BedokFunland JC's A Level H2 Chemistry Qns (Part 2)
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Forgive me posted :
How come the species with high boiling point will be more easily liquified?
Liquified means solid to liquid right? Or from gas to liquid? O.o Thanks in advance ! :)
"Liquified" means "to become liquid" (from either gas or solid).Ask yourself : What's a liquid? It's the state of matter when the molecules are so close together that we call it a liquid instead of a gas (but still moving fluidly rather than vibrating about fixed positions, as per the solid state). Then ask yourself : why or how do molecules of a liquid manage to stay so close together? It's because of the intermolecular van der Waals /or hydrogen bonding that holds them close toegther?
So you realize that the stronger the intermoleculer van der Waals or hydrogen bonding, the easier it is to keep the molecules together in the liquid state (as opposed to weaker van der Waals which allow the molecules to fly apart into the gaseous state), which is to "liquify" (or "liquefy") it.
And if the intermoleculer van der Waals or hydrogen bonding is strong, what can you say about the boiling point? Higher, of course! Because it takes more energy to overcome such intermolecular attractions in order to boil the species (eg. H2O, NH3, etc).
Now do you see the link between boiling point and how readily a particular species is liquified? Chemistry is all about seeing links. If you can see the links between every topic of Chemistry (physical chemistry, inorganic chemistry, organic chemistry), then you would not only truly understand and appreciate Chemistry, you will enjoy it so much that getting an A grade will be (relatively) effortless.
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Forgive me asked :
Which following sew give best yield of 2 bromo-4- nitro methyl bezene?
A) alkylation -nitration - bromination
B) alkylation bominaton nitration
Ans is a becos it's major product. Bt how do I know it's major product . And after I add nitro into my methyl benzene , should I do 2-4 or 3-5 ??which one should i follow?
Alkyl groups are electron donating by induction (actually hyperconjugation, but this term is not required at H2 level), hence they are mildy activating and are ortho-para directing.
Halogen groups withdraw electrons by induction more strongly than they donate by resonance, hence they are mildly deactivating but because they donate by resonance, they are still ortho-para directing.
Nitro groups withdraw electrons by both induction (due to a +ve formal charge on N) as well as by resonance (due to the N=O double bond allowing the nitro group to withdraw by resonance without violating N's octet). Accordingly, nitro groups are strong deactivators, and are meta directing.
Activators win deactivators, and ortho-para directors win meta directors.But both option Also form the same product . How u know Which is one gt major whic one dun have?
If you carry out option B : alkylation bominaton nitration, notice that because Br is large and poses significant steric hinderance, the major product will see the Br+ electrophile substituting away the para (rather than ortho) proton. In contrast, in option A : alkylation -nitration - bromination, because the nitro group is also fairly large and will more likely take up the para position, the subsequent Br+ electrophile will be forced to take up the ortho (relative to the methyl) position.
Q2) which pair of reagent from
Carbocations
A) CH3COCH3 + HCN
B) CH2=CHCH3 + HBr
C) CH3CH2Cl + NaOHIsnt it all suppose to form carbocation? Why only b?
Q2.
A) CH3COCH3 + HCN, upon nucleophilic addition of the CN- nucleophile, the intermediate generated is an alkoxide ion (ie, the -ve formal charge is on the O atom, the C atom does not have a +ve formal charge).
C) CH3CH2Cl + NaOH, because this is a primary alkyl halide, it undergoes SN2 rather than SN1. (Due to both sterics and electronics reasons).Q4) if the decay rate of a radioactive isotope z decease from 1800 coun per
Min to 180 count per min in one week , wad is the half life?Q4.
Molarity (desired) / Molarity (initial) = (1/2)^n
n = 3.322 half-lives = 1 week
Hence half-life = 0.301 weeks -
Originally posted by SgStudentStressed:
is Na2O an alkali or not? cos an an alkali is a soluble base so is NaOH an alkali along with Na2O im veey confused with it. pls help me clarify what is what
An alkali ( noun ) is any solution ( adjective : "alkaline solution" ) which contains a higher concentration of OH- ions over H+ ions.Bases which are able to undergo hydrolysis ( hydrolysis = "chemical reaction with water" ) to generate soluble products are called "soluble bases" because the solid base appears to dissolve in water (it is really undergoing a chemical reaction with water, ie. hydrolysis, to generate soluble products).
For instance, Na2O or sodium oxide, contains the strongly basic O2- dinegative oxide ion, which in a bid to stabilize itself, abstracts protons H+ ions from water, generating 2OH- ions.
O2- + H2O ---> 2OH-
Since NaOH is soluble (unlike say, MgO), the resulting solution is alkaline, ie. has a higher concentration of OH- ions over H+ ions, and has pH greater than 7 (note that pH = 7 holds true for neutral solutions (only) at 25 deg C; therefore at 25 deg C, any pH greater than 7 is considered alkaline).
There are 3 different definitions of acids (and therefore, of bases). These are : Bronsted-Lowry, Lewis and Arrhenius. For 'O' levels, the Arrhenius definition "an acid dissolves in water to generate H+ ions" is often used, even though the other two definitions (Bronsted-Lowry and Lewis), used at 'A' levels, are more correct and useful.
To summarize :
A (Bronsted-Lowry) base is any species that accepts protons H+ ions. A 'soluble' base undergoes hydrolysis ('chemical reaction with water') to generate a solution containing a higher concentration of hydroxide OH- ions over protons H+ ions, which is called an "alkali" or "alkali solution" or "alkaline solution".
