on these modules.
|Ashley_ann Tucker 11/1/00 7:21:09 PM:
Module 26: Question #1 Dr. Ferede, Can you please
give a couple of scenarios that would be helpful
in understanding the subtle nature of partially
charged atoms. Also, when finding the partial
charge on every atom within a molecule; is the
only benefit obtained by doing so the
determination of the polarity of the molecule or
is there other determinants we can arrive at with
this information? Module 26: Question #2 What
happens in a formal charge situation if the
electrons are not shared equally between
covalently bonded atoms? Is this situation
possible or would it just be considered a partial
charge if the electrons are not shared equally
between the covalently bonded atoms? Please if
time permits, discuss this concept in more detail.
Thank you. Module 26: Question #3 How can we, just
by observation determine which resonance
structures are better? I really am rather lost
here. Okay, I understand that we determine the
formal charge on each of the elements in both of
the structures. Then we observe to see which has
the fewest number of atoms with the formal charge.
Then whats the next step other than stopping here
and guessing? I seem to be missing something. Can
you please help?
||#1) If two nonidentical element
share a pair of electrons, this pairs of electrons are not
going to be shared equally. To describe the unequal
sharing of electrons we can use formal charges, oxidation
numbers or partial charges. All these three ways are
electron book keeping method to just give us informations
on how much the probability of finding electron density is
on one atom compared to the other atom in a shared bond.
Each of these three methods have their own use and
application for example when we are trying to balance
redox equations we use oxidation number, and when we want
to see which is a better resonance structure then we use
formal charge and when we want to see the amount of
polarity in a bond we use partial charges.
#2. Please read above the answer I gave for #1.
If you can write different Lewis dot strucutres for a
molecule or a polyatomic ion, then these Lewis dot
structures are called resonance structures. One way to
determine which is the best resonace structure (the
highest contributor to the actual structure) is by
assigning formal charges. Like you said in your question
the Lewis dot strucuture that has the fewest formal charge
is the best resonace strucutre. for example if we have two
resonace strucutres (resonace structure A and resonance
strucutre B) for one moleucle and we want to decide which
one is the best then we assign formal charges for each
element in the two resonance sturctures. Let us say in
this example the molecules has 3 atoms. and we assign
formal charges for each atom (using the formula given for
finding formal charges). Let us say in resonance strucure
A all the atoms have 0 formal charges and resonance
structure B has 0 on atom #1, -1 on atom #2, and +1 on
atom #3. then we would choose resonance structure A to be
the best resonace structure since it has less charge
|Kevin_kwanghoon Ha 10/31/00 8:28:35
Hello professor, I still haven't received the
study guideline for module 26 which would be due
this week. I'm not sure if we have to do them or
not this week. Do you want us to do module 26 this
week? or is it just me who didn't receive it?
||I sent it to everbody last
saturday. Anyway I will send it to you again today. Yes,
it is due this week.
|Teresa_chanton Brown 10/30/00 4:17:01
Professor Ferede I do not understand this question
from the study guide. can you help me? ? The
partial charge for an atom A bonded to only one
other atom B = group #A -# of lone pairs –
(?A)/(?A + ?B) where ? is the electronegtativity
(which you look up in the periodic table of
||There are different types of
electrongenegativity tables. However the most common one
and the one we will use is the one constructed by Linus
Pauling which is shown on page 423 of your textbook. just
replace the "?" with the electrongegativity
values for each element in the compound. for example if
the molecules is HCl then to find the partial charge on H
= group #H - #lone pairs of H - (electronegativity of
H)/(electronegativity of H + electrongegativity of Cl).
|Shahar Kalev 10/27/00 12:23:27 AM:
Can i say that IE is exactly the same like delta
H? Why? Shahar
||You can have delta H (which
just means the change in heat for a reaction at a constant
pressure) for any reaction.
The change in heat of a reaction can be measured under
constant pressure then it is called delta H or qp or it
can be measured under constant volume then it is called
qv. So when you report the heat given off or absorbed in a
reaction you report delta H (qp) or qv depenending how you
run your reaction. Most reactions in an ordinary lab are
done under constant pressure (like the calorimetric
experiments you did in your lab) so if we measure the heat
of the reaction then we report delta H. Ionization energy
is the change in heat when an element in a gaseous state
goes to a positive ion in a gaseous state.
M(g) ---> M+(g) + e
This reaction is an endothermic reaction and therefore the
sign will be positive, and sometimes the heat needed is
refered as the enthalpy of the reaction.
Most reactions involving gases are carried out under
|Shahar Kalev 10/27/00 12:22:23 AM:
Which element has larger IE, Atomic radii, afinity
and ionic radii and why? Se or Sb? How can you
determine it when you are asked to do so with
elments that are diagonaly to one another? Shahar
||You can only use the general
trends to prodict some properties such as electron
affinitity and atomic radius for elements in the same
group or the same period. Even for that there are lot of
excetpion to the general trends especially for electron
affinitiy, However one general guideline you can use when
you are comparing elements that are not in the same group
or the same period is, change in property within a period
is more dramatic than a change with in a group. As you
notice with in the same period there are elements that are
metals, nonmetals and metalloids. whereas you can find
groups that are exclusively one or the other (all metals
or all nonmetals). for example all group IA and IIA are
all metals and all group VIIA and VIIIA are non metals.
Therefore you can use this general guideline if you have
to predict properties for elements that are not in the
same period and or same group. Even then you really need
experimental data to say what the trend would be to get
Now let us look at Se and Sb if you look up their electron
affinity Se = -194.97 whereas Sb = -103, this is what you
would have expected since usually nonmetals have higher
electron affinity than metals (or metalloids). and Se is a
nonmetal and Sb is a metalloid. The trend in the
ionization energy is also as expected Se = 941 and Sb =
834, this is like expected because of the metallic
property of Se and metalloid property of Sb. their radius
is Se = 119 and Sb = 140. you can only attribute this to
more d electrons.
In conculsion there is not a very reliable trend to
predict the properties of elements that are not in the
same period or group.
|Shahar Kalev 10/27/00 12:19:34 AM:
When we talking about empirical formula in module
24, is it always the same like the basic ionic
compound? (i.e. Na2O) Shahar
||the empirical formula is the
smallest whole number ratio of elements in a compound. The
empirical formula formula is also represent the actual
formula (called the formula unit for ionic compounds) for
ionic compound. for example the emprical formula for the
following compounds is the same as the formula unit: NaCl,
MgBr2, Na2O. Only in few cases is the empirical formula
different than the formula unit in ionic compounds eg for
Merculary(I) bromide its formula unit is Hg2Br2 but its
empirical formula is HgBr.
However for covalent compounds usually the actual formula
(refered as molecular formula for covalent compounds) is a
higher number multiple of the empirical formula for
example for the compound benzene, its molecular formula is
C6H6 whereas its empirical formula is CH.
|Kevin_kwanghoon Ha 10/26/00 11:00:00
Could you explain about affinity. I am not quite
sure what it is, yet.
||Electron affinity that is
reported for different elements and ions is the amount of
energy given off (in most cases) or absorbed when an
electron is added to a species in a gaseous state. for
example if we measure the energy when
Cl (g) + 1 e' --> Cl- (g) then we call that amount of
energy electron affinity. When an electon is added to an
element in a gaseous state usually energy is given off
(specially for the halogen groups) that means it is a
favorable process. But in some cases it is not a favorable
|Kevin_kwanghoon Ha 10/26/00 10:57:44
Ions are held together by electrostatic force
between them. So, is there no chemical bonds
between them at all? Does the Lewis structure not
apply to ions?
||there is a chemical bond
between oppositly charged ions and it is called ionic
bonding (or electrostatic attraction).
we can also write Lewis dot structure for ionic compounds
but it is not discussed in details since it is just
writting the Lewis structure of each ion involved
remember when we write the Lewis dot structures for
elements we write the symbol of the elmenent and the
valence electrons represented as dots. we do the same for
ions. We write the symbol of the elements and write the
valence electrons (after they have lost or gained) and
then we put a bracket around the symbol of the element and
write their charge.
for example if I want to write the Lewis dot structure for
NaCl, I will write the Lewis dot structure for Na ion
(Na+)and Cl ion (Cl-).
Na element has one valence electron but Na+ has none and
therefore it will be written as [Na+] and for chlorine
element it has seven valence electrons then when it
becomes Cl- then it has eight valence electrons and
therefore its Lewis structure will show [Cl]- (there are
eight dots around the Chlorine but I am not able to show
it here). Overall it will be shown as [Na+] [Cl]- (with
eight dots written around the Cl
Another example if I was showing the Lewis dot structure
for MgCl2 I will show [Mg2+] 2[Cl]- (showing the eight
dots around the chlorine.
|Kevin_kwanghoon Ha 10/26/00 10:54:52
I learned that that there are different ways of
drawing the Lewis electron dot structure between
the same elements. But would it be right on the
exam if I just draw one of them that fits the
||You are right, there are
differnt ways and guidelines you can use any one that you
understand. The point is that you end up with the correct
Lewis structure. After you finish drawing the Lewis dot
structure (with what ever guideline you have followed)just
make sure that you have done the following.
1) used the right number of valence electrons(and
remember to add 1 electron for a negative charge and
subtract one electron for a positive charge).
2) make sure each element has the correct number of
electrons around it. Remember in a molecule or a
polyatomic ion the following need to be satisfied
a) hyderogen must have 2 electrons around it.
b) C, N, O, F must always have 8 electrons around them
(the octet rule)
c) there are some elements that are satisfied with less
than eight and you are responsible for knowing that Be is
satsfied with 4 electrons, and B, Sn can be satsfied with
d)elements that appear in periods 3, 4, 5, 6, 7 can have 8
or more. eg S can have 8 in compounds like SCl2 but has 12
in compounds like SF6.
the most common elements you will see for this are S,P,
|Quynh-mai_tu Nguyen 10/26/00 3:45:49
Dr. Ferede, 1) I'm not clear about the differences
between the enthalpy formation and the latice
energy. Could you explain them? 2) What is the
equation for enthalpy formation and the equation
for the latice energy? 2) Could you explain the
answer for the following self test problem #5 on
module 23? if the EA increase down group, then why
is Cl > Br? Copy of problem #5, module 23. In
which of the following pairs is the element with
the highest electron affinity listed second? F and
O F and Cl - Correct Answer Cl and Br I and Te
Explanation: Electron affinity generally increases
across a period, and generally decreases down a
group. The F atom is the exception because its
electron affinity is lower than that for the Cl
atom. F > O F < Cl Cl > Br I > Te
Information that supports this question can be
found in the lesson titled "Atomic
Properties", on the screen titled Electron
||for answers for #1 and #2
questions please read the answer I gave for Teresa_Chanton
Brown on 10/24/00 at 6:25:22 pm down below.
2) Electron affinity generally increases across a
period (horizontal), and generally decreases down a group
(vertical). So by this general trend Cl should have a
higher electron affinity than Br, since Br is lower in the
group than Cl
There are quite few exception to this rule, one of
these exception is the halogen family (group). From the
general rule F should have higher electron affinity than
Cl but it does not. the reason for this being the unusal
small size of F. However the other halogens follow the
general trend. i.e. Cl>Br>I>At.
Since there are so many exception for electron affinity
trend and different explanations for the exceptions just
remember the general trend and these three exceptions.
|Eric_m Yoshida 10/26/00 10:51:37 AM:
Dr. Ferede, When talking about the size anamoly of
second and third row transition metals we have the
"lanthanide contraction", but is there
an anamoly for the actinide series as well?
||The 5f orbitals of the
actinides are not effective in repelling the 7s electrons
like in the case of in the lanthanides, however we would
not out of a trend since since this phenomonon is already
seen by the lanthanides. In the case of the lanthanides it
is anomoly since we are comparing it the 5th period that
do not have any f orbitals.
|Ashley_ann Tucker 10/24/00 8:05:22
Module 25: Question #7 Will we be required to draw
Lewis electron dot structures for molecules and
polyatomic ions for this Friday's quiz? Question
#8 Besides, the fact that these dot structures can
help us to predict the structure of the molecule
or ion; what are some other benefits they might
||Yes, this friday quiz will
include modules 23, 24, 25. and since drawing Lewis
electron dot structures for molecules and polyatomic ions
is big part of module 25, you will have questions on
those. Remember the quizzes come directly from the Self
Tests, Addional Study and FAQs of these modules. So work
out these problems before the quiz and you will know
exactly what to expect.
also I will like to point out that I have given you a good
guideline on how to write Lewis dot structures in the
study guideline I sent you for module 25.
|Ashley_ann Tucker 10/24/00 7:58:48
Module 25: Question #6 Are there exceptions to the
octet rule we sholud know about? If so can you
please advise. Also, is there any other helpful
tips available to help predict the results for the
most common reactions?
||exceptions to the octet rules
are discussed in module 26, (for next week). Therefore I
will wait until next week to answer this question. If
after doing the module next week if you still have this
question then let me know. I am also sending you outlines
for this module.
for the second part of your question, please refer to the
answer I gave to Quynh_Mai_tu on 10/5/00 at 2:47pm down
below. In the middle of my answer I have listed the common
type of reaction and how to predict their product.
|Ashley_ann Tucker 10/24/00 7:52:12
Module 25: Question #5 Dr. Ferede, is it safe to
say that valence electrons work to bond to other
atoms and core electrons do not?
||Yes, it is the valence
electrons that are lost, or gained or shared in a chemical
|Ashley_ann Tucker 10/24/00 7:47:20
Modlue 24: Question #4 Why is the group number of
a main group metal related to the number of
valence electrons for that element?
||that is how it is numbered so
that it is convenient.
|Ashley_ann Tucker 10/24/00 7:41:06
Module 24: Question #3 Dr. Ferede, Why is it that
a compound such as NaCl2 not form? Does it have
anything to do with the lattice energy for NaCl2
being twice the amount of that of NaCl. If so, can
you please explain the concept in further detail.
||for two elements to come
together to form ionic solid there three energies
. electron affinity,Cl going to Cl- eg
lattice energy, eg Na+(g) + Cl-(g) going to NaCl(s)
and all these three energies must add up to give a
negative energy change, a more stable state (lower
energy), i.e. the ionic solid must be at a more stable
state state than the individual elements involved.
