ATAR Notes: Forum
VCE Stuff => VCE Science => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Chemistry => Topic started by: Rod on December 20, 2013, 10:36:06 am
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Hi guys,
I just wanted to know how difficult chromatography (chapter 6), spectroscopy (chapter 7) and combining and choosing analytical techniques (chapter 8) are on a scale of 1 to 10. I'll be starting these topics in about one week's time. Obviously, as you can see I'm going ahead, and I just wanted to know whether I should do the review questions. A lot of smart people are telling me I should, while others have told me I shouldn't. I personally like doing questions, as it solidifies my knowledge, but these questions are taking a lot of my study time away.
Thanks
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I think that learning the theory of chromatography and spectroscopy techniques is mainly just memory work- consult both textbooks, others' notes etc. So in that sense it is not too bad. What people find difficult is the interpretation of chromatograms, and readouts from the various instruments- which is what is generally given more marks on the exam. So, in order to maximize your marks on the application side of these topics, I would suggest doing selected review questions (some of them are badly written and not likely to appear on an exam) in conjunction with Checkpoints, Neap or Lisachem books that give you really good targeted practice. Personally I found holidays the best time to cover the most content, and leave more application practice to during the term. Hope this helps
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Hi guys,
I just wanted to know how difficult chromatography (chapter 6), spectroscopy (chapter 7) and combining and choosing analytical techniques (chapter 8) are on a scale of 1 to 10. I'll be starting these topics in about one week's time. Obviously, as you can see I'm going ahead, and I just wanted to know whether I should do the review questions. A lot of smart people are telling me I should, while others have told me I shouldn't. I personally like doing questions, as it solidifies my knowledge, but these questions are taking a lot of my study time away.
Thanks
Chapters 1-4 are really all covered in year 11, so if you had a strong year 11 foundation in volumetric & gravimetric analysis, as well as stoichiometry, then it shouldn't be too difficult.
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I think that learning the theory of chromatography and spectroscopy techniques is mainly just memory work- consult both textbooks, others' notes etc. So in that sense it is not too bad. What people find difficult is the interpretation of chromatograms, and readouts from the various instruments- which is what is generally given more marks on the exam. So, in order to maximize your marks on the application side of these topics, I would suggest doing selected review questions (some of them are badly written and not likely to appear on an exam) in conjunction with Checkpoints, Neap or Lisachem books that give you really good targeted practice. Personally I found holidays the best time to cover the most content, and leave more application practice to during the term. Hope this helps
So are these three chapters mainly memory and interpretations ? And not many calculations?
Thanks
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So are these three chapters mainly memory and interpretations ? And not many calculations?
Thanks
Yeah predominantly. That said,UV-vis calibration curves for example, are often used as a part of a larger stoichiometry question. This often happens in chem (combination of several topics into one question), which is why review questions aren't always the best, as they are not always exam-style. Most people find interpretation the hardest because , eg. in HNMR, it requires logic and referring to the data book , which is definitely higher-level thinking.
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Thanks guys;
Can someone also please help me with gravimetric analysis. Last week I did all the gravimetric analysis questions, but not in the best of ways (kept making mistakes). So I did some more questions today to make up, and had a bit of trouble. My method of answering these questions are;
- Read the question carefully, find the precipitate, write down a fully balanced equation
- After writing down the equation, write down everything you have
- Use mole ratios to convert whatever you have to what you need
Is that method alright? I don't know why but I'm getting heaps of grav-questions wrong. Any help/tips appreciated!
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Thanks guys;
Can someone also please help me with gravimetric analysis. Last week I did all the gravimetric analysis questions, but not in the best of ways (kept making mistakes). So I did some more questions today to make up, and had a bit of trouble. My method of answering these questions are;
- Read the question carefully, find the precipitate, write down a fully balanced equation
- After writing down the equation, write down everything you have
- Use mole ratios to convert whatever you have to what you need
Is that method alright? I don't know why but I'm getting heaps of grav-questions wrong. Any help/tips appreciated!
When doing gravimetric analysis, there are a number of different analyses covered. For instance, you can determine water content, find the composition of a compound or find the mass of a particular substance by forming a precipitate. The methods I approach for gravimetric analysis:
1. Read the question.
2. Write out the equation; balanced, with states.
3. Apply any given data to my equation (e.g. we might be told that after a silver chloride precipitate is formed, dried and weighed, a mass of 0.754g is obtained. So, over the AgCl(s) on my equation, I'll just take note of the fact that I have that much of it.
4. Use stoichiometry to answer relevant questions.
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When doing gravimetric analysis, there are a number of different analyses covered. For instance, you can determine water content, find the composition of a compound or find the mass of a particular substance by forming a precipitate. The methods I approach for gravimetric analysis:
1. Read the question.
2. Write out the equation; balanced, with states.
3. Apply any given data to my equation (e.g. we might be told that after a silver chloride precipitate is formed, dried and weighed, a mass of 0.754g is obtained. So, over the AgCl(s) on my equation, I'll just take note of the fact that I have that much of it.
4. Use stoichiometry to answer relevant questions.
I see, that's a great method. Thanks Yacoubb :)
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I think that learning the theory of chromatography and spectroscopy techniques is mainly just memory work- consult both textbooks, others' notes etc. So in that sense it is not too bad. What people find difficult is the interpretation of chromatograms, and readouts from the various instruments- which is what is generally given more marks on the exam. So, in order to maximize your marks on the application side of these topics, I would suggest doing selected review questions (some of them are badly written and not likely to appear on an exam) in conjunction with Checkpoints, Neap or Lisachem books that give you really good targeted practice. Personally I found holidays the best time to cover the most content, and leave more application practice to during the term. Hope this helps
I think it would also be helpful to know what exactly goes on in these techniques. Like, for instance, how NMR actually works. They could certainly ask questions on that.
Interpretations of chromatograms, I think, is relatively straightforward in comparison as doing enough practice questions should help you with that. Most of my practice came through Checkpoints and I think it helped.
Thanks guys;
Can someone also please help me with gravimetric analysis. Last week I did all the gravimetric analysis questions, but not in the best of ways (kept making mistakes). So I did some more questions today to make up, and had a bit of trouble. My method of answering these questions are;
- Read the question carefully, find the precipitate, write down a fully balanced equation
- After writing down the equation, write down everything you have
- Use mole ratios to convert whatever you have to what you need
Is that method alright? I don't know why but I'm getting heaps of grav-questions wrong. Any help/tips appreciated!
I think the easiest way of thinking about gravimetric analysis questions is to keep a mental track of where the heck the ions are going. For instance, in last year's chemistry exam, there was a question that asked to find the % by mass of calcium ions in a CaO ore. We were given the mass of calcium oxalate precipitated after treating the calcium oxide with oxalic acid. The simplest way to go about a question like this is to realise that all of the calcium ions in CaO appear in CaC2O4, so you can then easily work out mole ratios. Gravimetric analysis is essentially extended stoichiometry IMO.
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Hey guys,
So I've been stuck in a fairly easy question for a while now:
In this question I have to balance the reaction Cr2O7^(-2) ------- Cr^(3+)
So here is what I did:
Cr207^-2+14h+ +9e-------- Cr^3+ +7h20
What am I doing wrong here??
Thanks!
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Cr is not balanced
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Cr is not balanced
Yeah sorry that was silly, but apparently it's 6e- and not 9e-. That's the mistake that I've done.
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Hey guys,
So I've been stuck in a fairly easy question for a while now:
In this question I have to balance the reaction Cr2O7^(-2) ------- Cr^(3+)
So here is what I did:
Cr207^-2+14h+ +9e-------- Cr^3+ +7h20
What am I doing wrong here??
Thanks!
Something that I rarely do personally, but always helps to help check, is to confirm that the number of electrons balances the total change in oxidation number. Even if you hadn't noticed the glitch of having two chromiums, you should be wondering, where are these nine electrons coming from?
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if you balance the Cr, you'll have +6 charge on the right hand side of the equation, so you would need 6e-(instead of 9e-) on the left hand side to balance the equation
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Something that I rarely do personally, but always helps to help check, is to confirm that the number of electrons balances the total change in oxidation number. Even if you hadn't noticed the glitch of having two chromiums, you should be wondering, where are these nine electrons coming from?
Still don't get it, sorry.
So the reactant side should have a total charge of -2+ 14= 12
And the products should have a total charge of +3
12-9=3
So shouldn't we add 9 electrons to the reactants?
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if you balance the Cr, you'll have +6 charge on the right hand side of the equation, so you would need 6e-(instead of 9e-) on the left hand side to balance the equation
Ohh I see. Thank you homer and lzxnl ! :D
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Hey guys just an easy question here,
For a redox reaction to occur, must all elements in the reaction change in oxidation number? Or at least one?
Thank you
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Hey guys just an easy question here,
For a redox reaction to occur, must all elements in the reaction change in oxidation number? Or at least one?
Thank you
Just at least one. As an example, consider the reaction you gave with dichromate. The hydrogen ions, when forming water, still have oxidation number +1.
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Hey guys, just needed some help on this question;
Q22.
The thermite process can be used to weld lengths of railway track together. A mould
is placed over the two ends of rails to be joined and it is filled with a charge of
aluminium powder and iron(III) oxide. When the mixture is ignited, a redox reaction
occurs to form molten iron which joins the rails together.
c Write the overall equation for the thermite process.
So lets start with the half equations;
Fe03 +2e-
Fe2+ 3O2-
What would the other half equation be? Aluminium powder was used...
Thank you
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I was trying to do an arrow after 2e- but I'm not sure what I did. Anyway I tried to write Fe03 + 2e0 arrow Fe2+ + 3o2-
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Another question here sorry, in this one I'm confused as fk
I won't type up the whole question, it's pretty long but it's about titrating solid iron with potassium premanganate to produce Mn2+ and Fe3+.
So after doing the whole, long question, I got everything wrong because my equation was wrong; here is the equation I did:
2Fe + Kmn04+ 8H+ ------- Fe3+ + Mn2+ + 4h20
Here is the answers:
5Fe2+(aq) + MnO4(aq) + 8H+(aq) → 5Fe3+(aq) + Mn2+(aq) + 4H2O(aq)
So as you can see that Mn04- stuffed me up. Where does that Mn04 come from? Why did they just suddenly take out the potassium out of it? Where did the potassium go?
Thanks guys, sorry again for all the questions and confusion
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Potassium is merely a spectator ion in the reaction. It doesn't react with any other species and hence can be omitted from the equation. There's no problem with putting the potassium in, but you'll just have potassium ions as one of your products, which you're missing in your equation.
You also seem to have 2 Fe on the left hand side of your equation, but only one Fe on the right hand side of your equation.
Also, the answers suggest you're not titrating solid iron, which makes sense as it is difficult to obtain accurate measurements from a titration of a solid.
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Hey guys this isn't a chemistry question but I just need some advice.
I've been doing many chemistry and physics questions these holidays, but something new that I've been doing is that I've kept the solutions in front of me on my laptop (although not visible). It's not like I look at the solutions before doing the question, I do the question first, show my working and answer, and then IMMEDIATELY look at the solutions. I've noticed that I've stopped double checking my answer like I did before to check if I've done any silly mistakes. Will this new habit cause me any damage to my studies? Should I stop? Should I only look at it if I am completely stuck? Any advice appreciated.
Thank you
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Hey guys this isn't a chemistry question but I just need some advice.
I've been doing many chemistry and physics questions these holidays, but something new that I've been doing is that I've kept the solutions in front of me on my laptop (although not visible). It's not like I look at the solutions before doing the question, I do the question first, show my working and answer, and then IMMEDIATELY look at the solutions. I've noticed that I've stopped double checking my answer like I did before to check if I've done any silly mistakes. Will this new habit cause me any damage to my studies? Should I stop? Should I only look at it if I am completely stuck? Any advice appreciated.