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Q1. You never realized, all these years, that the so-called "displacement" reaction is actually redox?!?!? Cl2 is reduced to Cl-, while oxidizing Br- to Br2!!!
Q2. Acid anhydrides are electrophilic. That's something which you should be able to figure out by analyzing its structure. Guys like girls, nucleophiles like electrophiles. Only nucleophiles have the 'balls' (the 'balls' usually refer to a lone pair, but can refer to a pi bond in the case of alkene nucleophile and benzene ring nucleophile) to attack electrophiles. A nucleophile such as phenol, can attack acid anhyrdide electrophiles in a similar fashion to attacking acyl chloride electrophiles, to generate a phenate ester.
Q3. Enthalpy of neutralization is defined as the enthalpy change per mole of water generated. So multiply by the coefficient of water in the equation.
Q4. Not required for H2 syllabus. If Cambridge wants you to apply it directly, the question will provide the formula.
Q5. If you keep adding excess strong base, then the pH will continue to rise, approaching (but never reaching) the pH of the strong base used. The OH- ion is a stronger base compared to the basic salt generated, which means if you're asked to calculate the pH at a point beyond the equivalence point (equivalence point = stoichiometric neutralization between acid and base, or oxidizing agent and reducing agent; end-point = point when a suitable indicator changes colour, it only approximates equivalence point; but for 'A' levels both terms may be used interchangeably), just consider the molarity of the stronger OH- ion from the excess NaOH added in, and ignore the contribution of OH- from the hydrolysis of the salt, because as predicted by Le Chatelier's principle, the hydrolysis of the salt is suppresssed by addition of extra OH- ions.
Q6. Yellow. Fe3+ is a clear-cut case. The case of Cr3+ is more probematic. For 'A' levels, you can just state the colour of the hydrated (ie. hexa-aquated) Cr3+ as green, even though strictly speaking and in theory, if all 6 ligands are water, the colour of (hexa-aquated) Cr3+ is pale violet. Problem is, in a percentage of any given sample, the counter anion (eg. Cl-, NO3-, SO4 2-, etc) will always substitute one or more of the water ligands, generating the stronger colour green, which will mask away the weaker pale violet colour. For more info on complex ions and their colours, check out Jim Clarke's website, and Rod Beavon's website.
Q7. Do not use the term "acylation" on its own, like some JCs badly teach. Because the term when used alone is ambiguous : it could refer to either Friedal-Crafts acylation, which involves first the reaction between a suitable Lewis acid catalyst such as AlCl3 and the acyl halide, followed by electrophilic aromatic substitution (recommended you always include the word "aromatic" to remind yourself that the benzene ring is the nucleophile in these reactions, but for 'A' levels Cambridge won't penalize you if you leave out the word "aromatic" in the term) where the benzene ring is the nucleophile. Or the term "acylation" could refer to nucleophilic acyl substitution, the mechanism for which is addition-elimination, and is the reaction responsible for generating esters, generating amides, hydrolyzing esters, hydrolyzing amides, and the reaction of nucleophiles with acid anhydride electrophiles, amongst other reactions.
If Cambridge asks you "what type of reaction is this? acid + alcohol ---> ester + water", the exam-smart candidate will give all 4 of these correct answers, to ensure you secure the full marks : esterification (because an ester is generated), condensation (because a water molecule or small alcohol is eliminated to join two molecules together intermolecularly, or two parts of the same molecule together if intramolecularly), nucleophilic acyl substitution (because it is nucleophilic substitution occuring on acyl group rather than an aliphatic group), and addition-elimination (which is the mechanism for nucleophilic acyl substitution, just as SN1 and SN2 are the mechanisms for nucleophilic aliphatic substitution).
To answer your question : with a Lewis acid catalyst, the reaction proceeds at room temperature, no heating is necessary.
Q8. Do not blindly follow Markovnikov's rule. It may fail sometimes (eg. in addition of HX onto 3,3,3-trifluroprop-1-ene). Rather, understand why it usually works (ie. comparative stability of the two alternative possible carbocation intermediates during the mechanism, which you may be asked to draw out in the 'A' level exam) and you can work out when it may fail.
Q9. Hydrolysis can be acidic (ie. proton-catalyzed) or alkaline (ie. base-promoted). Since esterification (using alcohol and carboxylic acids, instead of acyl halides) employs acidic conditions, therefore acidic hydrolysis of esters are also reversible (ie. it's actually the same reaction under the same conditions! it's only a matter of initial molarities of LHS vs RHS; ie. Qc vs Kc : at initial conditions, does position of equilibrium lie to the left or right?). Accordingly, alkaline hydrolysis of esters is not reversible.
Originally posted by JC2011:Hi, as i’m doing my revision, I realised that I am unsure of the qns below. Thanks.
1) Redox
I understand that Cl2 and Br ion can undergo displacement reactions, but can they undergo redox as well?2) Organic Compounds
Are we required to learn all the reactions of acid anhydrides in H2 chemistry? I only know that it reacts with alcohols & amines.3) Chemical Energetics
Why do we need to multiply the stoichiometric coefficient of the limiting reagent for enthalpy change of reaction and not the enthalpy change of neutralisation?4) Reaction Kinetics
I understand that the Arrhenius equation means that a lower Ka would result in a larger rate constant, k, hence faster reaction. However, how do we apply this equation to a kinetics question?5) Ionic Equilibra
In a titration curve between a weak acid (CH3COOH) and Strong base (NaOH), I understand that salt hydrolysis occurs at the equivalence/end point. However, what happens after end point?6) Transition Elements
I’m unsure of the real colour of [Fe(H2O)]3+, is it yellow or very pale violet?7) Does the process of acylation (Electrophilic substitution) reaction requires heating? other than anhydrous AlCL3?
