Chemistry: Post your doubts here!

[hassam]

if Hg bonds to COOH how can it disrupt the hydrogen bonding the COOH group was involved in b4? since CO part is there even in COOHg

 

[abcde]

if Hg bonds to COOH how can it disrupt the hydrogen bonding the COOH group was involved in b4? since CO part is there even in COOHg

The -C=O bond does not play a role in hydrogen bonding. -OH is responsible for hydrogen bonding between molecules and with H being replaced by Hg, that role diminishes. (Hydrogen bonding is possible only when H is attached to a very electronegative element such as O.)

 

[abcde]

http://www.xtremepapers.com/CIE/International A And AS Level/9701 - Chemistry/9701_w09_qp_11.pdf
Q21) How do we figure out that the total no. of C=C double bonds is 5?

 

[JD REBORN]

Can somebody explain to me Q)20.Plz explain it with drawings of isomers.

 

[JD REBORN]

Also Q)21.Explain it fully

 

[hassam]

The -C=O bond does not play a role in hydrogen bonding. -OH is responsible for hydrogen bonding between molecules and with H being replaced by Hg, that role diminishes. (Hydrogen bonding is possible only when H is attached to a very electronegative element such as O.)

yea thanks.....bt one thing abt the the bracketed thing u mentioned:see in DNA and protien's alpha helix structure th H-bonding is b/w CO and NH groups......so th criteria u mentioned is only necessary to be fulfilled by one of the molecules amongst the two involved in H-BONDING

 

[hassam]

FOR part(ii) ......it seems rather counter-intuitive to me that F groups repel the water molecules......can't they form H bonds with water.....bt markscheme says CF3 part can't form H-BONDS wid water

 

[omg]

thanks bro

if dat thanks was for mee... dn im nt bro! u may call me a sis

 

[abcde]

yea thanks.....bt one thing abt the the bracketed thing u mentioned:see in DNA and protien's alpha helix structure th H-bonding is b/w CO and NH groups......so th criteria u mentioned is only necessary to be fulfilled by one of the molecules amongst the two involved in H-BONDING

I don't get what you mean by the second line. Hydrogen bonding is still possible if hydrogen is attached to fluorine or nitrogen. (F > O > N are the three most electronegative elements in order of decreasing electronegativity). Can you please rephrase your question?

 

[abcde]

FOR part(ii) ......it seems rather counter-intuitive to me that F groups repel the water molecules......can't they form H bonds with water.....bt markscheme says CF3 part can't form H-BONDS wid water

It IS rather counter-intuitive. But I guess the application part simply requires the use of common sense, which dictates that if fluorine-containing groups prevent water from flowing through, they would obviously not form hydrogen bonds with water and this is what we're asked to identify here. On the contrary, oil molecules would be removed by the formation of van der Waals dispersion forces.
http://www.chemguide.co.uk/CIE/section113/learningf.html this simply asks us to accept the fact without questioning.

 

[larina]

Assalamo Alaikum

can i have some notes for p5!!?!

 

[katie]

Can anyone readin this comment please send me Chemistry SL May, 2011 Paper1(multipule Choice) and Paper2 and Paper 3 (options) + marcking scheme to this email address: [email protected]

thank you!
P.s I Need them sent ASAP!

 

[OakMoon!]

AoA!
The answer to the volume used of the acid in the practical's ms was 31 something! There was nothing wrong with the solutions as far as the concentration is concerned so is it possible that the concentration of indicators play a role in titration?

The indicator doesn't makes difference to the titration. Usually, 3-4 drops can easily bring a change, but if it doesn't adding more of it won't be a problem and you titration values will remain the same.

 

[OakMoon!]

http://www.xtremepapers.com/CIE/International A And AS Level/9701 - Chemistry/9701_w09_qp_11.pdf
Q21) How do we figure out that the total no. of C=C double bonds is 5?

First of all remember that one ring acts as the same as on double bond in calculations. If this compound, C20H28O, was an alkane how many Hydrogen atoms would it have? 20*2+2=42, right?
42-28=14 For each double bond 2 hydrogen atoms are reduced. In this case there are 14/2=7 double type of bonds. 1 is the cyclic ring, 1 is the C=O and then there are 5 C=C bonds. I hope you get it.

 

[OakMoon!]

Can somebody explain to me Q)20.Plz explain it with drawings of isomers.

 

[abcde]

First of all remember that one ring acts as the same as on double bond in calculations. If this compound, C20H28O, was an alkane how many Hydrogen atoms would it have? 20*2+2=42, right?
42-28=14 For each double bond 2 hydrogen atoms are reduced. In this case there are 14/2=7 double type of bonds. 1 is the cyclic ring, 1 is the C=O and then there are 5 C=C bonds. I hope you get it.

That's a nice explanation. Thank you.

 

[XPFMember]

That's a nice explanation. Thank you.

Assalamoalaikum wr wb!
Yeah true that...
Jazakallah Khairen
When abcde posted it, I tried long doing that but I couldn't get it...thanks a lott

 

[VelaneDeBeaute]

AoA!
In what range our answers to the practical titrations must fall within? 0.1 cm^3, is it? If some readings are more than, lets say, 0.2 cm^3 apart from the original, must they be recorded or not? Is there enough time to take more than 3 set of readings?
JZK for replies!

 

[abcde]

AoA!
In what range our answers to the practical titrations must fall within? 0.1 cm^3, is it? If some readings are more than, lets say, 0.2 cm^3 apart from the original, must they be recorded or not? Is there enough time to take more than 3 set of readings?
JZK for replies!

W.S!
Yes, the titration experiment requires us to obtain at least two titres that are within 0.1 cm^3 of each other. In some cases, a difference of 0.2 cm^3 is accepted, too but don't give yourself that leverage. You should record all three 3 readings but ensure that two of them are within 0.1 cm^3 of each other. Usually, no. But if you're confidently done with all of your paper correctly, you may perform one more for greater accuracy. That would be a rarity worth applause.

 

[VelaneDeBeaute]

^AoA!
Thankyou abcde!