Thank you
I think maybe only look at solutions if you're stuck. Finish the questions, and then check your answers, using the worked solutions if you've gotten something wrong and you don't know how to correctly solve it.
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I think maybe only look at solutions if you're stuck. Finish the questions, and then check your answers, using the worked solutions if you've gotten something wrong and you don't know how to correctly solve it.
That sounds good. Will be doing that from now on.
Thank you Yacoubb !
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Hey guys!
I've been going through the spectroscopic techniques (Flame tests, AES, AAS, UV, IR, NMR) and I was looking at the questions, it seems as if VCAA like combining many techniques into one question, making us read of several graphs usually for finding the structure of a compound.
I've been doing alright in that, but I am struggling to fully cement all the theory of these techniques.
So firstly, I was fine with flame test and AES, struggling in AAS, can someone please help me?
Here is what I know;
So we use AAS as it is more versatile and can be used for several elements. Unlike the first two techniques, where light is emitted, this technique focusses on the absorption of light. We know that different elements absorb different wavelengths and energies of light in order for their electrons to become excited, this is because of differing electron arrangements and different numbers of protons and neutrons. So THEREFORE, we can shine light onto an element, see how much it absorbs, and by looking at the amount it absorbs we can find how much and what the element is. Here is an anology I have created, I have friends named Mary and Matthew. They both look identical so I can't distinguish. Mary likes eating dog food, while Matthew hates it, Matthew likes chicken but Mary is vegan so would never eat chicken. So one day I through a piece of chicken on the floor, one of them pick it up and eats it, that person must be Matthew and the other person cannot be Matthew.
So the steps in AAS:
1. Element is placed in lamp
2. Element is vaporised WHY DO WE VAPORIZE IT DONT UNDERSTANT
3. A specific wavelength of light is shone onto element
4. Remaining light goes into monochromator and I have no idea what the purpose of this monochromator is
5. A detector counts the light left. And by taking this away from the original, we know how much has been absorbed and thus can find out WHAT the element is and also.. apparently how much - how the heck do they now how much??
THANK YOU!!
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Your very own thread eh? ;)
In the flame the solution that the metal ions are in is evaporated and the ions become atomised, so reduction occurs for example. Na+(aq) + e- -> Na(g) I never really understood how it becomes vaporised in the gaseous state, BUT the function is to allow the atoms to maximise their exposure to the characteristic electromagnetic radiation, think about a block of metal or little pieces and then shine a light, the little pieces will more accurately absorb what's being emitted. Have a look on chem guide uk for explanations on everything! :D
Mono(one)chromator(colour) it selects one wavelength, usually the strongest one possible for accuracy
You can't find out what you have (if that's what you were mentioning not sure?) because you need to know what's in the solution to put in the right cathode lamp made of the same metal!!
You construct a calibration curve (it's pretty much a linear relation, but deal with curve for now) with absorption vs. concentration. You put known standard solutions of that metal ion in and record the absorbance, from there you can plot the corresponding points on your graph (the absorbance and concentration_, then you put in your unknown sample, you get an absorbance reading and from there you can see the resulting concentration. The calibration line establishes the relationship between absorbance and concentration, they're unrelated variables until you connect them like this :) drawing up a calibration curve, or reading one is key for most spectroscopic techniques ;) You'll see heaps of these in the textbook :)
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Your very own thread eh? ;)
In the flame the solution that the metal ions are in is evaporated and the ions become atomised, so reduction occurs for example. Na+(aq) + e- -> Na(g) I never really understood how it becomes vaporised in the gaseous state, BUT the function is to allow the atoms to maximise their exposure to the characteristic electromagnetic radiation, think about a block of metal or little pieces and then shine a light, the little pieces will more accurately absorb what's being emitted. Have a look on chem guide uk for explanations on everything! :D
Mono(one)chromator(colour) it selects one wavelength, usually the strongest one possible for accuracy
You can't find out what you have (if that's what you were mentioning not sure?) because you need to know what's in the solution to put in the right cathode lamp made of the same metal!!
You construct a calibration curve (it's pretty much a linear relation, but deal with curve for now) with absorption vs. concentration. You put known standard solutions of that metal ion in and record the absorbance, from there you can plot the corresponding points on your graph (the absorbance and concentration_, then you put in your unknown sample, you get an absorbance reading and from there you can see the resulting concentration. The calibration line establishes the relationship between absorbance and concentration, they're unrelated variables until you connect them like this :) drawing up a calibration curve, or reading one is key for most spectroscopic techniques ;) You'll see heaps of these in the textbook :)
Thank you so much Edward !! :D
I've understood everything you have said! Now with that all out of the way, a little more questions have popped up unfortunately :(. My first question is that I see how we can use this method quantitatively, but how can we use it qualitatively? Also, what's the point of using and learning flame tests + AES when we have AAS, which is basically 100x better than both combined.
Thanks again Ed! :)
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Well I don't really think it is qualitative?? I mean to use it, you have to have the cathode lamp fitted with the same metal you want to analyse, unless you want to put a solution in there and see a green flame from Cu2+ ions haha. I've never really considered it qualitatively due to the specificity of the technique.
Well AES is a glorified flame test to start with :P this is just basic qualitative technique, you'd use to it try and guess/observe the flame colour so if you saw green fire for a sec then you'd go OK there's Cu2+ probably, so you fit the AAS with the copper lamp and go from there.
I guess AES is just a cheap qualitative technique then you go to AAS; the expensive, specific quantitative one. :)
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Well I don't really think it is qualitative?? I mean to use it, you have to have the cathode lamp fitted with the same metal you want to analyse, unless you want to put a solution in there and see a green flame from Cu2+ ions haha. I've never really considered it qualitatively due to the specificity of the technique.
Well AES is a glorified flame test to start with :P this is just basic qualitative technique, you'd use to it try and guess/observe the flame colour so if you saw green fire for a sec then you'd go OK there's Cu2+ probably, so you fit the AAS with the copper lamp and go from there.
I guess AES is just a cheap qualitative technique then you go to AAS; the expensive, specific quantitative one. :)
Understood! Pretty stupid question haha! Thank you ;D
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Hey guys,
Just a question on the unit 4 course here, in chapters 18-22 we would learn about the production of ammonia, ethane and etc. Apparently those four chapters are like a detailed study? So we only pick one out of the four and only need to know the one of those chapters for the exam?
Thanks
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Detailed studies have been removed from the exam (from the 2013-2016 study design) but your school will still have a SAC on it. Your teacher will probably tell you which one your class will be studying as the SAC will be on that.
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Detailed studies have been removed from the exam (from the 2013-2016 study design) but your school will still have a SAC on it. Your teacher will probably tell you which one your class will be studying as the SAC will be on that.
So we don't necessarily need to know a single chapter from 18-22 for the exam? But will need to study one for a SAC during school? That's pretty good!
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So we don't necessarily need to know a single chapter from 18-22 for the exam? But will need to study one for a SAC during school? That's pretty good!
Yeah you can look past those chapters :) Forget ethene completely, it's not existent on the study design now, you'll either study sulfuric or nitric acid or ammonia production. If you study sulfuric, contact me and I can send you the questions I had on my SAC to practice ;) ALSO try and get your hands on checkpoints 2010,11,12 [any of those unit 4 ones] (not 2013-14) it has a whole section in the Unit 4 area for these detailed studies (question banks) that the later combined 3/4 versions from 2013 don't have..
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Yeah you can look past those chapters :) Forget ethene completely, it's not existent on the study design now, you'll either study sulfuric or nitric acid or ammonia production. If you study sulfuric, contact me and I can send you the questions I had on my SAC to practice ;) ALSO try and get your hands on checkpoints 2010,11,12 [any of those unit 4 ones] (not 2013-14) it has a whole section in the Unit 4 area for these detailed studies (question banks) that the later combined 3/4 versions from 2013 don't have..
Awesome! Thank you :D!
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Hey guys, can I please get some help on UV-visible spectrometry;
So I'm just going to jot down some stuff I know, please let me know if any of my explanations are incorrect, and please add any other theory that I have not included.
I am finding this part of chemistry very hard to get my head around, and extremely boring.
Okay, so UV-visible spectrometry is one of six types of spectrometry techniques we learn in the chemistry course. It can be used for qualitative analysis, but mainly quantitative analysis. The first thing I was thinking of before reading 'mainly used for quantitative analysis', I was just like 'for f*** sake, why can't we just use AAS then, why do we have to learn this useless stuff?' - But the reason why we use UV instead of AAS because apparently an AAS machine costs a million while a UV costs 600 dollars, so that's one difference, and the other because AAS cannot analyse samples that are organic and non-metal, but a UV-visible can.
In simple steps, when a scientist is to use a UV-visible spectrometer, he would first analyse the sample to find out what it is, and also what colour and wavelength of light the sample best absorbs. He would then, by using many samples create a absorbance versus concentration curve, which the scientist can then use to find the unknown. Beer's law (A=ecl) can also be used in this situation.
In complex steps, when a scientist is to use a UV-visible spectrometer, he shines the light source into a monochromator, the monochromator chooses the best wavelength and colour of light to go into the sample solution. The absorbance of the light/wavelength is then measured my a detector which records the results.
Ok so that's all I know. I wasn't able to incorporate what a 'reference cell' is into it. And it felt terrible when I was typing it up, I didn't use my notes/text book but I had to really take my time in writing it up.
Thanks everyone.
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A reference cell is basically the cell that is being analysed, except without the actual analyte. So this includes the glass test tube and whatever solvent was used to dissolve the organic analyte, but NONE of the analyte itself. The chemist usually runs the reference cell through the spectrometer before the analysis of the actual analyte so that the absorbance of the glassware and the solvent can be recorded. Then the computer does some weird stuff with this knowledge, so that when you actually analyse the analyte, the absorbance readings would be solely attributed to the absorbance of the substance you are analysing!
So basically, the purpose of the reference cell is to factor in and remove any interferences from the testtube and the solvent in which the organic analyte being held in :) We call these undesirable interferences - background interferences.
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when a scientist is to use a UV-visible spectrometer, he would first analyse the sample to find out what it is, and also what colour and wavelength of light the sample best absorbs.
You need to know how he finds out the identity of the chemical! Basically, the spectrometer scans the substance with the entire UV- Visible spectrum of light, producing a characteristic graph of Absorbance vs Wavelength unique to the compound. The graph looks something like this http://commons.wikimedia.org/wiki/File:UV-vis_spectrum_of_host-guest_binding.jpg
The computer can then match the graph to its database of graphs and match the shape to its couterpart molecule.
Also, through this graph, you will be able to find out which wavelength of light (eg. colour - both the same thing) the compound absorbs best. This would be the x-value of the maximum turning points on the graph I linked you above.
Hope that helps :)
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A reference cell is basically the cell that is being analysed, except without the actual analyte. So this includes the glass test tube and whatever solvent was used to dissolve the organic analyte, but NONE of the analyte itself. The chemist usually runs the reference cell through the spectrometer before the analysis of the actual analyte so that the absorbance of the glassware and the solvent can be recorded. Then the computer does some weird stuff with this knowledge, so that when you actually analyse the analyte, the absorbance readings would be solely attributed to the absorbance of the substance you are analysing!
So basically, the purpose of the reference cell is to factor in and remove any interferences from the testtube and the solvent in which the organic analyte being held in :) We call these undesirable interferences - background interferences.
Understood!!! You are really good at explaining stuff :D, thank you so much!
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Understood!!! You are really good at explaining stuff :D, thank you so much!
I do try.. X-) haha jks, thanks! I appreciate it :)
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You need to know how he finds out the identity of the chemical! Basically, the spectrometer scans the substance with the entire UV- Visible spectrum of light, producing a characteristic graph of Absorbance vs Wavelength unique to the compound. The graph looks something like this http://commons.wikimedia.org/wiki/File:UV-vis_spectrum_of_host-guest_binding.jpg
The computer can then match the graph to its database of graphs and match the shape to its couterpart molecule.