Does Electrophilic addition of Br2 (aq) (E.A.) follow the markovnikov’s rule?9) Why is the hydrolysis if ester an irreversible rxn and esterification a reversible rxn?
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To secure full marks for this qn (a common A level exam qn), you have to explain :
why formation of CaF2 is more favourable than formation of CaF.
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why formation of CaF2 is more favourable than formation of CaF3.The Universe is a shrewd businessman. It will automatically favour the reactions that generate the most profit, thermodynamically speaking, for itself.
While ionizing Ca to Ca2+ requires more investment compared to ionizing only to Ca+, but the returns when forming much stronger ionic bonds between Ca2+ and F- are much higher (compared to weaker ionic bonds between Ca+ and F-), and therefore significantly increasing the proft margin.
The investments refer to the endothermic ionization energies, the returns refer to the exothermic lattice enthalphies, the profit margins refer to the formation enthalpies.
Hence, the universe finds that it's a far more profitable business venture to manufacture CaF2 instead of CaF.
In the case of CaF3, even though it's true that the ionic bonds between Ca3+ and F- would be even stronger than Ca2+ and F-, and thus the returns would be higher, but because ionizing Ca2+ to Ca3+ requires removing and electron from an inner quantum shell, the endothermic investment is crazy expensive! So much so that the overall formation enthalpy would actually be endothermic! It's an unprofitable, ill-advised business venture, and (unlike say, Temasek Holdings) Mr Universe is way too smart to be duped into such wasteful, loss-making dealings.
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snowflyssb asked :
why does BrCl4- exist but not FCl4-? is it due to difference in electronegativity?
why phosphorus does not occur naturally as P(triple bond)P, as opposed to nitrogen?UltimaOnline answered :
There are several reasons, the most important being the lack of availability of vacant, energetically accessible orbitals required to expand F's octet.
If F had 4 bond pairs and no lone pairs (ie. non-violated octet), then being in Grp VII, it would consequently have a tripositive formal charge, which is unacceptably destabilizing on any atom, let alone F, the most electronegative element in the entire periodic table.
P4 is more stable than P2, while N2 is more stable than N4.
In the hypothetical triply bonded P2 molecule, the pi bonds would be formed by the side-on or sideways overlap of the unhybridized 3p orbitals of the P atoms. Because 3p orbitals are highly diffused (compared to 2p orbitals), the side-on or sideways overlap would be highly ineffective, resulting in extremely weak pi bonds, that fail to effectively stabilize the hypothetical triply-bonded P2 molecule.
In contrast, the sideways or side-on overlap of the unhybridized, less diffused 2p orbitals of the N atoms is effectively, resulting in strong pi bonds that effectively stabilize the triply-bonded N2 molecule.
This is the reason why molecular nitrogen exists as triply-bonded N2, while molecular phosphorus exists as singly bonded P4. In point of fact, this is also the reason why CO2 is simple molecular, while SiO2 is giant covalent. -
forgiveme asked :
1.RBr->R+ + Br-(fast)
R+ +oh- -> ROH ( slow)
Does the rate eqn need to include br-?2) why ch3ch2ch=chcl
Is inert towards hydrolysis?
Q1. No, the rate equation is 1st order wrt OH- and 1st order wrt R+, but since R+ is an intermediate rather than a reactant, you have to substitute R+ with RBr in the rate equation.
Q2. The resistance to hydrolhysis is due primiarly to the partial double bond character between C and X (where X = any halogen), making cleavage of the bond difficult under normal conditions. There are two alternative explanations for this (Cambridge will accept either explanation).Resonance explanation : The lone pair on the halogen is delocalized by resonance to form a pi bond with the sp2 hybridized C (which it can do so without violating the C atom's octet, thanks to the existing pi bond which can delocalize away to form a lone pair on the neighbouring sp2 hybridized C atom), thereby giving the C-X bond, partial double character.
Orbital overlap explanation : Due to the overlap of the pi orbital of the double bond and the p orbital of the halogen, the C-X bond has double bond character.
An additional contributing factor or explanation, which has nothing to do with the strength of the C-X bond, has to do with the fact that in Chemistry, guys only like girls, similarly nucleophiles only like electrophiles. Nucleophiles hate other nucleophiles (since they're not homosexual), and therefore the electron-rich double bond of the alkene nucleophile repels away the electron-rich OH- nucleophile.
Both factors above (ie. partial double bond character of the C-X bond, due to resonanc and/or orbital overlap; as well as the repulsion between the two electron-rich nucleophilic species) result in a significantly increased activation energy required for nucleophilic substitution (including "hydrolysis", if the nucleophile is water or the OH- ion) to occur.
A temperature of approximately 350 deg C and a pressure of approximately 200 atm, is usually employed to hydrolyze halogenobenzenes (aka aryl halides) into phenol.
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'O' and 'A' Levels : Chemistry Calculations
4) Calculate the concentration of a solution of sodium hydroxide given that 25.0 cm3 of it required 18.8 cm3 of 0.0500 mol dm-3 H2SO4.
