Chemistry: Post your doubts here!

[Farhad221]

in AS don't worry about it.

But what types of questions do you think will come up on the hybridization

 

[nehaoscar]

Can anyone tell me the products of oxidation of But-1-ene by Hot, Conc, Potassium Manganate(VII)?

Will it be Carboxylic Acid + CO2 + H2O?

Yes
CH3CH2COOH + CO2 + H2O

 

[Abdul Hanan]

But what types of questions do you think will come up on the hybridization

In AS i have seen no question of hybridization in past papers...

 

[Farhad221]

In AS i have seen no question of hybridization in past papers...

what a relief ok thanks . With maths and chem tomorrow i was freaking out if i had to remember hybridization .

 

[Mahnoorfatima]

Polymerisation of Alkanes is not in the syllabus.

i mean alkenes. Stupid autocorrect.

 

[BhaiArshad]

i mean alkenes. Stupid autocorrect.

Go to the bottom of this page:
www.chemguideforcie.co.uk/section102/learningd.html

 

[nehaoscar]

http://maxpapers.com/syllabus-mater...hment/9701_may-june-2011-all-question-papers/

Variant 22 - Question 1 - c - i
In part b iv
The moles of NaOH that reacted with acid are found to be 0.04 moles
So in part c i
Shouldn't the moles at equilibrium be 0.06 moles for the acid and ROH and 0.04 mol for ester and H2O (Since 0.04 reacts and originally 0.1 is present so left over = 0.1-0.04 = 0.06)
The mark scheme says other way round... how?

 

[Turki AbdulAziz]

For the substitution of alchohols to form halogenoalkanes, Sulphur dichloride oxide (SOCl2) can be used as a reagent. But the problem is what are the conditions for the reaction?? I've tried searching the internet but i got nothing. Can someone please help.

 

[princess Anu]

n

For the substitution of alchohols to form halogenoalkanes, Sulphur dichloride oxide (SOCl2) can be used as a reagent. But the problem is what are the conditions for the reaction?? I've tried searching the internet but i got nothing. Can someone please help.

no conditions..

 

[princess Anu]

why is the observation sol turns yellow and not that yellow green gas ie chlorine disappears?

 

[nehaoscar]

n

no conditions..

For the substitution of alchohols to form halogenoalkanes, Sulphur dichloride oxide (SOCl2) can be used as a reagent. But the problem is what are the conditions for the reaction?? I've tried searching the internet but i got nothing. Can someone please help.

Reflux

 

[asadalam]

why is the observation sol turns yellow and not that yellow green gas ie chlorine disappears?

What year is this?Shouldnt it be that the iodine is displaced from by Cl2,and that would give a brown color to the solution?Maybe if all the I is displaced that would lead to Cl2 going in excess and causing solution to turn yellow maybe?Still what year is this from?

 

[NIM]

View attachment 53071

Thank you soooo much!!

 

[Studydayandnight]

Can anyone show me to how to solve this??

 

[NIM]

Need help !!
9701/43/O/N/13
Q#5b

 

[The Sarcastic Retard]

Guys in the paper will we have to explain sp sp2 and sp3 hybridization?

You should have a rough idea. Its very useful for knowing shape and bond angles.

 

[Mahnoorfatima]

http://studyguide.pk/Past Papers/CIE/International A And AS Level/9701 - Chemistry/9701_w05_qp_2.pdf
In question 5 part f, shouldn't the aldehyde change into OO-C-R as the reaction of Tollen's reagent woth aldehyde yeilds:
CH3CHO+(Ag(NH3)2)+----> CH3COO- ?? Anyone?

 

[The Sarcastic Retard]

http://maxpapers.com/syllabus-mater...hment/9701_may-june-2011-all-question-papers/

Variant 22 - Question 1 - c - i
In part b iv
The moles of NaOH that reacted with acid are found to be 0.04 moles
So in part c i
Shouldn't the moles at equilibrium be 0.06 moles for the acid and ROH and 0.04 mol for ester and H2O (Since 0.04 reacts and originally 0.1 is present so left over = 0.1-0.04 = 0.06)
The mark scheme says other way round... how?

Moles of NaOH used is 0.045
Moles of NaOH that reacted with HCl is 0.005
Remaining NaOH that will react wid acid = 0.045-0.005 = 0.04
Hence 0.04 0.04 0.06 0.06

 

[nehaoscar]

Moles of NaOH used is 0.045
Moles of NaOH that reacted with HCl is 0.005
Remaining NaOH that will react wid acid = 0.045-0.005 = 0.04
Hence 0.04 0.04 0.06 0.06

But then 0.04 is the amount of acid with which the NaOH reacted
So would the equilibrium remaining mixture be 0.1 - 0.04 = 0.06 ?
Like I know it's not but please could you explain why

 

[nehaoscar]

But then 0.04 is the amount of acid with which the NaOH reacted
So would the equilibrium remaining mixture be 0.1 - 0.04 = 0.06 ?
Like I know it's not but please could you explain why

Moles of NaOH used is 0.045
Moles of NaOH that reacted with HCl is 0.005
Remaining NaOH that will react wid acid = 0.045-0.005 = 0.04
Hence 0.04 0.04 0.06 0.06

Oh wait nevermind! i just got it after reading the question! Thanks!