Qualification > Sciences

ALL CIE CHEMISTRY DOUBTS HERE !!

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Deadly_king:

--- Quote from: Ari Ben Canaan on October 09, 2010, 05:17:13 am ---Ah, damn. I didnt think of it in that way.  :-X

+rep.

--- End quote ---

It's alright dude  ;)

Deadly_king:

--- Quote from: moon on October 08, 2010, 09:24:05 pm ---plz help in Nov.2009 ppr22 Q2(v) &

also Nov.2009 ppr42 Q8(c) (ii), d(ii),(iv)

thanx in advanc.

--- End quote ---

Nov 09 P2 Q2(v)

In most cases, Hydrogen has an oxidation number of -1.

Therefore in the first 3 cases. take oxidation number of H to be -1. Since they are all neutral molecules, resultant charge should become to zero. This principle is applied to ionic compounds only.

MgH2 :
Let oxidation number of Mg be x.
x  + 2(-1) = 0 ---> x = +2

Apply same principle for AlH3 and you'll get +3.

As for the other two, they are covalently bonded such that oxidation state of hydrogen is +1

PH3
Let oxidation no of P be x.
x + 3(+1) = 0
x = -3

same method for H2S.  ;)

nid404:

--- Quote from: Deadly_king on October 08, 2010, 07:50:11 am ---Hehe......yeah I guess so  ;)

Anyway it would be nice if you could confirm the method since I didn't obtain the exact answer  ???

--- End quote ---

Your method is not wrong really...

what I would have done instead

at 298K 1 mole occupies 24dm3
so At 596 K 1 mole occupies 48 dm3

1g of steam= 1/18 moles
volume of steam= 1/18 X 48= 2.67 dm3

Deadly_king:

--- Quote from: Garfield on October 09, 2010, 06:54:45 am ---Your method is not wrong really...

what I would have done instead

at 298K 1 mole occupies 24dm3
so At 596 K 1 mole occupies 48 dm3

1g of steam= 1/18 moles
volume of steam= 1/18 X 48= 2.67 dm3

--- End quote ---

Oooh.........yeah.
I guess I chose the long way :P

It's far easier to understand like you described  ;)

Thanks :)

nid404:

--- Quote from: Deadly_king on October 09, 2010, 07:04:13 am ---Oooh.........yeah.
I guess I chose the long way :P

It's far easier to understand like you described  ;)

Thanks :)

--- End quote ---

Considering it's an MCQ , you wouldn't want to do it the long way ;)

My pleasure :)

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