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BedokFunland JC's A Level H2 Chemistry Qns (Part 2)

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    H2 Chemistry @ BedokFunland JC (near VJC & TJC)
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    • Originally posted by Flying grenade:


      Pretentious packaging?  Is the common person suppose to infer, an amino acid that is branched, is very good?

      Anyway, what's so good about branched chain amino acid (BCAA)?


      Yes, BCAAs are useful specifically for sports and bodybuilding purposes, to prolong stamina and reduce DOMS after workouts.



    • Originally posted by MightyBiscuits:

      May I ask a general qns? My school tutor claimed that for energy lvl diagram when the 3d and 4s orbital is filled, 3d orbital will have a higher energy than 4s orbital but when unfilled, the 4s orbital will have a higher energy than 3d orbital. Is that true?

      Yes, this is true for the transition metals, as the electrons in the inner 3d orbitals will repel the electrons in the outer 4s orbitals, raising the 4s orbital to a higher energy level than 3d. Consequently, during ionization process, electrons are removed from the outer 4s orbital before electrons are removed from the inner 3d orbitals.

      Edited by UltimaOnline 08 May `16, 9:07PM
    • Originally posted by Flying grenade:

      Hi ultima, 

      I read cs toh book page 108 and understood *why* O2(g) is a *better choice* than F2(g) as an oxidiser.

      But i thought of comparing oxidising strength between F2(g) and O2(g)

      I know F2(g) is the strongest oxidising agent of all known chemical compounds  (or is it just the strongest, common oxidising agent?)

      Question : how to determine oxidising power of O2? In the data booklet, there isn't any equation for O atom , e.g. O2 + 4e- -> 2O2- , unlike F, which has a eqn F2 + 2e- -> 2F- , with a given E°value. 

      Is oxidising power related/affected to electronegativity? Are there other factors? 


      2. Came across this "Why is F2 a stronger oxidising agent than Cl2 even though F2 has a less exothermic E.A than Cl2? "


      But don't really understand fully

      Why is F2 a stronger oxidizing agent compared to Cl2 even though F2 has a less exothermic EA compared to Cl2?

      Oxidizing strength is primarily related to electronegativity, but secondarily to hydration enthalpy, because in the natural world, redox reactions occur mostly in the aqueous environments (ie. water being the solvent that brings the reactants together and allowing chemical reactions to occur), hence the redox potentials that are relevant are in the aqueous state.

      Yes, F2 a stronger oxidizing agent than Cl2 even though F2 has a less exothermic EA compared to Cl2. First, understand why Cl2 has a more exothermic EA compared to F2 : it's because even though F is more electronegative than Cl, but the charge density of F- is too high due to small ionic radius, concordantly the inter-electron repulsions within the small F- ion are highly destabilizing. Consequently, the EA for F isnt as exothermic, as F- isn't as stable as might be imagined solely due to electronegativity (as would be the case with Cl to Cl-), as there is also a destabilizing (ie. endothermic) component to the EA process for F (but not for Cl), ie. the inter-electron repulsions of the high anionic charge density F- ion.

      Next, if the EA of Cl is more exothermic compared to that of F, shouldn't Cl2 be a stronger oxidizing agent (ie. more positive reduction potential) compared to F2? Yet the Data Booklet clearly shows F2 having a more positive reduction potential compared to Cl2, and F2 is thus the stronger oxidizing agent. This is because as I mentioned earlier, we're interested in redox reactions occurring in the aqueous solvent or phase, and thus concordantly by Hess Law, reduction enthalpy = electron affinity enthalpy + hydration enthalpy ; and likewise oxidation enthalpy = ionization enthalpy + hydration enthalpy.

      Because F- has a much higher anionic charge density compared to Cl-, it's hydration enthalpy is also that much more exothermic (ie. much stronger ion - permanent dipole interactions formed). And after Hess Law is applied, the overall reduction potential of F2 to F-, is still more positive compared to the reduction potential of Cl2 to Cl-, and thus in aqueous environments, F2 is still a much stronger oxidizing agent compared to Cl2.

