All midterm questions (content etc..) can be directed to this post. Any comments recieved before 5pm Oct 14 (24 hours before the exam) will be replied to.
The midterm will be in ECS 123 from 5-6 pm Friday Oct 15.
All students whose surnames begin with A or B should write their test in Mac A144 (David Lam Auditorium) with the other section.
The Chem 101 webpage has the official midterm information.
http://web.uvic.ca/~chem101/midterm.html
Sample Midterm : http://web.uvic.ca/~chem101/C101Midterm1A2009.print.pdf
Sample Midterm Key: http://web.uvic.ca/~chem101/C101Midterm1A2009.key.pdf
Things you need
-student ID
-soft pencil
-pen
-Sharp EL-510R calculator
Answers to the Last Comments
Anonymous said...
Have we covered all the information necessary to complete quiz 4?
October 14, 2010 10:51 AM
Yes
Anonymous said...
Hi, for the 2008 midterm question, the answer gives that the maximum number of electron in an atom with a principle quantum number of 4 is 32. How is this possible? Wouldn't the answer be 36 because that is how many electrons are in Kr?
Thank you.
October 14, 2010 10:55 AM
Kr has electrons in the 1s, 2s, 3s, 3p, 4s, 3d and 4p orbitals.
The electrons with a principle quantum number of 4 are in the 4s,4p,4d,4f
Steven said...
Hi Dr. Sirk,
I was wondering if a completely filled d-subshell is still considered part of an atom's valence electrons? For instance, if Zinc is 1s2,2s2,2p6,3s2,3p6,4s2,3d10....are the valence electrons 4s2,3d10?
Nope, a completely filled d orbital counts as core, so the valence electrons are 4s2
Along the same lines, how do we know which electrons of an atom are considered valence electrons? Do we just assume all electrons added after the preceding noble gas are valence?
Yes, except for a completely filled d orbital
How do we know if K+ or Ar has a larger atomic radius?
They are isoelectronic, so the one with the greatest number of protons is smaller
Finally, are we required to calculate lattice energies on the midterm? I don't remember this ever being covered in class.
No direct calculations, but you should be aware of the formula (coulomb’s law) and how charge/radius would affect the lattice energy (E=kQ1Q2/d) Q1=charge on cation, Q2=charge on anion, d=radius of anion+radius of cation
Thanks in advance!
You’re welcome
October 14, 2010 11:02 AM
Anonymous said...
2009 midterm mc question 12. How do you determine the lattice energies of the RbCl, RbBr, and RbF, which are not given on the data sheet? thanks!
You just need to determine the largest lattice energy. They all have the same charge (+1)(-1) and the same cation (Rb+) so the one will the SMALLEST radius will have the greatest lattice energy (see above question)October 14, 2010 11:47 AM
Anonymous said...
3. Which of the following electron configurations is CORRECT for the ground state of the given atom?
A. S: [Ne] 2s2 2p5
B. P: [Ne] 3s2 3d3
C. Br:[Ar] 4d10 4s24p5
D. Rb: [Kr] 5s1
E. Ga: [Ar] 4s2 4p1
how do you know do you know that the answer is D?
The correct electron configurations are:
A. S: [Ne] 2s2 2p6
B. P: [Ne] 3s2 3p3
C. Br:[Ar] 3d10 4s24p5
D. Rb: [Kr] 5s1
E. Ga: [Ar] 3d10 4s2 4p1
So only Rb is correct
October 14, 2010 3:54 PM
Anonymous said...
Question 2c on midterm 2009:
What are the n values for the transition (labeled or unlabeled) that corresponds to the ionization energy
for the hydrogen atom.
answer:From n = 1 to n = ∞
What does that mean? Thanks
October 14, 2010 4:12 PM
In ionisation the electron is completely removed from the atom. Complete removal corresponds to the n=infinity energy level. The n=infinity corresponds to no attraction to the nucleus.(Note the electron is removed from the highest ground state energy level. For H this is 1. For Li this would be 2s)
Anonymous said...
a) A red laser pointer emits light at a wavelength of 650nm. What is the frequency of this light? b) what is the energy of one mole of these photons? c) The laser pointer emits light because electrons in the material are excited (by a battery) from their ground state to their excited state. When the electrons return to the ground state they lose the excess energy in the form of 650nm photons. What is the energy gap between the ground state and the excited state in the laser material?
How do I calculate part c) ?
