BedokFunland JC's A Level H2 Chemistry Qns (Part 2)
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Hoay posted :
To UltimaOnline
you mentioned the following generalization regrading the spliting pattern:
Strong ligands are not the same as strong field ligands.
Weak ligands are not the same as weak field ligands
Would you please elaborate it some examples?
First of all, this aspect is totally beyond the H2 Chemistry syllabus. And it's unlikely, even when Cambridge on occasion go beyond the syllabus in their questions, for Cambridge to ask directly on this. In other words, H2 Chem students really needn't concern themselves regarding this.
As an example, consider the water molecule versus the hydroxide ion, as ligands. Obviously, the hydroxide ion is a far stronger ligand, because it's a lot more electron-rich (the O atom in OH- has a negative formal charge, while the O atom in H2O only has a partial negative charge), and therefore it's a far stronger ligand (and nucleophile, and base) compared to the water molecule.
(A Lewis base is a dative bond donor. If you're a Lewis base donating dative bond to a metal ion, you're called a ligand. If you're a Lewis base donating dative bond to a proton, you're called a Bronsted-Lowry base. If you're a Lewis base donating dative bond to an electrophile, you're called a nucleophile. Notice that NH3, functions well as all 3 types of Lewis bases.)
Experimental evidence have shown, however, that when complexed with water ligands, the magnitude of the energy difference of the non-degenerate d orbitals (of the metal ion) after d-d* splitting, is greater than when complexed with hydroxide ligands. In other words, we say that the water ligand is a stronger-field ligand compared to the hydroxide ligand.
What determines whether a ligand is a strong or weak ligand? The strength of a ligand parallels it's basicity and nucleophilicity (even though occassionally for some species, basicity and nucleophilicity do not parallel each other, eg. F- is a stronger base compared to I-, but I- is a stronger nucleophile compared to F-. H2 Chem students should be able to figure out why).
What determines whether a ligand is a strong field or weak field ligand? Several factors, including the electronegativity of the coordinate dative bond donor, the charge density of the ligand molecule/ion, etc. To understand this fully requires a thorough understanding of crytal field theory, ligand field theory and molecular orbital theory, much of which are *way* beyond the H2 syllabus.
http://en.wikipedia.org/wiki/Crystal_field_theoryhttp://en.wikipedia.org/wiki/Ligand_field_theory
http://en.wikipedia.org/wiki/Molecular_orbital_theory
For those interested (ie. not required for H2 Chem syllabus), the most relevant content has been reproduced below :The size of the gap Δ between the two or more sets of orbitals depends on several factors, including the ligands and geometry of the complex. Some ligands always produce a small value of Δ, while others always give a large splitting. The reasons behind this can be explained by ligand field theory. The spectrochemical series is an empirically-derived list of ligands ordered by the size of the splitting Δ that they produce (small Δ to large Δ; see also this table):
I− < Br− < S2− < SCN− < Cl− < NO3− < N3− < F− < OH− < C2O42− < H2O < NCS− < CH3CN < py < NH3 < en < 2,2'-bipyridine < phen < NO2− < PPh3 < CN− < CO
It is useful to note that the ligands producing the most splitting are those that can engage in metal to ligand back-bonding.
The oxidation state of the metal also contributes to the size of Δ between the high and low energy levels. As the oxidation state increases for a given metal, the magnitude of Δ increases. A V3+ complex will have a larger Δ than a V2+ complex for a given set of ligands, as the difference in charge density allows the ligands to be closer to a V3+ ion than to a V2+ ion. The smaller distance between the ligand and the metal ion results in a larger Δ, because the ligand and metal electrons are closer together and therefore repel more.
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Organic Chem
>>> on the basis of bond energies, what could be the products of the following reaction? <<<
CH3*+CH3CH2Cl ---> ???
A. CH4+CH3CH*Cl
B. CH3CH2CH2*+HCl
C. CH3CH2CH3+Cl*
D. CH3CH2CH2Cl+H*This is a poor qn, as it doesn't specify if multiple steps or only one step is involved. If multiple steps are involved, then more than one answer is eventually possible, except for option D (as far as the H2 syllabus on free radical substitution is concerned, you *never* get a H radical). But assuming only one step, the best answer here would be C. The C-Cl bond is weaker than the C-H bond. The methyl radical can therefore induce the homolytic cleavage of the C-Cl bond, and combine with either the resulting ethyl radical or the resulting chlorine radical. Since the C-C bond is stronger than the C-Cl bond, enthalpy considerations will favour the methyl radical combining with the ethyl radical, rather than the chlorine radical.
>>> why does the reaction between propane and chlorine not give a high yield of 2 - chloropropane? <<<
There are two factors that determine product distribution where free radical substitution mechanism is involved : 1) No. of H atoms substitutable 2) Stability of alkyl radical intermediate. The first factor is simpler, and compulsory to know for the H2 syllabus. The 2nd factor is more complicated, and is optional to know for the H2 syllabus (but if you're gunning for a distinction, you better find out about it). To get 1-chloropropane, you can substitute any of the 6 hydrogens on the 1st and 3rd carbon. To get 2-chloropropane, you can only susbtitute either of the 2 hydrogens on the 2nd carbon. Therefore, in terms of this factor (No. of H atoms substitutable), you expect 1-chloropropane to be the major product, with 75% yield (2-chloropropane with 25% yield).
(The following is optional for the H2 syllabus)
However, based on the other factor (Stability of alkyl radical intermediate), because 2-chloropropane involves a more stable secondary alkyl radical intermediate (compared to the primary alkyl radical intermediate required to form 1-chloropropane), and recall that the more stable the intermediate, the lower the activation energy required to reach the intermediate, hence the major product (ie. lower Ea = faster reaction = major product) based on this factor, would be 2-chloropropane. In practice, because both factors are often contradictory, the actual experimental yield will be somewhere between the two extreme predictions based on either factor alone, ie. a weighted average. There have been a number of prelim paper qns which require you to consider both factors. Because Cambridge is increasing the difficulty of the H2 Chem papers, students gunning for a distinction in H2 Chem should be prepared to go deeper than the basic requirements of the syllabus, and explore optional areas such as these.