    • Question : how to determine oxidising power of O2? In the data booklet, there isn't any equation for O atom , e.g. O2 + 4e- -> 2O2- , unlike F, which has a eqn F2 + 2e- -> 2F- , with a given E°value.

      I told you before : in the natural world, these redox reactions occur mostly in aqueous environments / solvents / phases. Do you think the highly unstable O2- ion is able to exist in aqueous environment? Of course not! It will kena immediately hydrolyzed to OH-, and (depending on pH) can be further protonated to H2O.

      So for the oxidizing power of O2, look in the Data Booklet for the reduction potential of O2 to either OH- or H2O (depending on pH). (Ignore obscure & rare reduction products like peroxides, super oxides, ozonides, etc, unless otherwise specified by the question).

    • Originally posted by theophilus:

      Hi, is being able to solve A level questions sufficient? I am doing well on the TYS but prelim papers are ridiculously tough, isit unneccessarily tough? 

      Being able to solve A level questions is necessary but not necessarily sufficient. If you stick to doing A level papers only, without Prelim papers, you may or may not get A grade. Of course, just by doing Prelim papers (with or without A level papers) is by itself no guarantee that you'll get A grade either. Every year, there are dozens (sometimes over a hundred) of unfortunate students who do worse (eg. B grade) for their A levels compared to their Prelim exams (eg. A grade).

      My advice would be to focus on the style of A level papers, but at the same time broaden your understanding by learning from the "ridiculously" difficult questions and mark schemes from the JCs. Unless of course, you already find A level papers difficult, in which case confusing yourself with the even tougher Prelim papers would be counter-productive.

    • A BedokFunland JC original H2 Chemistry question

      Source : https://pixabay.com/en/photos/rain%20forest/

      A drop of water hangs from a leaf. Over the next few hours, depending on environmental conditions, the drop of water either increases or decreases in size (ie. mass and volume). Identify, describe and explain 2 of these environmental conditions (excluding wind and rain). Your answer must include all of the following terms : "Le Chatelier's principle", "enthalpy", "entropy" and "Gibbs free energy".

    • Originally posted by Flying grenade:

      A classmate of mine asked a qn in class in which the tcher was stumped : why electron shell closer to nucleus, has lower energy level?

      For a simplified version (which will suffice for A level purposes), you can think of it as : to maintain your orbit further from the nucleus, you (as an electron in the outer electron shells) naturally require more energy to sustain your further orbit and resist the electrostatic pull of the positively charged nucleus.

      Going any deeper like the classic scientific enigmas of "Why don't electrons fall into the nucleus?", and "Why don't the protons in the nucleus repel each other and cause the nucleus to fall apart?" (also see Cordus physics) that goes significantly beyond A levels (so we we're *not* going discuss it here, and it's a safe bet that neither your school teacher nor your private tutor will entertain you on this), and at University levels, physicists and chemists have come up with a few different theories or explanations or ways of looking at these enigmas. You may come across some of these ideas in the future if you choose to study Physics or Chemistry in the University. For now, if you're still curious (though you should of course focus on your A levels), you can google out some preliminary reading for your entertainment and satisfaction.

      Edited by UltimaOnline 16 May `16, 6:19PM
    • Originally posted by Ng.keebin:

      Why is 4-aminobenzenesulfonic acid a much weaker acid than benzenesulfonic acid?

      2 separate points :

      The amino group is electron-donating by resonance, hence destabilizing the negatively charged sulfonate group in the conjugate base. In addition, 4-aminobenzenesulfonic acid exists as a zwitterion in aqueous and solid states, and because ammonium groups (ie. conjugate acid of amines) are very weak acids with small Ka values, hence zwitterionic 4-aminobenzenesulfonic acid is a much weaker acid than non-zwitterionic benzenesulfonic acid.