In the case of NaCl2 the amount of energy to form Na2+
is so high that the lattice energy is not enough to
compasate for that. where as the formation of Na+ is not
very high and the amount of lattice energy is enough to
compasate for that.
|Ashley_ann Tucker 10/24/00 7:27:52
Module 23: Question #2 Dr. Ferede, Since
ionization involves moving electrons from a
particular electron shell to a position outside of
the atom,(n=infinity), is energy always required
to overcome the attraction of the nuclear charge?
And is the sign of the ionization energy; if
required always going to be positive?
||Yes, energy is required to
ionize, and since energy is added to the system, then the
sign for the energy is positive.
|Ashley_ann Tucker 10/24/00 7:17:23
Module 23: Question #1 Dr. Ferede,when an ion has
one unpaired electron is it always considered to
be paramagnetic or is it ever a time where it
would be considered diamagnetic? Can you please
give an example of this in the form of configuring
a transition metal ion.
||Any substance (usually we are
referring to nuetral compounds rather than ions) that has
unpaired electrons is paramangentic. Write the electronic
configuration of a transition metal first Then inorder to
know if it is a diamangetic or paramangentic substance you
really need to write the box notation in which you do one
box for s orbitals, 3 boxes for p orbitals, 5 boxes for d
orbitals and 7 boxes for f orbitals.
for example the electronic configuration for Cr is
1s2 2s2 2p6 3s2 3p6 4s1 3d5. If we look just look at the
valence electrons (for transition metals considered the
highest energy s orbitals and d orbital) then it would be
4s1 and 3d5. In this case for the box notation there will
be 1 box for s orbital (holding 1 e) and 5 boxes for d
orbitals (each holding 1 electron). Then as you can see
this will be a paramangetic substnace.
if you want to do it for Cr+3 then take away three
electrons from the the Cr and then do the same thing as
you did for Cr above.
|Teresa_chanton Brown 10/24/00 6:25:22
PRofessor Ferede You said that if we do not
understand the difference between the enthalphy
equation for NaCl and the lattice energy for NaCl
to tell you. Well I don't understand.
||I just want you to understand
the difference between the meaning of enthalpy of
formation and lattice energy. the enthlpy of formtion is
the energy given off when elements in their standard state
form 1 mole of a substance. whereas the lattice energy is
the energy given off when ions in a gaseous state come
togehter to form solid lattice.
eg the equation for enthalpy of formation for NaCl will
Na(s) + 1/2Cl2(g) -- > NaCl(s)
whereas the equation for the lattice energy of NaCl will
Na+(g) + Cl-(g) -- > NaCl(s)
|Teresa_chanton Brown 10/24/00 5:00:50
Professor Ferede COuld you give another example of
main group elements that electron configurations
that are of the nearest noble gas. When you do
noble electron configuration , does it depend on
which noble gas you are near? How do I determine
how to do it.
||the main group elements are
group 1A to VIIIA
metals (group 1A, IIA, IIIA) will lose 1, or 2, or 3
electrons to have an electron configuration like the noble
gas that is behind them.
Na will loose 1 electron (Na+) to become like Ne.
group 1A have S1 as a valence electron and
therefore lose 1 valence electron to aquire a noble gas
eg Li 1s2 2s1 when it becomes Li+ will lose 1 electron to
have electronic configuration of 1s2 (which is like the
electronic configuration of the noble gas He)
another example is Na has 1s2 2s2 2p6 3s1 and when it
becomes Na+ it will have 1s2 2s2 2p6 (this is the same
electronic configuration as the noble gas Ne) for group
IIA their valence electrons are S2 and
therefore they loose 2 electrons to aquire a noble gas
structure eg Mg has 1s2 2s2 2p6 3s2 when it looses its 2
electrons it becomes 1s2 2s2 2p6 (which is the same
configuration as the noble gas Ne)
group IIIA, have s2 p1 as their valence electrons but
only Al in this group looses 3 electrons to aquire a nobel
gas configuration like Ne
for nonmetals: they gain 1 (for group VIIA) , 2 (for
group VIA), or 3 electrons (for group V, but only N, P in
this group) to aquire a noble gas configuration that
appears after them.
eg F has an electron configuration of 1s2 2s2 2p5 and when
it becomes F- then it will have an electron configuration
of 1s2 2s2 2p6 (which is the same electronic configuration
|Teresa_chanton Brown 10/23/00 8:37:16
When answering this question should I do the
eletron confihuration to help me? Write the number
of electrons and the number of protons for the
following species and prove to yourself that they
are isoelectronic species? o O2-, F-, Na+, M2+
||No you really do not need to
write down the electronic configuration.
all you need to know is the atomic number of each
element from the periodic table then to find out the
number of electrons in each ion, add one electron for each
negative charge and subtract one electron for each posive
O atomic # = 8, O2- = 10e'
F atomic # = 9, F- = 10e'
Na atomic # = 11, Na+ = 10e'
Mg atomic # = 12, Mg2+ = 10e'
As you can see O2-, F-, Na+, M2+ all have the same
electrons. and the definition of isoelectronic species is
species (can be elements or ions) that have the same
number of electrons.
Notice these species have different # protons. Remember
the # of protons is determined by the atomic number and
they do not lose or gain protons when they become ions.
only electrons are lost or gained to become ions.
|Teresa_chanton Brown 10/23/00 4:41:35
professor Ferede What do you mean when you ask the
question : ? Is what is expected from the general
trend and the experimental result for ionization
energy of nitrogen and oxygen the same? This
question is from the study guide.
||the general trend in ionization
energy is that it increases from left to right across a
period that is it takes more energy to remove an electron
off of an element that is on the right side of a period
than on the left side. for example the ionization energy N
is more than C.
following this general trend we would expect O to have
higher ionization energy than N. But the experimental data
shows the other way round, where N has a higher ionization
energy than O.
the reason for N having higher IE than O is clearer to
see if you first draw the electronic configuration of O
and N in a box notation. N has has only unpaired electrons
in the 2p orbitals, where as O has one of its p orbital
paired. because these two electrons in O are in the same
space there is electron-electron replusion making it
easier to remove an electron from O than N.
|Teresa_chanton Brown 10/23/00
Professor Ferede Why is it that the outermost
orbitals are the orbitals with the highest energy?
I thought it was the other way around.
||The outermost electrons are the
orbitals with the highest energy because they are the
least stabilized by the attraction of the positive
charge(s) i.e. the protons of the nucleus since they are
so far from the nucleus. The inner electrons are more
attracted by the nucleus since they are near the positive
charge(s)and they are more satisfied and therefore more
stable than the outermost electrons and therefore less
|Teresa_chanton Brown 10/22/00 7:46:16
Professor Ferede I don't understand how to answer
this question from the study guide. What kind of
answer are you looking for? ? Which orbitals
contain the electrons in s-block elements p-block
elements d- block elements f-block elements
||the question should read which
orbitals contain the "valence electrons" and the
the s-block elements are group 1A and group 2A elements
and therefore their valence electrons are in the
the p-block elements are group 111A to V111A elements
and their valence electrons are in the s and p orbital
for example Boron's valence electrons are2s22psup>1
and Carbon is 2s2 2p2
for d-block (the transition metals) contain d-orbitals
f-block (lanthanide and the actinde and also referred
as the inner transitin elements) contain f orbitals.
|Eric_m Yoshida 10/20/00 12:04:41 PM:
Dr. Ferede, How do we determine the amount of an
element present using spectroscopy? I know that we
can identify it easily by comparing the
wavelengths present, but how do we know how much
is there? Will the lines be thicker when more
element is present?
||There are different kind of
spectropscopic method used in identifing the identity of
compounds (what elements, how much of each elements in the
compound) the concentration of a compound in solution.
examples of common spectroscopic methods are called
Infrared spectroscopy, UV/vis spectroscopy and NMR
spectroscopy. You will learn about these methods in a
chemistry course such as chem 21 and 22. These topics are
not discussed in chem 11.
|Shahar Kalev 10/20/00 12:34:02 AM:
Why does an element with more protons has more
energy? lets say cr6+ has more energy than cr 3+.
We know that electrons are the responsible for
||Saying element with more
protons have more energy is not a correct statement. for
examples for elements in the same period, the ones that
have more protons have their valence energy at a lower
energy state than the ones that have less protons and
therefore are smaller atoms. for example F is a smaller
atom than Li. However with in a group elements with more
protons have their valence energy at a higher state than
the one that have less protons and this is just because
there are more electrons and the attraction of the protons
is not so effective.
Looking at Cr+6 and Cr+3, both of them have the same
number of protons, the difference is the number of
electrons, cr+6 has 3 less electron than Cr+3.
Remember when cations and anions form the number of
protons do not change it is the number of electrons that
change. That is what most chemistry is about, what happens
to the electrons not what happens to the protons. Special
chemistry that deals with protons and nuetrons (or the
nucleus of the atom) is nuclear chemistry which is not
discussed in chem 11
|Shahar Kalev 10/20/00 12:30:16 AM:
What is the difference between electromagnetic
radiation emmited by heated bodies and heated
gases? module 20. Shahar
||The difference in the emitted
radiation arises from the difference in the electronic
structure of atoms and solid matter. Gas-phase atoms have
the electronic energy level structure of isolated atoms.
Heating a gas thermally or with an electric discharge
excites an electron in an atom to a higher energy level.
The atom emits light of a wavelength that corresponds to
the energy difference between its initial and final
electron states (remember E = hc/). Since the transition
is between two very specific energies, the light emission
occurs at a specific wavelength. Since there are multiple
electronic energy levels in an atom, there is a multiple,
but finite, number of emitted wavelengths. A solid
consists of a very large number of atoms that are bonded
together, producing a very large number of closely spaced
energy states. The electronic structure of the solid
determines its appearance- for example, whether the solid
is transparent, colored, or metallic looking. Just as in
an isolated atom, heating a solid promotes electrons to
higher energy levels. A heated solid emits light because
these electrons in excited states return to lower states
to release energy. Transitions occur to all of these
closely spaced energy states resulting in a continuous
emission spectrum, which we call blackbody radiation.
(copied from FAQ of module 20)
|Shahar Kalev 10/20/00 12:27:07 AM:
Can you please send us outlines to the next
modules also? I think they were helpfull. Shahar
|Stephen Kim 10/19/00 8:26:14 PM:
could you explain mod 21 self test 10: Some d
orbitals are easily recognized because they have
||Let us first describe the p
orbitals. the p orbitals can be described as being like a
figur 8 or a dumb bell shape or having two lopes.
Now for the d orbitals: there are five d orbitals, four
of which are similar and the fifth one is different.
the four that are similar can be described as clover leaf
shape or having four lopes each and oriented at different
axis. the fith one looks like a p orbital (so it only has
two lopes)with a ring around it
|Teresa_chanton Brown 10/19/00 8:09:59
Professor Ferede Could you explain "ground
state" in electrons?
||The ground state of an atom is
the most stable state for the atom, Where all the
electrons are at the lowest energy possible. for example
the ground state of H is 1s1. that is the electron of
hydrogen would be in the first energy level in an s
orbital. If enough energy is added this electron could go
to the 2nd, 3rd, energy levels. It could have an
electronic structure of 2s1, or 2p1, or 3s1 etc depending
how much energy it has aquired. Then it is in an excited
state and therefore it is not in its most stable state
that is it is not in its ground state.
giving another example the ground state of Li is 1s2 2s1.
that is when the electrons are in the most stable state
(the ground state) but if they acquire energy then they
can go to a higher (excited state. Usually in textbook and
in your module the ground states of different elements are
reported not the excited states.
|Teresa_chanton Brown 10/19/00 8:06:22
Professor Ferede when using Planck's constant h
should I use the value 3.0 E8 , 2.998 E8, or
2.9979 E8. I am thinking that that is one of my
problems, when working out the problems at the end
of the chapters in the book.
||This are values for c (speed of
light) not h (Planck's constant) and the number of
signigicant figures really should not make that much
difference. it is just that the first one has only 2
significant figures and the second one has 4 and the third
one has 5. If in the question you are asked if the values
of other measurments (such as wavelength or frequency) are
given in 2 sig. figs then use the first c value but if the
other measurments are given in 4 sig fig use the 2nd one
and if they are given in 5 then use the third one. But
like I said it should not make much difference except the
number of significat that you will obtain will be
different (the integrity of the number will be the same).
The value for h (Planck's constant) is 6.626 x 10-34j.s.
I think that is where the problem lies, you must be using
the wrong constant.
|Teresa_chanton Brown 10/19/00 8:00:49
Professor ferede I am having a problem working out
problem # 26 in the text book chapter 7. could you
explain it in a breif way?
||When you are referring to a
particular question please copy or type the question.