H2SO4 + 2 NaOH ® Na2SO4 + 2 H2O
5) Calculate what volume of 0.05 mol dm-3 KOH is required to neutralise 25.0 cm3 of 0.0150 mol dm-3 HNO3.
HNO3 + KOH ® KNO3 + H2O
6) A 250 cm3 solution of NaOH was prepared. 25.0 cm3 of this solution required 28.2 cm3 of 0.100 mol dm-3 HCl for neutralisation. Calculate what mass of NaOH was dissolved to make up the original 250 cm3 solution.
HCl + NaOH ® NaCl + H2O
7) What volume of 5.00 mol dm-3 HCl is required to neutralise 20.0 kg of CaCO3?
2 HCl + CaCO3 ® CaCl2 + H2O + CO2
8) 3.88 g of a monoprotic acid was dissolved in water and the solution made up to 250 cm3. 25.0 cm3 of this solution was titrated with 0.095 mol dm-3 NaOH solution, requiring 46.5 cm3. Calculate the relative molecular mass of the acid.
9) A 1.575 g sample of ethanedioic acid crystals, H2C2O4.nH2O, was dissolved in water and made up to 250 cm3. One mole of the acid reacts with two moles of NaOH. In a titration, 25.0 cm3 of this solution of acid reacted with exactly 15.6 cm3 of 0.160 mol dm-3 NaOH. Calculate the value of n.
10) A solution of a metal carbonate, M2CO3, was prepared by dissolving 7.46 g of the anhydrous solid in water to give 1000 cm3 of solution. 25.0 cm3 of this solution reacted with 27.0 cm3 of 0.100 mol dm-3 hydrochloric acid. Calculate the relative formula mass of M2CO3 and hence the relative atomic mass of the metal M.
11) A 1.00 g sample of limestone is allowed to react with 100 cm3 of 0.200 mol dm-3 HCl. The excess acid required 24.8 cm3 of 0.100 mol dm-3 NaOH solution. Calculate the percentage of calcium carbonate in the limestone.
12) An impure sample of barium hydroxide of mass 1.6524 g was allowed to react with 100 cm3 of 0.200 mol dm-3 hydrochloric acid. When the excess acid was titrated against sodium hydroxide, 10.9 cm3 of sodium hydroxide solution was required. 25.0 cm3 of the sodium hydroxide required 28.5 cm3 of the hydrochloric acid in a separate titration. Calculate the percentage purity of the sample of barium hydroxide.
Partial Solutions :
Q4. Calculate moles of H2SO4. Since it's a diprotic acid, hence calculate moles of protons H+ involved, which is also the same as the moles of hydroxide ions OH- involved, which in turn is the same as the moles of NaOH involved. Now that you've the moles of NaOH, and the Qn gives you volume of NaOH, hence determine the molarity of NaOH, commonly written as [NaOH], using the formula : Molarity = Moles / Volume (in dm3).
Q5. Calculate moles of HNO3 present. Since both acid and base are monoprotic, hence determine moles of KOH required. Since molarity of KOH is given, hence determine volume required, using the formula : Molarity = Moles / Volume (in dm3).
Q6. Calculate moles of HCl neutralized, which would be the same as moles of OH- present in the 25cm3 NaOH(aq) aliquot, hence calculate the moles of OH- present in original 250cm3 of NaOH(aq) solution (from which the aliquot was obtained), thereafter multiply by the molar mass of NaOH to determine the sample mass of NaOH required by the qn.
Q7. Let x be the volume (in dm3) of HCl required. Given the molarity of HCl, calculate moles of HCl involved, in algebraic terms. Given sample mass of CaCO3, calculate the moles of CaCO3 (simply divide the sample mass by molar mass). Bearing in mind that the CO3 2- ion is diprotic, hence calculate the moles of HCl required for acid-base neutralization. Equate this value to the moles of HCl in algebraic terms found earlier, and solve for x.
Q8. Let x be the molar mass of the monoprotic acid. Given sample mass, find moles of the acid (simply divide sample mass by molar mass), in terms of algebraic variable x. Given 250cm3 volume, determine molarity of acid in the 250cm3 solution. Since a 25cm3 aliquot was used for titration, calculate moles of H+ (still in terms of x) involved in the titration. Given molarity and volume of NaOH(aq) used in the titration, calculate moles of NaOH involved, and equate to moles of H+ found earlier (in terms of x), and solve for x.
Q9. Find molar mass of hydrated ethandioic acid, in terms of algebraic variable n. Given sample mass, find moles of acid present (in terms of n). Given volume of 250cm3, find molarity of acid in 250cm3 solution, hence determine moles of diprotic acid neutralized in titration (ie. in 25cm3 aliquot). Since ethandioic acid is diprotic, hence determine moles of H+ neutralized in titration (in terms of n). Equate this expression to the moles of OH- neutralized in titration (use the formula : moles = molarity x volume, for the NaOH), and solve for n.
Q10. Let the molar mass of metal M, be x (grams). Accordingly, determine moles of M2CO3 present, in terms of x (use the formula : moles = sample mass / molar mass; note that sample mass is given, and molar mass will be in terms of x). Given 1dm3 solution, calculate molarity (still in terms of x). Given 25cm3 aliquot involved in neutralization, determine moles of CO3 2- ions involved in neutralization (use the formula : moles = volume x molarity). Bearing in mind the carbonate(IV) ion CO3 2- is diprotic, hence calculate moles of H+ neutralized (still in terms of x). Equate this expression to the moles of HCl used (use the formula : moles = molarity x volume), and solve for x.