    • Hi Nelexus, glad to hear you chose to read Chemistry in the University. Btw, while both NUS and NTU Chemistry are (approximately) equally good, the Science bookstore at NUS is better stocked compared to the Science bookstore at NTU. Fyi.

      Generally (for everyone reading this), please do not PM me Chemistry questions (PM me only if you're interested in joining my BedokFunland JC tuition), but post Chemistry questions on the forum for discussion so that everyone can participate, learn and benefit together.

      Next, for Uni level Chemistry questions, please do not post them here, as my professional interest (ie. my job) is specifically A level Chemistry.

      But no worries (this suggestion is for all JC graduate students taking Chemistry or Chemistry related courses in the University), I'll recommend you an excellent Chemistry forum (probably the best on the internet) where you can get help for all your Chemistry questions, all the way to Masters, PhDs and professional Chemical industry questions.

      Presenting.... the Chemical Forums (yeah that's the name, strictly functional and down-to-business, nothing flowery or fanciful).



      Nelexus PMed me :

      Hi Ultima, I have a few Chemistry questions that I would like to clarify with you. Am a graduated JC student but intending to read chemistry in uni. So I have been reading up on Chemistry before Uni starts. Some of the questions I am asking will be covering advance concepts far beyond the h2 chem syllabus.


      1. Resonance Vs Conjugation


      What is the difference between resonance and conjugation? From what I've read online, a continuous overlapping of adjacent p orbitals forms a conjugated system. The overlapping of p orbitals allows electrons to "move" in the conjugated system, resulting in delocalisation of electrons in that conjugated system. This is illustrated by the diagram below I've picked up from the internet.

      Therefore, from this scenario above, conjugation allows resonance to occur. 


      Now consider the conjugate base of benzoic acid and pyruvic acid. I believe they are all conjugated system. The thing is, for the benzoate ion, the electrons on oxygen in the carboxyl group cannot be delocalised into the benzene ring no matter how I draw, thus no resonance structure forms arises for the benzoate ion. Likewise for the pyruvic acid, the electrons cannot be delocalised into the ketone group.  So while they are conjugated, there are no resonance structure for . This sounds abit strange, considering that for a conjugated system with p orbitals overlapping with each other, electrons can "move" around the system, yet this is not possible for the electrons on the oxygen atoms of benzoate and pyruvate ion. My point is ilustrated below (pardon the big image)

      For this amine below that I've drawn, there's a conjugated system and the delocalisation of electrons can happen.  My confusion is tha tlogically,  conjugation which is the overlapping of p orbitals, should allow electrons to delocalise, and thus entails resonance. Yet this is not the case for some system. Why?

      2. pH of equivalenve points of polyprotic acids


      The diagram below shows the titration curve of sulfurous acid with sodium hyroxide.

      For he second equivalence point where only SO3 2- is present, the equivalence point can simply be obtained by obtaining the Kb of SO3 2-, and then find [OH-] using Kb = [OH-]^2/ [SO3 2-] am I right?

      Now the tricky part is the first equivalence point. The species present in the first equivalence point is HSO3- which is an amphoteric species. Meaning HSO3- can act both as a base and an acid as follow

      HSO3 - + H2O  --> SO3 2- + H3O+ Ka
      HSO3 - + H2O  --> H2SO3 + OH- Kb

      Is it possible to calculate the pH of the first equivalence point? 


      I understand that for amino acids, theres this formula pH = 1/2 (pka1+pka2) which is used for calculating the pI and the so called equivalence point as shown below.

      Logically amino acids are polyprotic acids as well, so would the same formula work? Or do I have to physically inspect whether Ka or Kb is larger, and then proceed to using the method I used for finding the second equiv point?