Right now let me write the question down then I will
chapter 7, #26. The most prominent line in the spectrum
of magnesium is 285,2 nm and other are found at 383.8 and
a) in what region of the electromangetic spectrum are
these lines found/
b) which is the most energetic line?
c)what is the energy of 1 mole of a photon of the most
d) How much more energetic is a photon of this light
compared with a photon associated with the least energetic
a) 285.2 nm is uv light, 383.8 is a border between uv
and visible light and 518.4 nm is visible light.
b) the one that has the shortest wavelength is the most
energetic, i.e. 285.2 nm.
c) first find the energy of 1 photon using the equation
E = h x nu (and remember h = 6.626 x 10-34) and
c= speed of light and lambda is the wavelenth in meters
(so you have to change nm to meters) . then after finding
it for one photon to find it for one mole multimply the
answer by Avogadros # (6.02 x1023.
d) calculate the energy of the wavelength of 285.2 nm
(which was done in answer c) then calculate the energy of
the wavelength of 518.4 nm then subtract one from the
other to get the difference in energy.
|Stephen Kim 10/19/00 7:28:39 PM:
Could you explain mod 20 question 4 on self test
please: How long did it take for radio signals to
return from the Explorer landing craft on Mars,
when the distance was 100 million km?
||The speed of any
electromagentic radiation is 3.00 x 10 8 m/sec
that means it will take it 1 sec to travel 3.00 x 10
Now change 100 million km into a meter and then divide
it by the speed of light and you will get your answer.
|Sang_hoon Chung 10/19/00 3:03:18 PM:
1)Is there any difference between heated body and
black body? 2)In one of the points in matter wave,
it talks about electron density. When we're
talking about electron density, does it mean the
number of electrons present? (p.315) 3)If so what
is the unit?
||All objects emit a range of
electromagnetic radiation at all times. This called
As temperature increases, the intensity of blackbody
radiation increases and the peak wavelength shifts to
shorter wavelengths. We can see the blackbody radiation
from a hot object such as an electric stove burner or the
tungsten filament in a light bulb because some of the
blackbody radiation is in the visible region of the
electromagnetic spectrum. A stove burner at several
hundred degrees appears red hot, and a light bulb, which
operates at several thousand degrees, glows white hot.
Unheated objects are at room temperature, which is
approximately 300 K. The blackbody radiation at this
temperature is in the infrared portion of the
electromagnetic spectrum, which we can not see visually.
However, infrared light can be detected with special
infrared-sensitive detectors or video cameras. Infrared
cameras are used for night security. They work on the
principle that intruders can be detected because a person
has a higher temperature, and therefore a blackbody
spectrum that is brighter and shifted (to a higher
frequency range), compared to the cooler surroundings. The
blackbody spectrum of an object can also be used to
determine its temperature (this was copied from FAQ of
2) when it talks about electron density, it does not
mean the number of electrons present, it is talking about
the probablility of finding electron is more or less
around a certain region. For example let us take hydrogen
atom at its ground state. The probablity finding the
electron is highest nearer to the nucleus at n = 1 than
when it is further from the nucleus let us say n = 4, or
etc. then we say the electron is density highest at n = 1
for an electron in a hydrogen atom at a ground state. The
propbablity of finding the electron gets smaller and
smaller as it goes further and further i.e. the electron
density gets thinner and thinner.
|Eric_m Yoshida 10/18/00 4:47:52 PM:
Dr. Ferede, I was confused at the last question of
the studyguide questions for module 20. It asks
"What does promoting electrons to n=infinity
indicate?" Is it referring to the orbitals or
shells? Is it referring to adding energy? Or is it
just saying that we can have an infinite number of
||Sorry about the confusion. it
is saying when an electron goes from n = 1, or 2 or 3 etc
to n = infinity by adding energy, what has happened to the
the answer I was looking for is: the atom has actually
lost the electron and the atom has been ionized.
|Kevin_kwanghoon Ha 10/18/00 6:31:28
In module 20, we use symbol "v" for
frequency. But I learned that the symbol for
frequency is "f" in my Physics class.
Why is it different in Physics and Chemistry? One
more thing. For First Law of Thermodynamics, delta
E = q + w in Chem, but in Physics delta U = q - w.
One last thing. For work done on its surrounding,
w < 0 in Chem, but w > 0 in Physics. I know
they all mean the same thing. But why do they use
different symbols or notations.
||for your first question, it is
just a matter of using different symbols. but for your
second and third question both U and E are used in
chemistry and physics. and the sign for w being different
is a matter of having different definition. Chemist define
this terms in terms of the system where as the physists
define it interms of the surrounding and hence a
difference in the sign. +w and -w
|Kevin_kwanghoon Ha 10/18/00 6:21:05
Is there a simple rule for electron
configurations? Could you outline the rule for
||the first simplest order is n=1
is filled first then n=2 then n=3 etc. as you know this
simple order is really not followed after n =3 since 4s is
filled before 3d and so on. Therefore a different order
than discribed above is used for elements that have more
than 18 electrons. One simply way to remember the order of
filling is shown on page 365 of your text book and I will
also explain it below.
3s2 3p6 3d10
4s2 4p6 4d10 4f14
5s2 5p6 5d10 5f14 "5g20"
6s2 6p6 6d10 6f14 "6g20" "6h24" etc
some orbitals are written in quatation " " since
there are no known elements that fill these orbitals.
then draw a diagonal arrow going from top of
"1s" to bottom of "1s" with the arrow
pointing down (it is like a diagnonal arrow you will draw
if you have a square and you are drawing an arrow down
from the top right corner of the square to the bottom left
corner of the square with an arrow going down)
then draw another diagnal arrow from top of "2s"
to the bottom of "2s" with an arrow down
then draw anther one from the top of 2p to 3p with an
then another from 3p to 4s with an arrow down and another
from 3d to 4p and 5s. continue drawing these arrows until
you have at least 7 parralel arrows. After you finish
drawing your arrows then start with the top arrow and
follow each arrow from the bottom of the arrow to the tip
of the arrow .
one difference with my explanation above and the one
diagram shown on page 365 of your book is that in the book
the "1s" is shown at the bottom.
Then the order of filling you will find will be
1s 2s 29 3s 3p 4s 3d 4p 53 4d 5p 6s 4f 5d 6p 7s 5f 6d
one important thing to notice is the order of filling
is ns, np, (n-1)d, (n-2)f
for example let us take n=6 then we have 6s and 6p (the
same n value), but we have 5d (1 less n value) and 4f (2
less n value)
there are some exception to this order specially when
you go to elements with higher atomic number since the
differecne in energy levels becomes smaller as you go
higher and orbital energy mixing occurs.
and there is also exception such as in Cr and Cu, where
all half filled degenerate 3d orbitals are half occupied
(in the case of Cr) or fully occupied (in the case of
Cu)before the 4s is filled
there is a good summary of noble gas electronic
configuration of elements on page 367-368 of your text
book. As far this class is concerned just know the general
order of filling and do not worry about the exceptions
except for Cr and Cu.
|Kevin_kwanghoon Ha 10/18/00 2:59:02
Module 20 / Self Test Q#9 An electron in a
hydrogen atom undergoes a transition from the n =
1 state to the n = 16 state. The result of this
transition will be: 1 A continuous spectrum. 2 An
emission line of relatively high frequency. 3 An
emission line of relatively low frequency. 4 An
absorption line of relatively high frequency. 5 An
absorption line of relatively low frequency.
||the two energy lever are far
apart so it is going to take a lot of energy to go from n
=1 to n = 16. therefore it is going to be high frequency
(short wavelength). it is going to be absorption since it
is going from low energy state to high energy state. so
the answer should be 4.
|Ashley_ann Tucker 10/17/00 11:35:19
Question/Request #1 Module 20: Dr. Ferede can you
please give some tips that would be beneficial in
aiding us with remembering the relationships
between the energy, frequency and wavelength of
radiation? Question #2 Module 20: Is wavelength
and frequency always inversely related? If so can
you please further explain the concept and why it
is that this relationship is always the way it is.
Also, will waves with the shortest wavelength
always have the highest frequency and those with
the longest wavelength, will they have the lowest
frequency? Question #3 Module 20: How does
electromagnetic radiation have electric and
magnetic fields oscillating simultaneously in
planes that are mutually perpendicular to each
other as well as in the direction of propagation
through space. Can you please explain this concept
in futher detail.
||wavelength and frequecy are
inversly proportional that means that long wavelength
corresponds to short frequency (and vise vesa). Let me
give you an analogy. Let us say you are standing in the
middle of the ocean and you are counting the number of
waves that pass you per minute. If the waves have long
wavelength only few of them will pass you by compared to
if the waves have short wavelength. Therefore the
frequency of the waves that have long wavelength is
frequency is directly propotrional to energy i.e. waves
that have higher frequency have higher energy (or vise
vera). going back to our water analogy more of the shorter
wavelength waves will pass you by per minute than a longer
wavelenth wave. and therefore since each wave has an
energy associated with it more of them will have more
#2)Yes , wavelength and frequency are always invesely
proportional. and waves with high frequency will always
have higher energy. This are relationships that have been
empirically determined. For electromangetic radiation the
relationship is summerized in this two equations.
one that relates frequency(nu) to speed of light (c)
and wavelength of a wave (lambda)
nu = c/lambda
the second eqaution is relates energy to frequency and
a proportionality constant, Plack's constant (h)
E = h x nu
|Teresa_chanton Brown 10/17/00 8:44:52
professor Ferede I was wondering, when an atom is
in an exicted state, is the molecule that the atom
occupies more likely to bond with other molecules.
Or will an element more likely donate or give an
atom to another element?
||When an atom is in an excited
state it has more energy than when it is in the ground
state. Therefore less energy is needed to pull an electron
off (ionize) of an excited atom than an atom that is in
the ground state.
|Teresa_chanton Brown 10/15/00
Professor Ferede could you explain to me how to do
the test question problem number 6 from module 19.
My understanding of this problem is not clear.
||In module #19, questions #6. 7,
8,and 9 are similar questions. I answered question #9 for
Stephen Kim on 10/13/00 at 12:34:21 down below. The
difference between #6 and #9 is that #9 is a longer and a
little more complicated than #6.
I will outline what you should do for all these types
of questions and then I will guide you how to do #6.
We are using Hess's law, which simply states that if
individual equations can be added up to give a final
equation then their delta H can also be added to give the
delta H of the final equation
step 1) look at the individual equations (you can
number them #1, 2, 3 etc if you want) and see if they are
written in the same direction, i.e. the reactants and the
products are on the same side as the final equation.
step 2) Then look to see if the number moles of
reactants and products in these individual equations (eqn
1, 2, 3 etc ) are the same as the final equation.
step 3: once all the individual eqautions (equation 1,
2, 3 etc ) are in the proper format as the final equation
then add them (as well as their delta H) up.
In order to add up individual equations to give us a
final equation, we add up all the reactant side of all the
equations and all the product side of all the equations
and then cancel any species that shows up the same on both
Now let us do #6
Step 1) look at the individual equations (in this case
we will call them eqn 1, 2) and see if they are written in
the same direction as the final equation.
question 1 is notwritten in the same direction as the
final equation but question 2 is . and therefore we need
to reverse question 1 and write it as 2NO2 -- >
N2O4delta H = -58 notice since we reversed the equation
then we need to reverse the delta H to make it -58 kJ
step 2) Now let us see if the number moles of reactants
and products in these individual equations (eqn 1, 2) are
the same as the final equation.
for equation 1, there is 1 N2O4 in this equation and there
is 1 N2O4 in the final equation. therefore we leave this
equation alone (i.e. we do not multiply it by a
equation 2: there is 1 NO in this equation but there are
2NO in the final equation and therefore we need to
multiply this equation (as well as the delta H) by 2
giving us "NO + O2 --> 2NO2 delta H - 112
step 3: Now all the individual eqautions (equation 1,
2) are in the proper format as the final equation and
therefore we can add them (as well as their delta H) up.
In order to add up individual equations to give us a final
equation, we add up all the reactant sides of all the
equations and all the product side of all the equations
and then cancel any species that shows up the same on both
side. in this case it becomes:
2NO2 + 2NO + O2 --> N2O4 + 2NO2 delta H -170 kj
in this case the 2NO2 cancel out since they appear on
the reactant and product side of the equation to give you
the final equation
N2O4 delta H - 170 kJ let me know if any of the steps are
|Kevin_kwanghoon Ha 10/13/00 1:18:09
Module 18 Question #3, The heat capacity of liquid
water is 4.184 J/g K and the heat of fusion and
vaporization of water are 333 J/g and 2260 J/g,
respectively. Determine the heat absorbed or
released when 75.0 g of water vapor at 100.0 ¡ÆC
solidifies to ice at 0.0 ¡ÆC.
||you need to do this steps
the gerenral steps are
steam at 100 oC --> water at 100 oC --> water 0
oC --> ice at 0 oC
step 1: steam at 100 oC to water at 100 oC. here there
is a phase change, to do this:
mass x heat of vaporization
step 2: water at 100 oC to water at 0 oC. Here there is
no phase change, to do this
mass x SH x delta T
step 3: water at 0 oC to ice at 0 oC. Here is a phase
change. to do this
mass x heat of fusion
then you add up all the heats that you have for the
Let me know if you do not understand any of these
|Kevin_kwanghoon Ha 10/13/00 1:11:34
Module 18 Question 10 The internal energy change
of a chemical reaction is –205 J. If 300 J are
evolved during the reaction, then the system has
done ___ J of work. I answered -95, but was
incorrect according to the Archipelago. I did get
95 for work since -205 = -300 + W, and W is 95.
But W > 0 means surroundings did work on
system, and the question is asking for how much
work the system has done. So, I answered -95. Is
my answer right or wrong?
||Yes the answer is +95, and yes
you are correct if is (+) then work was done to the system
not the reverse
|Kevin_kwanghoon Ha 10/13/00 1:03:26
Could you explain what is heat of fusion and
vaporization of water? Although the values are
given as 333 J/g and 2260 J/g, respectively, I
don't really understand what they are compared to
just heat capacity of water.
||Heat of fusion is just the heat
needed to melt (to turn it from a solid to a liquid) a
substance. The heat of fusion can be given per gram or per
mole. In this case where it says the heat of fusion is 333
j/g, it is saying to melt one gram of this substance it
takes 333 j.
heat of vaporization is the amount of heat to vaporize
(to turn it from a liquid to a gas) a substance. It can
can be given per mole or per gram. In this case it is 2260
j/g, which means that 2260 j of heat is needed to vaporize
one gram of this substance.
