Chemistry: Post your doubts here!

sidbloom1995

sidbloom1995

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mai jinn hun 2 said:
Can anyone help me with this questions?:unsure::unsure:
View attachment 37561
Click to expand...
in this one, H2PO4- ats like a base and donated electron and HPO4-2 acts as an acid
the trick is that when a buffer reacts with an acid or a base, the product include the reactent of the other (this is some thing that i have noticed and is not a given rule, watch in the example)

H+ + HPO4-2 ----------> H2PO4-
now
OH- + H2PO4- ---------> HPO4-2 + H2O

as for the part (iii)
you need to use the formula for acidic buffer solutions to find it's pH, which is
pH = pKa + log ( [salt]/[acid])
for pKa u use the value of 7.2, (sorry about this part im a little confused as to how to explain it, form what i remember, i took out the no. of moles of [H+] and used them again in pKa= (-log[H=]) equation and got 7.2 as answer so that's why you have to use 7.2, thought im still not 100% on that, sorry,)
can you give me the paper no. or this, i'll try solving it once more and see if i can answer it
 
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sidbloom1995

sidbloom1995

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sadiaali said:
Please can anyone help me to explain the structure of ice. I had a bad chemistry teacher:(
Click to expand...
structure of ice is very simple
it only have 2 bondings in it, covalent and hydrogen bonding
when water is freezing, these Hydrogen bondings arrange themselves in a hexagonal structure and form rings,giving ice an expanded volume and a fixed shape
 
sadiaali

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sidbloom1995 said:
structure of ice is very simple
it only have 2 bondings in it, covalent and hydrogen bonding
when water is freezing, these Hydrogen bondings arrange themselves in a hexagonal structure and form rings,giving ice an expanded volume and a fixed shape
Click to expand...
Please one more question :) Why it is tetrahedral while in liquid it is bent??
 
sidbloom1995

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it is not tetrahedral, it is non- linear and it is due to the lone pairs of electrons on the Oxygen atom
there are two lone pairs and they both act on bonded pairs to make them bend to the angle of 104.5*
remember when shaping a molecule, the number of bond pair and lone pair all count!!!
(this is just my idea of it and is not very logical but i tend to think about it as a family, the brothers get married and the "nands" cause repulsion and does not allow the "bhabis" to take up all the space and the space is equally divided depending on who is more or if they are equal in number :p ...............it's a sort of joke that helps me remember to consider both...u can forget it if it confuses u :p)
 
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sidbloom1995

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Abdel Moniem said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w05_qp_2.pdf
Question 5e, f and g
Click to expand...
your answers are kind of long to answer and i cant draw it's shapes over here, but i'll tell you the method
(e) with Na,
the OH group in the molecule will give it's H away ad form [O-Na+] and the rest of the structure remains same
with ethanoic acid
it forms ester linkage, the OH group again gives it's H away and the ethanoic acid gives it's OH away from -COOH and they form H2O. the O from a sort of bridge andlinks with the C of -COOH.
it forms this sort of bonding R- C-O-CH(=O)CH3, [RCOOCHCH3] (please write down as a displayed formula so that u get a clear understanding of the structure)
(f) with tollen's reagent
R-CHO + 2[Ag(NO3)2]+1 + 3OH- -----------> R-COOH + 2Ag + 4NH3 +2H2O, (this is the general equation of tollen's reagent!!)
with 2,4-NDPH
you see the NHNH2 group on top of the benzene ring right!!
that is the only structure that is involved with the reaction with Aldehyde
when u write it it's something like this :NH-:NH2
the -CHO group comes and give it's C=O's O to the 2H from NH2 to form H2O
the c from C=O bonds with N from NH2 and the resultant is N=C-R bond arangement
if we were to write it in an equation form,
C6H3(NO2)2N2H3 + R-CHO --------> C6H3(NO2)2N2H=CR + H2O
(g) in order to sow -cis -trans isomerism, there should be no two groups on the same carbon of C=C.
so the structure will be something like this
(CH(OH))H-C=C-H(CO2H)
again try to draw it as displayed formula and you will understand it better!!
 
mai jinn hun 2

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sidbloom1995 said:
in this one, H2PO4- ats like a base and donated electron and HPO4-2 acts as an acid
the trick is that when a buffer reacts with an acid or a base, the product include the reactent of the other (this is some thing that i have noticed and is not a given rule, watch in the example)

H+ + HPO4-2 ----------> H2PO4-
now
OH- + H2PO4- ---------> HPO4-2 + H2O

as for the part (iii)
you need to use the formula for acidic buffer solutions to find it's pH, which is
pH = pKa + log ( [salt]/[acid])
for pKa u use the value of 7.2, (sorry about this part im a little confused as to how to explain it, form what i remember, i took out the no. of moles of [H+] and used them again in pKa= (-log[H=]) equation and got 7.2 as answer so that's why you have to use 7.2, thought im still not 100% on that, sorry,)
can you give me the paper no. or this, i'll try solving it once more and see if i can answer it
Click to expand...

Still I didnt get even a word of it :(
How we'll know that what is acid and what is base from the data given in this question ?
I think so it was from 06m/j or from 09 nov var (41)
 
sidbloom1995

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mai jinn hun 2 said:
Still I didnt get even a word of it :(
How we'll know that what is acid and what is base from the data given in this question ?
I think so it was from 06m/j or from 09 nov var (41)
Click to expand...
about the acid or base part we have to see if the molecule is giving protons or accepting protons, H+, so it have to do with bronsted lowery base and acid definitions!!
HPO4-2 have only 1 H and can accept one more (since we have H2PO4-) so it will act as a base and accept one more H+ to become H2PO4-.
since there are two salts in our sample, that means that they both exist in our sample. so when an acid is added, HPO4-2 reacts with that acid and accept it's proton H+ and when a base is added, H2PO4- reacts with is and donate is its H+ proton!!
and your paper no. was not right!!
i checked both papers but the question is not there!!
 
ZaqZainab

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ZaqZainab

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Faaiz Haque said:
ok what about 7a i
Click to expand...
The general formula for ester is there in the pic It is always suppose to be there in an ester if it isn't then that is not an ester View attachment 37732
So they have given me C5H10O2

CH3-CH2-CH2-CH2-COOH----->C4H9-CO2H

CH3-CH2-CH2-COO-CH3------->C3H7CO2CH3

CH3-CH2-COO-CH2-CH3------>C2H5CO2C2H5

I am just moving the COO group If you notice in the first one the COO group is in the end and then in second one its in the second last place and in the third one its is in the middle You have to draw this with all the Bonds showing
 

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