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A pure sample of N2O4 (l) is introduced into an evacuated vessel. The vessel, of constant volume, is heated to a constant temperature such that the equilibrium below is established : 2NO2 (g) <---> 2NO (g) + O2 (g)
The value of the pressure p is then found to be 20% greater than if only NO2 (g) were present. Calculate the mole fraction of oxygen in this equilibrium mixture.
Solution :
2NO2 (g) 2NO (g) + O2 (g)
1 | 0 | 0
-2x | +2x | +x
1-2x | 2x | x
Total pressure = 1+x
Since (1+x)/(1) = 120/100, hence :
1 + x = 1.2
x = 0.2
Therefore mole fraction of oxygen = (0.2)/(1.2) = 0.16666 = 0.167 (to 3 sf) -
Animizer asked :
Anyone knows how to determine the suitability of the substance of salt bridges for half cells? Thanks!!
UltimaOnline replied :
Choose inert ions that are virtually never oxidized or reduced under aqueous conditions, and do not precipitate out with the other ions present. eg. Na+ or K+ and NO3-.
Animizer continued asking :
Ooh thanks! One more ques,
Why is oxygen and hydrogen produced in electrolysis of aq calcium nitrate?
The hydrogen part i understand. But the oxygen...
O2 + 4H+ + 4e- ---> 2H2O E=1.23
NO3- + 2H+ + e- ---> NO2 + H2O E = 0.81
In this case, NO2 should be preferentially oxidized at the anode.. But since oxygen is produced, it shows H2O is preferentially oxidized.. I don't understand.
Oh, and what about Hg(NO3)2? Hg is not inert.. But I see other inert ions being used as salt bridges too...UltimaOnline continued replying :
Try not to use mercury salts, as they are toxic, and as you say, they're not as inert as Na+ or K+. If you're referring to a particular setup that uses Hg(NO3)2 as a setup, scan and upload the material and post it here, so I can have a look at it and comment further on it.
NO3- ions are inert under aqueous conditions, because being anionic, it will be electrostatically attracted to the anode. But only oxidation occurs at the anode, and it is not possible to oxidize NO3- ions any further (since N is in Grp V, and it's OS here is already +5), therefore it remains unreacted. Similarly for SO4 2- ions. These ions can only be reduced and not oxidized, but being anionic, are attracted to the wrong electrode (the anode, where only oxidation occurs) instead.seller_john felt obliged to interject with a random, extra remark :
what a difficult forum session .
Animizer concluded the thread with :
Ah. Thanks alot! I didn't realise NO2 wasn't in the solution hahaha oops.
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Nish asked :
Since the questions for Chemistry at A levels are becoming more ''out of the box" type, I would like to know why Aluminium, not being categorised as a Transition Metal(as far as the A level syllabus is at), is able to form [Al(OH)4]- complex.
Thanks in advance!
P/s I searched via google and nothing much came up.
UltimaOnline replied :Aluminium is NOT a transition metal, because it does NOT form one or more ions with partially filled d orbitals.
Your misconception is that only transition metals can form complex ions. In reality, *any* metal can form complex ions, as long as its charge density is high enough to polarize the electron charge clouds (we visualize this as a lone pair) from the ligand (ie. donor atoms of the ligand molecule or anion) towards itself, forming what we call coordinate dative covalent bonds with the central metal ion (which may or may not be a transition metal).
Transition metals, due to their capacity or feasibility to exist in various oxidation states (with comparable stabilities), commonly have ions with charge densities high enough to form complex ions, which is why we say one property of transition metals is that they commonly form complex ions.
But just as we could say "Asian cultures traditionally have a closer affinity to martial arts, compared to Western cultures", it doesn't mean all Asians know martial arts, or that there's no such thing as a Caucasian man who knows martial arts.
Al3+ is just one of many non-transition metal ions, which are capable of forming complex ions. Another would be Be2+ (which has a similarly high charge density as Al3+).
Truth be told, it's a spectrum, rather than binary black and white, the transition from ion-dipole interaction to dative bond formation in complex ions. We can (more or less) confidently say the charge density of Be2+ is sufficiently high to form complex ions with actual dative bonds from the water ligands, while the charge density of Sr2+ is not high enough to form complex ions, but only high enough to have ion - permanent dipole interactions with water molecules.But what about Mg2+ and Ca2+? You could say that the true state of these ions in solution, have the charactersitics of both complex ions, as well as ion - permanent dipole interactions. In other words, somewhere in between. It's only in human minds that we go "huh? how can be somewhere in between?", coz the Universe couldn't care how humans think : to the Universe, Mg2+(aq) is Mg2+(aq) regardless of puny human concepts about "complex ions" or "ion - permanent dipole interactions".
Just as you were lied to at O levels that there's only "covalent bond" versus "ionic bond", now at A levels you realize that oh, it's actually "covalent bond with ionic character", and "ionic bond with covalent character". Similarly, there's a lot of lies (or at least, half-truths) that you're still being taught at A levels, a necessary evil because of the oversimplification process of education at various levels.
Notice that the higher level you go, the less black and white that chemistry, and life in general, becomes. At PSLE you thought you knew everything in the Universe. At O levels you thought you knew a lot about the Universe. At A levels you realize there's a lot more to the Universe that you don't know. At University levels you start to realize the Universe is infinitely more beautiful, complex, astounding and wonderous than humans can even begin to comprehend. -
Daniel798 asked :
The oxidation number of phosphorous in PCL5 is +5.
But PCL5 is formed as phosphorous expands it's octet into the 3d subshell and forms 5 covalent bonds with the 5 chlorine atoms.
But it did not lose and electron so why is it's oxidation number here +5? (I think i have a conceptual misunderstanding of what the oxidation number means).i thought that oxidation state/number showed how many electrons the atom has lost because i didn't know the definition of oxidation state/number so i got confused since PCL5 has number +5 so i though it lost 5 electrons when in actual fact, it did not lose any electrons.
UltimaOnline replied :At much higher levels of Chemistry, there is indeed a technical difference between Oxidation State (OS) and Oxidation Number (ON), but for A level H2 Chemistry purposes, OS and ON are to be considered identical.
The true meaning and formula for OS (or ON) is Formal Charge + Electronegativity consideration.
OS (or ON... from here on I'll just use the term OS) is a type of charge on the atom, one that considers electronegativity.
Formal charge is another type of charge on the atom, one that doesn't consider electronegativity.