Heat of fusion and heat of vaporization are heat needed
to change one state to another , i.e. from solid to a
liquid and from a liquid to a gas. However the heat
capacity is the amount of heat needed to raise the
temperature of a substance with in one state for example
water from 17 oC to 20 oC. heat capacity can also be given
per gram or per mole. If it is per gram: then it is
defined as the amount of heat needed to change 1 gram of
substance 1 oC (or K) and if it is given per mole than it
is defined as the amount of heat needed to change 1 mole
of a substance 1 oc(K)
|Stephen Kim 10/13/00 12:34:21 AM:
Self Test Question 19.9 The enthalpy change for
the reaction: 2 B5H9(g) + 12 O2(g)--> 5 B2O3(s)
+ 9 H2O() given the following information is _____
kJ. 5 B(s) + 9/2 H2(g)-->B5H9(g) H = 62.8 kJ 2
B(s) + 3/2 O2(g)-->B2O3(s) H = –1263.6 kJ 2
H2(g) + O2(g)-->2 H2O(l) H = –571.5 kJ I am
clueless to this one
||We are using Hess's law, which
simply states that if individual equations can be added up
to give a final equation then their delta H can also be
added to give the delta H of the final equation
step 1) look at the individual equations (in this case
we will call them eqn 1, 2, 3) and see if they are written
in the same direction, i.e. the reactants and the products
are on the same side as the final equation.
question 1 is not but but question 2 and 3 are. and
therefore we need to reverse question one and write it as
B5H9(g) --> 5B(s) + 9/2 H2(g) delta H = -62.8
notice since we reversed the equation then we need to
reverse the delta H.
step 2) Now let us look at if the number moles of
reactants and products in these individual equations (eqn
1, 2, 3) are the same as the final equation.
for equation 1, there is 1 B2H9 in this equation
however there are 2 B2H9 in the final equation and
therefore we need to multiply this equation (as well as
the delta H)by 2 giving us:
2B5H9 --> 10B +9H2 delta H = -125.6
equation 2: there is 1 B2O3 in this equation but there
are 5B2O3 in the final equation and therefore we need to
multiply this equation (as well as the delta H) by 5
10B + 15/2O2 --> 5 B2O3 delta H = -6816.5
equation 3: there are 2 H2O in this equation but there
are 9H2O in the final equation and therefore we need to
multiply this equation (as well as the delta H by 9/2)
9H2 + 9/2 O2 --> 9H2O delta H -5143.5
Now all the individual eqautions (equation 1, 2, 3) are
in the proper format as the final equation and therefore
we can add them (as well as their delta H) up.
In order to add up individual equations to give us a
final equation, we add up all the reactant sides of all
the equations and all the product side of all the
equations and then cancel any species that shows up the
same on both side. in this case it becomes:
2B5H9 + 10B + 15/2O2 + 9H2 + 9/2O2 --> 10B + 9H2 +
5B2O3 + 9H2O.
in this case the 10B's and the 9 H2's cancel out to
give you the final equation
2B5H9 + 12O2 --> 5B2O3 + 9H2O
and therefore add up all the delta H to give you the
delta H of the final equation
-125.6 (from eqn 1, obtained by reversing and myltiply
+-6816.5 (from eqn 2, obtained by multiply by 5)
-2571.5 (from eqn 3, obtained by multiplying by 9/5)
total = -9503.6Kj (Please check the maths, I did not use a
calculator and let me know if any of the steps are not
|Eric_m Yoshida 10/12/00 10:34:45 AM:
Dr. Ferede, Would you explain the concept of when
a reaction is thermodynamically favored. The
modules explanation with energetics doesn't seem
like it is enough. Is there more to it?
||Yes, it has to do with the
product(s) being at a lower energy than the reactant(s).
Please read for more information down below the answer I
gave to Ashley Tucker's question on 10/8/00 at 2:17:35
|Stephen Kim 10/10/00 5:21:48 PM:
Prof. Ferede, I am still confused what the
difference between a product favored reaction is
compared to a reactant favored reaction. Does it
mean product favored when just more products are
formed from the reactants or am i wrong?
||Product favored reaction means
when the reaction reaches equilibrium the products are at
higher concentration than the reactants.
eg if A + B ---> C + D.
at the end of the reaction there will be more of C and D
and little (or none) of A and B left.
|Teresa_chanton Brown 10/9/00 1:29:53
Professor ferede IS it acurate to say, that a
glass of cold water when palced in a room can
lower the temperature in a room? I was reading
that in the textbook. It ststed that the cooler
temperature in the glass can lower the temperature
in a room so that the room and everthing in it is
the same temperature.
||Yes, it is true. However if the
glass is small the effect of it will be so small it will
be neglegible. But let us say there is a big bathtub full
of cold water and the room is a small room you, the
cooling of the room will be noticable. Or vise vera, if
the bathtub is full of hot water, the room will get
warmer. In general if two system are at a different
temperature heat will flow from the hotter system to the
cooler system so that both systems will have the same
temperature. the hotter system will get cooler and the
cooler system will get hotter until both are at
|Ashley_ann Tucker 10/8/00 2:17:35 AM:
Question # 4 Module 18: Dr. Ferede: Is it
mandatory to use a fraction coeffienct for O2 when
writing an equation for the decomposition of one
mole of H2O. Please explain if this is the case or
not. Question #5 Module 18: Are coefficients
always taken to mean moles rather than molecules?
Question #6 Module 19: Why are exothermic
reactions considered product-favored when they are
at room temperature? Question #7 Module 19: W What
is the difference between the enthalpies in
regards to reactants and products. Please explain
||#4) When we write a balanced
equation, we usually use whole number integers. for
example for water it would be 2H2O --> 2 H2 + O2.
However we also sometimes use fractions (or decimals) as
long as it is balanced> For example in this case we
need to write the equation for 1 mole of water and
therefore we divide each coefficient by 2, leaving 1/2 on
H2O --> H2 + 1/2 O2.
We can not write
H2O --> H2 + O since oxygen does not exist in nature
as O but exist as a diatom, O2.
#5) remember moles are numbers and molecules are
substances, they completely reperesent a different thing.
confusing moles and molecules is like confusing dozen and
eggs. a dozen is just a number and eggs are substances.
The coefficient in a balanced equation represent numbers,
the numbers could be 1, 2, 3 or avogadros number (6.02 x
10 -23) which is a mole.
for example for the equation
2H2O --> 2 H2 + O2, the ratio of numbers of H2O;H2:O2
is 2:2:1, that could mean that if 2 molecules of H2O react
they will give 2 molecules of O2 and 1 molecule of O2 or
if 4 molecules of H2O react they would give 4 molecules of
H2 and 2 molecules of O2. Or if 2 X 6.02 x 10 -23
molecules of H2O react they will give 2 x 6.02 x 10 -23
molecules of H2 and 1 x 6.02 x 10 -23 molecules
or in anther words 2 moles of molecules of H2O react to
give 2 moles of molecules of H2 and 1 mole of molecules of
H2. We usually say 2 moles of H2O, rather than saying 2
moles of molecules to make it shorter.
#6)There are two factors to consider to see if a
reaction is product favored (spontaneous) or not. One is
enathalpy and one is entropy. You will learn how entropy
can affect if the reaction is spontaneous or not in chem
12. But enthlapy is discussed here. When reactants give
off energy (heat) to go to product they have lower
enthalpy which would says they are at more stable state.
Lower energy is a more stable state than a higher energy.
You can use the analogy of a very active boy and an old
man. the boy has a lot of energy and is less stable where
as the old man has less energy and at a more stable state.
Anyway when reactant go to products by giving off heat
then they are going to a more stable state and therefore
it is product favored.
#7) Each reactant and product has its own enthalpy
(energy related to strength of chemical bonds). We do not
measure the enthalpy of each reactant and product but we
measure the enthlapy differece (delta H) between reactants
and products and find out if they are exothermic and
endothermic and by how much by carring out experiments
such as calorimetric experiments.
|Ashley_ann Tucker 10/8/00 12:43:47
Question #1 Module 17: Dr. Ferede, When we are
required to make calculations that ask for us to
find the difference in temperatures, can we use
either Celsius or Kelvin temperatures or must we
use Kelvin. Question #2 Module 17 Is there a
difference between molar specific heat and
specific heat if so what is the difference and how
do you determine which is which? Question #3
Module 17 Is the main differences between
potential and kinetic energy is that kinetic
energy is moving energy and potential energy is
stored energy that can become kinetic energy was
it starts moving. Also, thinking about the reverse
of this process; if I had an object say a ball and
I dropped it(It would be an example of potential
energy going to kinetic energy)(right?)once it
stops falling and hits the ground would it return
to potential energy or something else? Is there a
transformation that takes place or is that energy
||#1) If you are doing the
difference in temperature, delta T, it would not matter if
you use the difference in celsius or Kelvin. Remember 1
unit change in celsius is the same as 1 unit change in
kelvin. for example the freezing point of water is 273 K
and boiling point of water is 373 K, the difference
between the two is 100. Also the freezing point of water
is 0 oC and the boiling point of water is 100
sup>oC . the differece between the two is 100. So a
unit change in Kelvin is the same as a unit change in
2) Ths specific heat is described as the amount of heat
per degree change per gram i.e. S.H. = J (or cal.)/g.
however the molar heat capcity is the amount of heat per
degree change per mole i.e. Molar heat capacity (usually
given the symbol C) = J (or cal.)/g.delta T
So as you can see one is describing the amount of heat
per gram of a substance and one is describing per mol of a
d) Your descripition of kinetic and potential energy is
fine. Your last statement where you say that energy is
recylced is put in a general form "energy is
conserved". There is a transformation of one into
another but the total energy is conserved.
|Alana_diaz Ayuyao 10/6/00 4:46:25 AM:
Module 14, Self Test #8: Can you explain why there
is only 5 moles of product and not 10?
||A + 2B --> 2C, it says 30
moles of A and 10 moles of B wrere reacted.
step 1: find the limiting reagent.
the mole ratio of A:B is 1:2 therefore if we take 30
moles of A then we need 60 moles of B. But only 10 moles
of B are available and therefore B is the limiting
2) Once we know the limiting reagent then we see that
the ratio of B:C is 2:2. Therefoe if the reaction is 100
%, 2 moles of B produce 2 moles of C, or in this case 10
moles of B produce 10 moles of C. However since the
reaction is only 50 %, 10 moles of B produce only 5 moles
|Kevin_kwanghoon Ha 10/5/00 10:04:43
Module 16 Self Test #10 Calculate the formula
weight of a monoprotic acid, HW, if 0.365 g of the
acid is dissolved in 100 mL water and titrated
exactly with 25.0 mL of 0.200 M Ba(OH)2. ---I
cannot get the balanced equation for this problem
||think of this monoprotic acid
as HCl. then write the balanced equation for HCl with
Ba(OH)2 to give water and BaCl2. after you write the
balanced equation then replace the HCl with HW. there you
will have your balanced equation. after you balance it the
ratio of HW to Ba(OH)2 should be 2:1 ratio.
After you have done your balanced equation then do the
usual 3 steps in any stoichiometric calculation (please
look at the question down below yours) to solve the
|Kevin_kwanghoon Ha 10/5/00 10:03:26
Module 16, Self Test #8 Vinegar is a dilute acetic
acid solution. 40.11 mL of 0.100 M NaOH were
required to completely neutralize a 6.00 g sample
of vinegar. What percent by mass of this sample is
||Steps to take:
First you need to write the balanced equation to find out
the mole to mole ratio of the acetic acid and the NaOH
(hint:acetic acid has the formula HC2H3O2 and acetic acid
is a monoprotic acid like HCl). when you write the
balanced equation you will see it is a one to one ratio.
After you write the balanced equation, you do the usual
3 steps in any stoichiometric calculation.
1) amount of known (in this case NaOH) --> moles of
known 2)moles of known --> moles of unknown (in this
case acetic acid) 3) moles of unknown --> amount of
1) find the number of moles of NaOH by multimplying M x L
2) find the number of moles of the acetic acid using the
mole ratio from the balanced equation.
3) find the amount (grams) of acetic acid by multipling
the moles by the molar mass.
Once you find the mass of acetic acid to find the %,
all you need to do is: mass of acetic acid calculated/6.00
g x 100%
let me know if you do not understand any of these
|Alana_diaz Ayuyao 10/5/00 9:27:05 PM:
The archipelago system is down right now. Do you
know when it will be back up?
||No, I do not know when the site
is down and when it is up ((I only find out when I tried
to log in, like you). They inform me about it only when it
is going to be down for an extended time.
Like I told Mai and Theresa down below please make copies
of all the documents you need way ahead of time (about
three weeks ahead) and not wait until the end to hand in
your assignments. In this way you will minimize any damage
caused by this problems.
|Teresa_chanton Brown 10/5/00 4:27:08
professor Ferede Is any one else having trouble
connecting to the archdl web site. In the
multimedia presentation of module 15. why is it
when we have 250mL of 0.140M NiCl2 *6H2O that the
6 moles of H2O are not calculated in the addition
||Sorry the Archiplelago site was
down for some times. Like I pointed out to Mai down below
please make copies of all the documents you need way ahead
of time (about three weeks ahead) and not wait until the
end to hand in your assignments. In this way you will
minimize any damage caused by this problems.
to find the molar mass of NiCl2.6H2O, the 6 H2O have to
be included with the NiCl2 and since molarity is mole of
solute/liter of solution they also have to be included.
|Quynh-mai_tu Nguyen 10/5/00 2:52:36
Dr. Ferede, This morning I was able to connect to
Archipelago homepage, but I couldn't this
afternoon (Thursday). Does anyone in the class
have similar problem?
||Sorry the Archiplelago site was
down for some times. I just want to remind you to please
make copies of all the documents you need way ahead of
time (about three weeks ahead) and not wait until the end
to hand in your assignments. In this way you will minimize
any damage caused by this problems.
|Quynh-mai_tu Nguyen 10/5/00 2:47:02
Dear Dr.Ferede, 1) In Module 15, slide #12, step
3, where did 4.27 came from? 2) In Module 16 and
Ch.5 in the book, please explain the chemical
processes that cause red carbage juice to change
color and act as indicator when a solution's
acidity changes. 3) Study Guide question #4,
module 16, will the chemical reaction will be
given on the exam or are the students suppose to
||1) pH = -log [H+] and therefore
[H+]= 10-pH. In this case pH = 3.37 and
therefore the [H+]= 10-3.37. we could leave it
that way, but the common way of writting it would be 10 to
the power of a whole number integer, not decimals or
fractions. Therefore we raise it to the next whole number
integer which is -4 and find the log of .37 which is 4.27.
therefore overall it would be 4.27 x 10-4. Or
another (or easier) way of doing it would be to enter
antilog of -3.37 in your calculator, and you will find the
2) The compound that is responsible for the color
change has a different structure when it is in acidic
solution since it is protonated (a proton is added to it)
and when it is in basic solution since it is deprotonated
( a proton is removed from it). We will talk a little more
about the specific type of structure that is resposible
for color in a later module. At this stage this is
3) Most of the time if a question is asked about a
reaction, the equation of the reaction will be given to
you, unless it is one of the reactions in which you can
easily predict the products. then you are supposed to
write down the products and then balance the equation then
work with it. At is stage you will ask which of the
reaction can we predict the products easily?
