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Originally posted by Flying grenade:
Originally posted by UltimaOnline :
I'll leave it to Flying Grenade (don't disappoint me) to show you how to solve it.
5.00g of an anhydrous Grp 2 metal nitrate loses 3.29g in mass when heated strongly
2M(NO3)2 => 2MO+4NO2+O2
What is metal M?
Flying grenade's suggested solution
mol = mass/mr
let mr of M be x
ηM(NO3)2 = 5/(x+124)
ηMO=1.71/(x+16)
since stoichiometric ratio of M(NO3)2 to MO is 1:1
we can equate ηM(NO3)2 = ηMO
solving two equations simultaneously, we obtain x=40.1 (3sf)
Metal M is Calcium.
Correct. Well done, Flying grenade.Fyi, my own working is [ 5g / [ 3.29g / (molar mass of 4NO2+O2) ] x 2 ] - [ (molar mass of NO3) x 2] = 40.1g
But your working will of course get full marks as well. Cambridge only marks the final answer, as long as your working is reasonable, you've some creative freedom to work it out as you wish.
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can write cold, alkaline kmno4 (aq)?
or must write cold, kmno4 (aq), NaOH(aq)?
for formation of diol fron alkene
can write acidified K2Cr2O7 (aq), heat
or must write H2SO4 (aq) , K2Cr2O7 (aq), heat?
my sch say must write the compound out
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Originally posted by Flying grenade:
can write cold, alkaline kmno4 (aq)?
or must write cold, kmno4 (aq), NaOH(aq)?
for formation of diol fron alkene
can write acidified K2Cr2O7 (aq), heat
or must write H2SO4 (aq) , K2Cr2O7 (aq), heat?
my sch say must write the compound out
In the A level exams, the reagents and conditions for each reaction, should be written in exactly the same way that either CS Toh, or Chan Kim Seng, or George Chong writes them (ie. above & below the equation arrow) in the reaction equations. Cambridge will accept any of these alternative ways of writing the reagents and conditions. All JC students are advised to closely follow these 3 sources, over their own school lecture notes. -
why U.O. say H- can't exist? can exist with o.s. -1 as in hydride, but cannot exist with mononegative formal charge is it?
can H have 2 electrons around it?
I'm confused, help !!
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for Haemoglobin, two N atoms are ionically bonded to Fe2+ central atom, the other 2 N atoms are datively bonded to Fe2+?
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Originally posted by Flying grenade:
why U.O. say H- can't exist? can exist with o.s. -1 as in hydride, but cannot exist with mononegative charge is it?
can H have 2 electrons around it?
I'm confused, help !!
Don't be silly. I never said H- can't exist. -
Originally posted by Flying grenade:
for Haemoglobin, two N atoms are ionically bonded to Fe2+ central atom, the other 2 N atoms are datively bonded to Fe2+?
Obviously. -
Originally posted by UltimaOnline:
Don't be silly. I never said H- can't exist.why mechanism the H atom cannot take a lone pair of e- uh?
H cannot have uninegative formal charge uh?
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Originally posted by Flying grenade:
why mechanism the H atom cannot take a lone pair of e- uh?
Because your mistake (a common error for Singapore JC students) is that you wrongly eliminated a H- when the reaction required the elimination of a H+. H- is extremely unstable, and thus will attack right back. It is therefore a non-viable leaving group. H- does exist, but must always be carried by a more stable, larger, lower charge density species, the carrier species known as the reducing agent which delivers the H- ion. -
THANKSSSS ULTIMAAA
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Originally posted by Flying grenade:
THANKSSSS ULTIMAAA
Also, if you're observant, you'll realize that whenever you mistakenly eliminate a H- instead of a H+, the O or N or C atom that was bonded to the H atom will consequently lack a stable octet, which will alert you to the error. When H+ is eliminated, it leaves its electrons behind, allowing the O or N or C atom to keep its stable octet. -
page 130 cs toh advanced guide
for electrolysis of brine, why when Hg is used as cathode, Na(s) is liberated, instead of H2(g)?
redn potential of H+/H2 =0.00V shld be more +ve than Na+/Na
from data booklet, no Na+/Na redn potential value, but have for Li+/Li, which is very -ve.
is it becos if there is no electrode(as in without the Hg electrode) , the Na(s) want to deposit but no where to deposit? (and hence H2 is liberated instead? lolol)
for electrolysis of Na2SO4, the H+ at the cathode come from water H2O uh?