October 14, 2010 4:15 PM
E=hv or E=hc/lambda with lambda equal to 650 nm (650*10-9m)
Anonymous said...
In the 2008 midterm, question number 3 asks for correct order of Zeff. The correct answer is
Na > Be > Li
However, I thought that Zeff, for the purpose of this course, was = # of electrons - the # of core electrons, which would, in effect, make the answer Na (1) > Be (2) > Li (1)
Does this mean that we are to assume Na has a much larger Zeff than is indicated by the formula above?
October 14, 2010 4:16 PM
Good question: If we want to estimate the value of Zeff , then Zeff = # of protons - the # of core electrons is reasonable to get at a number. But that is a very rough estimate and note that Li, Na, K, Rb Cs would all have the same value (1). Recall the trends in Zeff. It increases a lot across the row and slightly down a column. So you need to use the trends for this question.
Anonymous said...
what sort of unit conversions will we need to know?
nanometers, micrometers? what else?
October 14, 2010 4:19 PM
Any of the values in Table 1.5 are testable.
Anonymous said...
can you please explain the answer to number 6 on the 2008 midterm?: What is the maximum number of electrons in an atom that can have the principal quantum number n = 4? thank you
See above
October 14, 2010 4:36 PM
First, a question asks: Arrange the bonds in order of increasing
polarity ( C--S, B--F, N--O). I know that the larger difference in
electronegativity causes a stronger polarity. For this reason, I had
ordered it as N--O, C--S, B--F. However, the answer provided explained
that because S is down a row and to the right, it has almost an
identical electronegativity to C... are we going to be expected to
know this subtle difference for the exam?
Already answered in the blog questions
Similar to this, the practice exam asks: Of the compounds below,
______ has the largest lattice energy. A) KF B) KCl C) RbCl D) RbBr E)
RbF
Would the answer to this be RbF, as it is both less electronegative
than K, and F has the highest electronegativity?
Ionic compounds, therefore you should be aware of the formula (coulomb’s law) and how charge/radius would affect the lattice energy (E=kQ1Q2/d) Q1=charge on cation, Q2=charge on anion, d=radius of anion+radius of cation. Rest explained above.
Finally, in the Prep 101 session, it was mentioned that the intensity
of a photon affects the number of electrons ejected from a surface
during the photoelectric effect, but the frequency affects the kinetic
energy of the released electron. Do we need to know the part about
intensity, as I do not recall it being mentioned in the lectures.
It was in the notes under a subtitle:observations (also in the text)
From email:
ReplyDeleteHi Dr. Sirk,
Two Questions:
1. I know that formal charges will be on the midterm, but will anything regarding formal charges on resonance structures be on the midterm?
2. How do you know what the electronegativity of the atoms are (not on data sheet)so you can determine what the bond between them is by knowing the different bond ranges.
1) Resonance structures (section 8.6) will not be tested on the first midterm.
ReplyDelete2)You will need to know the difference between a metal and a non metal. Without the values of electronegativity given, you can can determine ionic (as it's between a metal and non metal) and covalent/polar covalent (between non metal and non metal)
Note that H is a non-metal.
Beyond that you can qualitatively determine the degree of polarity of a bond (dipole moment) by the trends. eg: H-X, where X is a halide. HF>HCl>HBr.
So the questions can be:
Is the bond ionic or covalent?
Which bond is most polar?
Question posted on Introduction:
ReplyDeleteI'm confused about lattice energy. How does it 'drive' ionic bonding? I understand that lattice energy is the energy required to break up the ionic compound when it forms, which shows the stability of the solid, but how does that make up for the net positive enthalpy of the ionic reaction (ie. that Na + Cl requires 147kj/mol more energy than the electron affinity of Cl provides)?
The energy that is required to break up the solid (energy in-so a positive value) is given off when the solid is formed (energy out-a negative value-spontaneous reaction).
ReplyDeleteSo the energy of the reaction Na +Cl2 --> NaCl
is
ionisation energy of Na (+)
electron affinity of Cl (-)
lattice energy (large -)
(plus some other smaller values that we do not concern ourselves with)
= a large negative value, therefore energy out, therefore spontaneous.
From email: Will the Born Haber Cycle be on the midterm?
ReplyDeleteThe Born-Haber cycle will not be on the midterm. The energy of salt formation( for our purposes) is simplified to the three main contributors: enthalpy of ionisation, electron affinity and lattice energy
ReplyDeleteCan you please explain the difference between radial probability and density probability functions?