Remember you learnt about five type of reactions, in
the lab last week as well as in your modules
Out of this five, some of them you can predict the
products easily. Let us look at each of them.
1) double displacement (exchange) reactions. You can
easily predict the products.
2) single displacement. You can also easily predict the
3) combination reaction. You can predict the products
of simple reaction of this kind for example H2 + O2 -->
4) decombosition reaction. You can also predict the
products of simple reaction of this kind for example HgO
--> Hg + O2
5) oxidation reduction reaction. You can predict few
reaction of this type if they are simple reaction. but you
should be able to predict the produts of combustion
reaction of hydrocarbons. Remember the product of
combustion reaction of hydrocarbons are always CO2 and
|Eric_m Yoshida 10/5/00 10:38:20 AM:
Dr. Ferede, What is the difference between
molarity and molality? I have heard of this term
before, but forgot the difference.
||Molarity is the # of moles of
solute in a liter of solution. However molality (which is
not discussed in your module) is the # moles of solute in
a liter of solvent. Remember a solution is a solute +
|Teresa_chanton Brown 10/5/00 6:25:28
Professor Ferede. I am sorry but I do not find the
comments in your last e-mail as a source of
encouragement. I don't think it helps any student
to berate them , because of the progeress of
others. I know it does not encourage me to do any
better. Although I do understand that this is a
class of active learning. Yet you cannot determine
the potential of other student by two students.
||I am sorry you felt hurt when I
singled out the two highest grades and congratulated them.
I know most of you did the best and studied hard. Like you
pointed out all of you there have different background and
different circumstances to deal with. I should have
included all the people that have done their personal best
in my congratulations. Thank you for pointing it out to
me. Teresa, However one thing I want you to understand is
that in a college class, or any class where a teacher
would not know every body’s circumstances, study habits
and other personal responsibilities, the teacher mostly
depends on the grade earned in quizzes and exams to assign
a grade or in this case singled out some for praise. I
know it is not fair, since every body background and
circumstances is different but mostly that is how it goes.
I will try to be more sensitive next time. As I said in my
last email, I was also going to send personal emails to
each of you and make comments individually. I apprecaite
all your comments and suggestions.
|Ashley_ann Tucker 10/5/00 12:45:47
Question #1 Module 14: When 6.25 mg of ether is
subjected to combustion analysis and 14.84 mg of
CO2 &7.56 mg H2O are obtained; What is the
percent of Hydrogen? Question #2 Module 15: Can
you please explain how in the lecture presentation
when they worked out the molarity of AlCl3 to be
.387M for 25.8g After having determined the
concentration of Cl3 in the solution by dividing
.387g by 1.0L and multipling that by the ratio of
3 mol Cl to 1 mol AlCl3. Where did the .387 g come
from? Isn't .387 the molarity of AlCl3? Question
#3 Module 16: What exactly is a standard solution?
||question #1, module #14;
. Please refer to the answer I gave to Nicole_marie Duncon
on 3/13/00 at 7:39:41 down below
question #2, module #15
please refer to the answer I gave to Diana_Jean Croft on
3/17/00 at 4:42:00 down below.
question #3 module 16:
A standard solution is just a solution whose concentration
is known to at least four significat figures. for example
an HCL solution that has 0.4352 M is a standard solution
|Shahar Kalev 10/5/00 12:24:33 AM:
Can you determine Molarity of a gas? Are gases
measured only by their volume? Shahar
||Yes you can measure the
Molarity of a gas if they are in solution for example if
HCl is a gas when it dissolves in water we talk about the
molarity of the solution. Howeve in a pure form we refer
to their the pressure, volume and temperaure.
|Ashley_ann Tucker 10/5/00 12:22:55
Question #1 Module 13 Is there always a limiting
||There is a limiting reactant if
the other reagent(s) are used in excess, but sometimes you
could use the exact amount of each needed and this is
called stoichiometric equivalent, in that case both
reagents are limiting reactants
|Shahar Kalev 10/5/00 12:19:18 AM:
What is Titration? How are you supposed to know
exactly the Molarity from using it?
||Titration is a procedure used
to calculate the concentration of an acid by reacting it
with a known volume and condentration of a base. this can
also be done for unknown concentration of bases or other
type of reactions.
during this procedure you put the unknown acid in a
flask and add a base that has a known concentration and
see how much base it takes to reach the end point (where
the acid has been nuetralized by the base) then using
stoichiometric calculation, calcualte the concentration
(molarity) of the acid.
To calculate the molarity you use the usual 3 steps in
stoichiometric calculation. First write the balanced
amount of unknown --> mole of unknown --> mole of
unknown --> amount (in this case molarity) of unknown
|Shahar Kalev 10/5/00 12:15:58 AM:
Calculate the molecular weight of a diprotic acid,
H2X, if 0.270 g of the acid dissolved in 250 mL
water are neutralized by 54.32 mL of 0.154 M KOH?
self test Module 16. Shahar
||steps to take:
1) write the balanced equation to find the mole to mole
ratio (you can think of H2X as H2SO4).
2) then find the number of moles of KOH used by
multipling M xL
3) from the mole ratio find the moles of H2X used.
4) form the number of moles and grams of the acid
calculate the molar mass using the equation
mole = mass/ Molar mass.
Let me know if any of these steps are not clear.
|Teresa_chanton Brown 10/4/00 9:10:39
Professor Ferede In the study guide , you asked,
When an acid base titration is carried out , the
indicator is added to the acid solution and then
the basic solution is slowly added to this
solution. At first, what is the color of the
indicator? Is it an indication of a basic or
acidic solution? Well when you ask the last
question I don't understand how to tell if it is
basic or acidic when the color of the indicator
has not changed color. If the indicator has not
changed color how do you knoew if it is acidic or
basic? Is it Neutral?
||the color of the indicator in
the acid solution is colorless, so as long as there is an
excess of acid in the flask the color will remain
colorless until it reaches the equivalent point and then
when a single drop of an excess base is added the color
will turn pink. the color of phenolphthalein indicatore is
colorless in acidic and neutral solution and pink in basic
|Kevin_kwanghoon Ha 10/3/00 7:46:24
Since you said you would like a comment on the
exam, I think it was a much harder exam than the
spring chem 11 online exam #1. I solved the
previous exam before our exam to practice, and I
didn't really have any problem. But our exam was
much more detailed than the previous one. One more
thing... We were not given a solubility table, and
Dr. Goueth wasn't even aware that it was supposed
to be given to us. Anyways, I hope the next exam
would be a little easier than this one.
||Thank you for your comment. I
am sorry you thought it was much harder than the spring
chem 11 exam. I tried to make it as comparable to it as
One thing I know that was diffenent was the balancing
the redox equation (which I emailed you and told you will
be on the exam). The spring class had this question
towards the end of the semester rather than in their first
exam. I changed it this time because I thought this was
really the right place to put it instead of revisiting the
topic towards the end of the semester.
Also most of the questions I brought from your
self-test, additional assignments, and faqs or similar
questions to them, and I did not think there would be any
problem there. Any way next exam I would give you a copy
of the spring semester exam again to practice with and
also some type of outline for the exam. I would try to
make the exams as fair as possible, but also all of you
need to do your part by reading, doing and studing all
Also remember doing your practice exam at home with all
the information around you and all the time you can take
is not the same as taking an exam in class with a limited
amount of time. When you do your homeworks and practice
exams I advise you to do them more than once. One time you
do them in a relaxed manner (taking as much time as you
need with all the information around you) but you have to
do it again in kind of a exam situation (with only limited
time, with no breaks in between and no added information,
like an open book around).
You really did not need the solublity table during your
exam that is why I did not ask Dr. Goueth to give it to
you. The question that needed the solublity table was # 13
in which you were asked to write the net ionic equation. I
gave you all the information you needed on the exam by
putting (s) for solid or (aq) for dissolved in water on
the equations. That is all you you could have gotten form
the solubility table and you were given this information
on the exam
|Eric_m Yoshida 10/2/00 8:38:46 PM:
Dr. Ferede, Has the exam been graded? Will our
scores be posted on the website or will we have to
wait till Friday?
||Some parts of the exam has been
graded. I have just sent you an email commenting on the
exam. I will try to email you your scores by thursday.
|Teresa_chanton Brown 9/28/00 8:55:11
professor ferede when the compounds in a net ionic
equation are all soluble, is that when there is no
||This is almost the same
question as Theresa Brown asked down below so I will
answer it the same way with little modification.
The answer to your question is yes and no. Remember there
are three requrements for this type of a reaction to take
place. 1) precipitate formation
2) gas formation
3) weak electrolyte and nonelectrolyte formation.
Therefore for a reaction that does not form a gas or weak
and nonelectrolyte, if all the compounds are soluble then
we can say all the ions are spectator ions. and there is
But for reaction that form weak or nonelectrolyte the
compounds might still be soluble and they might not be a
spectator ions and therefore there might be a reaction
|Teresa_chanton Brown 9/28/00 8:22:33
Professor Ferede Q1. when I am doing the net ionic
equation, when there are compounds that are
soluble are those the compounds that become
spectator ions? Also , thanks for lightening the
load. the homework load.
||The answer to your question is
yes and no. Remember there are three requrements for this
type of a reaction to take place.
1) precipitate formation
2) gas formation
3) weak electrolyte and nonelectrolyte formation.
Therefore for a reaction that does not form a gas or weak
and nonelectrolyte, if all the compounds are soluble then
we can say all the ions are spectator ions. But for
reaction that form weak or nonelectrolyte the compounds
might still be soluble and they might not be a spectator
you are welcome.
|Alana_diaz Ayuyao 9/28/00 5:45:46 PM:
Are you going to post the answers to the practice
test before the exam? Also, could I also get a
copy of this solubility chart? Thanks.
||No, that is just for practice.
Try to find the answers for the practice test on your own.
|Kevin_kwanghoon Ha 9/28/00 9:39:25
Thank you for sending me the solubility table. But
I could not open the file because when I tried to
open the attachment, it referred me to the
faculty.smc.edu site and I had to enter a faculty
password. Could you send it to me one more time?
as a Word document format, but not as an internet
||I will ask Theresa to give you
a copy during your lab
|Shahar Kalev 9/27/00 10:39:19 PM:
CAN YOU SEND ME PLEASE SOLUBILITY TABLE? I DID NOT
RECEIVE IT. TAHNK YOU. SHAHAR
||I will ask Theresa to give you
a copy during your lab
|Shahar Kalev 9/27/00 10:37:23 PM:
DO WE HAVE TO KNOW TYPES OF MOLECULES SHAPES FOR
THE EXAM?(LIKE LINEAR, TRIGONAL PYRAMID) SHAHAR
||No, that would be in your exam
|Shahar Kalev 9/27/00 10:35:17 PM:
WHAT IS THE DIFFERENT BETWEEN LIMITING REACTANT TO
||they mean the same thing
|Stephen Kim 9/22/00 11:33:55 AM:
Dr. Ferede, for the homework assignments, there
are no questions in the book numbered 37b and 57a?