      3. Amino Acids

      For amino acids, why is it that amino acids always exist in its protonated form? For example, in the titration of amino acids, the major species in the beginning is always NH3+ -R - COOH at a low pH. Should it be the case that say in the laboratory, the amino acids you work with are powders in the form of zwitterions NH3+ - COO- and then when you add it into water, the carboxylate group gets protonated and a basic solution is formed. Yet this is not the case for the examples I've seen on the internet, where the amino acids exist as NH3+ - R - COOH at pH < 7?


      Secondly, to say the isoelectronic point of an amino acids, does it mean to have equal concentrations of both the cation and anion, or the maximum concentration of the zwitterions? Some textbooks obtain the formula of pH = 1/2 (pka1+pka2) using the assumption that [cation] = [anion], which I think does not make sense.

      I quote from Essentials of Organic Chemistry by Dewick


      "The pH at which the concentration of the zwitterion is a maximum is equal to the isoelectric point pI, strictly that pH at which the concentrations of cationic and anionic forms of the amino acid are equal. With a simple amino acid, this is the mean of the two pKa. 

      in the derivation, how is the statement " cation = anion " true? Doesn't the dissociation of amino acids occur stepwise, meaning cation --> pka1 where cation = zwitterion --> equivalence where only zwitterion --> pka2 where zwitterion = anion --> 2nd equivalence where there is only anion. 

      I understand that these questions are far beyond the h2 chem syallbus which you may not be obliged to answer, but I really hope you do. I've scoured countless materials and resources on the internet to find no answer. Thank you in advance!


      UltimaOnline here again. I will give brief guidance & comments regarding Nelexus' questions. For deeper discussion, you're advised to discuss on the Chemical Forums.

      Conjugation (not to be confused with hyperconjugation) *allows* for delocalization of electrons to occur via resonance.

      In the benzoate ion, the negative formal charge on the O atom, cannot be delocalized by resonance into the benzene ring, as evidenced from the impossibility of drawing the required mechanism to delocalize the negative formal charge into the benzene ring (since resonance cannot involve the breaking or forming of sigma bonds).

      However, note that in the conjugate acid form, the pi electrons in the benzene ring can indeed be delocalized by resonance to the COOH group, ie. the COOH group is electron-withdrawing from the benzene ring by resonance, and is hence deactivating and meta directing. When deprotonated however, the COO- group, being electron-rich, ceases to be electron-withdrawing by resonance from the benzene ring, as the high electron charge density of the dinegatively charged COO2- resonance contributor is excessively destabilizing.

      The real reason why the benzene ring can help to stabilize the benzoate ion conjugate base (thus making benzoic acid a stronger acid than say, ethanoic or propanoic acid), is because the sp2 C atoms of the benzene ring have a higher % s orbital character, and is thus electron-withdrawing by induction (but not by resonance).

      Yes, the formula pH = 1/2 (pKa1 + pKa2) or pH = 1/2 (pKa2 + pKa3), is applicable for all solutions containing only an amphiprotic species, eg. at equivalence point for polyprotic or multiprotic acids, such as zwitterionic amino acids.

      A solution at its isoelectric point contains the highest possible molarity of its zwitterionic form (ie. no net ionic charge), and very low and equal molarities of both the cationic (ie. conjugate acid) and anionic forms (ie. conjugate base) forms. The molarities of the ionic forms at pI can be calculated using the relevant Ka values.

      Amino acids can (depending on the exact amino acid's R group) exist as dinegative anionic, uninegative anionic, neutral zwitterionic, unipositive cationic, and dipositive cationic. So if you were carry out a titration of an amino acid, you would usually start with either the fully deprotonated anionic form (if you're adding acid from burette), or the fully protonated cationic form (if you're adding alkali from burette).

      Edited by UltimaOnline 23 May `16, 3:45PM
    • Originally posted by Flying grenade:

      I don't have the photo of the qn, but i wrote it down, so I'll type it out

      Qn : write the reaction for titration of 100ml of 0.100M anilinium bromide (aminobenzene • HBr) with 0.100M NaOH. Sketch the titration curve for points Vb=0, 0.1Ve, 0.5Ve, 0.9Ve, Ve, 1.2Ve


      Species present at equivalence pt is phenylamine.

