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thanks ultima u are my savior i always so thankful when u reply my doubts
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Originally posted by Flying grenade:
thanks ultima u are my savior i always so thankful when u reply my doubts
Just to add :In CS Toh's update, the reason why it's incorrect to put either "slow" or "fast" is because the structure shown is the transition state, not the intermediate (there is no intermediate for SN2). Note that "slow" and "fast" must refer to actual steps converting reactant to intermediate, intermediate to intermediate, or intermediate to product. Transition states are simply human-imagined states in-between (ie. transitioning) actual steps, and thus cannot be either "slow" or "fast".
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for the factor of charge density, a charge is more favoured to be in a bigger atom? e.g. S compared to O?
because bigger atom , for the same charge, lower charge density, more stable?
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Originally posted by Flying grenade:
for the factor of charge density, a charge is more favoured to be in a bigger atom? e.g. S compared to O?
because bigger atom , for the same charge, lower charge density, more stable?
Varying stability of conjugate base (due to varying charge density) is the real reason (together with endothermicity of H-X bond dissociation enthalpy) why the strength of Bronsted-Lowry acidity is HI > HBr > HCl > HF.If Cambridge asks you why and you talk about electronegativity, you're shooting yourself in the foot, and will nullify any correct points you may have also written (because as long as you write contradicting points, Cambridge thinks you're trying to cheat your way through by using rainbow ruses and as such you'll get zero marks for the question).
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THANKSSSS ULTIMAAAAA !!!
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page 22 cs toh advanced guide
what does it mean by diluted 50 times?
Dilution factor. That is, the new, diluted solution will have a volume 50 times greater than the volume of the original, undiluted, solution.
in this case, 250cm3/5cm3 = 50.
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Originally posted by Flying grenade:
page 22 cs toh advanced guide
what does it mean by diluted 50 times?
Dilution factor. That is, the new, diluted solution will have a volume 50 times greater than the volume of the original, undiluted, solution.
in this case, 250cm3/5cm3 = 50.
You've defined it yourself liao. Moles of solute remain the same, but volume of solution now increased by 50 times. Hence molarity is decreased by 50 times. -
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omg
page 51 cs toh advanced guide
i have been learning wrong things throughout my life
cs toh say CO2 molecules have intermolecular pd-pd forces of attraction.
the next statement below "this is possible because in the solid state, CO2 molecules are fixed in position." does this statement mean only CO2(s) has pdpd? what is that statement for?
i thought only molecules with (net)dipole moment have intermolecular pd-pd.
i am now confused already. is CO2 a polar molecule? it's a symmetrical molecule
i always thought that in order for a molecule to be polar, the molecule must have net dipole moment
so CO2 has id-id (all molecules have) , does it have pdpd?
is CO2 a polar molecule without a net dipole moment,
or is CO2 a non-polar molecule without a net dipole moment but has pdpd?
so in order to have pdpd IMFs, just need two(or more) different atoms chemically combined tgt in a molecule, without a need for net dipole moment?
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Originally posted by Flying grenade:
omg
page 51 cs toh advanced guide
i have been learning wrong things throughout my life
cs toh say CO2 molecules have intermolecular pd-pd forces of attraction.
the next statement below "this is possible because in the solid state, CO2 molecules are fixed in position." does this statement mean only CO2(s) has pdpd? what is that statement for?
i thought only molecules with (net)dipole moment have intermolecular pd-pd.
i am now confused already. is CO2 a polar molecule? it's a symmetrical molecule
i always thought that in order for a molecule to be polar, the molecule must have net dipole moment
so CO2 has id-id (all molecules have) , does it have pdpd?
is CO2 a polar molecule without a net dipole moment,
or is CO2 a non-polar molecule without a net dipole moment but has pdpd?
so in order to have pdpd IMFs, just need two(or more) different atoms chemically combined tgt in a molecule, without a need for net dipole moment?