Do you mean do questions 37 and 57?
||Chapter 4 has those numbers. I
do not want you to hand in all 37 and 57. You can do all
if you want but hand in 37b and 57a.
|Stephen Kim 9/22/00 11:27:03 AM:
On Self Test 10.6, it asks for the balanced net
ionic equation for potassium carbonate and
aluminum sulfate, but in the reactants part of the
equation, potassium is not even on the equation!
||remember the net ionic equation
does not include the spectator ions, and patassium is a
spectator ion in this reaction. Inorder to write a net
ionic equation you have to first write the molecular
equation and then the total ionic equaion then your write
the net ionic equation.
molecular equation: 3K2CO3(aq) + Al2(SO4)3(aq) -->
3K2SO4(aq) + Al2(CO3)3(s)
total ionic equation: 6K+(aq) + 3CO3-2(aq) + 2Al+3(aq) +
3SO4-2(aq) --> 3K+(aq) + 2SO4-2(aq) + Al2(CO3)3(s) Net
ionic equation: (cancel out any spectator ions) 2Al+3(aq)
+ 3CO3-2(aq) --> Al2(CO3)3(s)
|Teresa_chanton Brown 9/22/00 6:26:29
Professor Ferede I am sorry for not being clear. I
still don't have the guidlines for moules 10, 11,
and 12. of the transcipt for module 7. I want to
apologize for not being clear, and thank you for
being patient. :)
|Roman Ferede 9/22/00 2:54:04 AM:
-----Original Message----- From: QUYNH MAI NGUYEN
To: email@example.com Sent: 9/21/00 6:03 PM Hi
Dr. Ferede, I tried to log on to VOH today, but I
couldn't. Are other students in our class having
similar problem? I did the self test for Module
10. One of my mistakes is problem #7. The
explanation given after the test is graded wasn't
clear enough. Could you work the problem out and
explain how to get the right answer? Thanks, I'd
really appreciate it. In your last email, you said
if the VOH is not working tomorrow, you will email
us an address that we can send our questions. I
haven't received any email from you today. Does
that mean VOH is working? Why can't I get through
it even though I'm using computer in school? My
printer at home is not working, is there anyway I
can save the transcript from the modules into a
diskette to print it somewhere else? Mai Nguyen
||Hello Quynh, I am sorry that
you are having problems with the VOH for the last 2 days,
but now it is working! I am going to copy your qusetion
and this anwer and paste it on the VOH so that the other
students also see it. to answer your questions first let
us first write the balanced molecular equation CaSO4(s) +
(NH4)2CO3(aq) --> CaCO3(s) + (NH4)2SO4(aq)and then
write the total (complete ionic) equation CaSO4(s) + 2
NH4+(aq) + CO3-2(aq) --> CaCO3(s) + 2 NH4+(aq) +
SO4-2(aq) Notice: CaSO4 and CaCO3 are written as molecules
since both are insoluble in water. Now cancel out all the
spectator ions (the species that appear the same on both
sides) and write the net ionic equation CaSO4(s) +
CO3-2(aq) --> CaCO3(s) + SO4-2(aq) If the SMC site is
not working you can send me an email to
A.MENKIR@CGIAR.COM. If you want to copy the transcript on
a diskette all you need to do is open the transcript, then
highlight it then copy it and go to your diskette, open a
microsoft word document and them paste it there. Good luck
and let me know if you have any more questions or concerns
|Kevin_kwanghoon Ha 9/21/00 10:51:43
How do I predict precipitation reactions for
elements that are not in the General Guidelines
for the Solubility?
||If they are not in the general
solubility guidline then you will have to be given more
information about it, i.e. you will be told if it is
soluble or not soluble.
|Kevin_kwanghoon Ha 9/21/00 10:46:35
What is the difference between oxidation number
and the charge on an atom? The numbers seem to be
||The charge show the # of
electrons lost or gained in ionic bonding but oxidation #
shows the number of electrons gained, lost or unequally
shared. the charge and oxidation numbers are the same for
ions. for example an ion that has a charge of -2 will also
have an oxidation number of -2. However covalently bonded
elements do not have any charge for example the C in CO2
does not have any charge so that we can not talk about the
charge on the carbon. Therefore we talk about the
oxidation number of the Carbon in CO2. The oxidation # of
the Carbon in CO2 is +4 (notice: it is not loosing 4
electrons but sharing four electrons but it is unequal
sharing in which the Carbon has less amount of the
|Kevin_kwanghoon Ha 9/21/00 10:44:23
According to the Common Oxidizing and Reducing
Agents table from the multimedia presentation, the
oxidation number for oxygen is usually -2, and -1
when in H2O2. But is it -1 or -2 when in H2O?
||the oxidation number for oxygen
is always -2 except when it is bonded to another oxygen
such as in H2O2 (then it is -1)or when it is bonded to F
(which is not at all common and you do not have to worry
|Teresa_chanton Brown 9/21/00 10:15:18
hello professor Ferede I fixed the problem with my
e-mail address, so go a ahead and send it and it
will come. The study guide for modules 10 11 , amd
12. I am very sorry for causing so much of a
problem. it was my computer's fault.
||I am glad your computer is
fixed. But from your question I do not understand if you
did or did not get study guidelines for modules 10, 11,
12. Also let me know if you are missing anything eles.
|Shahar Kalev 9/20/00 6:31:55 PM:
What`s the difference between H3O+ ion and H+ ion?
||actually H+ is a short hand way
of writting H+ in water. H+ in water is found bonded to a
water molecule and therefore it is written as H3O+.
|Shahar Kalev 9/20/00 6:29:07 PM:
in Acid-Base reactions, after balancing the
reaction, how do you determine whether or not to
break down the compounds on the reactant side to
ions? module 11. Shahar
||Look at the answer I gave to
Natalia on 5/20/00 at 5:21:38. There I explained how total
(complete) ionic should be written. Total ionic equations
are written showing species that dissociate or ionize as
ions but species that do not are written as molecules. For
strong acids, since they are strong electrolytes, then
they are written as ions but weak acids since they are
weak electrolytes are written as molecules.
|Shahar Kalev 9/20/00 6:26:20 PM:
Can you send us guideline for balancing redox? (As
you gave us for H.W. P. 995- text book)? in that
case, Are we supposed to study it by ourselves
from an outside resource? Shahar
||I sent you a detailed guideline
for balancing redox reaction at the bottom of study
guideline for module 12. Please read the guideline and if
you still do not understand it let me know what you
specifically do not understand and I will be glad to
explain it again.
|Ashley_ann Tucker 9/20/00 5:45:04 PM:
Module 10: Question #1 When molucules of solute
bump into one another can this type of occurence
be seen under a microscope? Module 11: Question #2
In what ways do specifing the definition of an
acid ease the work of practicing chemist. Please
give a practical example. Module 12: Question #3
How is it that when an acid-base reaction takes
place a positively charged hydrogen proton
transfers from one molecule to another molecule.
Why doesn't it stay where it is rather than
transfer. Please explain, is there something else
happening that I'm missing in this concept.
||1) no it can not.molecules or
atoms are too small to be seen under a microscope
2)defining a concept clearly lets a chemist know what they
are dealing with so that they can further explore or add
to that concept.
3)For an acid-base reaction or any reaction to take place
with out added energy usually the product(s) most be at a
more stable state than the reactant(s). You will learn
more about driving forces of a reaction in chem 12.
|Natalia Pokras 9/20/00 5:35:12 PM:
Hi Professor, 1)in Study quiz for module 11
question# 6. Why is it no reaction between
potassium nitrate and sulfuric acid? 2) I am also
concern how we going to take our first exam with
out all these tables ?
||Look down below at the answer I
gave. In this case there is no reaction since all the
species on the reactant and the product side are the same
and therefore theu camcel out to give no net ionic
this type have of exams have been taken with out any
tables before and I am sure you can do it too! I will be
sending you a short study guideline for your exam 1
|Natalia Pokras 9/20/00 5:31:28 PM:
Hallo Professor, I got very confused about module
10. 1) Is there are any way you can explain to me
how we can predict if the reaction will happens or
not.2) In the module self quiz question #8: The
balanced net ionic equation for a reaction that
occur when ammonium sulfide and iron (III) are
mixed. The right answere was 2FE + 3S = Fe2S2. Why
is it? What do we need to put on the product side
a product that is soluble or the one that or not
soluble?How to predict that its going to be no
||This is explained in details in
your study guideline but I will summarize it here.
The net ionic equation is written only when a reaction
takes place. If a reaction does not take place all the
species in the reactant side will be the same as the
species on the product side when we write the total
(complete) ionic eqaution, and therefore all the species
on the reactant side will cancel out all the species on
the product side leaving no "net ionic
There are three requirements for a reaction in aquiesous
solution to take place.
1) One of the products must be a solid(precipitate), i.e.
it is not soluble in water.
2) one of the products must be a an insoluble gas. Most of
the gases are simple gases you should recognize such as
H2, Cl2,CO2, H2S etc. But also you should know that three
common compounds decompose in aqueous solution to give
gases. These are a) H2CO3(aq) which decompose to give CO2
(g) + H2O b)H2SO3(aq) which decompose to give SO2 (g) +
H2O c) NH4OH(aq) which decompose to give NH3 (g) + H2O 3)
the third requirement is not so obvious. a weak
electrolyte or a nonelectrolyte must be formed. This weak
or nonelectrolytes are compounds that do not dissociate
into ions or dissociate to a very limited amount. examples
of these type of compoounds are covalent compounds such as
H2O or weak acids such as HC2O3O2 (acetic acid)
So in order for the reaction to take place one or more of
these requirements need to take place.
To answer the second part of your question, Fe2S3 is not
soluble in water and therefore it satisfies the first
requirement and therefore it is going to show in the net
|Teresa_chanton Brown 9/20/00 4:48:47
Professor ferede I am not sure if you have
recieved any of my emails, but I have not recieved
the study uidkines for modules 10, 11 and 12.
could you please send them to me.
||I just received your email
about not getting transcript for module 7 and study
guidelines for module 10,11, 12. I have been sending you
all those things with the email address you have provided
to the course website (Archimpelago). And this email
address seem to be the same as the one you just wrote, so
I do not understand why you are not getting them. Any way
I will send them again.
|Teresa_chanton Brown 9/20/00 4:42:54
Professor Ferede I am having trouble doing net
ionic equations. Can you give me a general
explanation on how to do them? And should I go
ahead and memorize the solubility/insolubility
||qustion 1) please refer to the
answer I gave to one of your class mates, Teresa Mico on
9/19/00 at 1:52:53, two questions below this.
question 2) please refer to the answer I gave to one of
your class mates, Eric Yoshida on 9/19/00 at 6:46:49, one
question below this.
|Teresa_chanton Brown 9/20/00 10:18:24
Professor Ferede can you further explain acids for
me? I want to know what happens to a componud that
has more than one H+ atom. After it loses it's
hydrogen atom, is it still an acid?
||More comprehensive definition
of an acid is the Lewis acid definition in which acids are
defined as an electron pair accepter. for example a
species that has an H+ such as HCl is
considered an acid since it is electron deficient and can
accept an electron. However species that do not have H+
can also be considered as acids, for example BF3.
However as far as chem 11 class is concerned you can use a
less comprehensive definition of an acid, which is the
Bronsted-Lowery definition (or Arrehnius definition) which
describe an acid as just a proton (H+) donor so
any species that has H infront of its formula can be
thought of as an acid.
A compound that has more than one hydrogen atom such as H2SO4
can still be an acid when it looses its proton. for
example H2SO4 becomes HSO4-
and since HSO4-still has a proton to
donate. However HSO4- is now also
considered to act as a base since it can also accept a
|Eric_m Yoshida 9/19/00 6:46:49 PM:
Dr. Ferede, Will we be required to memorize the
solubility table for our quizzes and exams? Is
there some easier way to remember what is on it?
||Yes, you are supposed to know
the general solubility table, look at the one I sent you
in your study guideline. Also I will email you a nice
solublity table one of your classmate, Teresa Mico has
|Roman Ferede 9/19/00 1:52:53 AM:
From: Teresa Micco To: Roman Ferede Sent: 9/18/00
10:20 AM Subject: Balanced net ionic equations I
am not understanding the solubility equations (net
ionic). When I am doing the test on module 10 I
got two answers incorrect and I still couldn't
understand why. Teresa
||In order to write net ionic
equation you first need to write molecular equation, total
ionic equation. I have described all of these types of
equation in details in study guidelines I sent you for
that module. Please refer to that study guideline. But
here I will summerize each of them: the molecular ionic
equation is the balanced equation written showing no ions,
as if all compounds are molecules. for example sodium
chloride is written as NaCl not Na+ Cl-. the total ionic
equation (also reffered as the the complete ionic
equation) is written in which compound that ionize and
dissociate are written as ions but the ones that do not
are written as molecules eg NaCl is written as Na+ Cl- but
water is written as H2O. to find out which one ionize and
dissociate, please refer to your study guidelines the net
ionic equation is written in which all the spectator ions
are cancelled out. To find out what spectator ions are
please refer to your study guideline. Teresa, I am going
to copy your question and answer and paste it in the VOH
so other students can also see it.
|Eric_m Yoshida 9/18/00 3:29:37 PM:
Dr. Ferede, Is there any other way to get the net
ionic equation? What is the real purpose for
having the net ionic equation? I tried following
the method described in the additional study, but
did not do well on the self test. Eric
||the real purpose of the net
ionic equation is to see the bare essential of a reaction.
In the net ionic equation only species (elements, ions,
compounds) that take part in the reaction are shown. The
spectator ions (this are ions that do not take part in the
reaction but are just there to accompany the ions that
take part in the reaction) are not shown.
Another student also emailed me with the same type of
question and I will refer you to the answer I gave her
just above this answer.
|Vattanasrisakulthai 9/16/00 11:48:05
Dr. Ferede- I have a question from a previous
module. I just wanted to know exactly WHY the
Lanthanide and Actinide series are seperate from
everything else on the periodic table. Does it
have to do with certain physical and chemical
properties? What exactly? Chai
||Yes, the lanthanide and the
Actinide elements have different properties than the other
elements but they are written at the bottom of the
periodic table because if they were included with the
other elements it would give the periodic table an odd
shape, so they are separated the way they are to give the
periodic table a neater, more compact appearance.
|Alana_diaz Ayuyao 9/15/00 12:13:35
Are all ionic compounds salts? Are there any
||all ionic compounds are salts
expcept acids, bases, acid oxides and basic oxides
|Alana_diaz Ayuyao 9/15/00 12:08:32
Is it possible that you can make our next quiz
another take-home quiz?
|Alana_diaz Ayuyao 9/15/00 12:07:15
Re: nomenclature & balancing chemical
equations, will we need to memorize the formulas
and charges of cations and anions?
||You should be able to predict
the charges on the ions by looking at the periodic table
for the main group elements for the other ones just
remember the most common elements which I gave you in the
|Sang_hoon Chung 9/15/00 3:19:26 AM:
Q3: Did you send out the numbers for the
homework(Assignment numbers) through the email?