For instance, consider this exam question :
In the case of carbon monoxide, if an electric field is generated, would the C atom or the O atom be attracted to the positive field? and negative field?
The typical JC student would assign partial charges based on electronegativity, ie. delta negative on O, and delta positive on C. Based on such imagined partial charges, the student would deduce that the O atom is polarized towards the +ve electric field, while the C atom is polarized towards the -ve electric field.
Wrong.
The reason for the typical JC student's error, is because he/she failed to consider formal charges.
Because the C and O atom both have 5 valence electrons from 1 lone pair and 3 bond pairs (ie. triple bond between the C and O atoms), and because C is in Group IV and O is in Group VI, the C atom has a -ve formal charge and the O atom has a +ve formal charge.
(In terms of stable octet though, both atoms have a full stable octet. "Stable Octet" and "Formal Charge" are two different ways chemists count valence electrons. A third way to count electrons, is the Oxidation State aka Oxidation Number).
Based on formal charges, you would expect the C atom to be attracted to the +ve electric field, and the O atom to be attracted to the -ve field.
Based on oxidation states (OS) however :
The OS on the C atom is (formal charge + electronegativity consideration) = -1 + +3 = +2
The OS on the O atom is (formal charge + electronegativity consideration) = +1 + -3 = -2
"Electronegativity consideration" refers to the concept that both electrons in the bond pair would be polarized away from the less electronegative atom and towards the more electronegative atom.
Based on OS, you would expect the O atom to be attracted to the +ve electric field, and the C atom to be attracted to the -ve field.
Notice that while the preceding statement appears, ostensibly or superficially, similar to the reasoning of the typical JC student, the critical difference is that, it is reasonable, valid and correct to deduce the polarity (of a polar molecule when exposed to an electric field) based on OS, and based on formal charges, but not (reasonable, valid or correct to base) on imagined partial charges on atoms that actually have formal charges.
Back to the CO molecule : since the two different models/concepts of Formal Charges and Oxidation States lead us to different (to be precise : opposite) conclusions as to the polarity of the CO molecule when exposed to an electric field, what are we to do? Which concept is more correct?
The concepts are both equally correct, because they are two separate and distinct concepts that do not, within their own framework, contradict each other; the more relevant question to ask is which concept outweighs which concept in the case of carbon monoxide being exposed to an electric field?
As it turns out, experimental evidence has shown that (Chemistry is a microcosm of real life, which is not black and white, so similarly for Chemistry : be aware it's often a case-by-case basis, but) in the case of carbon monoxide, the -ve formal charge and +ve oxidation state of the C atom, and the +ve formal charge and -ve oxidation state of the O atom, largely cancel each other out.
But ultimately, experimental evidence does indicate that the effect of the formal charges do indeed (slightly) outweigh the effect of the oxidation states, ie. the C atom in CO is slightly polarized towards the +ve electric field, and the O atom slightly polarized towards the -ve electric field.
Finally, in case any JC student is reading this and feels disturbed by what I've said in this post, do not worry too much : Cambridge will be reasonable. Cambridge will accept and award full marks to the student, regardless of which of the 3 explanations he/she gives : based on formal charges (the most correct), or based on oxidation states, or based on partial charges (the least correct).
As long as your explanation tallies with your prediction (ie. if you say the C atom has a -ve formal charge, but go on to say therefore the C atom would be attracted to the -ve field, you will get zero marks), Cambridge will give you full marks. This is perfectly understandable, because they just want to see if the candidate is able to make the correct conclusion based on the relevant arguement/concept (and since all 3 arguments/concepts are arguably, all somewhat reasonable, all 3 arguements/concepts will be accepted by Cambridge).
Edited to add :
In PCl5, the P atom has 0 lone pairs and 5 bond pairs. According to this structure :In terms of a "stable octet", we say it has an "expanded octet" of 10 valence electrons. But in terms of "formal charge", we say it has 5 valence electrons.
Oxidation State (OS) = formal charge + electronegativity consideration
Thus the OS of the P atom in PCl5 = (0 + +5) = +5
In terms of stable octet, it hasn't lost any electrons at all (in fact, it's gained another 5 electrons!)
In terms of formal charge, it hasn't lost any electrons at all.
In terms of oxidation state or oxidation number (OS or ON), it has lost 5 electrons to the more electronegative chlorine atoms.
So has the P atom actually lost any electrons or not?!? It depends on whether you're looking at it from the viewpoint of stable octet, formal charge, or oxidation state.
Have you lost or gained when you donate some cash to the poor? In material/monetary/financial terms, you've lost. In compassionate/ethical/spiritual terms, you've gained. Whether it's chemistry or real life, It's always a matter of perspective.
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Topic : Equilibria
Note that Kstab (stability constant) = Kf (formation constant) = the Kc value for the formation / stability of any complex ion. You should be able to write out the Kstab equation for any complex ion.For more info :
http://en.wikipedia.org/wiki/Stability_constants_of_complexes
1. The DiNitrogen TetrOxide Gas QnIf 0.50g of dinitrogen tetroxide is placed in a vacant 2dm3 container at 25°C, molecular dinitrogen tetroxide dissociates partially into molecular nitrogen dioxide. Calculate the partial pressure exerted by the dinitrogen tetroxide at equilibrium, given that the Kp for the reaction at 25°C is = 0.114 atm.
Ans : PN2O4 = 0.0349 atm2. The Cadmium Sulfide Ppt Qn
At 25°C, the Kc for the following reaction is 1 x 1081 :
2H+(aq) + 2NO3-(aq) + 3H2S(aq) ↔ 2NO(g) + 4H2O(l) + 3S(s)Given the dissociation constant of hydrogen sulfide is 1 x 10-19 and the solubility product of cadmium sulfide is 8 x 10-27, calculate the Kc for the following reaction :
3CdS(s) + 8H+(aq) + 2NO3-(aq) ↔ 2NO(g) + 4H2O(l) + 3S(s) + 3Cd2+(aq)
Ans : Kc = 5.12 x 10^593. The MonoHydroxoChromium(III) Ion Qn
At 25°C, calculate the pH of a 0.10 mol/dm3 solution of Cr3+(aq) ion, given that
the Kstab of the dipositive monohydroxochromium(III) ion = 1.0 x 1010.You may assume the counter anion does not undergo significant hydrolysis.