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pg 130
for electrolysis of H2SO4, why H2SO4 slowly become more concentrated?
for the anode reaction, Cs toh provided 2 equations
1st equation, more water form as product
2nd eqn, water as a reactant, then form O2 as product
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Originally posted by Flying grenade:
page 130 cs toh advanced guide
for electrolysis of brine, why when Hg is used as cathode, Na(s) is liberated, instead of H2(g)?
redn potential of H+/H2 =0.00V shld be more +ve than Na+/Na
from data booklet, no Na+/Na redn potential value, but have for Li+/Li, which is very -ve.
is it becos if there is no electrode(as in without the Hg electrode) , the Na(s) want to deposit but no where to deposit? (and hence H2 is liberated instead? lolol)
for electrolysis of Na2SO4, the H+ at the cathode come from water H2O uh?
Mercury thermodynamically (both enthalpically and entropically) stabilizes the reduced sodium metal in amalgam, and also on further condition that a concentrated (not dilute) solution is electrolyzed. -
Originally posted by Flying grenade:
pg 130
for electrolysis of H2SO4, why H2SO4 slowly become more concentrated?
for the anode reaction, Cs toh provided 2 equations
1st equation, more water form as product
2nd eqn, water as a reactant, then form O2 as product
At cathode, 4 H2O used (to supply the 4 H+), and O2 + 2 H2O + 4 OH- byproduct generated (not written in CS Toh's eqn, since he used 4 H+ instead of 4 H2O as reactant).At anode, 4 H2O used (to supply the 4 OH-), and 2 H2 + 4 H+ byproduct generated (not written in CS Toh's eqn, since he used 4 OH- instead of 4 H2O as reactant).
Hence, total H2O used = 8, but total H2O generated = 2 + 4 = 6.
Therefore, overall balanced redox equation is 2 H2O(l) --> 2 H2(g) + O2(g), and the solution becomes more concentrated (as H2O is electrolyzed away).
Note that Singapore JCs will insist you use H2O as the reactant for both cathode & anode, instead of CS Toh's H+ and OH- respectively.
Cambridge will accept both versions, unless the solution being electrolyzed is strongly acidic or strongly alkaline; which is why Singapore JCs have decided it's safer to instruct JC students to just always use H2O as the reactant.
And if you can't figure out why, or which cathode & anode equations are acceptable or not acceptable in such electrolytic setups of strong acidity or alkalinity, go ask your school teacher or private tutor.
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Originally posted by UltimaOnline:
Mercury thermodynamically (both enthalpically and entropically) stabilizes the reduced sodium metal in amalgam, and also on further condition that a concentrated (not dilute) solution is electrolyzed.and hence also a reason why low [H+]
AHHH I SEEEE THANKS ULTIMAAA !!!!
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Originally posted by UltimaOnline:
At cathode, 4 H2O used (to supply the 4 H+), and O2 + 2 H2O + 4 OH- byproduct generated (not written in CS Toh's eqn, since he used 4 H+ instead of 4 H2O as reactant).At anode, 4 H2O used (to supply the 4 OH-), and 2 H2 + 4 H+ byproduct generated (not written in CS Toh's eqn, since he used 4 OH- instead of 4 H2O as reactant).
Hence, total H2O used = 8, but total H2O generated = 2 + 4 = 6.
Therefore, overall balanced redox equation is 2 H2O(l) --> 2 H2(g) + O2(g), and the solution becomes more concentrated (as H2O is electrolyzed away).
Note that Singapore JCs will insist you use H2O as the reactant for both cathode & anode, instead of CS Toh's H+ and OH- respectively.
Cambridge will accept both versions, unless the solution being electrolyzed is strongly acidic or strongly alkaline; which is why Singapore JCs have decided it's safer to instruct JC students to just always use H2O as the reactant.
And if you can't figure out why, or which cathode & anode equations are acceptable or not acceptable in such electrolytic setups of strong acidity or alkalinity, go ask your school teacher or private tutor.
i see , thanks ultima
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Hi Ultima, please take a look at above qn, from 2007 (can't edit on same post cos can't find the cursor using phone hence cant type on the post above with pic
part V)
i understand can illustrate using the pair of same amino acids, but tcher instructed us to illustrate using pair of different amino acids to make it more challenging
https://www.dropbox.com/s/md4v22dtowaymp1/20160714_100436-1.jpg?dl=0
Question :
https://www.dropbox.com/s/amfgl1ifmblcm99/20160714_101102-1.jpg?dl=0 drawn three lys-ser R group interactions
do we have to draw full displayed formula when showing R group interactions?
need to show the linear H bonding between the R groups? is my drawing of (3) acceptable?
from the Qn, HSA is a globular protein, and has a roughly spherical shape in water, 67% of the amino acids are incorporated into an α-helix
if need show linear H bonding , then lys and ser is one shorter and one longer than the other,
if both R groups must originate from the same vertical starting position, then showing linear H bonding is quite difficult
Thanks, Ultima
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Originally posted by Flying grenade:
Hi Ultima, please take a look at above qn, from 2007 (can't edit on same post cos can't find the cursor using phone hence cant type on the post above with pic
part V)
i understand can illustrate using the pair of same amino acids, but tcher instructed us to illustrate using pair of different amino acids to make it more challenging
https://www.dropbox.com/s/md4v22dtowaymp1/20160714_100436-1.jpg?dl=0
Question :
https://www.dropbox.com/s/amfgl1ifmblcm99/20160714_101102-1.jpg?dl=0 drawn three lys-ser R group interactions
do we have to draw full displayed formula when showing R group interactions?