ReplyDeleteLook at the images in Figure 6.19 (a) and 6.23 (a) to help visualise this discription.
ReplyDelete6.23: This is the probability density function that shows the density of the electron (most dense at the center)
6.19: This is the radial probability (the probability density function multiplied by the surface area*).
The density is greatest at the center, but the volume is much greater as move away. Where this maxes out we get the radius.
If you imagine a dartboard with a hundred darts thrown it at it, this might make more sense. there may be the greatest density in the center (1 cm radius circle), but you have a greater proability of finding more darts if you took a 1 cm wide ring with and inner diameter of 50 cm and placed it over the dart board (simply because it has a greater volume)
*or more correctly the integrated volume; the small slice of volume as the surface area increases.
Do we need to know the oxidation numbers of atoms?
ReplyDeleteDo we need to be able to calculate effective charges on atoms using the u=Qr equation? (dipole moments)
ReplyDeleteFor electronegativity trends, does going up a column or going right have a greater effect?
ReplyDeleteFor a limited number, yes. You should know that the main group elements (s and p block) will have an oxidation state that gives them a noble gas configuration.
ReplyDeleteYou should know that the alkalis (group 1) all have oxidation states of +1 (and 0 the neutral atom, as all elements have)
Group 2 has +2
Group 8: Noble gases only have 0
Group 7: Halogens have -1
Group 6: have -2
etc..
Will we need to do calculations involving the formula u=Qr?
ReplyDeleteRe: u=Qr
ReplyDeleteIn general terms, anything in the assigned reading is fair game. However, an equation that may show up in the readings, but not covered in either the lectures or in any of the assignment questions or on the data sheet would most likely not be tested. Note that while the calculation may not be tested, the concepts behind the calculation are testable.
Re: Electronegativity across versus dowm
ReplyDeleteThe trend across is more pronounced than the trend down (This is true of I.E. and Zeff as well).
Question from Introduction:
ReplyDeleteFor lattice energy trends, how do you determine what d is for a compound such as MgF2? Would the lattice energy of MgF2 be less than that for MgO?
October 11, 2010 12:59 PM
ReplyDeleteAislinn Sirk said...
1) d is the radius of Mg^2+ plus the radius of F^1-. The stoichiometry does not affect d.
2) Yes, it would ( the increase in charge trumps the slight increase in radius). Note that lattice energies for some compounds are given in the data sheets.
Will Slater's law be on the midterm?
ReplyDeleteCalculations will not be. Reasoning behind it (core shield well, valence shield poorly) is testable.
ReplyDeleteThe rydberg formula isn't on the data sheet, do we have to memorize it?
ReplyDeletedo we need to know the hydrogen line spectra wavelengths and colours?
ReplyDeleteRe: Rydberg Formula
ReplyDeleteThe equation is given for both energy and frequency. By combining either with c=(wavelength)(frequency) (and E=h(frequency) for E) one can obtain wavelength.
Re Colours
ReplyDeleteNo,
But spectra and colourwise, you do need to know that red is lower energy than orange, yellow, green, blue, violet and the approximate wavelength for visble light (350nm-750nm)
You should also be aware that you can calculate any of the hydrogen lines (Energy, wavelength, frequency) (see question on Rydberg above)
Do we get a data sheet to take in with us for the the exam?
ReplyDeleteYes. It is very similiar to the one linked on the website.
ReplyDeletehttp://web.uvic.ca/~chem101/datasheet.pdf
The textbook says that you can determine if a bond is nonpolar covalent, polar covalent, or ionic by determining the difference between the electronegativity of two atoms, but it doesn't directly say what values would be considered great enough to form an ionic bond or a polar covalent bond - where is the distinction?
ReplyDeleteAlso, do we still have class on Friday?
The difference between covalent and polar covalent is straightforward. A bond between 2 nonmetals with an electronegativity difference of less than 0.4.
ReplyDeleteThe difference between ionic and polar covalent is fuzzier. Anything above 2 is ionic. Anything below 1.3 is polar covalent (You may find other values in different sources. This is not hard deadline). Between that a bond between a metal and non metal is ionic and a bond between nonmetals is considered polar covalent.
So both electronegativity and the type of element determine the naming of the bond.
And yes, Friday is a normal lecture class.
ReplyDeleteA question asks: arrange the bonds in each of the following sets in the order of increasing polarity: C--S, B--F, N--O.
