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Chemistry: Post your doubts here!

Amy Bloom

Amy Bloom

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Gémeaux said:
More exothermic means releasing more energy when the reaction occurs. This would be a more negative value, like -3400 kJ/mol. A less exothermic reaction would release lesser heat in comparison to a more exothermic reaction. This would have a less negative value, e.g. -2000 kJ/mol is less exothermic than -3400kJ/mol.
The - sign indicates that the reaction is exothermic, rest 2000kJ is a value lesser than 3400kJ mathematically.
Click to expand...
thanks! ;)
 
Kirabo Takirambudde

Kirabo Takirambudde

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Need some help with this:

In each of these two compounds, what is the strongest intermolecular force present?
ethanal CH3CHO
methoxymethane CH3OCH3

Thanks
 
messi10

messi10

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Please quick reply,

1. how alcohol is directly formed from alkAne.
2. how alcohol are directly reduced to AlkAne.
 
VelaneDeBeaute

VelaneDeBeaute

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NouranAyman said:
Al SAlam Alaikum! please i have few questions in chemistry AS pleaseeeeee help me asap exam is tomorrow :( ​
http://www.xtremepapers.com/papers/...d AS Level/Chemistry (9701)/9701_s07_qp_2.pdf which is may/june 2007 paper 2 btw, question 2(d) please for the first column , what is this is compound and what will be its product on oxidation??​
Next is http://www.xtremepapers.com/papers/...d AS Level/Chemistry (9701)/9701_s08_qp_2.pdf which is may/june 2008 paper 2 , question 5 (a)ii and 5(e) what structural formula should we write can someone explain!! :(​
and please some one post Good chemistry AS notes so i can revise from it tomorrow :(((( and guys hows your preparation going?? pleaseee reply me anyone. Im freaked out​
One more question was how can we find the bond angle when they ask us H-C-H for example and they say suggest bond angles ​
Click to expand...
For the first query, the compound will be a carboxylic acid because it's heated under re-flux.
For the second one, 5a(ii), it'll be oxidized to a carboxylic acid. For 5(e) it should have just -OH as the terminals. So that leaves us to fill up the bonds of Carbon. Hence C-C will be triple bonded, with OH at each of the terminals (as depicted by the marking scheme).
Go through your book, please! It has several chote chote important points that students tend to overlook!
You look for how many atoms are attached to the central one, and then how many lone pairs are there. Look in the Chp 3 or 4 closely in the Cambridge book, it has a table there! :p
 
NouranAyman

NouranAyman

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VelaneDeBeaute said:
For the first query, the compound will be a carboxylic acid because it's heated under re-flux.
For the second one, 5a(ii), it'll be oxidized to a carboxylic acid. For 5(e) it should have just -OH as the terminals. So that leaves us to fill up the bonds of Carbon. Hence C-C will be triple bonded, with OH at each of the terminals (as depicted by the marking scheme).
Go through your book, please! It has several chote chote important points that students tend to overlook!
You look for how many atoms are attached to the central one, and then how many lone pairs are there. Look in the Chp 3 or 4 closely in the Cambridge book, it has a table there! :p
Click to expand...
Thanks alot :) (Y) exam in 1 hr and 40 mins from now :0 May Allah make it easy for all of us, All the best
 
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freezingfires

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CAN ANYONE PLEASE SOLVE THIS QUESTION OF NOVEMBER 2008 PAPER 4 Q2 PART B

In the late 19th century the two pioneers of the study of reaction kinetics, Vernon Harcourt
and William Esson, studied the rate of the reaction between hydrogen peroxide and iodide
ions in acidic solution.
H2O2 + 2I– + 2H+ 2H2O + I2
This reaction is considered to go by the following steps.
H2O2 + I– IO– + H2O
step 1
IO– + H+ HOI
step 2
HOI + H+ + I– I2 + H2O
step 3
The general form of the rate equation is as follows.
rate = k[H2O2]a[I–]b[H+]c

(b) Suggest values for the orders a, b and c in the rate equation for each of the following
cases.
NUMERICAL VALUE
CASE a b c
step 1 is the slowest overall
step 2 is the slowest overall
step 3 is the slowest overall
[3]
 
Amy Bloom

Amy Bloom

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Can anyone help me in interpreting a two way paper chromatogram?
Explanations badly needed.
1.JPG
 
Gémeaux

Gémeaux

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freezingfires said:
CAN ANYONE PLEASE SOLVE THIS QUESTION OF NOVEMBER 2008 PAPER 4 Q2 PART B
http://www.xtremepapers.com/papers/...d AS Level/Chemistry (9701)/9701_w08_qp_4.pdf
Click to expand...
b) Each option gives a case choosing one of the three to be the slowest step, which the rate-determining step. The reactants of the equation in every case should appear in the rate equation if they affect the equilibrium. Taking step I, we have H2O2 + I– ---> IO– + H2O , this has H2O2 and iodide but no hydrogen ions. so the order for hydrogen would be zero, and that for the other two, 1.
a=1, b=1, c=0

Step II: (IO-) + (H+) ---> HOI here there is hydrogen so for c =1, but the other reagent is an intermediate that *never* appear in a rate equation. the intermediate was formed using H2O2 and I- so the instead of IO-, these would appear i.e.
a=1 and b=1, c=1

Step III: The equation is HOI + H+ + I– ==> I2 + H2O
there is again an intermediate HOI, this was formed using H+ and IO-. This means that there would be two sources of hydrogen ions, one the third equation and the other the second equation, giving c=2. IO- takes us back to the first equation, which uses H2O2 and I-, leaving us again with two sources of I- ( the third eq. and the first one) so b=2. H2O2 comes only once, so a=1.
a=1, b=2, c=2.

Hope it helps.:)
 
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freezingfires

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Gémeaux said:
b) Each option gives a case choosing one of the three to be the slowest step, which the rate-determining step. The reactants of the equation in every case should appear in the rate equation if they affect the equilibrium. Taking step I, we have H2O2 + I– ---> IO– + H2O , this has H2O2 and iodide but no hydrogen ions. so the order for hydrogen would be zero, and that for the other two, 1.
a=1, b=1, c=0

Step II: (IO-) + (H+) ---> HOI here there is hydrogen so for c =1, but the other reagent is an intermediate that *never* appear in a rate equation. the intermediate was formed using H2O2 and I- so the instead of IO-, these would appear i.e.
a=1 and b=1, c=1

Step III: The equation is HOI + H+ + I– ==> I2 + H2O
there is again an intermediate HOI, this was formed using H+ and IO-. This means that there would be two sources of hydrogen ions, one the third equation and the other the second equation, giving c=2. IO- takes us back to the first equation, which uses H2O2 and I-, leaving us again with two sources of I- ( the third eq. and the first one) so b=2. H2O2 comes only once, so a=1.
a=1, b=2, c=2.

Hope it helps.:)
Click to expand...

I got it!! Thank-you so much !! :)
 
M

memorypriest

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cowarrior said:
GUYS HELP NEEDED>>>>>>>>>

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s03_qp_1.pdf

Please help me solving questions:
2. 6. 8. 11. 13. 32. 33.
Click to expand...
Q2) B
1 of the side chain - 1 double bond gets saturated
2 of the side chain - 2 double bond gets saturated x 2 = total of 4 double bond saturated
1+4 = 5

Q6) C
Ideal gas equation work best for gases at High temperature and Low pressure

Q8) A

Q11) A
There is higher % of dissociation of H+ in NH3, thus more acidic.

Q13) C
Pick 1 acidic oxide and 1 alkaline oxide

Q32) C

Q33) A
 
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