Chemistry: Post your doubts here!

[syed1995]

when you make titration can we start with 50cm3 as an initial reading for each rough titration
in the MS it is written that do not give marks if start with 50cm3
plaese answer i am confused

DUDE if you post once .. we will help you.. don't be impatient.. Posting the same thing again and again makes people think that you are rude.

Well the initial value will be 0 or 10-15 or anything less than 50.. since the burette has 50 at the bottom and 0 at the top.

If you are starting with 50 means that you re-fill the burette.. or haven't set your apparatus correctly. HENCE 0 marks.

 

[ZainH]

no offence meant bro but its very basic
5bi) since carboxylic acid is removed in the first option Na can only react with alcohols hence the group is OH group
5bii use the equation 2Na + 2C3H5OH = 2C3H5O-Na + H2 so calculate moles of alcohol used and find your answer
5biii) it will show that 1 mole of the compond produces one mole of H2 hence it must contain two OH groups
hope it solves the problem

i) How come Na can only react with alcohols?
ii) How'd you find that equation?

Could you explain a bit more please?

 

[A star]

i) How come Na can only react with alcohols?
ii) How'd you find that equation?

Could you explain a bit more please?

well for your first question its the basic organic concept that Na(reactice metals) only reacts with alcohols and carboxylic acids to form salts sice the carboxylic acid is removed in first option alcohol was left
Q2 well they have molecular mass use the formulae n=(emperical formulae mass)/molecular mass and n(CH2O) then just react with sodium remove the two H and balance the equation

 

[hela]

you cannot start with 50 cm3 as thats the maximum reading a burete can have

I DID NOT GET IT

 

[hela]

What's ur problem!?!!

WHEN YOU MAKE TITRATION DO I HAVE TO REFILL THE BURETTE EACH TIME TILL 50 CM3 OR NOT

 

[hela]

DUDE if you post once .. we will help you.. don't be impatient.. Posting the same thing again and again makes people think that you are rude.

Well the initial value will be 0 or 10-15 or anything less than 50.. since the burette has 50 at the bottom and 0 at the top.

If you are starting with 50 means that you re-fill the burette.. or haven't set your apparatus correctly. HENCE 0 marks.

Umm yeah I thought it wasn't sending... lol... anyway I meant why can't we refill for each seperate titraion up to 50 since the burette fills completely up to 50 and the marking scheme says do not award mark if 50 cm3...I ALWAYS refill the burette why not... it makes the readings easier the calculate... Thanks for your patience but this is what i don't understand...

 

[syed1995]

Umm yeah I thought it wasn't sending... lol... anyway I meant why can't we refill for each seperate titraion up to 50 since the burette fills completely up to 50 and the marking scheme says do not award mark if 50 cm3...I ALWAYS refill the burette why not... it makes the readings easier the calculate... Thanks for your patience but this is what i don't understand...

Dude have you ever looked at a burette reading? It is 0 at top .. and 50 at bottom.. for value we always take the value at top ..

that's why it is final - initial .. cuz final is greater than initial.. if you completely filled the burette your starting reading will be 0. and the ending reading will be 15.1 for example..

so Final - Initial = 15.1 - 0 = 15.1 your reading.

 

[asd]

LOL. @ the burette guy.

 

[asd]

Oh, she's a girl.

 

[syed1995]

Oh, she's a girl.

lol.. happens with the best of us.

 

[hela]

right don't be rude dude I've seen a burette a couple of times I just have a bad teacher dude. Anyway good job!

 

[A star]

so your practical skills not up to the mark no problem the threshhold is low

 

[ZainH]

http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_s12_qp_22.pdf

Q1)c)i)

I understand how sulphur will have two lone pairs but how will nitrogen? It has 5 electrons in it's valence shell and it's only bonding with two other sulphur atoms. Could someone draw this for me and show me? Thanks in advance =)

 

[Muhammad Bin Anis]

the qu

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s12_qp_22.pdf

Q1)c)i)

I understand how sulphur will have two lone pairs but how will nitrogen? It has 5 electrons in it's valence shell and it's only bonding with two other sulphur atoms. Could someone draw this for me and show me? Thanks in advance =)

es


the question says assuming all bnds to be single,... so there is one lone pair and a single exra electron on nitrogen

 

[syed1995]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s12_qp_22.pdf

Q1)c)i)

I understand how sulphur will have two lone pairs but how will nitrogen? It has 5 electrons in it's valence shell and it's only bonding with two other sulphur atoms. Could someone draw this for me and show me? Thanks in advance =)

well N has 5 electrons .. 2 will be bonded that leaves 3 electrons.

1 Lone pair.. and 1 extra electron. SO 1 lone pair would be the answer.. unless 1.5 is possible

 

[littlecloud11]

Can anyone explain Question 2 (c)
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w07_qp_4.pdf

(i) i got, but for (ii) where did 4 come in the 4x^3 ( from markscheme)

littlecloud11 please sister, if possible.

The conc. oc Mg(OH)2 would be the same as the conc. of Mg2+ ions as the mole ratio is 1:1
Mg(OH)2 -------> Mg2+ + 2OH-
assume that the conc. og Mg2+ is x, then the conc. for OH- would be 2x

Ksp =[Mg2+] [OH-]^2
2*10^-11 = [x] [2x]^2
2*10^-11 = x* 4x^2
2*10^-11 = 4x^3
solve to get the value of x, which is the conc. og Mg(OH)2

Sorry for replying so late.

 

[syed1995]

right don't be rude dude I've seen a burette a couple of times I just have a bad teacher dude. Anyway good job!

I wasn't being rude. If I came across as rude I would like to apologize .. I was just pointing out the fact that the reading on the burette is upside down.

Sorry.

 

[19islandprincess96]

When do we use KCN as a reagent and when HCN?
In the paper 9701/41/M/J/10 Q4 part d it asks for the reagents and in the marking scheme it says KCN and that HCN negates.
So when do we use what?

 

[A star]

LOL. @ the burette guy.

Oh, she's a girl.

cyclone94 see what i mean

 

[A star]

When do we use KCN as a reagent and when HCN?
In the paper 9701/41/M/J/10 Q4 part d it asks for the reagents and in the marking scheme it says KCN and that HCN negates.
So when do we use what?

links please