Ans : pH = 2.5
4. The TetraamineCopper(II) Ion QnGiven that the mass solubility of copper(II) hydroxide in pure water is 3.315 x 10-5 g, what molarity of ammonia must be maintained at equilibrium in a solution in order to fully dissolve 0.10 mol of copper(II) hydroxide in 1dm3 of solution?
Given that the Kstab of tetraaminecopper(II) ion = 1.0 x 1012. All temperatures are at 25°C.
Ans : 12.7 mol/dm3 (since the molarity of NH3 must be slightly greater than 12.63 mol/dm3)
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A rant regarding how secondary schools and JCs teach poorly in the topic of electrolysis.
I teach my 'O' and 'A' level students to be able to understand the concept and to use the Data Booklet's standard reduction and oxidation potentials, to work out which species is reduced at the cathode, and which species is oxidized at the anode.At 'O' levels, the term "discharged" is often used (eg. "Cu2+ instead of Fe2+ is preferentially discharged at the cathode since Cu is less reactive than Fe" is a statement commonly taught at 'O' levels), which is not as comprehensive or as correct as "reduced" or "oxidized".At 'A' levels, this topic is very poorly taught in most JCs, which wrongly teach 'A' level students the use of the erroneous term "eeee-nought" or "eeee-naught" when it should be pronounced as "E-standard" and symbolizing either "standard cell potential" or "standard reduction potential" or "standard oxidation potential" or "standard electrode potential" all 4 terms actually having different definitions in electrochemistry.For instance, in the purification of copper :At the anode, Cu is preferentially oxidized to Cu2+, and less reactive metals such as Ag is not oxidized to Ag+, because the oxidation potential of Cu to Cu2+ is more positive (under standard conditions : -0.34V) than the oxidation potential of Ag to Ag+ (under standard conditions : -0.8V). Consequently, less reactive metals such as Ag, are left unoxidized at the anode and fall (due to greater density of metals compared to the aqueous electrolyte solution) to the bottom of the setup as anodic sludge, which is later processed to extract the more valuable unreactive metals.At the cathode, the Cu2+ is preferentially reduced to Cu, and the ions of more reactive metals such as Fe2+ (which were previously oxidized at the anode together with copper) are not reduced to Fe, because the reduction potential of Cu2+ to Cu is more positive (under standard conditions : +0.34V) than the redution potential of Fe2+ to Fe (under standard conditions : -0.44V). Consequently, ions of more reactive metals such as Fe2+, are left unreduced at the cathode, and remain in the electrolyte solution.In this way, impure copper at the anode is transferred to form pure copper at the cathode. -
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Originally posted by qdtimes2:
1. Below shows the electronic configuration of the three d-block elements in the Periodic Table.
Element X: [Ar]3d^7 4s^2
Element Y: [Ar]3d^8 4s^2
Element Z: [Ar]3d^10 4s^1
Which one of the following statement is incorrect?
A The electronic configuration of central metal ion for [Y(CN)6]4- is [Ar]3d^8
B Upon reduction from ZCl2(aq) to [ZCl2]-(aq), the solution turned colourless.
C The E* value of the X3+/X2+ is less positive than that of Z3+/Z2+
D X is likely to exist as K2X2O7.
Answer is D. I know B and C are out, but I chose A. I thought Y would lose 2 electrons to form Y2+, then forming the complex ion, which has the correct electronic config?
2. 2,2-dimethylpropane reacts with chlorine gas in the presence of ultraviolet light to give a mixture of products. Which one of the following statements is correct regarding this reaction?
A. Only one mono-substituted product is formed.
B. Only the propagation steps involves C-Cl bond formation
C. Only the initation step of the mechanism involves homolytic fission.
D. Both propagation and termination steps produce hydrogen chloride.
Answer is A. I chose C, because I thought the one in propagation step only produces 1 radicals; homolytic fission must produce 2 radicals right? Why is it A anyway? I thought they can be a lot of chlorine substitution going on in one molecule.
3. An organic compound X undergoes the following reactions.
i) It decolourises a solution of bromine in tetreachloromethane.
ii) It reacts with phosphorus pentachloride giving copious white fumes of HCl.
iii) It reacts with hot alkali to give a compound with two alcohol functional groups.
Which compound could be X?
A. HOH2CCH=CHCH=CHCH2Cl
B. Cl2CHCH=CHCH2COOH
C. BrCH2CH2CHClCH2COCl
D. ClCH2CH2Ch=CHCH(Cl)CH2OH
Answer is A. I know C and D are out, but gosh, A and B seems to be both correct!
4. (Choice A- statements 1,2,3 are all corect. B - statement 1,2 are correct. C - statement 2,3 are correct. D - Only statement 1 is correct)
GABA is a neurotransmitter released by red algae which encourages shellfish larvae to settle on the ocean bed.
GABA: H2NCH2CH2CH2CO2H
Which of the following statements are correct?
1. It is a 2-aminocarboxylic acid.
2. It is soluble in water due to zwitterion formation
3. It migrates to the anode of an electrolytic cell at pH2.
Answer is C (statement 2,3). I know 3 is correct, but how can it form a zwitterion like that? I thought there must be a so called alpha carbon, where the NH2 and COOH group must belong to the same carbon atom. What's a aminocarboxylic acid too? :/
sorry for the wall of texts, thanks for your replies man. these are from Vjc 2009 prelim paper.
Q1.
The electron configuration of an uncomplexed ion (ie. without ligands) and a complexed ion (ie. with ligands) is different : dative bonding means the electrons of the ligands enter into the vacant orbitals of the metal ion.Q2.
Initiation and propagation both involve bond homolytic cleavage (or fission, but cleavage sounds sexier). When only monosubstitution occurs (eg. excess alkane, limited halogen), then for 2,2-dimethylpropane, all hydrogens are equivalent (ie. gives the same product).Q3.
Option B generates a geminal diol, which readily dehydrates to form a ketone (or in this case, aldehyde).Q4.