ReplyDeleteI had thought that N--O would be the least polar, yet the study guide solutions manual said "since there is a big decrease in going from the second row to the third, this means that S is not only less than O, but essentially the same as C". Would we need to know that the shift downward of S trumps the closeness of N and O?
I know that we would need to know the general trend of electronegativity and how the increasing charges/decreasing distances between atoms creates a larger polarity, but would we need to know the minor differences as shown in this question?
Short answer: No.
ReplyDeleteLong answer: For this and other concepts, the assignment questions from the textbook will include questions that involve you looking up specific values in order to tell what the answers are.
Another note: You should be aware the EN changes more across a row than down a column ( as does IE, EA, radius).
ReplyDeleteFor question 1 d part II on the 2008 practice exam, why are B and Al included in the answer? Shouldn't the correct answer include only Ga, In and Tl?
ReplyDeleteCan you recheck the question, please? I can't find the one you are referring to.
ReplyDeleteDear Professor Sirk,
ReplyDeleteFor the probability function analogy you gave us in our notes, it states that there is a higher probability of landing in an area further away from the center. Also in the figure there is a higher density of dots closer to the center. In the 11th edition of our text book it gives us a similar figure stating that ,” the regions with a high density of dots correspond to relatively large values of psi2 and are therefore regions where there is a high probability of finding the electron.” Does this mean that there is a higher probability of finding an electron further away from the nucleus and that there is a higher density of electrons closer to the center?
Thank you
14. Which of the following descriptive combinations of atoms in a bond and type of bond is CORRECT?
ReplyDeletei. CsF, polar covalent
ii. F2, nonpolar covalent
iii. ICl, polar covalent
iv. CF4, ionic
A. i B. i and ii C. ii and iii
D. ii and iv E. iv
Is CsF ionic?
is CF4 non-polar?
I and Cl are very close together.. why is it polar covalent?
5. (a) [1 MARK] Arrange the following atoms in order of increasing first ionization energy (I1): (i.e. smallest to largest) Ne, Na, Mg, Al, Cl
ReplyDeleteNa < Al < Mg < Cl < Ne
Why does Al have a smaller first ionization energy than Mg?
Re: Probability
ReplyDeleteYes
Re
ReplyDeleteIs CsF ionic?
Yes-metal and non metal is an ionic bond.
is CF4 non-polar?
The bonds in C-F are very polar (The entire molecule is non polar, but that will not be covered, nor tested until later in the course)
I and Cl are very close together.. why is it polar covalent?
The electronegativity difference is greater than 0.4. Basicly (without numbers) any heteronuclear bond between two nonmetals is polar.
Re ionisation energy
ReplyDeleteThe first p electron is easier to remove than the 2nd s. (and the 4th p electron is easier to remove than the 3rd)
1)One of the textbook questions asked what bond between carbon atoms had the greatest bond enthalpy. There were three carbon atoms bonded together with a single bond, double bond, and triple bond.
ReplyDelete2)When writing out the electron configurations, does the orbital correspond to one of the boxes in which 2 electrons are located, and then is the subshell all of the boxes with the same principale quantum # (so the 3s, 3p, 3d, would be called a subshell)???
Thanks
I can't find the link to the midterm we went through on tuesday morning. Would you mind posting a link.
ReplyDeleteWill we need to know how to draw the exact drawings for the d-subshells (dxy,dxz...), or do we just need to know that there a four leaf clover shape?
ReplyDelete1) Triple bonds are the strongest and shortest. Single bonds are longest and weakest.
ReplyDelete2) principle quantum number (n) is a shell
azimuthal quantum number (l) specifies subshell
So 4 is a shell , 4s is a subshell, 4p is a subshell
Will there be questions on the midterm regarding lattice energies that we will have to calculate?
ReplyDeleteA link to the midterm can be found on
ReplyDeletehttp://web.uvic.ca/~asirk/overview.htm
Re: d orbitals
ReplyDeletePrevious midterms have asked for one example of a d orbital. No questions have asked for identification of a specific orbital. Knowing how to draw at least one would cover almost all of the questions on d orbital shape.
In my notes I wrote that theory predicted an infinite increase in intensity for blackbody radiation, but by assuming quantized energy, Planck was able to explain why that didn't happen. Why does quantized energy explain why it didn't increase infinitely?
