Chemistry: Post your doubts here!

[soumayya]

hey thnx this was where i got stuck: why didn't we take 2*+994 kJ/mol for N2??

when u break an N2 (N-N) molecule (bond energy), u will get 2 N atoms (zat's wat u need)..

 

[athaan]

Cyanohydrins can be made from carbonyl compounds by generating CN–ions from HCN in the presence of a weak base.

In a similar reaction, –CH2CO2CH3 ions are generated from CH3CO2CH3 by strong bases. Which compound can be made from an aldehyde and CH3CO2CH3 in the presence of a strong base?
A CH3CH(OH)CO2CH3
B CH3CO2CH2CH(OH)CH3
C CH3CH2CH(OH)CH2CO2CH3
D (CH3)2C(OH)CH2CO2CH3
this question is taken from O/N 2009, paper 11/12.

I only know that A is impossible, but how about B,C,D? they only differ in the arrangement of the carbon and amount of carbon atoms.
any help is greatly appreciated, and a fast reply is needed. I'll be having a school test on Monday, and this question bothers me

 

[soumayya]

hope zis hlps...

 

[athaan]

thanks for the post, i hope i can do well on paper 3 with that
has anyone had an explanation for my question in the previous page?

i hope i'm not rushing for an answer, haha.

 

[KurayamiKimmi]

Can anyone tell what we have to study for homogenous and heterogenous catalysis thank you to all the kind folk in advance

 

[autumnsakura]

For Q3, answer is B but I don't understand why D is wrong. Could someone explain and correct my facts?
For Q16, What does it mean when HCL has a negative enthalpy as oppose to HI positive enthalpy. What characteristic would it bring to HCL and HI?
Q18, I just plain don't know..

Would appreciate any help on explaining my 3 questions... thanks!

 

[autumnsakura]

I'm sorry i mis-interpreted the statement. The third option is *The rate of backward reaction increases*! Ofcourse it does. The rate of both the forward and backward reactions increases. Yes, the equillibrium shifts to RHS but because particles are more closer now its easier for them to be converted to products and reactants at the same time. So The more the pressure, the higher the rate.
Just as i made the mistake, i'd like you to remember the difference b/w altering the reaction itself and the rate of reaction. The rate of reaction is independent of the moles etc. and can be increased by increasing temperature or pressure.
Hope this clarifies!

Thanks for being sooo helpful

 

[autumnsakura]

Cyanohydrins can be made from carbonyl compounds by generating CN–ions from HCN in the presence of a weak base.

In a similar reaction, –CH2CO2CH3 ions are generated from CH3CO2CH3 by strong bases. Which compound can be made from an aldehyde and CH3CO2CH3 in the presence of a strong base?
A CH3CH(OH)CO2CH3
B CH3CO2CH2CH(OH)CH3
C CH3CH2CH(OH)CH2CO2CH3
D (CH3)2C(OH)CH2CO2CH3
this question is taken from O/N 2009, paper 11/12.

I only know that A is impossible, but how about B,C,D? they only differ in the arrangement of the carbon and amount of carbon atoms.
any help is greatly appreciated, and a fast reply is needed. I'll be having a school test on Monday, and this question bothers me

I did this question before so I'm going to copy to you the explanation I wrote to myself on the paper..haha! Not sure if you'll understand though..
-CH2CO2CH3 is a nucleophile that is formed. If you look at option C and draw it out, you'll find that after the CO2 it is followed by one CH3, which is exactly the same as the nucleophile given. The nucleophile here is suppose to act like CN-, so the C atom it bonds to should have OH group. Option C fullfills both conditions. Option D is wrong because it isn't an aldehyde (the OH is bonded to a C atom which has two CH3 groups attached thus ketone). Just draw everything out and compare. Break and join stuff and you'll see clearer.. Hope I helped..

 

[athaan]

thank you! the explanation was very helpful.

For Q3, answer is B but I don't understand why D is wrong. Could someone explain and correct my facts?
For Q16, What does it mean when HCL has a negative enthalpy as oppose to HI positive enthalpy. What characteristic would it bring to HCL and HI?
Q18, I just plain don't know..

Would appreciate any help on explaining my 3 questions... thanks!

Q3:
D is incorrect because: If you look by the fact of two elements with the same nucleon number, their mass are the same. As I remember, the mass of proton and neutron is considered to be equal. This is what I think. Correct me if I'm wrong. B is wrong, because: Atom, in fact, can be split into simpler parts. Atom consist of proton, neutron, and electron. Therefore, simpler parts of atom does exist (proton, neutron, electron).

I'll try to work out with the other questions..

 

[autumnsakura]

Glad I could help with your question. I had some head cracking to do too! Thanks for your help as well!

thank you! the explanation was very helpful.