Aminocarboxylic acid means both the amine group and the carboxylic acid group are present in the same molecule. In organic chemistry, the alpha carbon is desginated as the C atom directly bonded to the functional group. For amino acids of proteins, the highest priority goes to the carboxylic acid group, and the alpha C atom is the one bonded to the carboxylic acid group, and notice that (for amino acids of proteins) the amine group happens to be directly bonded to the alpha C atom as well, thus amino acids of proteins are called "alpha amino acids".
Any amino acid (whether alpha or beta or gamma, etc) are capable of forming zwitterions, because the amine group is basic and wants to accept a proton, and the carboxylic acid group is acidic and wants to donate a proton. When the amine group is protonated, the N atom gains a +ve formal charge. When the carboxylic acid group is deprotonated, the O atom gains a -ve formal charge. Thus forming a zwitterion. -
forgiveme posted :
Hey , I hope To clarify some Of my chem doubts . Hope u guys can help me me
Most qns are on inorganic chem .
1. Grp 7
How come reactivity of halogen with hydrogen must be explained with bond energy and not by their oxidizing strength ?when to use oxidingsing strength and when to use nod energy ?2How come complex silver diamine ion is colorless ? I tot complex is colored ? O.o
3.displacement rxn
State obv when Excess cl2 bubbled In kl right
So chlorine become clorine ion , does the ion has colour ?
And how come iodine gas form still need to react with iodine ion to form i3^-when cl2 is in excess . Isn't the iodine ion complete gone ?Last, sodium iodine and conc h2so4 will form iodine gas right?
Home come my lecture notes wrote violet fumes and black solid ?Tats about it . Pls help me . T.T
Q1.
2 reasons why bond energy is more directly relevant than redox potentals :
- the bond must first be broken before redox can occur.
- if u wanna quote data booklet redox potentials, the reactants or products must be in the aqueous state (as written in the redox equation), but the reaction of halogen with hydrogen does not involve the aqueous state.Q2.
Ag+ has completely filled d-orbitals. Not all complex ions have partially filled d orbitals, which is a prerequisite for colours arising from d-d* transitions.Q3.
Cl- ion is colourless. The I2 generated may be attacked by the I- nucleophile to generate the I3- ion, but unless the exam questions specifies you need to talk about the I3- ion, you can ignore its formation and just write the product as I2.Q4.
Black solid = solid iodine. Violet fumes = gaseous iodine.
forgiveme posted :Hey thanks alot
Is it true tat all trans element below cobalt Is fill 3d ?Eg Ag hereAnd is all halide ion colour colorless ? And for this type of qns we only need to state the colour change for the halide gas changed right ? The salt no need right ?
Q4:bt the equation suggests that the iodine formed
Is a gas . So I jus right violet fumes can alr right ? Or need to state both ?Thanks again
Q1. No, case-by-case basis. But barring further anomalous patterns for transition metals in period 5 and above (the anomalies for period 5 elements and beyond, akin to Cr and Cu of period 4, are beyond the A level syllabus), you should be able to work out the electron configuration based on the element's position in the periodic table.Q2. Yes all halide ions are colourless. Depends on the question. Sometimes, in trickier questions, you must consider the colours of both species present, eg. when unipositive VO2+ reacts with Fe2+, you get blue dipositive VO2+ and yellow Fe3+, the mixing of colours would result in a green solution. The exam smart candidate will state all 3 colours, with explanation (ie. specify that one product VO2+ is blue, the other product Fe3+ is yellow, and therefore the mixing of both will result in a green coloured solution), to secure full marks.
Q3. Being kiasu is a good trait for A level exams. Best to state both, but at a minimum, you're right that it's more important to say violet vapour of I2(g) is observed, rather than black solid of I2(s).
forgiveme posted :
Alright ! I understand alr !
Another qns is that
The redox potential for [fe(h2o)]^2+ is +0.77
redox potential for [fe(cn)6]^3- is +o.36v
Suggest a difference .
My tut and state that it's cos cn is a stronger ligand and hence more stable , les easily reduced
But , the ans in tys suggests that it's cause of the negative sign in cn complex,harder to accept electron .
Hmmm , so which one is right ?2. When comparing ionic radius between anion and cation right ,
Anion has a larger bond length becos is has one more quantum shell right ? Bt how come some ans says it's cos of electron-electron repulsion ?Haisss .
Sorry to bother u .
Q1. Both are correct, and contribute to the difference in the reduction potentials, and Cambridge expects the 'A' level candidate to know both answers. In the 'A' level exams, you should give both answers, to secure full marks.Q2. Again, both are correct. But the more direct & important reason is that the anion has one quantum shell more than then cation. The greater shielding effect (ie. inter-electron repulsion) only serves to [I]additionally[/I] increase the anionic radius by a little bit extra. (Eg. if each quantum shell measures an additional 1cm in diameter, then a certain cation would be 2cm, and it's anion would be 3.2cm instead of 3cm; the additional 0.2cm caused by additional shielding effect).
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Guys like Girls, Nucleophiles like Electrophiles.
Guys are nucleophiles, girls are electrophiles. Guys like girls, so nucleophiles like electrophiles. Only nucleophiles got the 'balls' to attack electrophiles (so the mechanism is always a curved arrow from the 'balls' of the nucleophile to the electron-deficient electrophile), the 'balls' (usually) referring a lone pair, though it can sometimes be a pi bond, as in the case with alkene guy (nucleophle) or benzene guy (nucleophile).
Water is more masculine (nucleophilic) than feminine (electrophilic), since the O atom in the middle can withdraw electrons inductively from both H atoms, resulting in the partial -ve charge on the O atom to be twice as great as the partial +ve charge on each H atom.
The Si atom of SiCl4 is in period 3, and thus has vacant d orbitals (remember : as long as you're in period 3, you have energetically accessible s, p and d orbitals, even if your p orbitals are empty). Because Cl is more electronegative than Si, and because there are 4 electron-withdrawing by induction Cl atoms, the Si becomes extremely feminine (electrophilic), and thus is readily attacked by water nucleophiles (guys can't resist such femininity). Isn't this the same scenario with CCl4? Then why doesn't CCl4 undergo hydrolysis readily, like SiCl4?