Q11. Let the % by mass of CaCO3(s) present in the 1g sample be x. Hence determine the sample mass of pure CaCO3(s) present, which is simply (x/100)(1g). Divide this by the molar mass of CaCO3(s) (refer to the periodic table) to obtain the moles of CaCO3(s) involved in neutralization (in terms of x). Find total moles of HCl present (moles = molarity x volume). Find moles of excess HCl by subtracting twice the moles of CaCO3, from the total moles of HCl present. (Why twice? because tghe carbonate(IV) ion CO3 2- is diprotic!). Equate this expression (of moles of excess HCl, in terms of x) with the moles of NaOH(aq) used in titration (use the formula : moles = molarity x volume), and solve for x.
Q12. Similar to Q11, but more involved and convoluted.
Let the % purity, which is the same as % by mass of Ba(OH)2 in the 1.6524 g sample, be x. Hence find sample mass of pure Ba(OH)2 present, in terms of x grams. Hence find moles of Ba(OH)2 present (formula : moles = sample mass / molar mass). Next find moles of OH- present (still in terms of x), bearing in mind that Ba(OH)2 is a diprotic base. Find total moles of HCl used (formula : moles = 0.2M molarity x 0.1dm3 volume). Subtract away moles of OH- present from total moles of HCl present, to obtain moles of HCl in excess (still in terms of x).Given molarity of HCl, find moles of HCl present in 28.5cm3. This would give you the moles of H+ involved in the other titration, which is the same as the moles of OH- involved in this other titration. Given that in this other titration, volume of OH- used was 25cm3, hence determine molarity of NaOH(aq). Hence determine moles of OH- present in 10.9cm3. Equate this value, to the moles of HCl in excess (in terms of x) found earlier, and solve for x.
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Hello.. I want to ask this.. want to double cfm..
for eg. H3C-MgBr
to find the electronegativity is take C-Mg = 1.24?
then CH3CH2OH , how to count the electronegativy? and how to determine whether is non -polar or polar? (Im freaking lost)
what the different btw Intra and inter molecular interaction?Electronegativity is regarding elements, not regarding molecules or bonds.
Polarity can be regarding bonds, or regarding molecules.
There are different versions of electronegativity ratings (eg. Pauling's versus Allen's versus .... etc), with some apparently contradictory results. For 'A' levels, you're not required to do any electronegativity calculations quantitatively. You're only required to do a simple qualitative assessment : eg. since C is more electronegative than Mg, hence the C-Mg bond is polar, and polarized towards the more electronegative C atom.
Alcohols are considered partially polar (the OH group) and partially non-polar (the alkyl group), which makes it a good solvent to use when one reactant is ionic (eg. Na+CN-) and the other reactant is relatively non-polar (eg. R-X). The shorter chain alcohols are considered polar enough to dissolve in water, while the longer chain alcohols are considered non-polar enough such that they do not dissolve in water.
Phenol is polar, but due to the hydrophobic nature of the large, non-polar benzene ring, it is only sparingly solule in water. Adding alkalis to increase the pH of the solution results in the deprotonation of phenol, generating the phenoxide ion, which is capable of ion - permanent dipole interactions with water, which is stronger than the limited hydrogen bonding the (previously protonated form) OH group had with water, and thus adding NaOH(aq) to a precipitate of phenol will dissolve it, since the phenoxide anion is more soluble in water compared to the molecular phenol.
Intra-school competition : competition within a school.
Inter-school competition : competition between school.
Intra-molecular bonds : covalent bonds within a molecule.
Inter-molecular interactions : hydrogen bonding or van der Waals (permanent dipole - permanent dipole aka Keesom forces, or permanent dipole - induced dipole aka Debye forces, or instantaneous dipole - induced dipole aka London dispersion forces) interactions.
Example ;
1,2-dihydroxybenzene has used up some of its capacity for hydrogen bonding, in forming intra-molecular hydrogen bonds. Hence it has less extensive inter-molecular hydrogen bonds compared to 1,4-dihydroxybenzene (which can only form inter-molecular hydrogen bonds), and consequently, 1,2-dihydroxybenzene has a lower boiling point compared to 1,4-dihydroxybenzene. Since to boil a species, you need to overcome it's inter-molecular (not intra-molecular) attractive forces (which in this case, refers to inter-molecular hydrogen bonding).hmm because
there is one question like..
H3C-Li H3C-K H3C-F H3C-MgBr H3C-OH
need to arrange from least polar to most polar..
So I did like C-Li , electronegativey is 1.57
C-K , 1.73
C-F : 1.43
C-Mg : 1.24
C-O : 0.89
so H3C-OH (least polar) , H3C,MgBr, H3C-F.H3C-Li,H3C-K (Most polar)
then 2nd question,
find the type of intra-molecular Bonds,
BeF2
CH4
CBR4
LIF
I use the electronegativity table to find out..
like more than 0.5 less than 2.0 = polar
less than 0.5 is non-polar
more than 2.0 is ionic...
3) CH4 , non polar
NH3 , 3.04-2.2 = 0.84 (polar)
Just that for CH3CH2OH --->iM SO lost
CH3NH2 also..
Ultima, I learn all this as above in just one lesson & they expect us to know all. we are not given resources also.
4) must determine the inter-molecular interaction
CHCL3 and C2H5OH just this I dk how?
