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Chemistry: Post your doubts here!

periyasamy

periyasamy

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Hai guys.Ok my question is the second n 3rd part.I understand how to get the no of c atom n there is a prescence of oh group.But unsure of how to get the no of h atoms...Thanks.Btw the answer is ch3c0ch3....
 

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Suchal Riaz

Suchal Riaz

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Browny said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w05_qp_1.pdf

Can anyone help me with questions 31,33,35,39?
Click to expand...
question number 31
the mass of molecule is 72 which means there are two isotopes of chlorine in it. chlorine-35 and chlorine-37. for these two isotopes the radius will be same as the number of electrons and number of protons are same. the isotopic mass and neucleon number are different.
question number 33
the only thing we need to know for this question is that graphite has hexagonal rings. so in a hexagons there are six points, three of them should be nitrogen and three of them should be B. therefore empirical formula is BN. second is true as we are told that it's structure is like graphite. and the next is again true as we are told that the structure is similar to graphite.
question number 35
all of the properties are correct. MgO is basic oxide. MgO will not conduct heat very well because of no free electrons like in metals which travel heat very quickly. and MgO has the highest boiling melting point in group 2. the reason is that it's size is very similar to oxygen so they fit it together very closely.
question number 39
in catalytic converter hydrocarbons are oxidised to carbon dioxide. CO is oxidised to CO2. but NO is REDUCED to N2. so 1 and 2 are correct.
 
sidbloom1995

sidbloom1995

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Anzuman turin said:
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,
Click to expand...

M/J 2012 v.11
Q9. they said that the total no. of moles are (2 + x)
that means that Q is 2 moles
at dynamic equilibrium, the rate of forward reaction is equal to the rate of backwards reaction so the no. of moles of P will not change!
thus the answer will be B because it has the same no. of moles in as in the start!!

Q22. the name is ethyl propanoate
it means that this ester was composed of propanoic acid and ethanol
since we reacted it with NaOH the ester will hydrolyse and form CH3CH2COO-Na+ and CH3CH2OH
there mass will be 96g for acid and 46g for alcohol
when we add there masses together we get 142g
divide each of these masses with the total mass and we will get the % of products formed!
so [96/142] x 100 = 67.6%
[46/142] x 100 = 32.4%

Q23. B
the structure suggest that it is an ester!
meaning the compound should have an OH group and COOH group
the structure has regular repeat pattern and that suggest that it is only made of a molecule that has an OH group on one side and COOH on the other side!
B is fits that criteria!

Q29. it is easier to remove Cl from C then it is to remove F form C
so when in UV light Cl wil ionize and damage the Ozone layer
when talking about most damage, take the one with most no. of Cl
that would be A in this case!

M/J 2012 v12.
Q26. 2,4-dinitrophenylhydrazine reagent is to test or the presence of aldehydes and ketones
it came negative so that means that non of these are present in out product
Na reacted with it to give H2 gas, that is a test for alcohol
with ethanol and H2SO4 , if it was an acid, a sweet smelling ester would have formed but this did not happen
all this suggest that it is not an aldehyde or keton or carboxylic acid!
the Na test suggest that it is an alcohol!
so the answer is C

im really sorry that it took me sooo long to answer your questions! im sorry if it caused you too much inconvenience!!
 
AbbbbY

AbbbbY

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sitooon said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s04_qp_4.pdf

6 e
Click to expand...

Aaah. Just did this 5 mins ago :D

Okay so basically, first things first, you gotta realize that 3 water molecules will be lost, so the Mr of the fragment[gly-ala-ser since equal amounts] will become (75+89+105-(18*3)) = 215.

Dividing 600,000 by 215 will give you the number of fragments present. I'm carefully choosing my words here. Fragments. Not residues. Each of these fragments will have 3 residues so the total fragments present * 3 = number of residues.

Mathematically,

600,000/215 = 2791 fragments
2791 fragments = 2791*3 residues = 8373 residues.
 
S

sitooon

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AbbbbY said:
Aaah. Just did this 5 mins ago :D

Okay so basically, first things first, you gotta realize that 3 water molecules will be lost, so the Mr of the fragment[gly-ala-ser since equal amounts] will become (75+89+105-(18*3)) = 215.

Dividing 600,000 by 215 will give you the number of fragments present. I'm carefully choosing my words here. Fragments. Not residues. Each of these fragments will have 3 residues so the total fragments present * 3 = number of residues.

Mathematically,

600,000/215 = 2791 fragments
2791 fragments = 2791*3 residues = 8373 residues.
Click to expand...
Thanks , :)
if you can little more expand " fragments and residues " ,
i mean whats the difference , we use the word fragments only in mass spectrometer :p
 
AbbbbY

AbbbbY

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Suchal Riaz said:
there are two c=c so two moles of Br2 are required to react completely. and (separately) 2 mol of h2 are required.
Click to expand...

Umm, you do realize you've answered 24 here, right? :p


Abdel Moniem said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w09_qp_11.pdf
Question 21
Click to expand...
quote="Suchal Riaz, post: 745892, member: 15977"]there are two c=c so two moles of Br2 are required to react completely. and (separately) 2 mol of h2 are required.[/quote]

A!

This is a pretty simple question which idk why a lot of students have trouble answering.

Tackle it like this.

If the compound was an alkane and had no rings or double bonds, number of H atoms present = 2n+2 = 2(20)+2 = 42

Now, every time there're 2H less, you'll either have a double bond or a ring. The question states that there's one aldehyde (C=O) and 1 ring structure which has 1 C=C so this accounts for 6 missing Hydrogens, thereby leaving you with 8 missing Hydrogens.

8/2 = 4 Hydrogens in the aliphatic side chain.

However, total C=C are 5 because you have to include the C=C from the cyclohexene + cis formation hence A
 
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