I told you liao : don't blindly trust Singapore JCs or your school lecture notes. The Distinction A grade student, will not blindly follow or memorize school lecture notes, but intelligently understand the Chemistry concepts in his/her own way. And the intelligent student will also sometimes disagree with his/her school teacher, because school teachers often disagree with each other as well.In CO2, the linear molecular geometry ensures that the 2 individual dipoles cancel out, hence it is an overall non-polar molecule. Being non-polar doesn't necessarily always mean only London dispersion van der Waals forces are present between molecules. Intelligently apply your own chemistry understanding case-by-case. For CO2, all 3 types of van der Waals forces are present : Keesom, Debye and London dispersion forces.
When comparing the intermolecular interactions for 2 different species separately, eg. when asked to explain the difference in melting or boiling point for 2 different species, you only need to state van der Waals forces versus hydrogen bonding or ionic bonding, and specify which is stronger. Only when both molecules have only van der Waals intermolecular forces, then it will be necessary to specify which type of van der Waals interactions are present between molecules of each species.
But be careful, contrary to what your school (ie. Singapore JCs) may have taught you, instantaneous dipole - induced dipole London dispersion van der Waals forces may not necessarily be weaker than permanent dipole - induced dipole Debye van der Waals forces which may not necessarily be weaker than permanent dipole - permanent dipole Keesom van der Waals forces, it depends on the total no. of electrons present and the molecular size, and hence the polarizability of electron charge clouds, and hence the magnitudes of partial charges and dipoles, and hence the strength of the electrostatic van der Waals attractions, regardless of which type of van der Waals is present.
You should be able to figure out such counter examples (which are contrary to the oversimplifications taught in Singapore JCs), including extreme counter examples where even instantaneous dipole - induced dipole London dispersion van der Waals forces may be even stronger than intermolecular hydrogen bonds! Shhhh! even if you know such counter examples, don't post them here and spoil the fun for others, let them figure these out for themselves! ;Þ
Lastly, for yourself, Flying Grenade, you can move on to another topic liao, don't keep flogging this dead horse here, and spamming multiple questions (I count 5 sub-questions in your post above) in a single post. If you still need further clarification, go ask your school teacher or private tutor.
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Hi UltimaOnline, for CO2, you mentioned that "In CO2, the linear molecular geometry ensures that the 2 individual dipoles cancel out, hence it is an overall non-polar molecule. Being non-polar doesn't necessarily always mean only London dispersion van der Waals forces are present between molecules. Intelligently apply your own chemistry understanding case-by-case. For CO2, all 3 types of van der Waals forces are present : Keesom, Debye and London dispersion forces.", is this only true for solid state?
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i am confident to say that london dispersion forces, debye forces, and keesom forces are present in CO2(g) and CO2(l) as well.
from google images "co2 solid state structure" , from looking at the structure, and since in a solid state, the molecules can only move about their plane of vibrations, i suspect that in CO2(s) , there is only london dispersion and keesom forces.
need savior ultimaonline to ascertain and give us his much revered imputs !
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in pg 51 of cs toh advanced guide, he wrote " This is possible because in solid state, CO2 molecules are fixed in position "
i Think he meant that, there is keesom forces in CO2(s)
But i think, keesom forces are also present in CO2(l) , and CO2(g)
because, there is permanent dipole arising due to the difference in electronegativity betwen O and C, and hence a dipole moment, and hence perm. dipole. there are debye forces too(except in solid state) , as an instantaneous dipole in one molecule can induce another dipole in any neighbouring molecule.
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but which one is the predominant(most extensive) IMFs between CO2(g) molecules?
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Originally posted by MightyBiscuits:
Hi UltimaOnline, for CO2, you mentioned that "In CO2, the linear molecular geometry ensures that the 2 individual dipoles cancel out, hence it is an overall non-polar molecule. Being non-polar doesn't necessarily always mean only London dispersion van der Waals forces are present between molecules. Intelligently apply your own chemistry understanding case-by-case. For CO2, all 3 types of van der Waals forces are present : Keesom, Debye and London dispersion forces.", is this only true for solid state?