      Oh use equation of dissociation of phenylamine ?


      Why cannot use amphiprotic formula in this case?since phenylamine is the cj base of phenylammonium cation 

      So not every time at equivalence point use amphiprotic formula, is it?

      Thanks for those guidance in thought process that you've written down !!

      Use amphiprotic formula only when the species present is amphiprotic. At the various equivalence points for acid-base titrations, the species present could be either only acidic (eg. H3PO4), only basic (eg. PO4 3-), or amphiprotic (eg. H2PO4- or HPO4 2-).

      Instead of blindly memorizing formulae without understanding (like the majority of pitiful Singapore JC students), you should intelligently understand what's going on, intelligently identify on a case-by-case basis what's the nature of the species present, and hence intelligently apply the relevant formula.

      And btw, phenylamine is a base, bases do not undergo dissociation (as erroneously taught in Singapore JCs), bases undergo hydrolysis.

    • Originally posted by Flying grenade:

      Hi ultima , i would like to ask,

      I have written down the kb value from the soln  book,


      I am confused by this part 

      In part i), the ka value is for phenylammonium cation, correct?

      For the part of vol base added = vol equivalence, i used kw/ka(phenylammonium) to get kb(phenylamine)


      So, when i use kw/ka is get kb of the cj base ah??

      Damnit, all the while i thought its ka and kb of the same molecule !


      When i apply kw/ka to find kb, is the kb the kb of same molecule or kb of its cj base!

      Ka of HA x Kb of A- = Kw

      Ka of BH+ x Kb of B = Kw

      Ka of conjugate acid x Kb of conjugate base = Kw

    • Originally posted by Flying grenade:

      Why is Mno4- a stronger oxidising agent compared to Cr2O72-?

      Because of a complex combination of reasons, which you needn't worry about for A levels. Most relevantly, the oxidation state of the heteroatom in the oxidizing agent (ie. +7 versus +6), but also the electronegativity, the charge density, the electron affinity enthalpy, hydration enthalpy, electron configuration, reaction mechanism, etc.

      So when you explore these issues at Uni level, you'll realize there are no longer any short, simple answers (those are the O level days of naivety), but a beautifully complex balance of separate (and oftentimes opposing factors). A levels is a transition state or intermediate between naively simplistic O levels, and beautifully complex Uni levels.

    • Originally posted by Flying grenade:

      One of my teacher  say that the reason why H- cannot reduce alkene is because alkene C=C bond is neutral(i.e. not polar)

      All along i was under this impression. 

      Today,  another different teacher say the reason is because the lone pair on H- repels with the pi e-cloud of C=C bond

      :/ i am flustered 

      See if you're intelligent enough to see how the following Singaporean link explains the answer to this H2 Chemistry question.


      Whether you get it or not, don't post further on this topic on this forum, don't spoil the fun for others! *evil laugh*

    • Originally posted by Flying grenade:

      Pang cheng cheong's book and my school cher say reaction of carbonyl cpds with 2,4dnph is at r.t. , but cs toh book say heat, and making sense's reagent and conditions say need warm!

      How??? These reagents and conditions makes me damn f**king agitated!

      Idk but this is one of my thought 24dnph reaction(condensation/nu sub) can be carried out under r.t. But perhaps warm, accelerates the reaction , right? So both are acceptable. My doubt, would cambridge mark both correct?? If i write heat will they know cos i know the reaction will be quicker? Or will they think i am dumb , the reaction requires heat to overcome Ea of reaction? So write 24dnph, r.t. more safe is it?

      Allow BedokFunland JC to enlighten thee :

      As far as 2,4-DNPH is concerned, Cambridge doesn't give a flying f**k whether you write "room temperature" or "warm" or "heat" or even if you don't specify the temperature at all.