Also, through this graph, you will be able to find out which wavelength of light (eg. colour - both the same thing) the compound absorbs best. This would be the x-value of the maximum turning points on the graph I linked you above.
Hope that helps :)
Damn I missed this post! Just got to read it!
Rightyo, so to sum up:
Reference cell:
A reference cell is used during UV-visible spectroscopy and infra-red spectroscopy. They are ultimately used in order to bring 100% accuracy of the analyte's absorption of the energy, either UV light or infra red radiation. A scientist can do this by first shining the light into the reference cell, which does not contain the analyte, the computer can then determine any disturbances that absorb the light as well as how much the cell absorbs. Then when calculating, the computer can take this away from the normal results so ONLY the analyte's absorbance is recorded.
Identifying a substance via UV spectroscopy:
In order to qualitatively determine the substance via UV spectroscopy, the scientist must expose all wavelengths and colours of UV light in the electromagnetic spectrum and then record it in a absorbance verses wavelength graph. He then can compare this to his database of graphs and determine what the substance is. He can do this because indentical substances will always have the same absorbances to the same wavelengths of lights.
Are you happy with that?
Thanks again
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Preeecisely :) Good work!
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So I was reading about some uses of infra-red spectroscopy, and one of them being that is is used for forensic science. Apparently in crime scenes they analyse carpet and find the molecular structure of the fibre, and I do understand how they can do that. But how the hell would that solve anything? It's not like the murderer/rapist had brought in carpet from his house to use during the crime? Have I missed something?
Thanks
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lols, ignore all the other crap about it is used in CSI lab or how it is used to convict a murderer XD, it will NOT be tested by VCAA obviously. Just focus on how IR works. Infra-red can be used to qualitatively determine the identity of a molecule and that has many uses in identifying organic compounds!
Btw, just a heads up - forensic investigations was removed from the course last year so you don't have to waste your time studying it in the textbook :P
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Hey guys,
So I'm just having some trouble with NMR. I know how to analyze and apply it, I know how to get the shift, split, size and environment, I know the functions of each and finally I know how to interpret a HNMR graph to find out the complete structure of a biomolecule. I also know how to interpret a CNMR graph, know the instrumentation well and have got all questions in the textbook correct. Is this all I have to know about it?
I have been struggling with the theory. I have been spending loads of time trying to understand all the processes of the NMR and how it actually works. Need some help. To be specific, I don't understand the spin stuff. So yeah, an atom has a spin and different molecules have different intakes of radiowave energy levels to change that spin. But how does that help us achieve the NMR graph? I don't understand the changes in magnetic fields, and overall how all this spin and energy stuff brings out the graph. Finally I don't understand the TMS, PB thought me what a reference cell is, and apparently a TMS acts as a reference cell, but how? And can someone please help me visualise what's actually happening when an atoms's particles 'spin', or 'change spin' or achieve magnetic fields.
Thank you
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Nobody was able to answer my question :'(
Well I have another, it's fairly straight forward.
How do you know what energy is required to stretch a bond? So for example HCL and HBR would both need different energies to stretch the bond. But how would you know which would need more energy, and which would need less.
Thank you.
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Hey Rod,
Apologies for not answering your previous question from yesterday, because, to be honest, I am not confident in answering it! I think you are going to have to wait for one of the 50 Chem boyz (lxznl, clippy, Edward21, thushan etc.) to give a crack at your question. Maybe PM them?
As for the HCl/HBr one however, I know for certain that less energy is required to stretch the bonds between heavy atoms. I know it seems slightly counter intuitive that less energy is required to move heavier objects, but be careful to take note that it is the bonds that infra-red radiation stretches - it doesn't move the atoms itself!.
So, in this case less energy would be required to stretch the bonds of? Yeah, you guessed it...HBr. (Br being heavier than Cl)
Regards,
PB
P.S. Apologies about the prev question I can only answer what I know :P + I don't want to mislead you.
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Hey guys,
So I'm just having some trouble with NMR. I know how to analyze and apply it, I know how to get the shift, split, size and environment, I know the functions of each and finally I know how to interpret a HNMR graph to find out the complete structure of a biomolecule. I also know how to interpret a CNMR graph, know the instrumentation well and have got all questions in the textbook correct. Is this all I have to know about it?
I have been struggling with the theory. I have been spending loads of time trying to understand all the processes of the NMR and how it actually works. Need some help. To be specific, I don't understand the spin stuff. So yeah, an atom has a spin and different molecules have different intakes of radiowave energy levels to change that spin. But how does that help us achieve the NMR graph? I don't understand the changes in magnetic fields, and overall how all this spin and energy stuff brings out the graph. Finally I don't understand the TMS, PB thought me what a reference cell is, and apparently a TMS acts as a reference cell, but how? And can someone please help me visualise what's actually happening when an atoms's particles 'spin', or 'change spin' or achieve magnetic fields.
Thank you
To be more specific, each nucleon has their own spin. In something with an even number of nucleons, the spins cancel each other out, so the nucleus doesn't really respond to a magnetic field and thus won't show up. This is why deuterium (hydrogen with a neutron as well) is used in substances that you don't want to show up in the NMR spectrum.
However, in something with an odd number of nucleons, the spins do not cancel. There is then a net magnetic moment (think of this as an attribute of the nucleus to interact with a magnetic field; the first case had a zero magnetic moment as all the spins cancelled so it doesn't interact with a magnetic field). In a magnetic field, this moment would tend to align with the direction of the field which is of lower energy. However, energy of the right amount is able to flip the alignment of the magnetic moment to perfectly oppose the direction of the magnetic field. This energy is what a chemical shift really is.
Nobody was able to answer my question :'(
Well I have another, it's fairly straight forward.
How do you know what energy is required to stretch a bond? So for example HCL and HBR would both need different energies to stretch the bond. But how would you know which would need more energy, and which would need less.
Thank you.
Erm...this is a weird question that you probably won't be asked in an exam. The thing is, there are a few factors affecting bond stretching. Firstly, the HCl bond is stronger than the HBr bond for a few reasons:
1. Chlorine is more electronegative than bromine, so the H-Cl bond is more polarised and there are stronger electrostatic attractions
2. Chlorine is smaller than bromine, so the H-Cl bond length is lower, resulting in a stronger bond
Secondly, bromine is more massive than chlorine. This would imply that more energy would be needed to displace the bromine atom.
Also how much are you stretching these bonds by? The bonds are originally of different lengths to begin with so how would you compare them?
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Erm...this is a weird question that you probably won't be asked in an exam. The thing is, there are a few factors affecting bond stretching. Firstly, the HCl bond is stronger than the HBr bond for a few reasons:
1. Chlorine is more electronegative than bromine, so the H-Cl bond is more polarised and there are stronger electrostatic attractions
2. Chlorine is smaller than bromine, so the H-Cl bond length is lower, resulting in a stronger bond
Secondly, bromine is more massive than chlorine. This would imply that more energy would be needed to displace the bromine atom.
Also how much are you stretching these bonds by? The bonds are originally of different lengths to begin with so how would you compare them?
Huh? Wasn't this question asked in like the 2010 chemistry exam?
Rod is just asking which bond requires more/ less energy to stretch...isn't it a pretty straight forward thing in VCAA terms? aka. heavier atoms = less energy to stretch bonds
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Hey Rod,
Apologies for not answering your previous question from yesterday, because, to be honest, I am not confident in answering it! I think you are going to have to wait for one of the 50 Chem boyz (lxznl, clippy, Edward21, thushan etc.) to give a crack at your question. Maybe PM them?
As for the HCl/HBr one however, I know for certain that less energy is required to stretch the bonds between heavy atoms. I know it seems slightly counter intuitive that less energy is required to move heavier objects, but be careful to take note that it is the bonds that infra-red radiation stretches - it doesn't move the atoms itself!.
So, in this case less energy would be required to stretch the bonds of? Yeah, you guessed it...HBr. (Br being heavier than Cl)
Regards,
PB
P.S. Apologies about the prev question I can only answer what I know :P + I don't want to mislead you.
Thank you so much PB! And don't worry about the previous question ! ;D Really appreciate all the help you give me!
So the sum up, the heavier an atom is the more weakly the bonds are attached. Therefore, the heavier it is the less energy is needed to break the bonds.
To be more specific, each nucleon has their own spin. In something with an even number of nucleons, the spins cancel each other out, so the nucleus doesn't really respond to a magnetic field and thus won't show up. This is why deuterium (hydrogen with a neutron as well) is used in substances that you don't want to show up in the NMR spectrum.
However, in something with an odd number of nucleons, the spins do not cancel. There is then a net magnetic moment (think of this as an attribute of the nucleus to interact with a magnetic field; the first case had a zero magnetic moment as all the spins cancelled so it doesn't interact with a magnetic field). In a magnetic field, this moment would tend to align with the direction of the field which is of lower energy. However, energy of the right amount is able to flip the alignment of the magnetic moment to perfectly oppose the direction of the magnetic field. This energy is what a chemical shift really is.
Erm...this is a weird question that you probably won't be asked in an exam. The thing is, there are a few factors affecting bond stretching. Firstly, the HCl bond is stronger than the HBr bond for a few reasons:
1. Chlorine is more electronegative than bromine, so the H-Cl bond is more polarised and there are stronger electrostatic attractions
2. Chlorine is smaller than bromine, so the H-Cl bond length is lower, resulting in a stronger bond
Secondly, bromine is more massive than chlorine. This would imply that more energy would be needed to displace the bromine atom.
Also how much are you stretching these bonds by? The bonds are originally of different lengths to begin with so how would you compare them?
Thank you so much Lzxnl! So all nucleons have a 'spin'. It just's even there are an even numbers of nucleons, they cancel out and thus no magnetic field is created. However, when there is an odd number of nucleons, the 'spin' is intact and the magnetic moment aligns with a magnetic field, and they tend to align in the lower energy part of the magnetic field. The amount of energy needed to flip over the nucleon in the higher energy direction is pretty much the chemical shift in a NMR graph. I have a much better idea of it now, thank you, I am going to save this for the end of March this year where I am going to learn it in school, I've found this the hardest concept to understand in AOS1.
Okay, so what detemines the strength of a bond is the size of the atom and the electronegativity of the atom. The lower in mass of the atom, and the higher electronegativity the higher amount of energy needed to break the bond. On the other hand when atoms are larger in mass, and lower in electronegativity, lower energy is needed to break the bond.
I'm stretching the bonds via infra red radiation.
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Huh? Wasn't this question asked in like the 2010 chemistry exam?
Rod is just asking which bond requires more/ less energy to stretch...isn't it a pretty straight forward thing in VCAA terms? aka. heavier atoms = less energy to stretch bonds
Well the problem is, there are two parts to bond-stretching. If your system has a higher mass, then the same force results in a lower acceleration. If your system has a stronger bond, it's like having a higher stiffness in a spring; same force results in less extension. It's not that straightforward.
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Well the problem is, there are two parts to bond-stretching. If your system has a higher mass, then the same force results in a lower acceleration. If your system has a stronger bond, it's like having a higher stiffness in a spring; same force results in less extension. It's not that straightforward.
Getting even more confused after reading this. Should I just save this for my teacher when school starts? Isn't even in the study design?
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lzxnl,
please use the airport wifi on something else other than confusing other people...
@Rod : in a sense, yes lzxnl is right. But the thing is that what he is trying to tell you is waaayyy out of the course. All you need to know is what I told you. Trust me :P. I know it sucks - but you really have to suck up to VCE's idea of chemistry, not real chemistry.
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Just stay superficial, stick to what the answers say in the textbook + practice exams and pay attention to the study design, that's my advice for VCE Chemistry. :P
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Just stay superficial, stick to what the answers say in the textbook + practice exams and pay attention to the study design, that's my advice for VCE Chemistry. :P
This is probably why I did relatively well. People post questions here and everyone else can provide in depth answers and I'm sitting here going "NMR that's the one with the lines right?"