need to show the linear H bonding between the R groups? is my drawing of (3) acceptable?
from the Qn, HSA is a globular protein, and has a roughly spherical shape in water, 67% of the amino acids are incorporated into an α-helix
if need show linear H bonding , then lys and ser is one shorter and one longer than the other,
if both R groups must originate from the same vertical starting position, then showing linear H bonding is quite difficult
Thanks, Ultima
You only need to fully display out the part of the R group which is directly responsible for the R group interaction. The rest of the R group can be condensed or skeletal.As far as possible, always show linear H bonding. Yes, your drawing of (3) is acceptable (except it can be improved, see next sentence below).
Regarding originating from the same vertical starting position : your drawing is actually misleading and incorrect.
Don't put a hypen between "lys" and "ser", it implies you wrongly think these are 2 directly adjacent amino acid residues (which would only be peptide bonded to each other, not H bonded by R groups).
The way you drew it out, from the R groups from same 'starting position', is incorrect and misleading. Because the amino acid residues with the R group interactions will be many residues apart, and because the polypeptide chain is non-linear (think of the polypeptide chain as curving and twisting convolutedly all over the place), hence you should have drawn the 2 amino acid residues as originating from different 'starting positions' on paper, either facing each other, or (more realistically) at an angle away from each other. See CS Toh's diagram on the bottom of page 381 (Advanced Study Guide).
Lastly, don't forget you're attempting to translate a 3D structure onto 2D paper. Each C-C bond is bent in 3D space, not linear. On 2D paper, it may not always be possible to illustrate perfect linearity (for all H bonds in all molecules), just do your best to reflect this. Cambridge will be reasonable.
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Thanks Ultima, for all the Details !!!
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if say, lys and ser is side by side joined by a peptide bond, is there H bonds between the R groups?
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Originally posted by Flying grenade:
if say, lys and ser is side by side joined by a peptide bond, is there H bonds between the R groups?
Usually not, due to steric factors. R group interactions are usually, and most importantly, between non-adjacent amino acid residues. If the only R group interactions are between adjacent amino acid residues, then there would be no meaningful tertiary structure, which is what gives proteins its biological function. In which case, proteins would be useless, and you would have died even before you were born.Now you understand why your drawing is misleading and incorrect? And your school teacher didn't point out this error?!? Or worse, your school teacher taught the entire class this wrong way?!?
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Originally posted by UltimaOnline:
Usually not, due to steric factors. R group interactions are usually, and most importantly, between non-adjacent amino acid residues. If the only R group interactions are between adjacent amino acid residues, then there would be no meaningful tertiary structure, which is what gives proteins its biological function. In which case, proteins would be useless, and you would have died even before you were born.Now you understand why your drawing is misleading and incorrect? And your school teacher didn't point out this error?!? Or worse, your school teacher taught the entire class this wrong way?!?
Yes Ultima, exactly. Whole class was taught wrongly. Ultima, that's why i am very very thankful ,of you, helping us type out, and more importantly, Elaborate out All(small and big, important and the nitty gritty ) the Details, and Most Importantly, teach and correct us to learn the correct and right education.
my only regret is i didn't discover BFJC earlier.
I Feel Very Frustrated and sad that I've discovered so much that I've learnt throughout my life, is wrong. Happy because at least i discover it.
i am very thankful that i receive education from BFJC, one of the only true educators and education.
Another important skill I've acquired from BFJC is to learn to suspect and question and double check if something is taught wrong, and i seek clarification via books, online, and forums.
some people don't even know they're learning the wrong things(like me in the past, receive without critically evaluating and discerning what I've learnt ), and once they discover,they'll be angry and happy.
for me, I haven't been doubting Chem stuffs I've learnt until this year, it's late, i wished i was more inquisitive and a critical thinker last time ,but better late than never
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Originally posted by UltimaOnline:
Usually not, due to steric factors. R group interactions are usually, and most importantly, between non-adjacent amino acid residues. If the only R group interactions are between adjacent amino acid residues, then there would be no meaningful tertiary structure, which is what gives proteins its biological function. In which case, proteins would be useless, and you would have died even before you were born.Now you understand why your drawing is misleading and incorrect? And your school teacher didn't point out this error?!? Or worse, your school teacher taught the entire class this wrong way?!?
he wrongly taught the entire class this way.
it was i, who sense there was something amiss, not right, wrong, erroneous, misleading, and incorrect, and thus, i seek clarification like a frantic chicken asap writing down typing down and uploading once i have my free period.