      However, don't mistake Cambridge's kind reasonable tolerance for weak bochup nonchalance : between PCl3 and PCl5, you *must* specify "heat" for 1 of these reagents, or Cambridge will penalize you. If Cambridge asks you to explain why this difference, do you know how to answer?

      (let's hope for more such out-of-the-box questions for this year's and all future Singapore H2 Chemistry papers, shall we? That will shut up those armchair critics who take perverse pleasure in criticizing the entire Singapore education system to be all about blind memorization without deeper understanding... such criticism applies more for PSLE, N and O levels, but less so for A levels and certainly not Uni levels; and among A level subjects, such criticism is even less valid for Chemistry than other A level subjects, hopefully Cambridge will continue this trend for Singapore's H2 Chemistry).

      Edited by UltimaOnline 24 May `16, 5:30PM
    • Originally posted by Flying grenade:

      Yeah okay, i shall stop asking u about reagents and conditions, too many differing R&C

      Cs toh advanced study guide pg 308 indicated halogen of benzene involves halogen carriers and at room temp. But it's summary at pg 315 indicate it needs heat. 

      Cs toh wrote friedel crafts reactions require heat,  but the small hci book written by pang peng cheong wrote r.t.

      Nvm, I'll just write one of them in the exams 

      Generally, just follow trusted sources such as CS Toh and the others (you know who they are). If there are any differences between these trusted authors, or even minor inconsistencies within each author, these differences are almost certainly going to be minor to the point of insignificance, and Cambridge will almost certainly accept all their (slightly differing) answers.

      Don't OCD about it lah. Be like me, lepak lepak, relak relak, anyhow do, anyhow write, anyhow skip, also get my A grade. *GuaiLanz LOL*

      Edited by UltimaOnline 13 Oct `17, 2:12AM
    • Originally posted by Flying grenade:

      Hi ultima, hope u can help me with this,

      Page 303 cstoh guide

      Halogenation of alkenes to haloalkanes

      Bottom of page, i don't understand why cs toh wrote , if bubbled a gas into aqueous Br2, if reddish-brown Br2 soln is rapidly decolorised, (without evolution of fuming HBr gas) indicates presence of c=c bond  .

      On top, the chemical equation has HBr as a product


      Is it because not fuming(means not concentrated) ah?

      Why did cs toh bring this up?

      Is it because of ethanoyl bromide ? 


      Checked(from bp and mp), it's a liquid at r.t.

      But it would hydrolyse readily in moist air to produce ethanoic acid and hydrobromic acid (HBr) ?


      Why cs toh mentioned *without evolution of fuming HBr gas* ? 

      Thank you

      CS Toh was deliberate and conscientiousness when he wrote that. If Br2 (in CCl4) was used on the analyte, and decolorization of the Br2, together with steamy fumes (indicating HBr) were both observed, that would be indicative of 2 other functional groups (covered in the H2 syllabus), but not the alkene functional group.

      Guess what these 2 functional groups are? I don't want to spoil the fun, so I won't spoonfeed you guys the answer on the forum. I'll encourage everyone reading this to figure it out for yourself.

    • Originally posted by Flying grenade:

      Hi ultima, hope u can help out on this !


      What product could it be?? Checked reactions, reagents and conditions for methylbenzene and alkenes, don't have that reagent and conditions! 



      Update : found soln

      Electrophilic Additions to Alkenes

      Reagents : HBr (gas) or 


      Elements Added :H-Br

      See, your problem is you (and most Singapore JC students) rely on memorizing from notes. If you understood mechanisms, you could have easily figured out the answer for yourself, even if you've never came across such a reaction in your notes.

      Sure, you can google out the answer now (ie. non-exam conditions), but what happens when Cambridge thinks up some new reaction that you've never come across before, to be used as an A grade exam question in the A levels?

      Well, I wouldn't be surprised if the majority of Singapore JC students wouldn't know the answer as well, in such a case, so I guess you needn't worry too much. It's a bell-curve, afterall.

