Chemistry: Post your doubts here!

[Nibz]

Where's it? I don't see it.

 

[Doctor Nemo]

In a titration between an acid (in burette) and an alkali, you may need to re-use the same titration flask.
Which is the best procedure for rinsing the flask?
A. Rinse with distilled water and then with the alkali.
B. Rinse with tap water and then with distilled water
C. Rinse with tap water and then with the acid.
D. Rinse with the alkali.

and another question is dat how come many carbonates react with acids when all carbonates are insolube except ammonium, potassium and sodium.......?

The answer is B because all you want to do is rinse the flask with water. You would start with tap water because it is less expensive than distilled water.

The carbonates might be insoluble in water but that doesn´t prevent the surface from reacting with acid. Metals are insoluble in water and they react with acid on the surface.

 

[ad98ify]

some mind-boggling A2 conceptual problems.

• why is methanoic acid a stronger acid than benzoic acid?
• why is the boiling point of silicon tetrachloride lower than carbon tetrachloride despite the fact that the only intermolecular force in group 4 tetrachlorides is van der waal forces and SiCl4 has greater molecular mass?
• why is melting point of lead greater than that of tin? even though lead has greater ionic radii and hence weaker metallic bonding within its element.

i know these questions are difficult but atleast please try to answer. it will mean alot

 

[MEGUSTA_xD]

The answer is B because all you want to do is rinse the flask with water. You would start with tap water because it is less expensive than distilled water.

The carbonates might be insoluble in water but that doesn´t prevent the surface from reacting with acid. Metals are insoluble in water and they react with acid on the surface.

I think it will be C, cause as you clean it with water you are hampering the concentration of acid, you need to clean it with acid after water to remove the water drops left inside

 

[omg]

is there is any site or sh from where i can get the ans of the SAQs of the chem applications????

 

[hassam]

The slight increase in melting point between tin and lead is due to the increased effective nuclear charge associated with the filling of f-orbitals and their relatively low screening effect . Morover, lead has a cubic close-packed structure which is tighter than the body-centred cubic arrangement in tin. this is what i found from a site....bt we dont need to know this for exam purposes

 

[hassam]

cie writes in one mS that condensation needs two different functional groups.....bt i have a problem with this cos PEG a molecule mentioned by cie in their application booklet plus in their exams ....this pmolecule's monomer is ethan1,2 diol....and PEG is a condensation polymer of this.....so where the hell are two different groups.....so i think CIE got it wrong on that ms,,,,,,

 

[hassam]

empirical formula of addition polymer is the same as that of its monomer.......this is a point mentioned by CIE in their ms....bt i never seem to understand it....i cnnt think wat wud be the end of molecule like.....i think it wud be like alkane bt this contradicts the ms point.....so kindly pls clarify this thing

 

[abcde]

empirical formula of addition polymer is the same as that of its monomer.......this is a point mentioned by CIE in their ms....bt i never seem to understand it....i cnnt think wat wud be the end of molecule like.....i think it wud be like alkane bt this contradicts the ms point.....so kindly pls clarify this thing

In polymers, we're considering very large numbers of monomer units, approximately ranging from 10 000 - 30 000. So even when you have two extra H atoms at the 'ends' that doesn't significantly alter the empirical formula. Consider a polymer consisting of say 15 000 ethene units.
Ethene: C2H4
=> Empirical formula: CH2
Poly(ethene): C30 000 H 60 002
=> Empirical formula: CH2 (roughly the same!)
So the empirical formula remains the same.
(P.S. That's just my line of reasoning.)

 

[ad98ify]

The slight increase in melting point between tin and lead is due to the increased effective nuclear charge associated with the filling of f-orbitals and their relatively low screening effect . Morover, lead has a cubic close-packed structure which is tighter than the body-centred cubic arrangement in tin. this is what i found from a site....bt we dont need to know this for exam purposes

Down the group, the shielding effect increases as the number of shells are increasing. hence, increase in shielding effect is a dominant factor here when it comes to determining properties of elements down the group. keeping this fact in mind, tin should have a higher melting point as its atomic/cationic radii is small and therefore stronger electrostatic attractions between delocalised electrons and the metal cation.
i m still confused sir

 

[jason234]

Can someone please upload the october november 2011 paper 22 markscheme

 

[Student12]

Can anyone PLEASE help me with Question number 3d part 2 ??? Its on Equilibrium !!