ReplyDeleteRe: Blackbody radiation
ReplyDeleteYOur notes are correct. The reasoning behind it is beyond the scope of this course but the Wikipedia entry for this is quite complete and accurate.
http://en.wikipedia.org/wiki/Black_body
Re: lattice energy
ReplyDeleteIn general terms, anything in the assigned reading is fair game. However, an equation that may show up in the readings, but not in any of the assignment questions or on the data sheet would most likely not be tested. Note that while the calculation may not be tested, the concepts behind the calculation are testable (ie charge is more important than radius).
Reposted (blogger misent a comment)
ReplyDeleteSo I don't fully understand the definition of a photon. Is it the lowest amount of energy required to make the photoelectric effect happen? or is it just the lowest measurement of energy in light? or is it something else
A photon is the smallest packet of electromagnetic radiation (light) allowed. It has energy (E=hv) v=frequency and travels at speed c.
ReplyDeleteIt is the smallest quantised packet of light, meaning it cannot be broken down into smaller pieces.
Despite the horrible flash animation ad, this link has a reasonable description of a photon.
http://physics.about.com/od/lightoptics/f/photon.htm
Reposted question:
ReplyDeleteWhat causes a different colour from the Line Spectrum to be emitted? Thanks!!
Hi Dr Sirk,
ReplyDeleteOn the 2009 Midterm, Written Answer Question 2, why are the n values for the transition that corresponds to the ionization energy for the hydrogen atom, from n=1 to n=infinity? When the electron is in the n=2 shell, for example, would it be considered ionized at that point then?
Thank you!
Different electron transitions between different energy level result in photons with different energies of light being emitted.
ReplyDeleteThe photons have specific energies and therefore have different wavelengths (E=hc/lambda) and therefore have different colours.
For the diagrams of the electron probability density for s orbitals (p. 231 in 11th ed), why is the probability density approaching infinity when r=0? Wouldn't there be 0 chance of finding an electron at the nucleus?
ReplyDeleteWould we be expected to sketch electron probability density graphs?
Re ionisation in a H
ReplyDeleteIn ionisation, the electron is entirely removed from the atom (n=infinity or no association with the proton)
The electron in the n=2 energy level is not considered ionised. It is considered to be "in an excited state"-meaning not in ground state, but still having excess energy (that can be released-see above question)
Do we have to know the Stern-Gerlach experiment?
ReplyDeleteCould you please look at the 2009 midterm, the written question #3a. I was wondering why there cannot be a double bond between Nitrogen and Oxygen? It would fill the octet rule and seems to make sense?
ReplyDeleteThanks
On the 2008 midterm, question 12, the key gives the answer that both P^-3 and Ga^+3 have completely full subshells, how are the orbitals of all occupied subshells of Ga^+3, completely full?
ReplyDeleteIs an orbital the way we illustrate electron probability density functions?
ReplyDeleteEven though you fill the s orbitals before the d orbitals, do you remove the electrons from the s orbitals first because it has a higher principle quantum number?
ReplyDeleteRe:p. 231 in 11th ed
ReplyDeleteYou are correct that there is a zero percent chance of finding the electron directly at the nucleaus, but if you look at the size of a proton versus the Bohr radius, you will see that it wouldn't show up on graph of this scale. (Also there is an uncertainty in the position of protons as well-not discussed in this course)
Re: Do we have to know the Stern-Gerlach experiment
ReplyDeleteYes
RE: 2009 midterm, the written question #3a. Why there cannot be a double bond between Nitrogen and Oxygen?
ReplyDeleteThe structure has 14 e- to place, 8 are used in bonds leaving 6 to put on N and O (H has a full shell with 2) Once those are placed, the octets are full.
Multiple bonds are added only when the octet can't be filled by just adding the electrons and they need to be shared.
RE: 2008 midterm, question 12,how are the orbitals of all occupied subshells of Ga^+3 completely full?
ReplyDeleteGa = [Ar]3d10 4s2 4p1
The electrons are removed from the 4p first, then the 4s, then the 3d. In this case, we can remove electrons from the 4s and 4p.
Ga3+ = [Ar]3d10
Re: Is an orbital the way we illustrate electron probability density functions?
ReplyDeleteYes
Re:Even though you fill the s orbitals before the d orbitals, do you remove the electrons from the s orbitals first because it has a higher principle quantum number?
ReplyDeleteYou do remove the electrons from the s orbitals first and it has a higher principle quantum number. This is a reasonable way to remember the rule.
However, we remove them from the s first because (in an ion) the 4s orbitals become slightly higher in energy than the 3d orbitals. In an atom the 4s orbital is lower in energyy.