      Edited by UltimaOnline 25 May `16, 11:28PM
    • Originally posted by Flying grenade:

      What does proticity mean? Can't find the meaning online


      Is it mean how many H it can donate?


      U.O. told me that formula of salt formed is based on proticity, based on the discussion of george chong's inorganic book having a error at page 40

      Proticity refers to how many protons a Bronsted-Lowry acid can donate, or how many protons a Bronsted-Lowry base can accept. Eg. H3PO4 is a triprotic acid, hence PO4 3- is a triprotic base.

      Since H3PO3 (Latin name phosphorous acid, Stock name phosphoric(III) acid, IUPAC name phosphonic acid) is a diprotic acid, hence upon complete deprotonation using excess NaOH(aq) in a Bronsted-Lowry acid-base reaction, the salt generated is Na2HPO3 (Latin name disodium phosphite, Stock name disodium phosphate(III), IUPAC name disodium phosphonate).

    • Stanton and Layard hypothesized in 1977–78 that toxicity of fibrous materials (particularly asbestos) is not initiated by chemical effects; they hypothesized that the effects of asbestos must be physical and that mechanical damage might disrupt normal cell activity—especially mitosis. There is experimental evidence that the very slim fibers (less than 60 nm, less than 0.06 μm in breadth) do tangle destructively with chromosomes (being of comparable size). This is likely to cause the sort of mitosis disruption expected in cancer.


    • Iodine (whether radioactive or not) accumulates in the thyroid gland, and with exposure to radioactive iodine (such as in the Fukushima nuclear disaster), what happens is that the radioactive decay of the radioactive iodine causes genetic damage to the thyroid gland (including activating proto-oncogenes and deactivating tumor suppressor genes), resulting in thyroid cancer. The only effective treatment (which works mainly as a preventive measure, meaning you have to take action the moment you hear news of a radioactive leak in your area) is to spam high doses of non-radioactive iodine (not as elemental iodine, which is readily reduced to iodide and is thus an oxidizing agent and thereby corrosive, but as an iodide or iodate salt), which works not as some magical bullet, but simply because of competitive inhibition, ie. the thyroid gland will have little propensity to absorb any of the radioactive iodine, if it is already saturated with non-radioactive iodide.

      Woman breaks silence among Fukushima thyroid cancer patients :

    • Originally posted by Flying grenade:

      Difference between acidity and strength of acid?

      Acidity of soln (molarity of H+) and the tendency for a acid to deprotonate is it?


      Acids do not necessarily have COOH functional group


      COOH functional group for carboxylic acid

      Measure of acidity (ie. pH) is referring to for solutions.

      Strength of an acid is measured by its Ka or pKa value.

      Proticity refers to how many moles of H+ one mole of the acid can donate, or how many moles of H+ one mole of base can accept.

      Obviously only those acids with the COOH group are called carboxylic acids. Obviously not all acids have that group. Why are you stating the obvious?

      Option D is the answer, because its conjugate base is most effectively stabilized (among the 4 options), by delocalizing the uninegative formal charge over 2 electronegative, electron-withdrawing O atoms (ie. similar acidic strength to carboxylic acids). You *have* to draw the mechanism for the resonance contributors to fully understand this.

    • Originally posted by Ng.keebin:

      For heterogenous equilibrium, why is pure solids and liquids not included when calculating Kc and Kp?

      Because it has been experimentally proven that the amounts of pure solids and liquids in heterogenous equilibria (that involve solvated reactants and products), do not affect the ratio of molarities of solvated reactants and products at equilibrium.

      Originally posted by Ng.keebin:

      Why when inert gas is added to a mixture of gases, the total pressure increases but partial pressures of each gas stay the same?

      Because partial pressure = mole fraction x total pressure. As inert gases are introduced, the mole fractions of the active gases decrease, while the total pressure increases, hence mathematically cancelling each other out.