Feels like I've already forgotten half the course, probably trying to make room for spesh.
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This is probably why I did relatively well. People post questions here and everyone else can provide in depth answers and I'm sitting here going "NMR that's the one with the lines right?"
Feels like I've already forgotten half the course, probably trying to make room for spesh.
Haha so true! And "relatively well" you smashed it ;) Haha I've been revising the course as I've started tutoring, it's coming back, slowly but surely :P
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lzxnl,
please use the airport wifi on something else other than confusing other people...
@Rod : in a sense, yes lzxnl is right. But the thing is that what he is trying to tell you is waaayyy out of the course. All you need to know is what I told you. Trust me :P. I know it sucks - but you really have to suck up to VCE's idea of chemistry, not real chemistry.
Point is, the question is dodgy. It's an unfair question IMO to ask as the course doesn't even talk much about factors affecting bond strength and stretching. You're better off devoting time and effort to more significant parts of the course.
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Point is, the question is dodgy. It's an unfair question IMO to ask as the course doesn't even talk much about factors affecting bond strength and stretching. You're better off devoting time and effort to more significant parts of the course.
Fair enough, the question was pretty dodgy but I don't like leaving anything out. I just can't start the next chapter knowing there are bits and pieces in the previous that I did not fully understand.
Also, a question similar to this was in the text book, so I kinda thought it was a part of the course.
Well thanks to you and PB I have got my head around it anyway, so let me concentrate on my hydrocarbons now :P. Chapter 10 (reactions of hydrocarbons) looks really hard!
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There are only so many reactions you'll have to go into. Here they are (from my memory)
Substitution of chlorine onto an alkane under UV light (C2H6 + Cl2 => HCl + C2H5 only if there is UV light),
Substitution of OH onto a chloroalkane (NaOH + C2H5Cl => NaCl + C2H5OH )
Addition of a halogen diatomic molecule/water over an alkene
(a. Br2 + CH2=CH2 => BrCH2CH2Br, CH2=CH2 + H2O => CH3CH2OH, only happens with acid catalyst)
Oxidation of an alkanol to an alkanoic acid (CH3CH2OH + O2 => H2O + CH3COOH, happens much faster with a better oxidant like dichromate or permanganate)
Esterification of an alkanol and an alkanoic acid (CH3CH2OH + CH3COOH => CH3CH2OOCCH3, requires acid or base catalyst and VCE only asks for the acid catalyst)
See, it's not that much is it? Someone add to this list if I've missed anything.
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Hey guys just a question here from the AOS revision in the heinemann text book
The amount of alcohol, CH3CH2OH, in your breath can be determined by blowing into a tube containing acidified potassium dichromate K2cr2o. Which substance has been oxidised and which substance has been reduced?
How would I know? I don't know how breath testing works...
Another question,
I've gone through all the organic reactions and in total there is 13. I'm already muddled by this! I can't do questions without looking at my organic reaction cheat sheet that I have made. I'm really want to do chemistry in university, my mate said apparently I'll have to learn 100s of other organic reactions !!! Is this true? :OOOOO
Thanks guys
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ethanol is oxidised and the dichromate ion is reduced. not sure why/how i know that, but it's just something that you need to know.
and yeah, much of chem is about memorisation, unfortunately.
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ethanol is oxidised and the dichromate ion is reduced. not sure why/how i know that, but it's just something that you need to know.
and yeah, much of chem is about memorisation, unfortunately.
All good, I'll try and find out a way to understand these equations though. I hate memorising.
Would I be asked that in a VCAA exam? The breath test question? It provides absolutely no information on what could get reduced or oxidised.
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Yeah, a questions like that in Heinemann can often require knowledge that is in the textbook pages before hand (page 48 to be precise) General chemistry knowledge should have you remember that dichromate is something that is usually reduced to Cr3+. You have probably balanced that half equation like 50 times by now!!
VCAA should give you a little more background info, at least products if not and unbalanced equation.
ethanol is oxidised and the dichromate ion is reduced. not sure why/how i know that, but it's just something that you need to know.
and yeah, much of chem is about memorisation, unfortunately.
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Yeah, a questions like that in Heinemann can often require knowledge that is in the textbook pages before hand (page 48 to be precise) General chemistry knowledge should have you remember that dichromate is something that is usually reduced to Cr3+. You have probably balanced that half equation like 50 times by now!!
VCAA should give you a little more background info, at least products if not and unbalanced equation.
Thank you!!!
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Hey guys what's the difference between amide and amino? Is it that NH is amine and NH2 is amino?
And when do we use 'amino' and 'amine'
Thanks!
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Hey guys what's the difference between amide and amino? Is it that NH is amine and NH2 is amino?
And when do we use 'amino' and 'amine'
Thanks!
First, you mention amide and amino. Then, you switch from "amide" to "amine". There is a difference.
For starters, amide has two meanings. In VCE chemistry, amide means a group where there is a carbon both double-bonded to oxygen and single-bonded to a nitrogen atom. Its other meaning is the NH2- anion, the conjugate base of ammonia (yes that exists; sodium amide, for instance, is a remarkably powerful base as you can imagine). Forget the latter
Now, "amine" and "amino" are equivalent when referring to the functional group. They are used differently when naming though; "amino" is used at the front, like "aminoethanoic acid), while "amine" is used at the back, like "butan-1-amine".
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Didn't understand sorry
So amide = NH
amino = NH2
When naming hydrocarbons, for amides we use;
1. Amino (at the start, so primary)
2. -amine (at the end, secondary)
We use primary ???when??? we use the secondary ??when??
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no, amide is C=ONH, so think of the bond you get when you join two amino acids in vce chemistry
if something has an amine group, and no other functional groups, then it will be in the form of _______-amine.
if the compound has multiple functional groups, theres sort of like, a hierarchy we use when it comes to naming. carboxylic acids are above amines in this hierarchy, so when we name something with an amine group and also a carboxylic acid group, we use "ic acid" at the end, and amino at the front. eg. 3-aminopropanoic acid.
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Hey guys,
I've just been having trouble with states.
When writing organic reactions, I seem to always get the states wrong. Eg, what states do I put in a reaction where an alcohol + carboxylic acid makes an ester? Or when alkene and water makes an alcohol? or when an alkane and bromine make a haloalkane??
All I know is that methane, ethane, are always gases. In alkanes carbons 5-16 are liquids, and above 17 are soft solids.
How do I understand states??
Also , are the reagents really important to in organic reactiosn??
Thanks!!
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Hey guys,
I've just been having trouble with states.
When writing organic reactions, I seem to always get the states wrong. Eg, what states do I put in a reaction where an alcohol + carboxylic acid makes an ester? Or when alkene and water makes an alcohol? or when an alkane and bromine make a haloalkane??
All I know is that methane, ethane, are always gases. In alkanes carbons 5-16 are liquids, and above 17 are soft solids.
How do I understand states??
Also , are the reagents really important to in organic reactiosn??
Thanks!!
Esters are generally liquid; they're not very soluble. Smaller alcohols are aqueous, just like with acids.
When alkenes and water react to form an alcohol...hmm...depends on the reaction conditions. Nowadays the hydration of ethene requires phosphoric acid at 250 degrees as a catalyst, so in that case water and alkene would both be dissolved in phosphoric acid. Alkenes don't react readily with water and need some catalyst.
For alkanes reacting with bromine, bromine is generally a liquid, while the alkane is either liquid or gaseous depending on size. Chlorine is always a gas, while iodine is generally solid.
States are related to the intermolecular forces that hold the molecules together. With increasing molecular mass, there are more electrons available that contribute to dispersion forces (more electrons, greater instantaneous dipoles, greater attractions), so the molecules are held together more strongly and boil/melt at lower temperatures.
Reagents are what cause the reactions to occur. They are just as important as the organic compound that you're trying to change.
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^^ Thanks Lxnl
Here's a question from the prac we did yesterday. How would using solid alcohols and liquid carboxylic acids when making esters affect an experiment? I see that using a solid, more yield will be given off the the alcohol, and less to the carboxylic acid, but how do these 'gross errors' affect the experiment?
Huh? More yield given off the alcohol? I have no idea what you're trying to say there.
If you're trying to react a solid with a liquid, your rate of reaction is going to be lower because not all of the solid can actually react. If the alcohol was dissolved in the acid, more of the alcohol molecules would be available for reaction.
Also, you need to ensure that your liquid carboxylic acid is dissociated enough as the esterification reaction using an alcohol and ester requires an acid catalyst (there are better ways of preparing esters :P)
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Huh? More yield given off the alcohol? I have no idea what you're trying to say there.
If you're trying to react a solid with a liquid, your rate of reaction is going to be lower because not all of the solid can actually react. If the alcohol was dissolved in the acid, more of the alcohol molecules would be available for reaction.
Also, you need to ensure that your liquid carboxylic acid is dissociated enough as the esterification reaction using an alcohol and ester requires an acid catalyst (there are better ways of preparing esters :P)
Thanks man!
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Hey guys just need some help on states for organic reactions, do you guys have any rules or anything that could help me?
This is what I know;
Most alcohols are liquid
Combustion reactions only have gases
Meth,eth,prop usually gases
Anything reaction with aq would usually produce aq
Anything reacting with water usually produces aqueous, btu sometimes liquids
anything non-polar is a gas
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Hey guys just need some help on states for organic reactions, do you guys have any rules or anything that could help me?
This is what I know;
Most alcohols are liquid
Combustion reactions only have gases
Meth,eth,prop usually gases
Anything reaction with aq would usually produce aq
Anything reacting with water usually produces aqueous, btu sometimes liquids
anything non-polar is a gas
Combustion reactions only have gases? I disagree. When burning fuel, which consists of hydrocarbons, the hydrocarbons are in liquid form when burnt. Similarly, sulfur dioxide is the product of burning sulfur, which is in solid form (I know this isn't organic). You just have to know what the states of the individual molecules are.
Meth, eth prop usually gases? Those alcohols are all liquid, propanamine is a liquid and none of the acids are gases (they're liquids). Propyl chloride is also a liquid. If it's non-polar, then the first three members of that homologous series are usually gases.
Aqueous reactions can form anything. Ethanoate ion reacting with chloroethane, all in water, can yield insoluble ethyl ethanoate; a similar product occurs with ethanoic acid and ethanol in acidifed water. Aqueous barium hydroxide reacts with sulfuric acid to form a precipitate of barium sulfate. Also, reacting sodium hydroxide with ammonium chloride, in water, can form gaseous ammonia if the reaction vessel is heated, while dehydrating ethanol in sulfuric acid in water can form gaseous ethene. Only assume that the product is aqueous if the product actually dissolves in water.
With your next one, if water is the solvent, you're generally correct on that.
Not all non-polar things are gases. Hydrocarbons aren't all gases, for instance. Stuff like decane is liquid normally. Similarly, molecular bromine is a liquid, while molecular iodine is a solid. Esters are quite non-polar (don't dissolve in water) but they're liquid.
As for rules, I don't have any hard and fast rules, but I think of them this way.
Most carboxylic acids are liquid (at least, the common ones)
Same with alcohols
Alkyl halides, if smaller, are gaseous, but at around three carbons they become liquid.
Butane is a gas, but pentane and higher are liquids (excluding isomeric effects on boiling points)
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Hi everyone,
Just been stuck on a question for a while ;
2.5 mol of an unsaturated fatty acid was converted to a saturated fatty acid by reaction with 15g of h2 gas in the presence of a catalyst. Calculate the number of carbon to carbon double bonds present in the unsaturated fatty acid.
My thoughts;
So the double bonded, unsaturated fat is obviously going through a hydrogenation reaction to become a single bonded fatty acid.
n(h) used was 15/1 = 15 mol
n(unsaturated fatty acid) = 2.5 mol
n(saturated fatty acid) = we don't know
hmm. maybe divide the total mole of hydrogen with the unsaturated fatty acid --- 15/2.5 = 6...