We are still putting e into the lowest energy orbital and taking them out of the highest energy orbital. The act of ionisation shifts the energies of the orbitals.
Have we covered all the information necessary to complete quiz 4?
ReplyDeleteHi, for the 2008 midterm question, the answer gives that the maximum number of electron in an atom with a principle quantum number of 4 is 32. How is this possible? Wouldn't the answer be 36 because that is how many electrons are in Kr?
ReplyDeleteThank you.
Hi Dr. Sirk,
ReplyDeleteI was wondering if a completely filled d-subshell is still considered part of an atom's valence electrons? For instance, if Zinc is 1s2,2s2,2p6,3s2,3p6,4s2,3d10....are the valence electrons 4s2,3d10?
Along the same lines, how do we know which electrons of an atom are considered valence electrons? Do we just assume all electrons added after the preceding noble gas are valence?
How do we know if K+ or Ar has a larger atomic radius?
Finally, are we required to calculate lattice energies on the midterm? I don't remember this ever being covered in class.
Thanks in advance!
2009 midterm mc question 12. How do you determine the lattice energies of the RbCl, RbBr, and RbF, which are not given on the data sheet? thanks!
ReplyDelete3. Which of the following electron configurations is CORRECT for the ground state of the given atom?
ReplyDeleteA. S: [Ne] 2s2 2p5
B. P: [Ne] 3s2 3d3
C. Br:[Ar] 4d10 4s24p5
D. Rb: [Kr] 5s1
E. Ga: [Ar] 4s2 4p1
how do you know do you know that the answer is D?
Question 2c on midterm 2009:
ReplyDeleteWhat are the n values for the transition (labeled or unlabeled) that corresponds to the ionization energy
for the hydrogen atom.
answer:From n = 1 to n = ∞
What does that mean? Thanks
a) A red laser pointer emits light at a wavelength of 650nm. What is the frequency of this light? b) what is the energy of one mole of these photons? c) The laser pointer emits light because electrons in the material are excited (by a battery) from their ground state to their excited state. When the electrons return to the ground state they lose the excess energy in the form of 650nm photons. What is the energy gap between the ground state and the excited state in the laser material?
ReplyDeleteHow do I calculate part c) ?
In the 2008 midterm, question number 3 asks for correct order of Zeff. The correct answer is
ReplyDeleteNa > Be > Li
However, I thought that Zeff, for the purpose of this course, was = # of electrons - the # of core electrons, which would, in effect, make the answer Na (1) > Be (2) > Li (1)
Does this mean that we are to assume Na has a much larger Zeff than is indicated by the formula above?
what sort of unit conversions will we need to know?
ReplyDeletenanometers, micrometers? what else?
can you please explain the answer to number 6 on the 2008 midterm?: What is the maximum number of electrons in an atom that can have the principal quantum number n = 4? thank you
ReplyDeleteOne type of sunburn occurs on exposure to UV light of wavelength in the vicinity of 325nm. a)what is the energy of a photon of this wavelength? b) what is the energy of a mole of these photons? c) How many photons are in a 1.00 mJ burst of this radiation?
ReplyDeletehow do i calculate c) ?
I was pretty confident going into this mid term, until i read the questions that were being posted here. Wow! good thing i read the blog. Looks like I have more studying to do.
ReplyDeleteThanks for the questions and the answers!
Are the conversions like m to A ( X10^-10) and constants such as 1.60 X10^-19C and 3.34 X10^-34 C-m going ot be on the data sheet or do we need to memorize them?
ReplyDeleteConstants will be provided. Conversions must be memorised.
ReplyDeleteRe Sunburn:
ReplyDeleteOne type of sunburn occurs on exposure to UV light of wavelength in the vicinity of 325nm. a)what is the energy of a photon of this wavelength? b) what is the energy of a mole of these photons? c) How many photons are in a 1.00 mJ burst of this radiation?
how do i calculate c) ?
-determine the energy in each photon. Divide 1.00 mJ by the energy per photon to get the number of photons.
All other comments have been answered in the main midterm post to tie the questions and answers together.
ReplyDeleteRe Thanks
ReplyDeleteYou are welcome. I am glad this is useful.
Unfоrtunatеlу, there is nο guaгantее.
ReplyDeleteThe mаjοг disadvantage of а 'Closed' sуѕtеm
іs thаt only thе most advanсed handsеt
avаilable, the iphone 5 yes, we're still calling it that and one in October for the iPad. He likened competition in the smartphone league tables.