I know that CH4 and CF4 is temporary dipole.. cos CH4 is non polar, CF4 is polar molecule, for cos of its shape, they cancel out each other , causing it to be non polar.
C-K isn't a covalent bond, it's ionic. Therefore you cannot apply electronegativity ratings, which are to determine the polarity of covalent bonds. But if your teacher wants you to go ahead anyway, then the C-K electronegativity DIFFERENCE (a keyword you kept missing in your posts) is approx 2.5 - 0.9 = 1.6, while the C-F electronegativity difference is approx 4.1 - 2.5 = 1.6. So based on these calculations, the bonds are approximately equally polar, which is misleading because one is more ionic in character, while the other is more covalent in character.
Yes, you're correct in your method to determine nature of intramolecular bonds.
CHCL3 and C2H5OH :
Between tricholoromethane molecules : permanent dipole - permanent dipole van der Waals (aka Keesom forces). Between ethanol molecules : hydrogen bonding. Between trichloromethane and ethanol molecules : permanent dipole - permanent dipole van der Waals (aka Keesom forces). -
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2010 'A' Level H2 Chemistry Paper 3.
Alcohol J, CxHyOH, is a volatile fungal metabolite whose presence when detected in air can indicate hidden fungal attack on the timbers of a house.
When 0.10 cm^3 of liquid J was dissolved in an inert solvent and an excess of sodium metal added, 10.9cm^3 of gas(measured at 298K) was produced.
When 0.10cm^3 of liquid J was combusted in an excess of oxygen in an enclosed vessel, the volume of gas (measured at 298K) was reduced by 54.4cm^3. The addition of an excess of NaOH(aq) caused a further reaction in gas volume of 109cm^3 (measured at 298K).
Use these data to calculate values for x and y in the molecular formula CxHyOH for J.
TPJC Chem teachers' solution :
In the first experiment, the gas evolved in the reaction with Na metal is hydrogen.
Moles of H2(g) = 10.9/24000 = 0.000454 molÞ moles of liquid J present = 0.000454 x 2 = 0.000908 mol
In the 2nd experiment, moles of CO2 evolved = 109/24 000 = 0.00454 mol
Þ moles CO2 : moles J (CxHyOH)
5 : 1 Þ x = 5
Moles of O2 used in the reaction = (109 + 54.4)/24000 = 0.00681 mol
Note that since CO2 was formed, it ‘masked’ the O2 that was used up
Þ Reduction in volume of gas = (vol O2 used up) – (vol CO2 formed)
Þ vol O2 used up = (reduction in vol) + (vol CO2 formed)moles O2 : moles J (CxHyOH)
7.5 : 1Þ C5HyOH + 15/2 O2 ® 5CO2 + ((y+1)/2)H2O
16 = 10 + (y+1)/2 Þ y = 11
ii) J is an alcohol that reacts with K2Cr2O7 Þ J is either primary or secondary alcohol.
J can be dehydrated to alkene Þ J contains the -CH2-CH(OH)- structure
Since alkene K is oxidized into ethanoic acid and propanone
Þ K is CH3CH=C(CH3)2Þ J is CH3CH(OH)CH(CH3)2
SAJC Chem teacher's solution :
When 0.10 cm3 of J is added to sodium metal, hydrogen gas is evolved.
CxHyOH + Na <-----> CxHyO-Na+ + (1/2) H2
No. of moles of hydrogen gas = 10.9 / 24000 = 4.542 x 10-4 mol
No. of moles of alcohol J = 2 x no. of moles of hydrogen = 9.084 x 10-4 mol
When NaOH is added to J, the carbon dioxide gas produced from the combustion is being absorbed.
No. of moles of CO2 = 10.9 / 24000 = 4.542 x 10-3 mol
Since all the C in CO2 is produced from combustion of J, we can find the value of x when finding out the ratio of the no. of moles of CO2 produced to that of alcohol J present initially.
No. of moles of CO2 / No. of moles of J = 4.542 x 10-3 / 9.084 x 10-4 = 5
Therefore, x = 5
When J is combusted in an excess of oxygen, the reduction in the volume of gas is due to the consumption of oxygen to produce water. This reduction in volume of gas is the difference between the volume of oxygen gas consumed in combustion and the volume of carbon dioxide gas produced.
Volume of gas being consumed= 54.4 cm3
No. of moles of gas being consumed = 54.4 / 24000 = 2.267 x 10-3 mol
Taking the ratio between the no. of moles of gas consumed and the no. of moles of alcohol J,
Ratio = 2.267 x 10-3 / 9.084 x 10-4 = 2.5 mol
Since we have determined x = 5 above, we can simplify the equation to,
C5HyOH (l) + ((19+y)/4) O2 (g) 5CO2 (g) + ((y+1)/2) H2O (l)
Hence, difference in no. of moles of gases on combustion = ((19+y)/4) – 5 = 2.5
y = 11
Alcohol J : C5H11OH
Observation : J reacts with acidified K2Cr2O7, J has molecular formula C5H11OH
Deduction : J is oxidised [1/2]. J is a primary or secondary alcohol.Observation : J dehydrated to form K, K reacts with hot KMnO4 to give equimolar mixture of ethanoic acid and propanone.
Observation : Oxidative cleavage
K has the structure (CH3)2C= and J has the structure CH(CH3)=
K is 2-methylbut-2-ene.
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In a reply to a post made by "forgive me" :
Regarding F- vs I-. your conclusions are wrong, and your reasons are also wrong. Further cryptic hints : why does a base behave as a base? what makes a nucleophile a nucleophile? What determines basic strength, and what determines nucleophilic strength?