So maybe 6 hydrogens added in during the hydrogenation reaction, so the fatty acid had 3 double bonds since each broken double bond equates to two hydrogens so 6/2 = 4
Can someone please help me? Even if I am right, can someone please explain to me how I did it lol! The last two sentences was just me randomly mucking around so I don't really know what I did.
Thank you!!
PS - No answers were supplied ;'(
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Hi everyone,
Just been stuck on a question for a while ;
2.5 mol of an unsaturated fatty acid was converted to a saturated fatty acid by reaction with 15g of h2 gas in the presence of a catalyst. Calculate the number of carbon to carbon double bonds present in the unsaturated fatty acid.
My thoughts;
So the double bonded, unsaturated fat is obviously going through a hydrogenation reaction to become a single bonded fatty acid.
n(h) used was 15/1 = 15 mol
n(unsaturated fatty acid) = 2.5 mol
n(saturated fatty acid) = we don't know
hmm. maybe divide the total mole of hydrogen with the unsaturated fatty acid --- 15/2.5 = 6...
So maybe 6 hydrogens added in during the hydrogenation reaction, so the fatty acid had 3 double bonds since each broken double bond equates to two hydrogens so 6/2 = 4
Can someone please help me? Even if I am right, can someone please explain to me how I did it lol! The last two sentences was just me randomly mucking around so I don't really know what I did.
Thank you!!
PS - No answers were supplied ;'(
(for some context) basically this is hydrogenation used in the evil food industry to create those bad trans fats!! :P
To explain try to visualise this.
1 mol of the fatty acid + (x amount) of mole of H2 --> 1 mol of saturated fatty acid
Each mole of H2 corresponds to one double bond, as you need 2 H atoms to saturated the C=C to form C-C.
So do a mole ratio of fatty acid : H2
2.5mol : (15/2)=7.5mol
1:3
Therefore if you need 3 mol of H2, for one fatty acid. It must contain 3 double bonds.
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(for some context) basically this is hydrogenation used in the evil food industry to create those bad trans fats!! :P
To explain try to visualise this.
1 mol of the fatty acid + (x amount) of mole of H2 --> 1 mol of saturated fatty acid
Each mole of H2 corresponds to one double bond, as you need 2 H atoms to saturated the C=C to form C-C.
So do a mole ratio of fatty acid : H2
Thank you so much man :D
2.5mol : (15/2)=7.5mol
1:3
Therefore if you need 3 mol of H2, for one fatty acid. It must contain 3 double bonds.
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Did you quote and forget to reply Rod haha?
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Did you quote and forget to reply Rod haha?
Hahaha didn't realise! But I did write some stuff down!
'Thanks man! really appreciate the help, I fully understand now'!
:)
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To what extent do we need to know about the three structures of DNA?
A lot of biology material in chemistry this year has been removed (according to my teacher) so I'm trying to keep it as chemistry as possible.
Primary:
- Covalent bonds form between the phosphate and the alcohol group of the sugar
- Covalent bonds form between nitrogen base and sugar
- Basically the nitrogenous base sequence makes up the primary structure
Secondary:
- Nitrogen bases form hydrogen bonds
- Adenine forms hydrogen bonds with thymine. Two bonds in total form, where hydrogen forms h bonds with either oxygen or nitrogen
- Cytosine form hydrogen bonds with guanine. Three bonds form in total, when hydrogen forms h bonds with either oxygen or nitrogen
- Double helix forms
Tertiary:
- Due to the negatively charge phosphate on DNA, they can coil around histone proteins
- These histone proteins allow DNA to become 'super coiled'. This allows DNA to be packaged into things like chromosomes
thank you;
My SAC is still more than 2 weeks away but I've predominantly finished all my prep. I still haven't got my results for my other one. Can someone give me some advice on what I should do for the next 2 weeks? I've done lots already.
Rod
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Hi guys,
a compound has the molecular formula C57H100O6
How many double bonds present in this compound? Explain.
What I'm thinking:
2n+1 is the formula for saturated compounds
So 57 x 2 + 1 = 115.
115-100 = 15 hydrogens missing. So 15/2 = 7.5, roughly 7 double bonds should be in this compound. What am I doing wrong here ;\.
Thanks
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isn't it 2n+2?
so 116-100=16 16/2=8?
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isn't it 2n+2?
so 116-100=16 16/2=8?
Nah man I'm pretty sure it's 2n + 1. And the answer is unfortunately not even close to 7, or 8. It's a tricky one ! :(
Edward21, Lxnl, PB, where art thou?
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wait.. is the answer 2?
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wait.. is the answer 2?
No 5 apparently ;\
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Hi guys,
a compound has the molecular formula C57H100O6
How many double bonds present in this compound? Explain.
What I'm thinking:
2n+1 is the formula for saturated compounds
So 57 x 2 + 1 = 115.
115-100 = 15 hydrogens missing. So 15/2 = 7.5, roughly 7 double bonds should be in this compound. What am I doing wrong here ;\.
Thanks
Assuming that there isn't a ring in this thing :P
The oxygens don't contribute to the double bond count. Just consider C57H100. It is known that C57H116 is a saturated alkane (the rule is 2n+2, not 2n+1 which leaves you with alkyl radicals. You can't have an odd number of hydrogens as then you have an odd number of electrons). You have 16 less hydrogens => should be 8 double bonds.
Now I think the answers may have considered this compound as a triglyceride composed of three fatty acids and they're not counting the ester carbonyl C=O bonds (three in total). That's my guess. If so...that's dodgy.
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How exactly does adding concentrated acid on an enzyme denature it?
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Another question here; this one is an urgent one, help appreciated (edward21 / lxnl / pb yess plsss
Okay, so I know that between cytosine and guanine there are three H-bonds, and between adenine and thymine there are 2 h-bonds. A h bond can be formed by hydrogen and O,N,F. I have each nitrogen base in my data booklet so I can just copy it out and draw the bonds. But at what nitrogen, hydrogen and oxygen atoms do the nitrogen bases form h-bonds? Do you guys understand me?
So how do I draw the bonds? Where to I start and at which atom for each nitrogen base? Is there a specific one?
Thanks
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Anyone ? ;\.
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Another question here; this one is an urgent one, help appreciated (edward21 / lxnl / pb yess plsss
Okay, so I know that between cytosine and guanine there are three H-bonds, and between adenine and thymine there are 2 h-bonds. A h bond can be formed by hydrogen and O,N,F. I have each nitrogen base in my data booklet so I can just copy it out and draw the bonds. But at what nitrogen, hydrogen and oxygen atoms do the nitrogen bases form h-bonds? Do you guys understand me?
So how do I draw the bonds? Where to I start and at which atom for each nitrogen base? Is there a specific one?
Thanks
Let's have a look at what happens with cytosine. The NH bonds to the deoxyribose sugar ring. Now, we have an amino group, a carbonyl and another nitrogen left. The hydrogens in the amino group as well as the oxygen on the carbonyl and the remaining nitrogen are able to hydrogen bond with the NH2 and the NH in guanine as well as the C=O in guanine.
Now let's look at adenine. The amino group (one of the hydrogens) as well as one of the nitrogens in the larger ring is able to hydrogen bond. In thymine, the C=O oxygen bonds with the hydrogen in the amino group while one of the N-H hydrogens bonds with one of the nitrogens in the adenine ring. The nitrogens with a hydrogen attached in thymine are chemically unable to participate in hydrogen bonding for reasons beyond the VCE chemistry course.
How exactly does adding concentrated acid on an enzyme denature it?
Enzyme tertiary structures depend in part on ionic interactions between NH3+ and COO- on amino acid side chains. If you add concentrated acid, the COO- groups are protonated, so the attractions disappear. Even worse, if you protonate extra NH2 groups, you may form repulsions between NH3+ groups. Overall, these repulsions will cause the finely designed active site of the enzyme to disappear, rendering it unable to fulfill its function.
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Hey guys,
can someone please clear up my redox terms;
Oxidation: Loss of electrons
Reduction: Gain of electrons
Oxidised: Has lost electrons
Reduced: Has gained electrons
Oxidant: Gains electrons from another substance and increases oxidation number of previous substance
Reductant: Loses electrons from another substances and reduces oxidation number
Thanks
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Oxidation - loss of electrons
Oxidised - compound that undergoes oxidation and loses electrons
Reductant - compound that loses electrons
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Reduction - gain of electrons
Reduced - compound that undergoes reduction and gains electrons
Oxidant - compound that gains electrons
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I think I've got it now. Got so confused argh.
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All the above is correct Rod. :)
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Need some help here;
So bonds that hold the tertiary structure of a protein are covalent bonds, hydrogen bonds, ion-dipole forces and dispersion forces. Why are covalent bonds apparently the 'most important' bond that makes the tertiary structure?
Thanks
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Covalent bonds, the strongest attractions in a protein, make up the primary structure. Without a primary structure, any discussion of tertiary structure is moot.
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Covalent bonds, the strongest attractions in a protein, make up the primary structure. Without a primary structure, any discussion of tertiary structure is moot.
Thanks heaps man :)
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Covalent bonds, the strongest attractions in a protein, make up the primary structure. Without a primary structure, any discussion of tertiary structure is moot.
Also another question;
I've been getting the starch vs cellulose to make bio ethanol questions wrong.
This is what I know;
In order to make bio ethanol (a biofuel) you need to start off with some simple sugars and then convert it to ethanol (can be done via fermentation). These simple sugars can be broken down from starch and cellulose.
Differences between is obviously starch would be easier to break down as plants use starch as storage, and would break starch down to get glucose for energy. On the other hand cellulose is used as structure by plants, so it would be harder to break down to simple sugars for ethanol production.
Apparently in refineries, all they have to do is add amylase to starch to break it down to glucose, then it can be used for fermentation to make ethanol. Cellulose is apparently much more complex. When cellulose is broken down, other sugar molecules excluding glucose are produced, they are harder to covert to ethanol and so makes the process slower and inefficient.
So obviously, due to efficiency and time starch would be better to use.
Could you please add any other information that I have missed? I've gotten three questions related to this stuff wrong.
Thanks
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Can anyone confirm the stuff I wrote above? ^^
Another question:
Apparently dilute sulphuric acid and magnesium oxide reaction is as follows;
Mgo + 2h+ -> mg2+ + h20
don't really get it, help appreciated,
Rod
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Can anyone confirm the stuff I wrote above? ^^
Another question:
Apparently dilute sulphuric acid and magnesium oxide reaction is as follows;
Mgo + 2h+ -> mg2+ + h20
don't really get it, help appreciated,
Rod
Is this got to do with redox? If yes, this is the half equation. Give me more detail and I'll try to help you.
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Can anyone confirm the stuff I wrote above? ^^
Another question:
Apparently dilute sulphuric acid and magnesium oxide reaction is as follows;
Mgo + 2h+ -> mg2+ + h20
don't really get it, help appreciated,
Rod
MgO is basic oxide. Hence, it is able to dissociate in water and form OH-, OH- + H+ -> H20. Correct me if I'm wrong.
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The oxide ion is a ridiculously powerful base; the pKa of its conjugate acid, hydroxide ion, is like 36 or something.
So oxide is protonated once to form hydroxide, which is itself still a strong base, so that reacts with more acid.
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Me needs some help pls
A haemoglobin has four iron atoms. When analysed, the haemoglobin has 0.33% of iron by mass. What is the molecular mass of a haemoglobin molecule?
Thanks
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Let's consider one mole of haemoglobin. There are four iron atoms, which means 55.8*4 = 223.2 g accounted for. This is 0.33% of the molecule's mass, so 100% of the molecule's mass is how much?
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Let's consider one mole of haemoglobin. There are four iron atoms, which means 55.8*4 = 223.2 g accounted for. This is 0.33% of the molecule's mass, so 100% of the molecule's mass is how much?
