Advanced Higher Chemistry 2012-2013

Ahem Ammonium* pardon my french

Hi I was wondering if anyone could help me out with 2012 Section A Q9? I just have no idea where to even start!

The formula mass is just there to trick a person ignore it !!!

To find out the solubility you would just square root the ka which is 1.80x10^-10
and you should get the answer as C

See for this question the [H+] would be 1x10-14 so the [OH-] would be 10^0
This means if they ask for [OH-] of pH 10 then it's 10^4?

10^-4

but it also forms a precipitate with most of the other ones too!

If this is the same question that I remember, the other ones AREN'T SOLUBLE.


The reactant needs to be soluble to form a precipitate as a product.

Could someone help me with question 33 and 38 the answers are c and d


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I can answer 33, The iodine comes off and leaves two CH3 molecules that combine to form C2H6.

So if you have CH3I and C2H5I and you lose the iodine, you have CH3 and C2H5 so the two CH3's can bind, the CH3 and the C2H5 can bind and the two C2H5's can bind, leading to the possibilities of ethane propane and butane.

This is probably a terrible explanation, but i made an attempt

In the flow chart, it says that Aldehydes oxidise to acids, but in the bright red book

it says that they react with Hydrogen Cyanide which is then hydrolysed to acids.
Which of these should I remember?

Does HCN not react with an aldehyde of carbon chain length n, which is then hydrolysed to form an acid of carbon chain length n+1?

How do I do this question?


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I can answer 33, The iodine comes off and leaves two CH3 molecules that combine to form C2H6.

So if you have CH3I and C2H5I and you lose the iodine, you have CH3 and C2H5 so the two CH3's can bind, the CH3 and the C2H5 can bind and the two C2H5's can bind, leading to the possibilities of ethane propane and butane.

This is probably a terrible explanation, but i made an attempt