As for your "amino acids : pH, pKa and pI" qn, to be honest, it'll be more effective to ask your JC teacher or tuition teacher (if you have one) about it. Students need at least one entire tuition lesson to properly understand and correctly apply "amino acids : pH, pKa and pI". It's unfeasible to expect a single forum post to thoroughly enlighten you on this topic.
Nonetheless, I won't leave you at the deep end of the pool. I'll give one (but just one) attempt at explaining this to you here. But it'll be less confusing if you have a tuition teacher to explain this to you in person ( JC teachers are so overworked and tired-out that they will mostly just brush you off with a "no need to know, don't ask so much", which is why there is a real need for tuition teachers).
So is there a need to know all about "amino acids : pH, pKa and pI"? Tis a grey area, in which Cambridge can argue that JC students should be able to deduce the answers to questions such as the one in the 2010 exam, even though JCs don't usually teach much about this topic. It always helps to be prepared to go a little (or if you've the aptitude for it, a lot) beyond the basic syllabus requirement, especially if you're committed to securing a distinction.
Peruse these links (open 3 browsing windows simultaneously).
http://themedicalbiochemistrypage.org/amino-acids.html
https://www.bio.cmu.edu/courses/03231/LecF04/Lec05/aa_titration.pdf
http://wwwchem.csustan.edu/chem4400/aapi.htmThe 3rd link above, notice that the isoelectric point must be such that the alpha ammonium group remains protonated, while the R group amine group remains deprotonated, hence the pH of the solution must be more acidic than 9.6 (the alpha ammonium pKa), but more alkaline than 6.0 (the R group ammonium pKa), hence the isoelectric point will be midpoint of the two pKa values, or (6+9.6)/2 = 7.8
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Originally posted by jlheartsfly:
Hi sorry to trouble again. can you explain the difference in TM and the s-block metals in their reactivity in water to me?thanks.
S-block metals are generally more reactive with water (with the reactivity increasing down the group) and react in a redox reaction with water (the metal being oxidized and water being reduced).D-block metals are generally unreactive with water, as the effective nuclear charge for the valence electrons are stronger (in turn because additional electrons enter into the d orbitals, which are more diffused than s orbitals and hence provide poorer shielding*), and therefore the oxidation potentials for oxidizing d-block metal atoms into their cationic states, are generally not as positive (ie. not as reactive) as that of s-block metals.
* Sometimes you need to say "d-orbitals provide poor shielding", while at other times you need to say "d-orbitals provide effective shielding", much like The Accountant Joke**, depending on the question. For instance, the former statement would be used to explain why the d-block metals generally have a smaller radius than s-block metals (d orbitals provide poorer shielding compared to s and p orbitals of the same electron shell). The latter statement would be used to explain why the atomic and ionization energies of the d-block metals remain similar (d orbitals provide more effective shielding compared to the s and p orbitals of the next electron shell). Why do we care about the next electron shell? Because unlike s-block metals of (say) period 4, for the d-block metals in period 4, together with the additional proton (as you go from left to right of the Periodic Table), the additional electron goes into the 3-d orbital rather than the 4-d orbital (for a period 4 s-block metal, the additional electron goes into the 4s orbital).
** The Accountant Joke :
A businessman was interviewing job applications for the position of manager of a large division. He quickly devised a test for choosing the most suitable candidate. He simply asked each applicant this question, "What is two plus two?"
The first interviewee was a journalist. His answer was, "Twenty-two".
The second was a social worker. She said, "I don't know the answer but I'm very glad that we had the opportunity to discuss it."
The third applicant was an engineer. He pulled out a slide rule and came up with an answer "somewhere between 3.999 and 4.001."
Next came an attorney. He stated that "in the case of Jenkins vs. the Department of the Treasury, two plus two was proven to be four."
Finally, the businessman interviewed an accountant. When he asked him what two plus two was, the accountant got up from his chair, went over to the door, closed it, came back and sat down. Leaning across the desk, he said in a low voice, "How much do you want it to be, boss?" The accountant got the job.Source :
http://www.accountingcoursesonline.com/accountantjokes.html -
Originally Posted by forgiveme :
Rate of decomposition of h202 can be determine by withdrawing fixed Vol of small sample of soln at known time into conical flask obtain cold dil h2so4 and titrating it with kmno4. Deduce rate of rxn of h2o2 is first order. I dun understand how come can use graph of kmno4 to find deduce. Ans write condition of h2o2 = mol of kmno4. Bt conc of h2o2 = to tat meh ?
Does HCOOH under give orange ppt On 2,4 dnph since the other side is a carbonyl?
Kinetics :
You're right that the mole ratio isn't 1 : 1. But the mole ratio is directly proportional. So plotting a graph of [KMnO4] is akin to plotting a graph of [H2O2], and a constant half-life is required to prove 1st order kinetics.
Organic :
Once the carbonyl group is bonded to a OH, it's considered a carboxylic acid group and not a carbonyl compound (ie. ketones and aldehydes), even though the C=O carbonyl group is present, as carboxylic acid groups have a different reactivity and hence takes priority. FYI, it is the resonance stabilization of the carbonyl group in carboxylic acids, esters and amides, that reduce the electrophilicity of the carbonyl C atom, to the extent that the weak nucleophile 2,4-DNPH (since it has two electron withdrawing by induction and by resonance nitro groups) is unable to react with carboxylic acids, esters and amides. -
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Originally posted by Plaxo:
2007 H2 Chemistry Paper 3 Qn 3 (b)(ii)
My question is what exactly is the information given in Data 1. Enthalphy change of formation of Ca2+ (aq)= -543 and f- (aq)= -333 ??