No, not only for solid state, but Keesom van der Waals forces predominate moreso for the solid state > liquid state > gaseous state, for obvious entropy reasons. If the reasons are not obvious to you, go ask your school teacher or private tutor. -
Originally posted by Flying grenade:
but which one is the predominant(most extensive) IMFs between CO2 molecules?
Flying Grenade, I'm not gonna entertain any more flogging of this dead horse. Go ask your school teacher or private tutor on this. I'll leave you with a final parting statement in reply to your (modified with state symbol removed) question quoted above : it depends on the state/phase of the CO2, and besides, Cambridge won't ask which predominates for the different states/phases, you need only state "van der Waals forces" (ie. all 3 types of van der Waals forces) as the intermolecular interactions between CO2 molecules, regardless of state/phase. -
a chem tcher from my sch say during lecture , non polar amino acids contains no electronegative atom
which is untrue right?
page 378 cs toh advanced guide, proline contains N atom, tryptophan contains N atom
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Originally posted by Flying grenade:
a chem tcher from my sch say during lecture , non polar amino acids contains no electronegative atom
which is untrue right?
page 378 cs toh advanced guide, proline contains N atom, tryptophan contains N atom
There's no such thing as a "non polar amino acid". What your school teacher is referring to, is the R group of the amino acid. And whether an R group is protic polar, aprotic polar, or non-polar, is a matter of common sense that should be instantly obvious to you from a 1-second glance, which you sure as hell shouldn't need to rely on your school teacher (or whatever he/she says) to figure out, ffs.I would like to give you the benefit of the doubt, and imagine you're asking the more intelligent questions on why tryptophan's R group cannot accept H bonds despite having an N atom, and is thus hydrophobic ; or why there isn't a 3rd pKa value for tryptophan despite a protonated N atom conjugate acid seemingly possible.
But then you cock this up by asking the coconaden question about why proline's R group is non-polar?!?
As to the more intelligent questions (that you didn't explicitly ask) : I'll give you a generous hint : RESONANCE.
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ooh, yeah ! regarding the intelligent tryptophan qn, yes , resonance! !! thanks, Ultima !!!
OOOH CRAP. looking at pg 378 cs toh advanced guide, at the non polar R group column, it does seem like proline and tryptophan's R group is polar. why is it non polar?? @.@ T-T
is it because by virtue it is hydrophobic, then it is non polar?
tyrosine and histidine also have bulky hydrophobic R groups
:x @.@
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ok it's visibly obvious that tyrosine is polar and uncharged.
no issue with tyrosine and histidine.
but R group of proline and tryptophan looks polar. definitely have net dipole moment.
is it because the magnitude of non polar characteristic is so large that it outweighs a small polar characteristic so overall non polar?
update : R group of proline and tryptophan is non polar, molecule is overall, polar, but it's polar characteristic is negligible so effectively, proline and tryptophan is non polar.
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Originally posted by Flying grenade:
ooh, yeah ! regarding the intelligent tryptophan qn, yes , resonance! !! thanks, Ultima !!!
OOOH CRAP. looking at pg 378 cs toh advanced guide, at the non polar R group column, it does seem like proline and tryptophan's R group is polar. why is it non polar?? @.@ T-T
is it because by virtue it is hydrophobic, then it is non polar?
tyrosine and histidine also have bulky hydrophobic R groups
:x @.@
Go figure this one out for yourself. If you can't, go ask your school teacher or private tutor. -
figured out.
any side chain that contains a hydrocarbon alkyl group, or a benzene ring, it's nonpolar.
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Originally posted by Flying grenade:
If you want me to confirm whether you've really correctly understood this, draw out yourself (ie. don't use an internet image of) the amino acid proline, circle its R group, and upload the image here for me to check. -
ultima can help check if i circle the correct R (in this case the alkyl ) group correctly, in my sketch of Proline, please !
https://www.dropbox.com/s/73pbse5uds7ib0j/20160705_181640-1.jpg?dl=0
Thank u !