The difference between the propensity for hydrolysis of SiCl4 versus CCl4 (the former is readily hydrolyzed while the latter is not), is due to 2 different reasons (give both reasons in the A levels if asked), which are :
Reason #1 - Si is able to accept the dative bond from the water nucleophile into its vacant, energetically accessible d orbitals without needing to simultaneously cleave its existing bonds with the Cl atoms (something which the C atom cannot do, as it does not have energetically accessible d orbitals). This means that the hydrolysis of CCl4 has a much higher activation energy (since it needs to first cleave its existing C-Cl covalent bonds in order to accept a new dative bond from the incoming water nucleophile) compared to the hydrolysis of SiCl4 (as the inter-electron repulsion between the new dative bond formed and the existing Si-Cl bonds will lower the activation energy required to cleave the old Si-Cl bonds). Bottomline : the lower the activation energy, the more readily hydrolysis can occur.
Reason #2 - The Si-Cl bond is longer than the C-Cl bond, since Si is in period 3 while C is in period 2. But be careful not to make the mistake of saying "[I]therefore the Si-Cl bond is weaker than the C-Cl bond[/I]" because if you check the Data Booklet, you may be surprised to note that the Si-Cl bond is actually stronger than the C-Cl bond, for reasons beyond the scope of the A level syllabus. So the bond strength isn't the reason for the difference in propensity for hydrolysis, but the bond length still is. How so? Using an analogy, if it's raining heavily in a windy thunderstorm and you don't want your face to get wet, do you put your umbrella near your face or far away from your face, to protect your face from the water nucleophile? Similarly, because Cl is more electronegative than both C and Si, the 4 Cl atoms arranged tetrahedrally about the central C or Si atom, are partial negatively charged, and thus can repel the incoming water nucleophile (remember : guys only like girls, guys don't like guys; the delta -ve Cl atoms repel the delta -ve O atoms of the water nucleophile) so the partially negatively charged Cl atoms function like 4 little umbrellas around your face, shielding your face from water nucleophile during a windy thunderstorm). And because the C-Cl bonds are shorter than Si-Cl bonds, the Cl atoms provide more effective shielding (like an umbrella held more closely to your face) against the incoming water nucleophile in CCl4 compared to in SiCl4.
Guys always like girls. It's Chemistry. -
Originally posted by Flaw ng:
Hi, i am a A level student and would like to ask this question about arenes.
In arenes, we have learnt about activating and deactivating groups when attached to a benzene ring can have various effects on the rate of electrophilic substitution.
My question is that are groups like -CH2OH, -CH2Cl , -CH2OR ( where R is a R group like an alkyl ) activating or deactivating to the benzene ring?
Can someone please tell me the answer and explain the reasoning behind the answer? please help :D thx
All 3 abovementioned groups are electron withdrawing overall, because they withdraw by induction (since the C atom is partially positively charged, due to the greater electronegativity of the O, Cl and O atoms respectively) and they neither donate nor withdraw by resonance (to donate by resonance the atom directly bonded to the benzene ring must have a lone pair; to withdraw by resonance, the atom directly bonded to the benzene ring must have a pi bond that can shift away when it accepts electrons from the benzene ring by resonance to avoid violating its octet for period 2 elements).Since these groups withdraw by induction (and neither donate nor withdraw by resonance), they are thus meta-directing.
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Nish asked :
1. When AgNO3 (aq) is mixed with MgCl2(aq), a white ppt forms which is soluble in NH3(aq). When AgNO3(aq) is mixed with FeCl2(aq), a grey ppt forms which does not dissolve in NH3(aq).
Which statements about one or both of these reactions are correct?
1 Ag+ ions act as an oxidising agent.
2 Ammonia forms a soluble complex with silver metal.
3 Fe2+ ions act as a catalyst in decomposing AgCl.---------
2. Given weighed samples of the same mixture of MgCO3 and BaCO3, how can the mole fraction of MgCO3 in the mixture be estimated?
1 Add a known volume of [HCl= 0.1], in excess, and back titrate the excess of the acid.
2 Add an excess HCl and measure, at known temperature and pressure, the volume of CO2 liberated.
3 Add an excess of HCl followed by an excess of H2SO4, filter, dry and weigh the ppt.---------
3. Two students separately have available equal volumes of 0.1moldm-3 silver nitrate, sodium ethanoate and KBr.
The first student, on mixing the sodium ethanoate and silver nitrate, obtains a white ppt. On adding KBr, the ppt turns cream.
The 2nd student adds AgNO3 to KBr and obtains a cream ppt. On adding CH3COONA, there is no further change.
Which statements about these observations are correct?
1 Silver ethanoate is insoluble.
2 Silver bromide is less soluble than silver ethanoate.
3 Ethanoate can oxidise bromide.For this question I'm not sure about statement 1 and 2.
Thanks! :qmarkshy:
Q1. Only option 1 is correct. Option 2 is wrong because NH3 ligands complex with Ag+ ion, not Ag metal. Option 3 is nonsense because Fe2+ ions are not regenerated, furthermore AgCl can't be decomposed if it isn't even formed in the first place (since the redox reaction between Ag+ and Fe2+ occurs instantaneously).Q2. All options 1, 2, 3 are correct. Solve simultaneous equations (algebraic variables x and y) in all 3 options, and the relative amounts of both carbonates can be determined.
Q3. Option 1 is poorly phrased, but can be considered correct if the word "insoluble" is replaced with "sparingly soluble". Option 2 is correct, as evidenced by the 2nd and 3rd paragraphs of the question. Option 3 is nonsense, as no redox reaction occurs during ionic precipitation reactions.
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Originally posted by Flaw ng:
However i checked several sources online, and it is mentioned that groups like
-CH2OH, CH2Cl, CH2OR are ortho and para directing which is really a contradiction to what you've said.
Could it be the "hyperconjugation" effect as mentioned in some of the sources i checked?
please help me out, im confused
If the OH, Cl and OR groups are directly bonded to the benzene ring, then because these substituents are electron donating by resonance (since they have a lone pair on the atom directly bonded to the benzene ring), even though they are electron withdrawing by induction (since O and Cl are more electronegative than C), be aware that resonance effects outweigh inductive effects unless the substituent is a halogen*, accordingly these 3 groups are ortho-para directing.Summary : as long as you donate electrons by resonance, you're ortho-para directing. Halogens are ortho-para directing even though they're deactivating (ie. deactivates the benzene ring making it a weaker nucleophile). Halogens are deactivating because they withdraw by induction more strongly than they donate by resonance*. But because halogens donate by resonance, they're still ortho-para directing.
