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Chemistry: Post your doubts here!

ZaqZainab

ZaqZainab

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♣♠ Magnanimous ♣♠ said:
ZaqZainab
well i don't remember from which paper they were repeated :( i can only say they were about from states of matter
Click to expand...
I didn't even know which ones are my friend said she realized 2 were
and remember guys don't wast time in calculations if they are straight forward do them if they are tricky(lengthy)but yet you know how to do it
come back to it after you finish the paper Thats what i learned from MCQ don't waste time you waste 10 minutes on a 1 mark hard question might make you lose 5 marks of easy questions as you will be in hurry again read the questions well after coming out of the exam hall my friend mentions that they asked NOT and i read wrong but trust me have faith like 2 days back i didn't know a thing properly today morning i just had this feeling that i have nailed every bit and the paper was good they say 'Whatever we plant in our subconscious mind and nourish with repetition and emotion will one day become a reality.' and that is true
and lastly BEST OF LUCK :)
 
S

sagar65265

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unique111 said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w11_qp_12.pdf
Q.4 C
Q.6 B (why not C?)
Q.10 A
Q.11 C ( why not A? Doesn't Br go from -1 to 0?)
Q.17 c
Q.18 D ( but question says made from water? :confused:
Q.26 A
Q. 29 D
Q. 35 B
Q.37 D
Q. 40 A
Please, somebody clear these doubts?
Click to expand...

26)

When a primary alcohol is heated with aqueous acidified potassium dichromate, it is, in general, oxidized.
With gentle warming, it is first oxidized to an aldehyde, and if the aldehyde product is further allowed to react in heated conditions where the oxidant (dichromate, in this case) is in excess, the carboxylic acid of that alcohol will be produced. So, in this case, it is not heated excessively AND the aldehyde is removed immediately by distillation after it is formed. Therefore, we can say that the product obtained in the reaction is, indeed, an aldehyde. If the yield of this is 70%, let's see what we can find out from the equation:

Ethanol produces Ethanal on reaction with dichromate, and the pure reaction with the oxidant removed to simplify things is

CH3CH2OH + [O] ==> CH3CHO + H2O

(Note that the equation is not balanced as in reality - this is only to show us that the ratio of Ethanol to Ethanal is ideal conditions is 1:1. This is not the right reaction, which is itself very complex, and can be found here).

So, we have 2.30 grams of Ethanol. The number of moles of Ethanol in this mass = (Mass)/(Mass of 1 mole) = (2.3)/(12 + 3*1 + 12 + 2*1 + 16 + 1) = 0.05 moles.

We expect an equal number of moles of Ethanal to be produced, so we can say that the number of moles of Ethanal produced should be equal to 0.05 moles. However, since the yield is only 70%, we can expect that the actual number of moles we get = (70/100) * 0.05 = 0.035 moles.

Ethanal has a Molar Mass of (12 + 3*1 + 12 + 1 + 16) = 44 grams per mole. Therefore, (Mass of 0.035 moles) = (0.035) * (44) = 1.54 grams = A.

Hope this helped!
Good Luck for all your exams!
 
GCE As and a level

GCE As and a level

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sagar65265 said:
26)

When a primary alcohol is heated with aqueous acidified potassium dichromate, it is, in general, oxidized.
With gentle warming, it is first oxidized to an aldehyde, and if the aldehyde product is further allowed to react in heated conditions where the oxidant (dichromate, in this case) is in excess, the carboxylic acid of that alcohol will be produced. So, in this case, it is not heated excessively AND the aldehyde is removed immediately by distillation after it is formed. Therefore, we can say that the product obtained in the reaction is, indeed, an aldehyde. If the yield of this is 70%, let's see what we can find out from the equation:

Ethanol produces Ethanal on reaction with dichromate, and the pure reaction with the oxidant removed to simplify things is

CH3CH2OH + [O] ==> CH3CHO + H2O

(Note that the equation is not balanced as in reality - this is only to show us that the ratio of Ethanol to Ethanal is ideal conditions is 1:1. This is not the right reaction, which is itself very complex, and can be found here).

So, we have 2.30 grams of Ethanol. The number of moles of Ethanol in this mass = (Mass)/(Mass of 1 mole) = (2.3)/(12 + 3*1 + 12 + 2*1 + 16 + 1) = 0.05 moles.

We expect an equal number of moles of Ethanal to be produced, so we can say that the number of moles of Ethanal produced should be equal to 0.05 moles. However, since the yield is only 70%, we can expect that the actual number of moles we get = (70/100) * 0.05 = 0.035 moles.

Ethanal has a Molar Mass of (12 + 3*1 + 12 + 1 + 16) = 44 grams per mole. Therefore, (Mass of 0.035 moles) = (0.035) * (44) = 1.54 grams = A.

Hope this helped!
Good Luck for all your exams!
Click to expand...
Could u help me plz with my doubts ^^
 
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