Because I have not yet gotten any of them.
||I just send them out in my last
email. However I will send them to you again tomorrow in
case you did not get them.
|Sang_hoon Chung 9/15/00 3:12:55 AM:
Q2: When elements bond together, how do I
determine what goes in the middle?
||Usually this must be determined
experimentally however there is a rule of thumb to predict
what goes in the middle. The single element, usually the
least electrongegative element goes in the middle. there
are quite a few excetion to this rule but works most of
|Sang_hoon Chung 9/15/00 3:10:31 AM:
Q1: How do I determine a charges of an element?
||it depends on the element
if it is a main group element (reperesentative
element)i.e. that is group A elementes 1A to VIIA, then
you can predict the charge on the element by just looking
at the group #. For metals it is the same as the group
number for nonmetals it is group #-8.
|Kevin_kwanghoon Ha 9/15/00 12:29:41
Some ions form regular lattices, but some ions
don't. What makes some ions form regular lattices?
||Ionic compounds have all
regular lattice. The unit cell in the lattice might be
different for different ions. Compounds like glass (it is
not ionic compound) form irregular shape and they are
called amporhous solids.
|Kevin_kwanghoon Ha 9/15/00 12:23:14
What is the difference between "theoretical
yield" and "actual yield"?
||The actual yield is the
experimental yield, the amount one obtains in the lab when
one carries the experiment. The theoretical yield is the
yield calculated from the balanced equation. the actually
yield is smaller or equal to the theoretical yield
depending on the condition of the reaction or the type of
|Kevin_kwanghoon Ha 9/15/00 12:10:41
What does "average weight of a molecule"
mean compared to atomic weight?
||remember each atom in a
molecule has an average weight calculated from the weight
isotopes of each atom and their abundance. For example the
average weight of carbon is 12.01. This weight is not for
C-12 or C-13 or C-14 but is an average weight of all the
three. Therefore the weight of a molecule must be
calculated from the average weight of each atom in the
|Stephen Kim 9/14/00 9:01:03 PM:
This question is 3.85 from the homework. I had a
very difficult time with this question. Chromium
is obtained by heating Chromium (III) oxide with
carbon. If you want to produce 850 kg of chromium
metal, what quantity of Cr2O3 (in kilograms) is
||stepts to take
1) write the balanced equation
2) find how many moles 850 kg of cromium oxides is
3) then calculate how many moles of hromium that would
give you in the balanced equation
4) then change the number of moles of chromium to grams of
try this steps and if you are still having problems I will
solve it in details tomorrow.
|Shahar Kalev 9/14/00 5:46:30 PM:
Can you send us a list of the most common ions
that would be good for us to memorize?
||You should be able to predict
the charges on the main group elements by just looking at
the group number in the periodic number. Look at the anwer
I gave to Alana Ayuyoa on 09/15/00 concerning this. For
the other common ions refer to the list I gave you in your
study guide for that module and also look at the list in
your lab book for experiment that you did last week.
|Shahar Kalev 9/14/00 5:43:58 PM:
Is there a common charge to the transmition group
||this is a similar question
asked by Ashley Tucker on 9/12/00. So please look at the
answer I gave her down below.
|Shahar Kalev 9/14/00 5:41:40 PM:
Why when multiplying 2 by 1.008 the result is
2.016? isn`t the answer supposed to contain the
least amount of significant figuers? (1)
||First let us talk about the
difference between exact numbers and measured numbers. the
number of significant figures in a measured number is
dependent on the precision of the instrument used. If you
use a more precise instrument you will have more
significat figure. However exact numbers are independent
of any instrument and therefore they have infinit
significant figures. three common type of exact numbers
1) counted numbers 2) numbers in a formula 3) numbers in
exact conversion such as metric unit to metric unit. In
the case of your question if 2 is a counted number then it
is considered to have infinit significat number and
therefore the measurement that has the least significant
number will be 1.008 (four significatnt number) and
therefore the answer will have four significant number. If
however 2 is a measured number then the answer will have
only one significant number. So to answer your question
the number of significat figure in the answer is going to
depend on if 2 is an exact number or a measured number so
the person that is asking that question has to supply more
|Eric_m Yoshida 9/14/00 4:04:48 PM:
Is there any problem in referring to the Law of
Conservation of Matter as the Law of Conservation
||They are sometimes used
interchangeably and some put it as the law of conservation
of mass and energy.
|Teresa_chanton Brown 9/14/00 3:26:49
I've called tech support about the downloading
problem on the site. yet I haven't heard any thing
back from then, will you be able to send
transcripts for mod 7?
||I did send you the transcript
for module 7 in my last email. I will send it to you
again. If you do not hear from the technical support site
call them again.
|Ashley_ann Tucker 9/14/00 5:40:12 AM:
This question is for everyone. Is anyone having
problems downloading the web site? I keep getting
a server error message everytime I go to the URL.
Did it change or is the server down? Someone
please advise. Thank you.
||You call the technical support
site if you still have problems
|Eric_m Yoshida 9/13/00 8:07:36 PM:
Has anyone seen the Multimedia Presentation for
module 7? After reading the transcripts, it seems
like it has a lot of examples of ions. Please tell
me if I missed any important info that was not on
the transcript. Thanks
||If you have the transcript it
should be enough
|Ashley_ann Tucker 9/12/00 3:08:00 AM:
Question#1 Module 7: Dr. Ferede, How can one
predict the number of electrons a transition metal
will lose when it forms a cation? Is it possible?
Question#2 Module 8: Can you please explain
further regarding percent composition and how one
determines the percent if the formula of the
compound isn't available. What would be the
logical way to calculate the percent? Is it
laboratory analysis and if so is there any other
way to get this information? Qusetion#3 Module 9:
When working with an aqueous solution, does one
insert the H2O portion into the equation. What
type of reaction would follow if this were done?
Is H2O ever considered to be a reactant or
product? If so in what cases. Please give an
example. Thank you.
||1) predicting the charge on the
transition metals is not as straight forward as predicting
the charge on the main group elements (representative
elements) . The rule of thumb for transition metals is
that they will loose their outer most s electrons and if
they have variable charge then they will lose one or two
of their d electrons. Therefore most transtion elements
when they form charges they will have +2 (from losing
their s electrons) and/or will have +1. +3 electrons from
losing their d electrons or s and d electrons.
#2) It can be done experimentally by first analyzing which
elements are present in the compound and how much of each
element is present in a given weight of the compound. This
is called elemental analysis. form this information you
can obtain the percent composition even though you do not
have the formula.
3) sometimes the water can be used just as a solvent in
that case it does not appear in the equation, we just
mention the compounds are aquesous, for example when
AgNO3(aq) react with NaCl(aq) water just act as a solvent.
However in some reaction water can take part in the
reaction then it has to be written as a reactant or
product in the equation. eg. Na + H2O to form NaOH + H2
|Natalia Pokras 9/12/00 1:13:03 AM:
Hallo Professor, would you be so kind and clearify
to me 2 things. I looked to the syllable that you
hand out in the class and it's says that we will
get a maximum of 25 points for the self test (the
one after each module), but on the other page you
have 35 modules to cover. So, what the deal for
that tests (you said earlier that if a student got
80% or higher, he will get a full point)? 2)It is
conserning the syllable also it's said 80 points
for 12 labs (one of it will be dropped)and for the
final lab its going to be 40 points. In the
syllable that you sent to us through e-mail on
8.18.00 you have 70 points for the labs. Could you
explain how much points will each lab? Thank you
||Sorry about the confusion. The
points are like it says in the email I sent you. I sent
Dr. Guoeth the copy that was not reviewed. therefore it is
25 points for the self tests, 70 points for the lab and 40
points for the lab final
|Natalia Pokras 9/11/00 11:24:57 PM:
Hallo Dr. Ferede, I see that some people from our
class having a problem with the down loding of
module 7. I downloaded all of them exept #7. Some
how it's does not work on my computer also. I did
very bad on the self quizez for module 8 and 9
because firsteful I thought that we can do quizes
after watching the actual module, but for module 8
and 9 it is not so. To succed on that quiz we
really need to know module 7. Moreover because
nobody does not have a transcript its even
imposible to get a good score for rest of this
week module. Moreover our prelab base on the
knowledge that we suppost to get from themodule 7
(and question number 10 from the quiz that you
gave us to home also). I think this module is
extremely important in term of understanding how
to read and write the formulas and its going to be
very sad if we will not be able to study with the
||I understand the problem. I
will give every body the total point on the self tests on
module 7,8,9 as long as they attempt it. Read down below
(Eric's question) about the solution for module 7 problem.
|Stephen Kim 9/11/00 8:28:19 PM:
eric, i am having the same problems as you are!
after the download is completed, when i try to
pick it up using my lesson player, the lesson that
i just downloaded is not there ! i have tried
several times in regards to this. Dr. Ferede,
could you please send the transcripts for modules
7, 8, and 9! Thanks
|Stephen Kim 9/11/00 8:09:50 PM:
dr ferede, are the homework questions that are due
this week the ones you have assigned to us for
chap 1, 2, and 3 in the recent email that you have
||Like I pointed it out in my
email only chapter 3 is due this week. I already given you
credit for chapter 1 and 2. Do chapter 1 and 2 for your
|Eric_m Yoshida 9/11/00 3:26:58 PM:
Dear Dr. Ferede, I am having trouble downloading
the multimedia presentation for module 7. Is
anyone else having difficulty with this module? I
think that when I saved it the first time, I sent
it to a different location. The second time I
downloaded it, I sent it to the Lessons folder of
my hard drive. When I try to load it on the lesson
player, "No lesson loaded" appears.
Would you send me a copy of the transcripts,
please. Thank you, Eric
||Two other students also told me
that they are having problem with it. I have called the
Archepelog technical support and asked them what is the
problem and they told me they will let me know by
tomorrow. Next time if you are having any problems like
this please call the techenical support at 888-273-0814
and ask them for help. I will let you know the answer as
soon as I hear from them. Also I will like to remind all
of you in the future try to download and print the
trascript of the multimedia presentatation, copies of the
additional study, faqs and self test way ahead of their
due dates about three weeks ahead. If there is no solution
for module 7 by tomorrow I will send you the copy of the
|Teresa_chanton Brown 9/11/00 1:37:52
Hello Professor, I am a little confused in which
Laboratory Book I should be buying, Is it just the
Lab book that all the Chem 11 classes are using or
is it one the comes with the software?
||it is the same lab book that
all the chem 11 use. Last friday the book store did not
have any that is why I gave copies of the labs you are
going to do for the next two weeks during the lab. If the
book store does not still have these lab books and you did
not get copies from me last friday you can ask Dr. Goueth
to give you a copy before your lab this friday.
|Eric_m Yoshida 9/11/00 10:07:10 AM:
Thank you for coming to visit on Friday. We all
forgot to ask you what you are doing in Nigeria.
Is it for some kind of research? Or personal?
||It was a pleasure to meet you
all. I am in Nigeria because of personall reason. Even
though last year I wrote a perposal asking if I could do a
research while I am staying there, my proposal was turned
|Natalia Pokras 9/10/00 2:01:29 AM:
Dear Professor, in the quiz #2 (the one that you
gave us in the class for modules 4-6) in the
question #10 you ask us about the formulas for
compounds. I understand that Cations and Anions
will be covered in the module#7. If this is
correct are you still want us to answere this
question in the quiz? Sincerely
||Yes please anwser the question
since that is the question from the outside source.
|Natalia Pokras 9/10/00 1:06:18 AM:
Hallo Professor, I got wrong answere for self test
question #7 module 8. In the question they ask to
find the sum of coefficients. According to the
Transcript of this module under the In Balancing
Chemical Equation Reactions That Forms Oxides in
the example about 10 Oxygen they sayd that number
5 is the coefficient. Would you be so kind to
claerify this moment? Thank you.
||Please copy and paste the
question you are referring to because I could not find it
that is why I can not answer it
|Eric_m Yoshida 9/8/00 10:00:08 AM:
Dear Dr. Ferede, On the quiz, question #10 Why is
the atomic mass of carbon 12.011 grams? That seems
rather a lot for one carbon atom to weigh? Isn't
atomic supposed to be molar mass? If the carbon
was in g/mole, this question should be for molar
mass. Which is it?
||that is for 1 mole of carbon
atoms. 12.01 g is the molar mass.
|Stephen Kim 9/6/00 8:25:54 PM:
What does it mean when it says that most metals
are malleable? Does this mean that most elements
can react to other elements? (from self test
question 7 on mod 4)
||it means that the metals can be
hammered in to thin flat surface like aluminum foil and
they are also ductible which means that they can be drawn
into a wire like copper wire.
|Stephen Kim 9/6/00 5:27:17 PM:
Dr. Ferede, How are the modules for the class
correspond to the chapters in the textbook? Does 3
modules per week cover 1 chapter of the book (ex.
mod 4, 5, and 6 to chapter 2) Also i realized that
you did not assign us homework questions from the
book to turn in on friday's during lab...
||I sent you an email that
describes which modules correspond to which chapters. If
you do not have the email you can get a copy of it in the
lab this week. Also I did not send you the homework
problems for last week and this week since I want you all
to have your CDs and have every thing you need. I will
start giving you the homeworks next week starting for
|Stephen Kim 9/5/00 11:53:02 AM:
Prof. Ferede I had a question in regards to study
guide questions. Did you want us to answer the
first 10 and the last 10 questions on the study
guides for all the modules from now on? I just
wanted to make sure what i needed to turn in.
||Yes, up to module 15 we will be
doing it that way. However after module 15 you will be
doing more questions depending on which module you do. If
the module was not covered in chem 10 I will expect you to
answer more questions, and I will let you know how many
when the time comes.
|Natalia Pokras 9/5/00 11:18:44 AM:
Hallo Professor, If during the Self Test I pushed
the incorrect buttom and than accidentally pushed
the safe bottom is it any way to cancel the
answere ( I mean the time before the correct
answere will appeares)?
||No, once you enter it you can
not there is no way to redo you answer
|Natalia Pokras 9/5/00 11:02:32 AM:
Hallo Professor, 1)In the Glossary of module 4 its
said that periodic table was developed in 1872,
but in the transcript of the modules its said
1869. Which one is right? 2) In the transcript
(module 5 section metals) it is not clear how many
atoms of group 2A element needed to form compound
with single Oxygen atom?
||1) Infact also it is not only a
1) scientist, Mendeleeve that is credited for developing
the periodic table but also an english scientist called
Meyer. Both did it around the same time independent of
each other. The exact time is not really known.