* To understand why halogens withdraw by induction more strongly than they donate by resonance, you have to consider two cases :
Case #1 : Fluorine.
F is the undisputed king of electronegativity. Hence, it is unsurprising that even though it donates well by resonance, it withdraws even more strongly by induction.Case #2 : Chlorine, Bromine and Iodine.
Cl, Br and I are not so strongly electronegative. Cl and Br are both less electronegative than Nitrogen; and Iodine is (something that most JC students don't realize) even less electronegative than Carbon. If their (electron-withdrawing) inductive effects are so weak, then why don't their (electron-donating) resonance effects outweigh their weak inductive effects? Answer : because their (electron-donating) resonance effects are even weaker! Why? Consider the sideways or side-on overlap of unhybridized p orbitals of the halogen and the C atom (of benzene) to form the pi bond (by resonance). Notice that in the case of Cl, it would involve the sideways or side-on overlap of the 3p orbital of Cl with the 2p orbital of C. Because 3p orbitals are significantly more diffused than 2p orbitals, the sideways or side-on overlap is ineffective and the pi bond formed is weak. Therefore, Cl is poorly electron-donating by resonance. And accordingly, Br and I, are naturally even worse electron-donaters by resonance.Hyperconjugation refers to a 3rd modality in which groups may donate electrons, in addition to by induction and by resonance. Specifically, this term is used to describe how alkyl groups donate electron density to stabilize carbocations. This is the more correct reason why a tertiary carbocation is more stable than a secondary carbocation is more stable than a primary carbocation is more stable than a (pop quiz : what's one level below primary? kindergarten? nope, its...) methyl carbocation. But for H2 'A' level purposes, you don't have to use this term "hyperconjugation", you can use the simpler concept of "induction" to explain why alkyl groups stabilize carbocations and why alkyl groups on a benzene ring are both (weakly) activating and also ortho-para directing.
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ForgiveMe asked :
1. Ticl3 is colored , TiF4(ionic bond) is not. How come TiF4 have no 3d electron?
2. Also, regarding the shape of complexes, when there are 4 ligands, how do u know it's sq planar and tetrahedral?
3. How do I know how many valence electron does. Nickel have , when it has variable oxidation state ? 0.o?
4. First order reaction with respect to halogenoalkane and zero order with respect to NaOH, suggests that the reaction is SN1. How come ah? Is there a link?Q1. Ti is [Ar] 4s2 3d2, hence Ti4+ is [Ar] 4s0 3d0, hence no d-d* transitions.
Q2. This topic on complex ions is very complex, most of it involves concepts and calculations beyond A levels (eg. crystal-field theory, ligand-field theory, Jahn-Teller effect, etc). For A levels, you have no choice but to memorize a list of complex ions and their shapes. You can check out Jim Clarke's A level Chem website, and Rod Beavon's A level Chem website, for a list of complex ions commonly tested at A levels.
Q3. You should be able to work out the electron configuration for ground state atoms and ions, based on the element's position in the periodic table.
Ni is [Ar] 4s2 3d8
Ni+ is [Ar] 4s1 3d8
Ni2+ is [Ar] 4s0 3d8
Ni3+ is [Ar] 4s0 3d7
Ni4+ is [Ar] 4s0 3d6
etcQ3. Yes of course there's a link. If you don't see the link, that means either you don't truly understand kinetics, or you don't truly understand organic chem mechanisms, or (most likely) both. (Unfortunate fact is, the vast majority of JC students don't truly understand chemistry even after taking it at A levels, they just memorize notes. So it's not your fault, it's just the Sg education system).
In the SN1 mechanism, the rate determining (ie. slow) elementary step involves only the electrophile (the alkyl halide), not the nucleophile (the OH- ion). Therefore, the rate equation is :
rate = k [electrophile]^1 [nucleophile]^0
which is same as
rate = k [electrophile]^1
which is same as
rate = k [alkyl halide]^1
which is same as
rate = k [alkyl halide]
Note : "alkyl halide" means "halogenoalkane". -
blacktoast94 posted :
i cant' do this super basic qn. :(
why CH3CH(CL)COOH and CH3CH(OH)COOH are capable of dissolving in an organic solvent?
i thought the interactions between them are id-id interactions hence don't release sufficient energy to overcome the H-bonding between the solute molecules. :/
First of all, organic doesn't mean non-polar. It means "carbon containing". Many organic solvents are at least slightly polar and/or capable of limited hydrogen bonding (eg. ethers cannot donate hydrogen bonds but can accept hydrogen bonds).Secondly, even if enthalpy change may be slightly unfavourable (for some solutions), but often enough, the favourable entropy change may outweigh the unfavourable enthalpy change, to give a favourable Gibbs Free Energy change for the dissolving process.
Thirdly, Both carboxylic acids mentioned above have an ethyl group that is capable of instantaneous dipole - induced dipole van der Waals interactions (do not use acronyms like id-id, it's a bad habit) with the alkyl groups of the organic solvent (which you didn't specify).
Remember that as the alkyl group gets longer, the overall polarity of the molecule decreases significantly, even with a polar group such as OH or COOH present.
Fourthly, the Cl atom of the chlorocarboxylic acid is large and has significantly polarizable electron charge clouds, allowing for significantly favourable instantaneous dipole - induced dipole van der Waals interactions with the organic solvent.
Fifthly, some of the hydroxycarboxylic acid's capacity for hydrogen bonding has been used up for intramolecular hydrogen bonding, consequently resulting in less extensive intermolecular hydrogen bonding. In effect, there is much less hydrogen bonding between molecules of the hydroxycarboxylic acid required to be overcome by the organic solvent.
Sixthly, it really does depend [B][U]a lot[/U][/B] on the exact organic solvent used. Did you know, that phenol, phenylamine and benzoic acid are a lot more soluble in ether solvent, than in either aqueous (water) solvent or CCl4 (non-polar) solvent? This is because both (ethers and the 3 compounds mentioned) are both less polar than water, but more polar than CCl4 solvent.
As a general rule, partially polar compounds (ie. capapble of some hydrogen bonding, but has a large, non-polar benzene ring or hydrocarbon chain, such as phenol, phenylamine and benzoic acid) are more soluble in ether than water, while ionic species (eg. phenoxide ion, phenylammonium ion, and benzoate ion) are more soluble in water than ether.