So it would be 223.2 x 303
=67636 g ......... I got up to here and was stuck :\. How is this even possible..
Another question Nliu!
What is the exact point of rinsing the burette and pipette with the solutions which they will deliver? And also rinse the volumetric flask and with distilled water?
Thanks again
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So it would be 223.2 x 303
=67636 g ......... I got up to here and was stuck :\. How is this even possible..
Another question Nliu!
What is the exact point of rinsing the burette and pipette with the solutions which they will deliver? And also rinse the volumetric flask and with distilled water?
Thanks again
Firstly you need to rinse them with water to clean the burette and pipette. Then, as there's water of a lower solute concentration than your solution, if you just immediately use that burette or pipette, you'll find that your burette or pipette solution delivered will be slightly diluted by this water.
You rinse all equipment with water to ensure their hygiene. Volumetric flasks, however, are going to have water added to them anyway; it doesn't matter if you rinse them with more water.
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When 1.193g of a hydrocarbon compoound was completely burnt in pure oxygen to produce 4.039 g of c02 and 0.826 g of water. What is the empirical formula?
It's a multi choice and I have no idea.
Know I need to find the mole of the hydrocarbon. Could start of finding mole of c02 and water, but then what do I do?
thanks
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When 1.193g of a hydrocarbon compoound was completely burnt in pure oxygen to produce 4.039 g of c02 and 0.826 g of water. What is the empirical formula?
It's a multi choice and I have no idea.
Know I need to find the mole of the hydrocarbon. Could start of finding mole of c02 and water, but then what do I do?
thanks
I got the answer CH. If that's right, let me know and I'll show you what I did. If it's wrong, I'll try again cause I was unsure of one step.
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I got the answer CH. If that's right, let me know and I'll show you what I did. If it's wrong, I'll try again cause I was unsure of one step.
My answer was CH aswell.
m(CO2)=4.039g
n(CO2)=0.0918mol
n(C) = 0.0918mol
m(H2O)= 0.826g
n(H2O)=0.0458mol
n(H)=0.0918mol
Empirical Formula
C H
0.0918/0.0918 : 0.0918/0.0918
1:1
.:. CH is the formula
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My answer was CH aswell.
m(CO2)=4.039g
n(CO2)=0.0918mol
n(C) = 0.0918mol
m(H2O)= 0.826g
n(H2O)=0.0458mol
n(H)=0.0918mol
Empirical Formula
C H
0.0918/0.0918 : 0.0918/0.0918
1:1
.:. CH is the formula
Ohh I get it, thanks. Ill check if it's right and let you guys know soon, but it looks as if it's the only way. We didn't use the mass of the hydrocarbon but ;\.
Thanks Rishi and Blondie :)
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Ohh I get it, thanks. Ill check if it's right and let you guys know soon, but it looks as if it's the only way. We didn't use the mass of the hydrocarbon but ;\.
Thanks Rishi and Blondie :)
n(c)=0.0918 x12
m(c) = 1.10g
m(h)= 0.0918g
1.10 + 0.0918 = 1.19g
.. Which is the mass of the hydrocarbon! ;)
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n(c)=0.0918 x12
m(c) = 1.10g
m(h)= 0.0918g
1.10 + 0.0918 = 1.19g
.. Which is the mass of the hydrocarbon! ;)
Yay! thanks heaps guys :)
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Hey guys I've got a couple of more mole questions, sorry :(
Someone precipitated the nickel present as a compound Ni(C4H7N2O2)2. A sample of ore was 4.248 g, it yielded 1.525g of precicpitate, what is the nickel content of this ore expressed as a percentage by mass?
What I'm thinking:
Mole of ore: 4.248/288.7 =0.0147
Mole of Ni: 0.0147
Precipitate = 1.525g, n(Ni) = 0.0147.. Not sure what to do next.
CaC03 ---------> CaO + CO2
a 1.839g sample of mixture of calcium carbonate and calcium oxide was heated and a mass loss of 0.572 g was recorded. What is the percentage by mass of calcium oxide in original sample?
What I'm thinking:
Not sure with this one ;\
And one more:
11g of a hydrocarbon was completely burnt in air and 18g of water was produced. What is the emperical formula of the hydrocarbon?
N(h20= 18/18 = 1
n(h) = 2x1 = 2 mol
m(h) = 2x1 = 2
so m(carbon of hydrocarbon) = 11-2 = 9
9/12 = 0.75 mol
so n(c) = 0.75
n (h) = 2
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Hey guys I've got a couple of more mole questions, sorry :(
Someone precipitated the nickel present as a compound Ni(C4H7N2O2)2. A sample of ore was 4.248 g, it yielded 1.525g of precicpitate, what is the nickel content of this ore expressed as a percentage by mass?
What I'm thinking:
Mole of ore: 4.248/288.7 =0.0147
Mole of Ni: 0.0147
Precipitate = 1.525g, n(Ni) = 0.0147.. Not sure what to do next.
Hey Rod
First of all, don't be sorry about asking questions. It's good that you are clarifying things you are unsure about
Here's what I would and if I'm wrong, sorry
n(Ore) =0.0147
n( Nickel) 0.0147 (Great start)
m( Nickel) = 0.0147 x 58.7
= 0.86289 g
(% of nickel) = 0.86289 / 1.525
= 0.5658295082 x 100 = 56.58 %
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CaC03 ---------> CaO + CO2
a 1.839g sample of mixture of calcium carbonate and calcium oxide was heated and a mass loss of 0.572 g was recorded. What is the percentage by mass of calcium oxide in original sample?[/b]
What I'm thinking:
Not sure with this one ;\
Having a complete guess with this one:
Since calcium carbonate + calcium oxide= 1.839
Heating it would get rid of the water including the oxygen. 1.839-0.572 = 1.267g
1.267/1.839 x 100 = 68.9 %
This is a complete guess as even I'm not sure why I did certain steps :-\ It would be a easter miracle if I got it right. lol
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N(h20= 18/18 = 1
n(h) = 2x1 = 2 mol
m(h) = 2x1 = 2
so m(carbon of hydrocarbon) = 11-2 = 9
9/12 = 0.75 mol
so n(c) = 0.75
n (h) = 2
Good start. First of all, is the answer C24H4O ? The answer doesn't look right so let me know if it's not and I'll try again
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Hey Rod
First of all, don't be sorry about asking questions. It's good that you are clarifying things you are unsure about
Here's what I would and if I'm wrong, sorry
n(Ore) =0.0147
n( Nickel) 0.0147 (Great start)
m( Nickel) = 0.0147 x 58.7
= 0.86289 g
(% of nickel) = 0.86289 / 1.525
= 0.5658295082 x 100 = 56.58 %
[EDITED]
Just realise I need to do all the calculations properly, Ni(C4H7N2O2)2 is the precipitate, hence the mole of Nickel is 5.282x10^(-3) mole. Mass of Nickel = 0.3100 (g). Percentage by mass 0.3100/4.248 x 100 =7.299%.
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I don't think you should divide the mass of nickel to the mass of precipitate, cause they are asking for percentage by mass in the ore sample. hence mass of 0.86289 g of nickel in ore is roughly around 20.31%.
Oh really? I always thought percentage by mass was dividing by the total amount. Also, why would they give you that extra info if you don't need it?
Ah well, we'll just wait for Rod's answer
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Hey guys, thanks for all your help.
The answer to the first question is 7.3%, second question is 29.3% and third C3H8
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[EDITED]
Just realise I need to do all the calculations properly, Ni(C4H7N2O2)2 is the precipitate, hence the mole of Nickel is 5.282x10^(-3) mole. Mass of Nickel = 0.3100 (g). Percentage by mass 0.3100/4.248 x 100 =0.07299%.
So is this what you did?
n(Ni(C4H7N2O2)2 ) = 0.0052823
n(Ni) = 0.005283
m(Ni) = 0.005283 x 58.7 = 0.31 g
Hence 0.31/4.248 x 100 = 7.3 % And that's correct. Thanks pal.
Now, how about the other two ? ;)
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[EDITED]
Just realise I need to do all the calculations properly, Ni(C4H7N2O2)2 is the precipitate, hence the mole of Nickel is 5.282x10^(-3) mole. Mass of Nickel = 0.3100 (g). Percentage by mass 0.3100/4.248 x 100 =7.299%.
oh yeah...makes more sense now. Sorry for the wrong answer Rod and nhmn0301 :(
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oh yeah...makes more sense now. Sorry for the wrong answer Rod and nhmn0301 :(
All good Rishi! Thanks heaps for trying :), you got us there. Lets try and figure out the other two!
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Question 2:
So because this is a mixture of both CaCO3 and CaO and if you look at the equation, there is CO2 presents, which is a gas. When you get this compound up, CO2 will escape, which accounts for the mass loss. Hence, 0.572 is the mass of CO2, hence mole of CO2 is 0.013 mole. Now the Carbon in CO2 would give us the mole of CaCO3, hence the mass of CaCO3 is 1.3013g. Substrate mass of CaCO3 from the sample we have the mass of CaO, which is 0.5277g. Hence, percentage by mass of CaCO3 is 29.3%
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Let me have a shot at the other two, let's try and figure them out.
2)
Still have no clue.
3)
n(h2O) = 18/18 = 1 mol
n(h) = 1x2 = 2 mol
m(hydrocarbon) = 11g
m(carbon in hydrocarbon) = 11 - 2x1 = 9g
n (carbon) = 9/12 = 0.75
C: 0.75/0.75 H: 2/0.75
= 1 : 2.67
x3 C3H8
Yay!
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Question 2:
So because this is a mixture of both CaCO3 and CaO and if you look at the equation, there is CO2 presents, which is a gas. When you get this compound up, CO2 will escape, which accounts for the mass loss. Hence, 0.572 is the mass of CO2, hence mole of CO2 is 0.013 mole. Now the Carbon in CO2 would give us the mole of CaCO3, hence the mass of CaCO3 is 1.3013g. Substrate mass of CaCO3 from the sample we have the mass of CaO, which is 0.5277g. Hence, percentage by mass of CaCO3 is 29.3%
Thanks man, I've just figured out question three if you want to take a look at it
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Thanks man, I've just figured out question three if you want to take a look at it
same !!!! Yay, first success of the night :)
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Question 2:
So because this is a mixture of both CaCO3 and CaO and if you look at the equation, there is CO2 presents, which is a gas. When you get this compound up, CO2 will escape, which accounts for the mass loss. Hence, 0.572 is the mass of CO2, hence mole of CO2 is 0.013 mole. Now the Carbon in CO2 would give us the mole of CaCO3, hence the mass of CaCO3 is 1.3013g. Substrate mass of CaCO3 from the sample we have the mass of CaO, which is 0.5277g. Hence, percentage by mass of CaCO3 is 29.3%
So:
Assume that 0.572 g was lost to CO2
n(c02) = 0.572/44 = 0.013 mol
n(c) = 0.013 mol
And then what did you do? Sorry.
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So:
Assume that 0.572 g was lost to CO2
n(c02) = 0.572/44 = 0.013 mol
n(c) = 0.013 mol
And then what did you do? Sorry.
Sorry for the crappy working out, I was in a rush, you got the mole of Carbon in CO2, because CaCO3 is the only compound in that mixture that contains Carbon, you can assume that this amount of Carbon is from the CaCO3, hence, mole of ( C ) = mole ( CaCO3 ). Hence, mass of CaCO3 = 0.013 = 1.3 (g).
mass of CaO = mass of compound - 1.3
then find percentage by mass
sorry for the confusions.
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Hey guys just confirming that colorimetry is no longer in the course?
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Hey guys just confirming that colorimetry is no longer in the course?
