Chemistry: Post your doubts here!

[sidbloom1995]

View attachment 38889
EMPERICAL

can you post the MS!
thanks

 

[♣♠ Magnanimous ♣♠]

can you post the MS!
thanks

WHAT IS your answer..
No answer

 

[♣♠ Magnanimous ♣♠]

 

[sidbloom1995]

help plzz

9701.mj.06 q, 25
9701.on.06 q,37,3
9701.mj.01 q 34,26,
9701.on.07 q 33,37,4
9701.mj.08 q 27 ,
9701.on.09 ( varient 11) 28,24,21
9701.on.10(varient 12) 39,27,13
9791.0n.10 (varient 11) q 38 ( what product will form if tollen reagent react with it ),35,29
9701.mj.10 (varient 11) q 12,
9701.mj.11(varient 12) q 38,29,28,24,11,
*9701.mj.11(varient 11) q 36,27,34,26,16
9701.on.11.(varient 12) q 9
9701.on.11 (varient 11) q 27,29,21,
9701.mj.12, (varient 12) q 24,23
9701.mj.12 (varient 11) q 29,23,22,9,
9701.on.12 (varient 11) q 26,

O/N 2010 -11-
Q29. (sorry i didnt see it before) the ans is C because we will have 3 structures
2 will be from -cis and -trans
one will be a simple carbon chain of 4 carbons, butene!

Q35. sorry i dont know for 3 in this i'll tell u about it if i find anything!

 

[sidbloom1995]

WHAT IS your answer..
No answer

im sorry but im not confident enough to post my answer without confirmation!
i'll tell you the answer and if it is right, i'll post the procedure! it's CHO just!

 

[daredevil]

View attachment 38889
EMPERICAL

the method's right... but cant seem to find the final answer... :/
basicaly i also think it should be CHO like sidbloom said or maybe CHO2 ...

 

[Namehere]

View attachment 38889
EMPERICAL

If you add the percentages 37.5 + 4.17 + 58.3 you get 99.97, so there is a 0.03% missing from somewhere...

 

[sidbloom1995]

the method's right... but cant seem to find the final answer... :/
basicaly i also think it should be CHO like sidbloom said or maybe CHO2 ...

i also used the same procedure!
but since it was asked, i cant help but think that there might be something more to the question??!!
so i want to verify my answer!

 

[Namehere]

View attachment 38889
EMPERICAL

However, if you dont take that into account, you get that the ratio of C: H: O is 1: 1.3344: 1.166
if you times all by 6 so you get integer numbers for all of them -> 6: 8.01: 6.996, which simplifies to 6:8:7
Thus, the empirical formula is C6H8O7, which is correct, since E330 is known for citric acid, which has that formula.

 

[Namehere]

Easy Tip: Never simplify until the last moment!

 

[daredevil]

If you add the percentages 37.5 + 4.17 + 58.3 you get 99.97, so there is a 0.03% missing from somewhere...

oohh ryt... never thought of that....

where did u get this question anyway magnanimous?

 

[DeViL gURl B)]

http://i.imgur.com/G2akAHB.jpg


Try zooming in to see the reactants/conditions. If they're not visible quote me or send me a VM and let me know I'll scan it and reupload. Idk if it's clear enough took a pic from my phone.

Also note of the of arrows is the other way around.. Primary Alcohol -> Aldehyde ->Carboxylic
The Aldyhyde -> Carboxylic arrows are the other way around.

Aldehyde -> Carboxylic reagents are K2Cr2O7 + Acid under reflux
and
Carboxylic -> Aldehyde reagents are LiAlH4 or NaBH4

THANK YOU!!! Soo muchhh!! Well it ain't that clear...so if u don't mind scanning.. It'd be great!
Well I didn't get the aldehyde part...cuz couldn't get the pic properly....
THANK YOU SOO MUCH btw! ^_^

 

[DeViL gURl B)]

Aoa,
And heyy people! Umm I'm lookin for an easy form of PERIODICITY! As most of the papers..have those questions..and in quite a tricky way! So any one who has like...an awesome summary...like the one which won't get outta yr brain...sorta thing! PLEASE SHARE!! Cuz I really sometimes struggle at that part of the paper!
Thank u

 

[sitooon]

Using E0 value ,
Find the products of each compound at Anode ,
AgF
FeSO4
MgBr2

Ans . Oxygen for first two , and Br for MgBr2

 

[♣♠ Magnanimous ♣♠]

If you add the percentages 37.5 + 4.17 + 58.3 you get 99.97, so there is a 0.03% missing from somewhere...

can u explain deeply...
i m not getting...
i have done almost the full some but at last--> i m not getting how to convert that decimal to whole number like u did of C6H8O7

 

[♣♠ Magnanimous ♣♠]

oohh ryt... never thought of that....

where did u get this question anyway magnanimous?

in my topical answers...
but i dont have the mark scheme of it.

 

[♣♠ Magnanimous ♣♠]

the method's right... but cant seem to find the final answer... :/
basicaly i also think it should be CHO like sidbloom said or maybe CHO2 ...

maine yahi steps kiye hai par age kuch baki he jo muje nahi ata hai.

 

[daredevil]

in my topical answers...
but i dont have the mark scheme of it.

see the deal is that if we are considering the PERCENTAGE of elements in the compound it is just like if we are taking 100g of the substance and then saying that e.g. 33.4g of it is C.
but the thing here is that if u add all the percentages we get 99.97% which means that a 0.03% part of the whole compound is unknown... it is prob something other than these three elements....
now if it is not an authentic source of cie papers (or even if it is) my guess wud be that they misprinted the question so dont fret over it all too much bcz ur method is definitely correct if u did it like u said u did...
other than that we can still be open for suggestions from other fellows and see if anyone has an explanation.

 

[A star]

wow i donot get half the things being said here . so not reading from the start :3

 

[daredevil]

please someone explain this quesiton...