      If you're asking why the position of equilibrium doesn't shift, it's because the molarities of the active reactants and products (of the reaction) doesn't change upon addition of inert gases, so neither Qc nor Kc changes, hence the position of equilibrium doesn't shift.

      Edited by UltimaOnline 11 Jun `16, 12:23AM
    • Originally posted by theophilus:

      Ultima, for A levels purposes can I simple assume that polar will always be able to be soluble in polar reagents and non polar solvents wil mix with non polar reagents ? Ionic salts will always dissolve because they can form ionic like bonds with polar water molecules. Anything that can form hydrogen bonds iwith water can dissolve in water.  as long as the bonds holdingh the solid together is weaker than the bonds that will form with the solvent, solid will surely dissolve. please correct any misconceptions ! im struggling with the whole solubility thing ( not the energetics kind but the bonding of solvent solution )

      Not so simple. At O levels, everything is simplistically 'always'. At Uni levels, nothing is simplistically 'always'. A levels is between O levels and Uni levels, expect anomalies and having to examine case-by-case.

      'Like dissolves like' is usually applicable for most cases, but the question can add in more variables to get you to discuss relative solubilities of a polar solute in 2 or 3 different solvents, all of which are polar (for example).

      Not all ionic salts are soluble, hence you have Qsp and Ksp. But species like phenylammonium ions and phenoxide ions, are indeed more soluble than their molecular conjugate acid/base forms, due to ion - permanent dipole interactions which are stronger than hydrogen bonds (provided the formal charge is on a small period 2 atom).

      Even if a molecule can form hydrogen bonds with water, it may not be very soluble, because of the non-polar parts of the molecule (eg. benzene ring) thermodynamically preferring to form van der Waals forces with the non-polar parts (eg. benzene ring) of the other solute molecules, while the snobbish water solvent molecules thermodynamically prefer to form hydrogen bonds with other water solvent molecules. Hence, aqueous solubility of alcohols decrease with increasing length of its non-polar hydrocarbon chain.

      So once you see it as a beautiful complex spectrum of solubilities, rather than an oversimplistic black-and-white binary system of just 'soluble' or 'insoluble', you can let go of your frustrations and it will finally make logical sense for you.

      Originally posted by theophilus:

      Also e.g say when an amine is added into HCl, the solution surely dissolves and forms a salt rite? but isnt it reacting with the acid which permanently changes its chemical formula and not dissolving ? compared to maybe amine being dissolved in water which only forms hydrogen bonds with water and does not react. 

      Also Hormones must be made up of 2-amino carboxylic acids like proteins or any kind of amino acid?

      Yes, it is reacting with the acid to be protonated, and it's formula changes from its conjugate base form to its conjugate acid form, and its conjugate acid form is more soluble than its conjugate base form, because ion - permanent dipole interactions are stronger than hydrogen bonding, as long as the formal charge is on a small period 2 atom.

      Hormones may or may not be protein in nature (H2 Bio students have an advantage over H2 Physics students for this topic). Only proteins are polymers of alpha-amino acids.

    • Originally posted by Flying grenade:

      when the Eθcell of an overall cell reaction is +ve , do we say the reaction is thermodynamically/energetically favourable/feasible?

      Best to use the phrase "thermodynamically feasible". Because a positive cell potential = a negative Gibbs free energy change, by the formula (tested in the new H2 Chem syllabus) : delta G = - n F cell potential

    • Originally posted by Flying grenade:

      george chong org chem page 7 

      how does valence bond theory and hybridisation theory help us to determine if an organic compound is e-rich or e- deficient?

      From electron geometry, you can work out orbital hybridization. The greater the % s orbital character, eg. sp > sp2 > sp3, the more electron rich that atom is, because the unhybridized p orbitals (not used for hybridization to form sigma bonds) can overlap sideways to form pi bonds. Eg. Triply-bonded nitrile sp > doubly-bonded imine sp2 > singly-bonded amine sp3.

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