Well it's more of less a cheap version of UV-Vis focussing only on the visible part haha. :P Doesn't hurt to know, but UV-Vis is what will be examined :)
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Well it's more of less a cheap version of UV-Vis focussing only on the visible part haha. :P Doesn't hurt to know, but UV-Vis is what will be examined :)
Thanks pal :)
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Some questions here;
Some questions on NMR spectroscopy;
How come CH3F would have a larger chemical shift than CH3I? Fluorine has a smaller molecular mass than iodine, but how does that relate?
Does NMR essential look at the arrangement of bonding electrons around the nucleus being investigated??
And one more question on mass spec;
I get how ions with smaller masses will deflect more as they pass through a magnetic field, but how come a small charge ratio contributes even more to the deflection?
Thanks :)
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Some questions here;
Some questions on NMR spectroscopy;
How come CH3F would have a larger chemical shift than CH3I? Fluorine has a smaller molecular mass than iodine, but how does that relate?
Thanks :)
It's because fluorine is more electronegative than Iodine. The more electronegative elements have a greater chemical shift so they are often towards the left side of the graph
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Almost. The fluorine atom is highly electronegative yes, which pulls negative charge away from the C atom. In turn, the C atom will tend to pull some negative charge away from the H atoms, deshielding the H atoms. Increased chemical shifts are associated with greater de-shielding.
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Some questions here;
Some questions on NMR spectroscopy;
How come CH3F would have a larger chemical shift than CH3I? Fluorine has a smaller molecular mass than iodine, but how does that relate?
Does NMR essential look at the arrangement of bonding electrons around the nucleus being investigated??
And one more question on mass spec;
I get how ions with smaller masses will deflect more as they pass through a magnetic field, but how come a small charge ratio contributes even more to the deflection?
Thanks :)
In physics, the magnetic force is given by qvB, where q is the charge, v is the speed and B is the magnetic field (only if the field is perpendicular to the direction of motion; don't worry about that here)
So using Newton's second law for uniform circular motion (yes, the magnetic force is always perpendicular to the direction of motion so it satisfies the direction requirement for circular motion), mv^2/r = qvB
r = mv/qB
See how m/q determines the radius of the circular motion?
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Crap, last post 50 days ago! Doesn't vce go fast!!
Having a hard time getting my head around the effects of changing pressure in a reaction;
So apparently, with any change in pressure, the reaction will oppose it by moving to the side with less gas particles. Not sure what it means by 'moving to the side with less gas particles'. Does it mean just move to the side of the reaction with less coefficients?
Don't understand the whole theory of pressure vs reaction
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Crap, last post 50 days ago! Doesn't vce go fast!!
Having a hard time getting my head around the effects of changing pressure in a reaction;
So apparently, with any change in pressure, the reaction will oppose it by moving to the side with less gas particles. Not sure what it means by 'moving to the side with less gas particles'. Does it mean just move to the side of the reaction with less coefficients?
Don't understand the whole theory of pressure vs reaction
So for the pressure factor (it's related to volume factor as well): according to Le..... (bleh, I never know how to spell his name, but whatever).
For eg, if I have PCl3 + Cl2 <=> PCl5
assume constant temperature, if we double the pressure, we halve the volume and hence we have increase the concentration of each gas (amount/volume). Think of it like the negative feed back system (since you did Bio :D), the system will try to partially negate the "stimulus" by partially decrease the concentration of each gas/unit (i.e decrease the pressure). For this particular equation, the RHS only has 1 molecule and the LHS and 2 molecules. As a result, the system responds by favouring the forward reaction in an attempt to partially reduce the concentration of the LHS species until equilibrium is regained.
You can also think of it by using CF value:
CF = [PCl5] / ( [PCl3] x [Cl2] ) = Kc
when you introduce a change (halve volume => double concentration)
CF = 2[PCl5] / ( 2 [PCl3] x 2 [Cl2] ) = 1/2 [PCl5]/[PCl3]x[Cl2] => which is half of the original Kc value, hence CF < Kc. Hence, the system will try to equilibrate by increasing the CF value, by producing more PCl5 at the expense of PCl3 and Cl2. Hence, forward reaction is favoured.
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So for the pressure factor (it's related to volume factor as well): according to Le..... (bleh, I never know how to spell his name, but whatever).
For eg, if I have PCl3 + Cl2 <=> PCl5
assume constant temperature, if we double the pressure, we halve the volume and hence we have increase the concentration of each gas (amount/volume). Think of it like the negative feed back system (since you did Bio :D), the system will try to partially negate the "stimulus" by partially decrease the concentration of each gas/unit (i.e decrease the pressure). For this particular equation, the RHS only has 1 molecule and the LHS and 2 molecules. As a result, the system responds by favouring the forward reaction in an attempt to partially reduce the concentration of the LHS species until equilibrium is regained.
You can also think of it by using CF value:
CF = [PCl5] / ( [PCl3] x [Cl2] ) = Kc
when you introduce a change (halve volume => double concentration)
CF = 2[PCl5] / ( 2 [PCl3] x 2 [Cl2] ) = 1/2 [PCl5]/[PCl3]x[Cl2] => which is half of the original Kc value, hence CF < Kc. Hence, the system will try to equilibrate by increasing the CF value, by producing more PCl5 at the expense of PCl3 and Cl2. Hence, forward reaction is favoured.
Ahhhh, makes much more sense nhmn. Just one more thing, how does decreasing the volume decrease increase concentration for gases?
Thanks :)
Oh also, where is your school up to for chem? :P. We finished unit 3 last week, and we're barely moving through unit 4, getting lazy. Sensing we are going to rush through it all in term 3.
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Oh silly me, we increase the volume, hence dilute it so concentration decreases, whereas if we take out some volume concentration increase. Is that right or am I still wrong? Need to get the basics right
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Or is it that decreasing the pressure results in less collisions so the reaction rate decreases and so does the concentration? But wouldn't everything decrese consequently?
far out aren't I confusing myself?? ahh
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Or is it that decreasing the pressure results in less collisions so the reaction rate decreases and so does the concentration? But wouldn't everything decrese consequently?
far out aren't I confusing myself?? ahh
For this example, you are changing the pressure of GAS PARTICLES. So there's no dilution here. If you have a container with gas particles inside, when you increase the volume of the container, the concentration of gas particle (I.e the amount of gas particle per volume) decreases, hence the PARTIAL pressure of each gas will decrease (they exert less force on the surface of the container). When I say increase or decrease pressure, I mean partial pressure of each gas here. To disrupt equilibrium, the change in pressure must effect the partial pressure of the gases in the reaction equation UNEQUALLY to necessitate a response, not total pressure of the whole thing. For this particular example, there is 2 molecule of gas on LHS and 1 on the RHS, hence, when I use CF calculation from above, you see that the 2 on the numerator and 4 on the denominator end up doesn't cancel down but just simplify each other? Hence, you got 1/2 of Kc. If I have something like
A + B <=> C + D
For this particular example, even if you change the pressure, the equilibrium position won't change since the partial pressure of each gas all increase EQUALLY.
CF= 2 [C] x 2 [D] / 2 [A] x 2 = Kc since 4 and 4 cancels out. Hence, the system won't trigger a response.
Alternatively, think about P= nRT/V. Pressure and volume is inversely proportional.
In short, decrease volume => increase partial pressure => system tries to decrease partial pressure => nett forward reaction is favored in this first example.
Increase volume => decrease partial pressure => system tries to increase partial pressure => nett backward reaction is favored in the first example.
Let me know of there is any errors in this explanation though.
Err.... My school is in the middle of nowhere now -,-. We are having aspirin SAC next week and only about to finish unit 3 right before term 2 ends. Meaning we only have term 3 for unit 4....comparing to where other schools are up to, my school goes the slowest lol.
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Oops sorry, I accidentally hit the bold button on my phone.... Didn't mean holding anything in my explanation.
-
For this example, you are changing the pressure of GAS PARTICLES. So there's no dilution here. If you have a container with gas particles inside, when you increase the volume of the container, the concentration of gas particle (I.e the amount of gas particle per volume) decreases, hence the PARTIAL pressure of each gas will decrease (they exert less force on the surface of the container). When I say increase or decrease pressure, I mean partial pressure of each gas here. To disrupt equilibrium, the change in pressure must effect the partial pressure of the gases in the reaction equation UNEQUALLY to necessitate a response, not total pressure of the whole thing. For this particular example, there is 2 molecule of gas on LHS and 1 on the RHS, hence, when I use CF calculation from above, you see that the 2 on the numerator and 4 on the denominator end up doesn't cancel down but just simplify each other? Hence, you got 1/2 of Kc. If I have something like
A + B <=> C + D
For this particular example, even if you change the pressure, the equilibrium position won't change since the partial pressure of each gas all increase EQUALLY.
CF= 2 [C] x 2 [D] / 2 [A] x 2 = Kc since 4 and 4 cancels out. Hence, the system won't trigger a response.
Alternatively, think about P= nRT/V. Pressure and volume is inversely proportional.
In short, decrease volume => increase partial pressure => system tries to decrease partial pressure => nett forward reaction is favored in this first example.
Increase volume => decrease partial pressure => system tries to increase partial pressure => nett backward reaction is favored in the first example.
Let me know of there is any errors in this explanation though.
Err.... My school is in the middle of nowhere now -,-. We are having aspirin SAC next week and only about to finish unit 3 right before term 2 ends. Meaning we only have term 3 for unit 4....comparing to where other schools are up to, my school goes the slowest lol.
The explanation is pretty good, although the system isn't trying to increase the partial pressure; the change is the number of gas particles per unit volume and the system is trying to increase that. Also, increasing the volume is a dilution; your first line is slightly off.
Or is it that decreasing the pressure results in less collisions so the reaction rate decreases and so does the concentration? But wouldn't everything decrese consequently?
far out aren't I confusing myself?? ahh
The reason why you're confusing yourself here is that yes, increasing the volume and decreasing the pressure will decrease reaction rates. However, it affects BOTH the forward and backward reactions; it affects them different in different scenarios.
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http://www.vtextbook.com/?p=watch&subject=5&unit=4&area=15&topic=39&video=1544069
Rod - for further understanding, check the video above; if you think you have a general grasp of equilibrium and Le Chatelier's Principle, then watch from 10.25 to help answer your question, but if you think you don't understand the concept at all, then watch the whole video (~30 minutes).
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http://www.vtextbook.com/?p=watch&subject=5&unit=4&area=15&topic=39&video=1544069
Rod - for further understanding, check the video above; if you think you have a general grasp of equilibrium and Le Chatelier's Principle, then watch from 10.25 to help answer your question, but if you think you don't understand the concept at all, then watch the whole video (~30 minutes).
Thanks so much Thushan!
Big fan of your and edward21's chemistry videos by the way ! :)
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Thanks Lxnl and nhmnhm :)
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Hey everyone,
Why do we assume, in acid base equilibria, that the concentration of hydronium ions is the same as the concentration of the conjugate base produced in all acid-base reactions?
thanks
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n = CV
Volume is the same, so,
n = C
So, you're actually using the mole ratio. If the stoichiometry is 1:1, you assume the concentration is the same. In all of those acid-water reactions the stoichiometry will be 1:1, so you can assume the concentrations of the hydronium with the conjugate base will be the same.
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n = CV
Volume is the same, so,
n = C
So, you're actually using the mole ratio. If the stoichiometry is 1:1, you assume the concentration is the same. In all of those acid-water reactions the stoichiometry will be 1:1, so you can assume the concentrations of the hydronium with the conjugate base will be the same.
Ahh I see, thanks buddy :)
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Hey everyone, just want to confirm this, I've been hearing different things from different people
application of equilibrium and rate principles to the industrial production of one of ammonia,
sulfuric acid, nitric acid:
factors affecting the production of the selected chemical
waste management including generation, treatment and reduction
health and safety considerations
uses of the selected chemical.
That's a point from the study design. Do we need to know this for just our SAC? Or both SAC and exam?