Why? Because ionic species can form ion - permanent dipole interactions with water, which are stronger than the hydrogen bonding between water molecules. (You should be aware however that hydrogen bonding is generally stronger than ion - induced dipole interactions).
Cambridge could very likely set such an exam quesiton on "Separation of Organic Compounds by controlling the pH of Ether and Aqueous Solutions" in the upcoming A level exams.
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Blacktoast asked :
when BECL2 and CH3NH2 form a dative bond, how many hydrogen bond it is capable of forming?
"How many H bonds" vs "How many types of H bonds" are 2 different qns.
Furthermore, whether these are H bonds between molecules of the same species, of H bonds between these species and another species, eg. a solvent (eg. water), are again 2 different qns.
The correct answer, to the qn "how many H bonds may be formed between the molcule that is generated when 2 methylamine molecules complex with 1 beryllium chloride molecule" is ZERO.
This is because the N atom would have gained a +ve formal charge and have no more available lone pair for H bonding, and therefore cannot have electrostatic attraction with the partial +ve H atoms of a neighbouring complexed molecule of the same species. The Cl atoms are indeed partial -ve, but Cl is not electronegative enough to qualify for H bonding (at least for the 'A' level H2 syllabus).So the only interactions between the complexed molecules are permanent dipole - permanent dipole van der Waals, not hydrogen bonding.
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Dixoria asked :
Im pretty confused between the terms hydrolysis, hydration as well as solvation.
Does hydration = solvation?
So far some hydrolysis reactions i've encountered includes
1)reactions with water (Inorganic chem- complete hydrolysis of SiCl4 to form acidic soln)
2)acidic/basic hydrolysis of esters/amides
3)Partial hydrolysis(polarisation of O-H bond in [Mg(H2O)]2+ to form [Mg(H2O)5(OH)] + H+)
Is hydrolysis a chemical reaction whereas solvation is not?
And is (3) a considered a chemical reaction?
Why are there so many different type of "hydrolysis" reactions?The use of the term "hydration" by organic chemists in "hydration of alkenes" is anomalous and technically improper, and physical and inorganic chemists close one eye to the improperiety of this usage by organic chemists.
Apart from "hydration" of alkenes, in all other branches of chemistry, the difference between "hydrolysis" and "hydration" is :
Hydration is a subset of solvation, and refers to physical interaction between the solute and solvent.
Hydrolysis refers to chemical reaction with water. However (as with hydration), the term hydrolysis is also often loosely used.
For instance, in the hydrolysis of alkyl halides, the nucleophile used is often the OH- ion rather than water (OH- is a stronger nucleophile since the O atom has a -ve formal charge, whilst in water the O atom only has a -ve partial charge).
While the early use (in some branches of chemistry) of the term "hydrolysis" stemmed from the "lysis" of either water or the compound (reacting with water), but the term "hydrolysis" has now evolved to a more functional meaning of "chemical reaction with water".
For the difference between solution and solvation (or hydration, if the solvent is water), read Jim Clarke's webpage on "Solubility of Group II compounds" which will be helpful for 'A' level H2 Chem students :
http://www.chemguide.co.uk/inorganic/group2/problems.html -
Originally posted by Xavolon:
I'm really confused with whether Potassium or Lithium is a stronger reducing agent.
I've read about Standard Reduction Potential(althought I don't understand much) and it says that Lithium with the lowest reduction potential is the stronger reducing agent.However, according to the Reactivity Series, Potassium is more reactive than Lithium and thus is a stronger reducing agent.
Can anyone clarify this with me?Thanks in advance.
It so happens that this particular comparison you've brought up (or to be precise : Lithium), is problematic. Yes, K is indeed more reactive than Li (due to less endothermic 1st ionization energy of K due to greater distance between the +vely charged nucleus and valence shell, and greater shielding effect from greater no. of intermediate electron shells).
However, it is also true that the standard oxidation potential of Li is more positive than that of K. This may indeed be correctly interpeted as an apparent contradiction, at A levels. As it turns out, the reason for this contradiction has to do with the high charge density of the Li+ ion, resulting in extraordinarily strong ion-permanent dipole interactions with water, in the aqueous state. This phenomenon additionally favours the oxidation of Li to Li+ in the aqueous state (over K to K+), even though the 1st ionization energy of Li to Li+ is significantly more endothermic (compared with K to K+) in the gaseous state.
Therefore, for 'O' level purposes, it would suffice to simply state that because K is indeed more reactive than Li, hence K is a stronger reducing agent than Li. For 'A' level purposes, the high achieving student should be able to describe this apparent contradiction and explain it both ways.
Other than this particular problematic case of Li, you (both 'O' and 'A' level students) can use the Standard Redox Potentials accordingly, as per the example below :
Because a reducing agent is oxidized during the reaction, thus it is more appropriate to compare oxidation potentials (rather than reduction potentials). Since you're taking the reverse direction to these equations given, accordingly you should take the -ve sign for these potentials.
Eg. Based on the image below, since the reduction potential of Ag+ to Ag is +0.8V, hence the oxidation potential of Ag to Ag+ is -0.8V.
Let's put aside Lithium for the moment, and compare Na versus K.
Since the oxidation potential of Na to Na+ is +2.71V, while the oxidation potential of K to K+ is +2.92V, we see that K thus has a more positive potential (ie. greater potential = greater tendency = greater inclination = greater propensity) to be oxidized, compared to Na.
Therefore K is a stronger reducing agent than Na.
Now try it out for yourself, say, is bromine or iodine a stronger oxidizing agent? This time round, since oxidizing agents are being reduced, you should compare reduction potentials, not oxidation potentials.
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A pure sample of N2O4(l) is introduced into an evacuated vessel. The vessel, of constant vol, is heated to a constant temp such that the equilibrium below is established.
2NO2 (g) <--> 2NO (g) + O2(g)
The value of the pressure p is then found to be 20% greater than if only NO2 (g) were present.
What is the mole fraction,x, of oxygen in this equilibrium mixture?
A) 0.17
B) 0.20
C) 0.33
d) 0.402008 Nov P1 Q7
CS Toh's website : http://www.post-1.com/step-by-step/?a=solutions