This data is given together with enthalphy change of atomisation of Ca(s), enthalphy change affinity of flourine atom and enthalphy change of formation of CaF2(s).
Do I have to use information from data booklet as well, as it is not stated in the question ?
2006 H2 Chemistry Paper 3Qn 5
Write a balance equation when KI(aq) is added to a solution of Pb2+(aq). A bright yellow precipitate is formed. Further addition of KI(aq) causes the ppt to redissolve.
My Answer:
2KI(s) + Pb2+(aq) ----PbI2(s)+ 2K+(aq)
Yellow ppt of PbI2 is formed.
PbI2 (s) + I- (aq)----[PbI3]- (aq)
Further addition of KI causes the formation complex ion which is soluble.
However, the formula of the complex ion given in the TYS was [PbI4]-. Does that mean my answer is wrong cause I wasn't taught about this in my school. Is there a way of coming up with the right complex ions.
Thanks In Advance :)
Qn 1. The data given in the question suffices (for drawing a Hess Law enthalpy cycle), you do not need any further info from the Data Booklet. But if any exam qn appears to give inadequate data, feel free to quote values from Data Booklet to help you answer the question. It's just that in this instance, further info is not required.
Qn2. Different JCs teach different content. If you intend to get a distinction, you cannot rely merely on notes provided by any JC (even so-called top JCs), but you must be prepared to peruse multiple sources. Your fictitious tri-iodo complex ion will not score the mark. You must give the formula for the tetra-iodo plumbate(II) ion.
No single source (eg. any single JC's notes) will give an exhaustive list of complex ions, their formulae, their geometries and their colours. Cambridge deliberately does not release any list, to reserve the right to ask on obscure complex ions to test the candidate's ability to extrapolate and deduce probable formulae, geometries and colours.
You can find a more complete list of complex ions, their formulae, their geometries and their colours on Jim Clarke's website and Rod Beavon's website.
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Originally posted by Plaxo:
Thanks for reply UltimaOnline.
If you don’t mind me asking, what is the equation of the enthalpy change of formation of
the Ca2+ ion or f- ion stated in the qn. Cause I tot enthalpy change of formation is only for substance and not ions. Maybe i am mistaken it is not even enthalpy change of formation, then what is it ?
By Hess Law, formation enthalpy for Ca2+(aq) ion = atomization enthalpy + 1st ionization enthalpy* + 2nd ionization enthalpy + hydration enthalpy.* There is a technical difference between enthalpy and energy, but for 'A' level purposes these may be used interchangeably, eg. bond enthalpy = bond energy. Enthalpy change is heat change under constant pressure, energy change is heat change under constant volume. If a reaction generates gaseous products, some of the energy is used to push away the atmosphere, and enthalpy will be slightly less exothermic than energy change.
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Amino acids :
The most acidic group (ie. largest Ka, or smallest pKa) is the alpha carboxylic acid group, followed by the 'R' group carboxylic acid group, followed by the alpha ammonium group, followed by the 'R' group ammonium group (weakest acid).
The most basic group (ie. largest Kb, or smallest pKb) is the 'R' group amine group, followed by the alpha amine group, followed by the 'R' group carboxylate group, followed by the alpha carboxylate group (weakest base).
You should be able to explain why, based on "electron-donating by induction 'R' group" and "electron-withdrawing by induction alpha carbon".
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NH3 is a stronger field ligand (not quite the same meaning as stronger ligand, there is some correlation although there are also exceptions, eg. H2O is a stronger field ligand than OH-, although OH- is a stronger ligand than H2O) compared to H2O, which means replacing H2O ligands with NH3 ligands, results in a larger magnitude of d-d* splitting of the metal ion's orbitals. Exactly what effect this has on color, is a case-by-case matter.
Case 1
[Cr(H2O)6]3+ absorbs longer wavelength green light to give a violet coloured complex (although in practice, the counter ion present often substitutes away one or more of the water ligands, and you get green colour).
[Cr(NH3)6]3+ absorbs shorter wavelength violet light to give a yellow coloured complex.
Case 2
[Co(H2O)6]3+ absorbs longer wavelength yellow-violet light to give a blue-green coloured complex.
[Co(NH3)6]3+ absorbs shorter wavelength blue-ultraviolet light to give a yellow-orange coloured complex.
Case 3
[Cu(H2O)6]2+ absorbs longer wavelength orange light to give a blue coloured complex.
[Cu(NH3)4(H2O)2]2+ absorbs shorter wavelength yellow light, to give a deeper blue coloured complex.
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Originally posted by pokepokee:
hello
may i know why is pcl5 solid while pcl3 is liquid? thanks
Because PCl5 at room temperature and pressure exists as an giant ionic lattice structure of [PCl4]+ cations and [PCl6]- anions.Test Yourself :
Draw the structure of both cation and anion, stating their electron and ionic geometries. Work out the Oxidation States of phosphorus in both ions.
You need to use the BedokFunland JC formula for Oxidation State = Formal Charge + Electronegativity consideration, in order to correctly answer the Singapore-Cambridge A level 2011 P3 Qn, "What is the OS of Br in RCONHBr?"
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