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Author Topic: ALL CIE CHEMISTRY DOUBTS HERE !!  (Read 119132 times)

Offline EMO123

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Re: ALL CIE CHEMISTRY DOUBTS HERE !!
« Reply #585 on: May 28, 2011, 07:53:31 am »
Quote from: ~~!$!HUSH!$!~~ on May 27, 2011, 06:30:46 pm
in Q15,the AgCl is insoluble,so a reaction with a more reactive metal causes Ag's displacement(from AgCl to NaCl).
Q20 : the compound has 3 double bonds,therefore you use 2 to the power of 3 which is equal to 8.


now i got you thanxx Hush
after i tryed it all the ways i got you what you were meaning so thanxx hush
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Offline HUSH1994

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Re: ALL CIE CHEMISTRY DOUBTS HERE !!
« Reply #586 on: May 28, 2011, 09:36:33 am »
Anytime ;)
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Offline Dania

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Re: ALL CIE CHEMISTRY DOUBTS HERE !!
« Reply #587 on: May 28, 2011, 04:26:45 pm »
What is the pH of the buffer that results when 11g of NaCH3CO2 is mixed with 85ml of 1.0M CH3C02H and diluted with water to 1.0 L? (Ka of CH3C02H = 1.8 x 10-5)

a. 2.91
b. 3.86
c. 4.55
d. 4.74
e. 4.94
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:)

Offline iluvme

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Re: ALL CIE CHEMISTRY DOUBTS HERE !!
« Reply #588 on: May 29, 2011, 07:18:55 pm »
Could answer this question and explain it for me please?

Which alcohol gives only one oxidation product when warmed with dilute acidified potassium
dichromate(VI)?
A butan-1-ol
B butan-2-ol
C 2-methylpropan-1-ol
D 2-methylpropan-2-ol

The answer is B
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Offline The Golden Girl =D

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Re: ALL CIE CHEMISTRY DOUBTS HERE !!
« Reply #589 on: May 29, 2011, 07:33:39 pm »
Quote from: iluvme on May 29, 2011, 07:18:55 pm
Could answer this question and explain it for me please?

Which alcohol gives only one oxidation product when warmed with dilute acidified potassium
dichromate(VI)?
A butan-1-ol
B butan-2-ol
C 2-methylpropan-1-ol
D 2-methylpropan-2-ol

The answer is B

I guess what they mean by one oxidation product is only one product from this oxidation ...if it's a primary Alcohol A  , then it has two products => Aldehyde and carboxylic acid ...but if it's a secondary Alcohol B then it will give ONLY a Ketone which is why It's B

I hope I helped :)
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Offline iluvme

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Re: ALL CIE CHEMISTRY DOUBTS HERE !!
« Reply #590 on: May 29, 2011, 07:57:25 pm »
Quote from: ~ The Golden Girl  ~ on May 29, 2011, 07:33:39 pm
I guess what they mean by one oxidation product is only one product from this oxidation ...if it's a primary Alcohol A  , then it has two products => Aldehyde and carboxylic acid ...but if it's a secondary Alcohol B then it will give ONLY a Ketone which is why It's B

I hope I helped :)

Got it. Thanks. :)
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Offline The Golden Girl =D

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Re: ALL CIE CHEMISTRY DOUBTS HERE !!
« Reply #591 on: May 29, 2011, 08:00:56 pm »
Welcome :)
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Offline $H00t!N& $t@r

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Re: ALL CIE CHEMISTRY DOUBTS HERE !!
« Reply #592 on: May 30, 2011, 02:09:06 pm »
« Last Edit: May 30, 2011, 03:22:27 pm by $H00t!N& $t@r »
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Offline $H00t!N& $t@r

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Re: ALL CIE CHEMISTRY DOUBTS HERE !!
« Reply #593 on: May 30, 2011, 03:05:55 pm »
hello?
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Offline The Golden Girl =D

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Re: ALL CIE CHEMISTRY DOUBTS HERE !!
« Reply #594 on: May 30, 2011, 03:15:29 pm »
You'll get more replies if you gave us a link.
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elemis

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Re: ALL CIE CHEMISTRY DOUBTS HERE !!
« Reply #595 on: May 30, 2011, 03:21:18 pm »
I'm watching a movie right now. I'll answer this in 30-40 minutes time.

EDIT : Answering it now.
« Last Edit: May 30, 2011, 03:25:45 pm by Ari Ben Canaan »
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elemis

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Re: ALL CIE CHEMISTRY DOUBTS HERE !!
« Reply #596 on: May 30, 2011, 03:28:19 pm »
The question says 1cm3 of ice is heated.

Using the formula density = mass/volume you should be able to determine the mass of ice heated.

Next, you must realise that ice is basically water... Can you calculate the no. of moles of water in 1 gram of ice ? Should be easy.

Now, use the formula PV=nRT.

Rearrange for Volume. Remember the volume you calculate will be in m3 NOT dm3.
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elemis

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Re: ALL CIE CHEMISTRY DOUBTS HERE !!
« Reply #597 on: May 30, 2011, 03:31:46 pm »
Dehydration results in the removal of a water molecule.

Look at the double bonds in each option... that is where the water molecule was removed from.

Try drawing each structure with an added hydrogen on one of the carbons in the double bond and a OH on the other carbon.

Do not forget this point in the question :

is unreactive towards mild oxidising agents.

Its a hint... Think about it.
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elemis

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Re: ALL CIE CHEMISTRY DOUBTS HERE !!
« Reply #598 on: May 30, 2011, 03:33:37 pm »
You should be able to eliminate two of the options straight away.

In this scenario, look for the carboxylic acid group.... COOH. Also, look for carbons with four different structures attached to them.

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Offline $H00t!N& $t@r

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Re: ALL CIE CHEMISTRY DOUBTS HERE !!
« Reply #599 on: May 30, 2011, 04:16:23 pm »
Quote from: Ari Ben Canaan on May 30, 2011, 03:31:46 pm
Dehydration results in the removal of a water molecule.

Look at the double bonds in each option... that is where the water molecule was removed from.

Try drawing each structure with an added hydrogen on one of the carbons in the double bond and a OH on the other carbon.

Do not forget this point in the question :

is unreactive towards mild oxidising agents.

Its a hint... Think about it.

Oh i see.. so its D because it is the result of dehydrating tertiary alcohol which is unreactive towards mild oxidising agents.  ;D
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