 

[rodent123]

I have too many queries please help!
1.is platinum electrode used for transitional metals and non metals only in electrochemistry? not for metals? please somebody clarify this electrode thingy
2.difference between hydrolysis.deprotonation and acid base reaction
3. standard hydrogen electrode is used only when we want to determine the electrode potential of any particular electrode and not the emf ?

 

[smzimran]

Can anyone PLEASE help me with Question number 3d part 2 ??? Its on Equilibrium !!

CO2 + H2 ---> CO + H2O
initial moles 0.50 0.50 0.20 0.20
equilibrium moles (0.50-x) (0.50-x) (0.20+x) (0.20+x)


equilibrium conc. = equilibrium moles (because volume = 1 dm3)

Kc = 1.44
Kc = [CO][H2O] / [CO2][H2]
1.44 = (0.20+x)(0.20+x) / (0.50-x)(0.50-x)
1.44 ={ (0.20+x)/(0.50-x)}^2
taking root of both sides
1.2 = 0.20+x/0.50-x
0.60-1.2x = 0.20 +x
0.40 = 2.2x
x = 0.18
Now, as
CO2 + H2 ---> CO + H2O
equilibrium moles (0.50-x) (0.50-x) (0.20+x) (0.20+x)
0.38 0.38 0.32 0.32

 

[abcde]

Can anyone PLEASE help me with Question number 3d part 2 ??? Its on Equilibrium !!

Which part of it is unclear? The mark scheme shows a very explicit calculation. Assuming that x moles of CO2 and H2 react, x moles of CO and H2O will be produced as the molar ration is 1:1. So, at equilibrium no. of moles of CO2 and H2 = (0.5 - x) since the reactants are being removed from the mixture while the no.of moles of CO and H2O = (0.2+x) since these are being formed. The rest is the application of the Kc formula.

 

[Student12]

Thanks 'abcde' and 'smzimran' i just realised where i was going wrong.
& can you give any suggestions on the calculations of chemistry ? what can i do to get better in it ? :/ i know practice but then from where.. ?
Do u knw any website for the calculations ? or hv any worksheets or questions to solve ?

 

[Shahroz95]

Assalamoalaikum, everyone!

Can anyone tell me about the Equlibrium constant, Kp, and the Concentration constant, Kc. I need the formula of the Kp, and I know pretty much of Kc, I just need the differnece between the two. Thanks in advance!

 

[abcde]

Thanks 'abcde' and 'smzimran' i just realised where i was going wrong.
& can you give any suggestions on the calculations of chemistry ? what can i do to get better in it ? :/ i know practice but then from where.. ?
Do u knw any website for the calculations ? or hv any worksheets or questions to solve ?

The CIE endorsed Chemistry book by David Acaster has plenty of practice questions related to each topic. So if you're not yet comfortable with the past paper ones, look it up there. Then move on to the past papers.

 

[abcde]

Assalamoalaikum, everyone!

Can anyone tell me about the Equlibrium constant, Kp, and the Concentration constant, Kc. I need the formula of the Kp, and I know pretty much of Kc, I just need the differnece between the two. Thanks in advance!

W.S!
Kp is simply the equilibrium constant in terms of partial pressures. Since gases can occupy any container, there volume is indefinite. The Kp provides a better measure of the equilibrium constant in reactions involving gases as conc. = moles/volume, which is difficult to determine.
To find partial pressure, use: p = n/N x P, where
n: no. of moles of the particular gas (whole partial pressure is being found)
N: total no. of moles (of all gases)
P: total pressure (of all gases).

Once you have the partial pressures, Kp is found in exactly the same way as Kc. Just use the partial pressures where you'd use concentrations.

 

[Student12]

Can anyone explain how to do question number 2, 15 and 31 from this paper i've attached ?