It's not explicit in the study design and my teacher has said we don't need to know it for the exam
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Oh also another question here; can someone please correct me:
An increase in pressure results in a decrease in volume. Hence there is less space for the particles, resulting in more collisions, increasing the rate of reaction. On the other hand, a decrease in pressure results in an increase in volume, hence there is more space for particles, resulting in less collisions. This decreases the overall reaction rate.
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Your studied chemical will NOT appear in the exam - don't stress. Once the SAC is gone, throw out the notes without fear. ;) But don't - the notes are useful to hold onto as possible examples for applications of equilibria.
You're close - I'm not sure that volume is actually related to collision theory. It's all about two things:
- orientation
- kinetic energy
Let's refer to the ideal gas law, PV = nRT. You see, when we increase pressure, we'll get a resulting increase in temperature. This means we'll see an increase in kinetic energy, and that's what's giving us more collisions, not our decrease in volume. That's what's increasing the rate of reaction, and vice-versa for a decrease in pressure.
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Your studied chemical will NOT appear in the exam - don't stress. Once the SAC is gone, throw out the notes without fear. ;) But don't - the notes are useful to hold onto as possible examples for applications of equilibria.
You're close - I'm not sure that volume is actually related to collision theory. It's all about two things:
- orientation
- kinetic energy
Let's refer to the ideal gas law, PV = nRT. You see, when we increase pressure, we'll get a resulting increase in temperature. This means we'll see an increase in kinetic energy, and that's what's giving us more collisions, not our decrease in volume. That's what's increasing the rate of reaction, and vice-versa for a decrease in pressure.
Understood! thanks pal :)
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Let's refer to the ideal gas law, PV = nRT. You see, when we increase pressure, we'll get a resulting increase in temperature. This means we'll see an increase in kinetic energy, and that's what's giving us more collisions, not our decrease in volume. That's what's increasing the rate of reaction, and vice-versa for a decrease in pressure.
Amazing explanation.
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Thanks, hahah. I was absolutely horrible at 3/4 chem, and I've rapidly picked up my game in uni. It's only when you suck at something and become good at it that you can explain it to others.
(welp, if you're good to begin with you can still teach people, but I digress... :P)
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Thanks, hahah. I was absolutely horrible at 3/4 chem, and I've rapidly picked up my game in uni. It's only when you suck at something and become good at it that you can explain it to others.
(welp, if you're good to begin with you can still teach people, but I digress... :P)
*ahem
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I added an adamendendum. I love our smart peoples, too. <3
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Hey everyone
Can someone please explain why catalysts have no effect at all at equilibrium? thanks
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Catalysts work by reducing the activation energy of a reaction. However, reducing the activation energy of the forward reaction also decreases the activation energy of the reverse reaction. As a result, both forwards and backwards reaction rates are sped up.
Or, if you want, a catalyst isn't mentioned in the equilibrium constant expression so its concentration shouldn't affect the equilibrium (a bit hand-wavy but hopefully it helps you remember)
Finally, catalysts aren't consumed in the reaction. Imagine you add a catalyst to an equilibrium mixture. The amount of catalyst hasn't changed. Therefore, there is no change to the system. Do you expect the equilibrium to shift?
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Catalysts work by reducing the activation energy of a reaction. However, reducing the activation energy of the forward reaction also decreases the activation energy of the reverse reaction. As a result, both forwards and backwards reaction rates are sped up.
Or, if you want, a catalyst isn't mentioned in the equilibrium constant expression so its concentration shouldn't affect the equilibrium (a bit hand-wavy but hopefully it helps you remember)
Finally, catalysts aren't consumed in the reaction. Imagine you add a catalyst to an equilibrium mixture. The amount of catalyst hasn't changed. Therefore, there is no change to the system. Do you expect the equilibrium to shift?
thanks lzxnl :)
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When nitric acid acts as an oxidant, it oxidises another species and is itself reduced. As a result of being reduced there will be a decrease in the oxidation number of nitrogen. In HNO 3 the oxidation number of N is +5 In NO 3 -, the oxidation number of N is +5 In N 2, the oxidation number of N is 0 In NO, the oxidation number of N is +2 In NO 2, the oxidation number of N is +4 Hence NO 3 - will not be produced when HNO3 acts as an oxidant.
Not following. If HN03 is to be reduced, then why does whatever HNO3 oxidises have to have nitrogen with less oxidation number than the nitrogen in HNO3?
Thanks
2;AM I know this is pissing me off
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The only reason for that is that the maximal oxidation state of nitrogen is +5. Hence, if HNO3 were to oxidise another nitrogen-based species (eg NO, N2, NH3), then that species needs to be able to be oxidised. This other species cannot have an oxidation number of +5, because it cannot be oxidised to attain a higher oxidation number. Therefore, the oxidation number of this other species needs to be less than +5.
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The only reason for that is that the maximal oxidation state of nitrogen is +5. Hence, if HNO3 were to oxidise another nitrogen-based species (eg NO, N2, NH3), then that species needs to be able to be oxidised. This other species cannot have an oxidation number of +5, because it cannot be oxidised to attain a higher oxidation number. Therefore, the oxidation number of this other species needs to be less than +5.
thanks man
Another question; here's a study design point:
Comparison of the renewability of energy sources including coal, petroleum, natural gas, nuclear fuels and biochemical fuels
I'm just planning to make a simple summary table for this. Our school is skipping it as well, apparently it doesn't really come up in the exam. Would a table be enough you reckon?
thanks
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Yeah, just know which are renewable. Knowing what 'carbon-neutral' is doesn't hurt.
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Yeah, just know which are renewable. Knowing what 'carbon-neutral' is doesn't hurt.
Would a reasonable explanation for carbon-neutral be that there is no net increase in the CO2 levels in the atmosphere through the burning of these fuels?
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Why does the accumulation of charge in one half cell in the galvanic cell stop electricity from being produced?
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Would a reasonable explanation for carbon-neutral be that there is no net increase in the CO2 levels in the atmosphere through the burning of these fuels?
VCAA would probably accept that
Why does the accumulation of charge in one half cell in the galvanic cell stop electricity from being produced?
That's because when charge is accumulated, you have an external electric potential that begins to counter the cell potential.
Also, it means you don't have a complete circuit.
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VCAA would probably accept that
That's because when charge is accumulated, you have an external electric potential that begins to counter the cell potential.
Also, it means you don't have a complete circuit.
Stilll not making sense, it should though I do physics ;'(.
What do you mean by electrical potential countering the cell potential?
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I think what he means is that, the accumulation of electrons in one half of the galvanic cell would make it highly negative (obviously...haha). So that means that it is harder for electrons to flow in the circuit because the negativity would start to oppose and repel the incoming electrons.
I believe that is what he means by an external potential countering the cell potential.
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I think what he means is that, the accumulation of electrons in one half of the galvanic cell would make it highly negative (obviously...haha). So that means that it is harder for electrons to flow in the circuit because the negativity would start to oppose and repel the incoming electrons.
I believe that is what he means by an external potential countering the cell potential.
Awesome, thank you PB :)
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Just copying and pasting some of my notes. If highlighted WHY? next to some of my notes because I need a bit more info, and the textbook/internet has not really provided me that info.
All galvanic cells have two half cells, with each containing an electrode. The type of electrode we use depends on some factors. If one member of the electrode is a metal, then that particular metal itself is used the electrode. Why?
If there is no metal present, an inert electrode such as platinum or graphite is used. Why?
In some half cells, if one of the half cells is a gas, in this case a special kind of electrode is used, and basically a glass tube is coated over the platinum or graphite electrode. Why?
cheers
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Study design question:
For primary cells, do we need to specifically know
- dry cells
- alkaline cells
- button cells
And for secondary cells:
- car batteries
- nickle based batteries
Thank you!
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Scrap the last question I asked, but still need a bit of help on the question I asked on the 7th of July.
But here's any easy one, help appreciated :)
Just with fuel cells, what makes them so good is that the waste products in fuel cells is always water. Unlike other cells such as galvanic cells, which produce ions like Pb2+ and Cu2+ etc. Why is it better to have water as a waste product than charged ions? Is it just so it's easier to remove?
And another question, the products from reactions of fuel cells start of as charged ions, like H+, how do these charges become water? Do they react to a specific electrolyte roaming around in the fuel cell? If so what electrolyte
thanks
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Question 12
Consider the following statements about alkanes.
I Alkanes react with oxygen to produce carbon dioxide and water.
II Alkanes react with hydrogen chloride to produce chloroalkanes.
IIIAlkanes react with sodium hydroxide to produce alkanols.
Which of the statements above are true?
A. I only
B. I and II only
C. I and III only
D. I, II and III
Answer is A.
Worked solution
A is correct because I is the only correct statement. Alkanes are a hydrocarbon and react with oxygen in a complete combustion reaction to produce carbon dioxide and water.
B is incorrect because II is an incorrect statement. Alkanes must react with chlorine in a substitution reaction to produce chloroalkanes.
C is incorrect because III is an incorrect statement. Alkanes react with water in a substitution reaction to produce alkanols.
D is incorrect because II and III are incorrect statements. Alkanes must react with chlorine in a substitution reaction to produce chloroalkanes and react with water in a substitution reaction to produce alkanols.
Waah? Why doesn't INSIGHT consider the reaction between alkane + cl a reaction? Sure it's a substitution, but why can't it be a reaction?
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Study design question:
For primary cells, do we need to specifically know
- dry cells
- alkaline cells
- button cells
And for secondary cells:
- car batteries
- nickle based batteries
Thank you!
I wasn't asked to know specific battery types for my year
Scrap the last question I asked, but still need a bit of help on the question I asked on the 7th of July.
But here's any easy one, help appreciated :)
Just with fuel cells, what makes them so good is that the waste products in fuel cells is always water. Unlike other cells such as galvanic cells, which produce ions like Pb2+ and Cu2+ etc. Why is it better to have water as a waste product than charged ions? Is it just so it's easier to remove?
And another question, the products from reactions of fuel cells start of as charged ions, like H+, how do these charges become water? Do they react to a specific electrolyte roaming around in the fuel cell? If so what electrolyte
thanks
Well...if you have charged ions as products, they may precipitate out of solution if their concentration goes too high. That's not a problem with the hydrogen oxygen fuel cell.
As for H+ becoming water, try the reaction O2 + 4H+ + 4e- => 2H2O
Question 12
Consider the following statements about alkanes.
I Alkanes react with oxygen to produce carbon dioxide and water.
II Alkanes react with hydrogen chloride to produce chloroalkanes.
IIIAlkanes react with sodium hydroxide to produce alkanols.
Which of the statements above are true?
A. I only
B. I and II only
C. I and III only
D. I, II and III
Answer is A.
Worked solution
A is correct because I is the only correct statement. Alkanes are a hydrocarbon and react with oxygen in a complete combustion reaction to produce carbon dioxide and water.
B is incorrect because II is an incorrect statement. Alkanes must react with chlorine in a substitution reaction to produce chloroalkanes.
C is incorrect because III is an incorrect statement. Alkanes react with water in a substitution reaction to produce alkanols.
D is incorrect because II and III are incorrect statements. Alkanes must react with chlorine in a substitution reaction to produce chloroalkanes and react with water in a substitution reaction to produce alkanols.
Waah? Why doesn't INSIGHT consider the reaction between alkane + cl a reaction? Sure it's a substitution, but why can't it be a reaction?
Your question explains my disdain for the VCE chemistry course in not teaching detail. Alkanes react with elemental bimolecular chlorine in the presence of UV light because UV light breaks the Cl-Cl bond, producing two chlorine atoms with 7 valence electrons that are pretty damn reactive. As a result, these chlorine atoms are able to collide with alkane C-H bonds and break them.
Chloride, however, is pretty unreactive. It already has a stable octet of electrons, so why would it need to react?
Btw, the given answers are inaccurate. Alkanes don't react with water; the answers have confused alkanes with alkenes.
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Thank you so much!