Q3:
D is incorrect because: If you look by the fact of two elements with the same nucleon number, their mass are the same. As I remember, the mass of proton and neutron is considered to be equal. This is what I think. Correct me if I'm wrong. B is wrong, because: Atom, in fact, can be split into simpler parts. Atom consist of proton, neutron, and electron. Therefore, simpler parts of atom does exist (proton, neutron, electron).

I'll try to work out with the other questions..

 

[Amy Bloom]

Hello there, can anyone respond to these 3 questions on chemical energetics and explain the answer to me please:

(1) Which of the following elements in its crystalline form will have the lowest enthalpy change of vapourisation? [ answer is A, why is it not B?]
(A) Argon
(B) Chlorine
(C) Phosphorus
(D) Silicon
(E) Sulfur


(2) The gaseous oxides of nitrogen have positive enthalpy changes of formation. Which one of the following factors is likely to make significant contribution to these enthalpy changes? [answer is D, why is it not E?]
(A) the tendency of oxygen to form oxide ions O2-
(B) the high electron affinity for oxygen atoms
(C) the high electron affinity of nitrogen atoms
(D) the high bond energy of the nitrogen molecule, N2
(E) the similarity of the electronegativities of oxygen and nitrogen.


(3) The standard enthalpy changes for two reactions are given by the following equations: [answer is D]

2Fe (s) + 3/2 O2(g) → Fe2O3 ▲H = -822 Kjmol-1​

C(s) + 1/2 O2 (g) → CO (g) ▲H=-110 Kjmol-1​

What is the standard enthalpy change for the following reaction?

Fe2O3 (s) + 3C (s) → 2Fe (s) + 3CO (g)​

(A) -932 Kjmol-1
(B) -712 Kjmol-1
(C) -492 Kjmol-1
(D) +492 Kjmol-1
(E) +712 Kjmol-1

 

[NouranAyman]

Hey I wanted to ask, what are we supposed to study in the chemical periodicity part, like group 2 and nitrogen and sulphur?? Urgent please some one tell me

 

[Amy Bloom]

Hey I wanted to ask, what are we supposed to study in the chemical periodicity part, like group 2 and nitrogen and sulphur?? Urgent please some one tell me

You have to see section 9 of inorganic chemistry
Nitrogen and sulfur is another part, not in periodicity but found in inorganic chemistry, refer to the syllabus (see attached).
If you are AS student: you gotta study what's NOT in bold.
If you are A2 student you gotta study the above plus the things in bold, that is everything
Hope my answer was helpful.

 

[leosco1995]

Quick question: In Electrochemistry (A2), if both of the reactions have negative electrode potential values, which one goes forward? The most negative or least negative?

 

[Amy Bloom]

Quick question: In Electrochemistry (A2), if both of the reactions have negative electrode potential values, which one goes forward? The most negative or least negative?

the least negative one goes fwd as it is more "feasible" than the more negative one.

 

[Soldier313]

Hello there, can anyone respond to these 3 questions on chemical energetics and explain the answer to me please:

(3) The standard enthalpy changes for two reactions are given by the following equations: [answer is D]

2Fe (s) + 3/2 O2(g) → Fe2O3 ▲H = -822 Kjmol-1 i)​

C(s) + 1/2 O2 (g) → CO (g) ▲H=-110 Kjmol-1 ii)​

What is the standard enthalpy change for the following reaction?

Fe2O3 (s) + 3C (s) → 2Fe (s) + 3CO (g)​

(A) -932 Kjmol-1
(B) -712 Kjmol-1
(C) -492 Kjmol-1
(D) +492 Kjmol-1
(E) +712 Kjmol-1

okay this is the best way i could come up with to explain this qn:



so basically the part i have denoted with i) is the oxidation of Fe and the part i have denoted with ii) is the oxidation of C, i have multiplied it by 3, coz in this reaction, there are 3 atoms of Carbon.
so, we are finding x, x + (i) = 3x(ii)
x + -822 = 3x(-110)
x = -330+822
x = +492 Kjmol-1

hope that helps
sorry couldn't do the first two qns:./

 

[Paparazzi478]

ok so here i have a chemistry p5 question 2 ....... so dont skip this question i really need the answer!!!!
http://www.xtremepapers.com/papers/...d AS Level/Chemistry (9701)/9701_s08_qp_5.pdf
so in Q2 part (f) they asked to draw construction lines
so can you tell me what these construction lines are and how to make them on the graph ????

 

[Amy Bloom]

Soldier313
Thanks mate!

 

[Muhammad Bin Anis]

Quick question: In Electrochemistry (A2), if both of the reactions have negative electrode potential values, which one goes forward? The most negative or least negative?

http://www.xtremepapers.com/community/members/amy-bloom.17629/
Another way of remembering :

(the more reactive the metal is , the more stable it is in the ionic form)

the one that has the lower electrode potential value is more reactive - so it prefers to remain in the ionic form.


Therefore it will not reduce from +2 or +3 (or any positive value) to zero.

 

[Soldier313]

Soldier313
Thanks mate!

Anytym buddy