2) Group 2A compounds when they form ionic compound have a
charge of 2+ therefore when they combine with oxygen that
has (2-) they will react in a one to one ratio. for
|Ashley_ann Tucker 9/4/00 6:31:57 PM:
Question #1 Module 4: How is it that some
molecules of elements can be considered polyatomic
and homoatomic at the same time? Can you please
explain this concept further. Also, if this is the
case will all the atoms be identical and be
represented by the same symbol? Question #2 Module
5: Since alkali metals are considered to be so
reactive that they are never found free in their
natural state; is it possible to break their bonds
and isolate them from the elements they are bonded
with? Question #3 Module 6: How does one determine
or recognize the diffences between the different
types of stereoisomers? Are there other ways to
accomplish this task other than the placement of
||#1) polyatomic just means that
it contains more than one atom. homonuclear means that it
contains the same type of atom. An example of a polyatomic
ion that is homonuclear is Hg22+
#2) Yes it is possible to isolate them from their
compounds. for example Na can be islolated from NaCl. #3)
stereoisomers such as geometric isomers (cis/trans
isomers) have different physical properties where as
stereoisomer such as enantiomers (nonsuperimposable mirror
images) have the same physical properties except by the
way they rotate light and the way they interact with some
molecule. You do not learn a lot about stereoisomers in
chem 11 so it would not be very appropriate for me to go
over isomerism in details right now.
|Natalia Pokras 9/3/00 12:44:28 AM:
Hi Professor, In the quiz under the #7 you asked
us how many ozone and oxygen atoms are contained
in 48.00, but you not specified the unit for
48.00. Would you be so kind and clearify this.
||Sorry, It is 48.00 grams of
ozone (O3). Remember almost all the questions on the
quizzes come directly from the Self Test, Additional
Excercises or FAQs.
|Natalia Pokras 8/30/00 11:56:19 PM:
Hallo Professor, Module 1 and 2 were easy to
understand, but I have many questions for module
3. How we determing if an element is neutral or
not? If an element is not neutral how we calculate
||Ask as many questions as you
want. That is one way that I would know that people are
doing what they are supposed to be doing. All elements are
neutral. They loose, gain or share electrons when they
combine with other elements to form compounds. The number
of electrons and protons in an element is equal and it is
equal to the atomic number. You can calculate the number
of neutrons by substracting atomic number from mass
number. When an element loose or gain to become a charged
particle, only the number of electrons change. The number
of protons and neutrons stay the same.
|Stephen Kim 8/30/00 9:23:20 PM:
I dont understand question number 13 in the self
test for module 3? There is no explanation for the
||Please copy and paste the
question that you are reffering to. Let me do that this
time. Which statement or combination of statements given
below are generally not correct concerning atomic weight?
1) Isotopic atomic weights and isotopic abundances are
generally known with the same accuracy. 2) Both quantities
must be known to calculate the elemental atomic weights.
3) Individual isotopic atomic weights are close to whole
number values. 4) Individual isotopic atomic weights are
exactly whole number values. As the answer says only 2 and
3 are correct statements. 1 is not correct because the two
values are obtained from different sources therefore they
can not be known to the same accuracy. number 4 is not
correct and to get the explanation refer to the answer ot
one of the FAQ in this module
|Natalia Pokras 8/29/00 11:21:55 AM:
Hallo Professor, Are we have any time frame (until
12pm for example)for the assignments that we have
to do at home(like Self test or answeres from
||you can hand it in any time of
the day as long as it is on that day
|Tohpun Haskul 8/29/00 12:48:25 AM:
I just wanted to be clear on something in Module
1. The question is: "At room temperature what
state do most elements exist as?" I couldn't
really find the answer.
||Most of them in a solid state.
Only two are in a liquid state and the noble gases and H2,
N2, O2, F2, Cl2 are in a gaseous state
|Linda_suzanne Gudz 6/5/00 7:39:27 PM:
How can we tell whether a "gas" is the
element alone or element2 ie one question asked
about Argon Gas & Chlorine Gas but the result
was for Ar (alone) and Cl2. Do we just have to
memorize the gases that form diatomic molecules?
||the gases that are inert, like
the noble gases they exists as monoatom and the other
elemental gases are reactive so they exist as diatoms (H2,
Cl2, N2, O2)
|Narmina_s Pashayeva 6/4/00 7:10:51
could you please remind how to find a bon order
||1. If you already draw the
Lewis dot structure then the bond order is the number of
bonds between the two atoms. that is if there is only one
pair of electrons then it is a single bond, a bond order
of one and if there are two pairs then it is a double
bond, a bond order of two and if there are three pairs
then it is a triple bond, a bond of three. If the molecule
has resonace structure the bond order would be one and a
fraction or two and a fraction depending on the molecule.
2. If you are asked to calculate bond order using
molecular orbital theory then you need to have the
molecular orbitals for the molecule (i.e. sigma 1s,
sigma*1s, sigma 2s etc..). Then you add up all the
electrons that are in the bonding orbitals and subtract
all the electrons in the antibonding orbitals and then
divide by two.
|Narmina_s Pashayeva 6/4/00 7:08:35
reg. polarity. in the module 29, additional study
they say that NF3 is assymatrical and therefore
polar. But in the last test we had there was the
BF3molecule that was as you said non-polar. Why is
||Remember it is the number of
lone pairs and bonding pairs of electrons around the
central atom that determine the geometry of a molecule.
Even though NF3 and BF3 have the same number of atoms
around their central atom (N and F), the central atoms do
not have the same number of lone pairs and bonding pairs
of electrons. N has one lone and three bonding pair
whereas B has only three bonding pairs. Therefore their
goemetry can not be the same.
This is a very important point to understand about the
VSEPR model. Make sure you understand it. If not let me
know about it and I will try to explain again.
|Linda_suzanne Gudz 6/1/00 12:12:28
THIS WAS QUESTION 6 of MODULE 38 ADDITIONAL
EXAMPLES: I have the following problems with it:
a. it asks re N2 & O2 but answers re Helium
b. I understand that effusion is related to speed,
ie the faster the speed, the faster the effusion.
I also understand that speed is inversely related
to mass, ie up mass down speed. But the solution
seems to put Oxygen effusing faster than Helium,
yet Helium is a far lighter gas & therefore
faster. Shouldn't helium, or the LIGHTER gas,
diffuse, effuse faster?
How fast do oxygen molecules effuse through a
barrier relative to nitrogen molecules?
We need Graham's Law to determine the rate
differential between oxygen and nitrogen
O2 on top
He on bottom
note: equation didn't copy
Oxygen molecules effuse through a barrier 2.83
times faster than He molecules.
||It is a mistake. The question
should have been O2 and He. And you are right the He
should effuse faster than O2
|Linda_suzanne Gudz 6/1/00 11:33:32
Module 39: London Force. The study outline says
that H2 & Cl2 are attracted only by London
Force: but aren't they covalently bonded, ie,
aren't they sharing an electron pair?
||The bond that hold the
molecules together in liquid or solid state is an
intermolecular force (H-bonding, dipole-dipole, and
London). it is true that each molecule has an
intramolecular bond i.e. covalent bond but when the
molecules go from solid --> liquid --> gas (all
physical reactions) what is being broken (or weakened) is
the intermolecular force. The covalent bond is broken when
a chemical reaction occurs. therefore since H2 and Cl2 are
do not have H-bonding or dipole-dipole force they can only
have London force.
|Roman Ferede 5/31/00 5:24:59 PM:
Question 8 Self Test: Suppose 40.0 mL of hydrogen
and 60.0 mL of nitrogen, each at standard
conditions, are both transferred to the same 125
mL container. The pressure of the mixture at 0 ?C
is ____ atm. I thought this question was
written/solved wrong. If the gases are at
standard, aren't they at 25 degrees C and 1 atm.
On transfer, don't they take a volume of 100 ml,
then reduce the volume to match the reduction
temperature which reduces atm, which is a 25/298
(or 8%) reduction? Linda Gudz
||No the question is
written/solved correctly. Standard condition is 0 oC and 1
atm. However standard condition for thermodynamic values
such as delta H are at 25 oC and 1 atm. I know it is kind
|Diana_jean Croft 5/31/00 8:10:43 AM:
Module 39 and outline: In the printed module 39
information it says to study the table to see the
factors that control viscosity and in your study
guidelines for module 39 it says to know what the
factors are that control viscosity. However, since
we can't see the table - what are these factors?
||You are right! The two main
factors that influence viscosity are length of molecules
and shape of molecules.
Two molecules with the same type of intermolecular force
but one is longer than the other, the longer one will have
also if two molecules that have the same intermolecualr
force but have different shape in which one is more
spehrical shape and one is more elonated shape the one
that has the elongated shape will have higher viscosity.
|Narmina_s Pashayeva 5/30/00 10:48:06
in the self test, module 38 question 7 reads: the
volume of oxygen required to burn 75 L of hydrogen
sulfide according to this reaction
2H2S+3O2=2H2O+2S02, where all the substances are
gases under the same condition of temperature and
pressure, is ? They take the given volume and
multiply it by 3/2 I don't understand why do they
||This is based on Avogador's and
Guy Lussac,s law, which states the same number of moles of
any gas at the same temperature and pressure occupy the
same volume. for example if we have the same number of
moles of gas O2, or N2, or CO2 they will occupy the same
volume. Remember at STP one mole of any gas will occupy
Now to answer your question, for a reaction that involve
only gases as reactants and products the number of moles
of these gases is directly proportional to the volume.
that is if we have one mole of each gas then they will
occupy the same volume howerweve if we have one mole of
gas A and two moles of gas B then gas B will occupy twice
as much volume as gas A.
Now if look at the mole ratio of the gases in this
equation there are 2 moles of H2S for every 3 moles of O2
and therefore their volume is going to be proportional to
that ratio 2:3. Therefore if there are 75 L of H2S then
there would be 75 L x 3/2 = 112. 5 L of O2
|Narmina_s Pashayeva 5/29/00 2:16:47
I can not link to the table again, so could you
please send it to me again or tell me how can i
find it somewhere else, and also my
"balancing redox reactions" e-mail is
missing too, so could you please send me that one
as well. Thank you.
||To get the table you can go to
my home page, click on chem 11, click on lecture notes,
then click functional group.
I will send you the email again. It is in the middle of
the page so you have to scroll down to find it.
|Diana_jean Croft 5/24/00 3:02:16 PM:
Module 36: Self test question #4 reads "In
polyethylene, what is the geometry and bond angles
around a carbon atom?" I figured since
ethylene is C2H2 and there is a double bond
between the C atoms, it should be trigonal planar.
So when another ehylene comes and bonds with first
ethylene, it breaks the double carbon bond and
forms single bonds with all other atoms? So what
type of reaction is this called?
||alkenes undergo addition
reaction. When they react they loose a double bond to add
molecules of Cl2 or H2 or H2O but they can also add
another molecule of alkene by a different mechanism called
free radical addtion polymerization. In this case the
double bond dissappers to form a free radical (an odd
electron species) which then adds another alkene. When the
two alkene bond to each other both of them lose the double
bond and become a single bond. Therefore the carbons will
not have a geometry of trigonal planar but they will have
a tetrahedral geometry.
|Diana_jean Croft 5/24/00 1:03:20 PM:
Module 37: On the self test several of the
questions use the equation PV=nRT which is fine,
but then they state that n=mass of compound/molar
mass of compound. I'm a little confused what
they're trying to figure out here. For example, in
question number 8 "A .340 mol sample of
oxygen gas occupies a volume of 40.0 L at 1.05 atm
and 298K. The weight of this sample is....g."
Why did they need to figure out the m/M business
when the number of moles is given?
||You are right since n = the
number of moles, and they have already given you the
number of moles for question #8, all you need is just plug
it in the Ideal gas equation. However for question #9, the
question asks you for the mass then what you have to do is
first calculate n using the Ideal gas equation and then
calculate mass by using the relationship that n = m/M i.e
|Julia_m Sian 5/22/00 11:59:26 PM:
Which is stronger, London force or dipole
interaction or H-bonding? Which should we consider
in predicting relative B.P. and F.P. ? And also in
relation to atomic size?
||These are the intermolecular
forces I was talking about for your previous question.
Right now I will not go into more details about each of
them. I will wait until we get to module 29. But to answer
your question simply: If the molecules have similar
molecular weight then the one that has hydrogen bonding
will have stronger intermolecular force than the one that
has only dipole- dipole interaction which in turn would
have higher intermolecular force than the one that just
has London force.
however as the molecular weight increases London force
becomes important. for example a large molecule with only
London force could have higher intermolecular force than a
small molecule that has hydrogen bonding.
|Julia_m Sian 5/22/00 11:57:11 PM:
I think butane is a gas and pentane is a liquid at
room temp. or something like that? So why is
butane a gas and pentane a liquid at room temp?
||this has something to do with
the forces that hold this molecules together to keep them
in liquid or solid state. these type of forces are called
intermolecular forces. and there are different type of
intermolecular forces which depend on the structure and
the molecular weight of the substance. We will talk about
these intermolecular forces in module 39 and determine why
butane has a lower boiling point than pentane (or why
butane is a gas and pentane is a liquid at room
|Moshe Golan 5/22/00 11:47:29 PM:
Module 37: What are the presure and temperture
limits in which we can consider all gases as
||Real gases behave like Ideal
gases at high temperature and low pressure.
|Moshe Golan 5/22/00 11:45:12 PM:
Why are free radical produced? Is their formation
||remember there is three steps
for a free radical reaction
even though the initiation step is an energy consuming
process (endothermic), the propagation step is an
exothermic for these reactions. that is where the driving
force for these reactions comes from. also the termination
step is an exothermic reaction
|Diana_jean Croft 5/22/00 1:53:17 PM:
Module 36: In "common addition polymers"
it says that teflon is an inert polymer. what does
||it means it is not reactive.