Chemistry: Post your doubts here!

[strangerss]

can u answer plz ...
i posted it several times but no one answered

in the first question , where there is NH2 that mean that's where the Br was because this reaction is a nucleaphilic substitution reaction so the nucleaphile which is the NH3 will displace the weaker nucleaphile which is the Br and a hydrogen will replace the other bromine , so from the diagram given if you count from left to right the NH2 is present in the 5ht position , which means in it's place there was a Br so the only answer with a Br in the 5th carbon atom is D, that's how I figured it out , but maybe there's another way , but I hope you understood , you should start by drawing the displayed formula of the options given I suppose. In the second question an increase in pressure in the left container will push the mercury so it's level rises to the right, right?when we increase the temp. for the first reaction the equilibrium will shift to the left hand side where there is more number of moles so more pressure , in the second reaction the number of moles on both sides of the equation is the same so where ever the position of equilibrium goes the same number of moles will occupy the container , thus there is no increase in pressure and obviously in the third reaction it will make the mercury level rise to the left which is not what we want. the third question the kc of the reaction was 2:1 right?(it's given in the question)so in the second reaction they switched the reactants so the kc becomes 1:2 PLUS the number of moles is half so it should be the square root of the Kc so it would be the square root of 1/2 which is A . note: if the kc is equal to 2 this mean it's 2/1 beacuse kc is the ratio right? you might ask why we should put the square root of the answer and not divide the new kc by 2 (because the mole number in the second equation became half) and the answer for that I really don't know hehe , may be it's sort of a rule in ratios or something. hope you understood

 

[daredevil]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w06_qp_1.pdf
Q37
daredevil

oo my no ...thts a tough one -__- :/

 

[daredevil]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w06_qp_1.pdf
Q37
daredevil

oo my no ...thts a tough one -__- :/

 

[biba]

can u help plz

sorry for the late reply!!!
i m confused with the organic question cuz i havnt revised organic yet
for equilibria:
q34: 1 is only correct because at the initial stage ,R contains twice as much gas particles as S. If this ratio is maiintained at different temperatures the pressure in r and S wil change but they wil stil be equal to each other i.e. the Hg level will not change!! in 2 and 3,N2 and the given mixture satisfies the condition i mentioned,the amount of gas in bulb remains the same!
whereas in 1 as temperature increases, the equilibrium shifts to the right since the forward reaction is endothermic. The number of gas particles increases and therefore the pressure in R will be higher than that in S.

q9: answer is A bcuz:
for reaction 1: Kc =[X2Y]^2 / [X2]^2 x [Y2] = 2 mol/dm^3
for reaction 2 : Kc = [X2][Y2]^0.5 /[X2Y] ,this becomes,
1/{[X2Y]^2 / [X2]^2 x [Y2]}^0.5 = 1/(2)^1/2

 

[unseen95]

please please please help me with question number 11 of http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_w06_qp_1.pdf with reason.

 

[h4rriet]

please please please help me with question number 11 of http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w06_qp_1.pdf with reason.

You have to make an ICE chart, i.e. initial concentration, change in concentration, and final concentration. The initial concentration of the acid and alcohol is 1 and 1. The change in concentration of the acid and alcohol is -x and -x, and the change in concentration of the ester and the water +x and +x. So the equilibrium concentration of the acid and alcohol is 1-x and 1-x, and the equilibrium concentration of the ester and water is x and x. Now you can find the Kc.

 

[h4rriet]

can u answer plz ...
i posted it several times but no one answered

34. Use Le Chatelier's principle. An increase in temperatue will favour the endothermic side of the reaction. This side produces twice the number of moles. Equation 2 has the same number of moles on each side, so the temperature won't affect the number of moles produced, because the mole ratio will still be 1:1.
9. The reaction is reversed, so the Kc is the reciprocal of the forward reaction's Kc. All the moles are divided by 2, so the square root of the Kc is taken.

 

[Rahma Abdelrahman]

yeah you're right .. so you got it ? great to hear that that question 29 was exhausting

Yeah I got it , thanks

 

[Alool]

does anyone know what's the GT for O/N 2005?

 

[A star]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w06_qp_1.pdf
Q37
daredevil Alice123https://www.xtremepapers.com/community/members/daredevil.11280/its D or 1 only try to break the compounds

 

[daredevil]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w06_qp_1.pdf
Q37
daredevil Alice123its D or 1 only try to break the compounds

didn't understand it

 

[Aymen Ezazi]

in finding the mass why did you multiply by 1*10^-3 ?

density = mass/ volume , so mass will be density multiply by volume.. and volume was in cm^3 so convert it to dm3... but if u dont want to, anyways u'll have to divide it by 1000 to change grams into kilograms..

 

[rawaaaan]

this one please
How does concentrated sulfuric acid behave when it reacts with sodium chloride?
A as an acid only
B as an acid and oxidising agent
C as an oxidising agent only
D as a reducing agent only

 

[messi10]

http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_w09_qp_11.pdf

Can anyone please explain, question number 5 of the above paper.

Regards!

 

[Omar99]

http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_w11_qp_11.pdf
some one in question no. 4

 

[Acelife]

Solve this question please
A student calculated the standard enthalpy change of formation of ethane, C2H6, using a method based on standard enthalpy changes of combustion.
He used correct values for the standard enthalpy change of combustion of ethane (–1560 kJ mol–1) and hydrogen (–286 kJ mol–1) but he used an incorrect value for the standard enthalpy change of combustion of carbon. He then performed his calculation correctly. His final answer was –158 kJ mol–1.
What did he use for the standard enthalpy change of combustion of carbon?
A –1432 kJ mol–1
B –860 kJ mol–1
C –430 kJ mol–1
D –272 kJ mol–1
Ans is C...i cant get the right answer :/

 

[daredevil]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w09_qp_11.pdf

Can anyone please explain, question number 5 of the above paper.

Regards!

B is ur answer bcz in A and D there are no poles bcz similar elements on both sides of the compound cancel the polarity of the compound. with those options eliminated, C has O and Cl at the two oppoite sides while B has Cl and H. the greater difference of electronegativity or polraity is between H and Cl so B is the sure answer.

 

[messi10]

B is ur answer bcz in A and D there are no poles bcz similar elements on both sides of the compound cancel the polarity of the compound. with those options eliminated, C has O and Cl at the two oppoite sides while B has Cl and H. the greater difference of electronegativity or polraity is between H and Cl so B is the sure answer.

Hay there,

Thanks for helping out, but its still not solved
Option B has H and H on both sides.. (not H and Cl, if that would been, then it would be a sure answer, but its not like that)
Option C has Cl and Cl on both sides..


And I'll be grateful if you explain the question number 10 of the same paper. Why its answer is C, whats wrong in B ??\

Regards

 

[1597.2217]

And I'll be grateful if you explain the question number 10 of the same paper. Why its answer is C, whats wrong in B ??
Regards

Because there is always some activation energy required. Option (C) has the smallest activation energy, which is the leap in the curve after the initial straight line. For option (B), the quick initial leap represents the activation energy. Hope this helps.

 

[darknessinme]

oo my no ...thts a tough one -__- :/

Hot KMnO4 causes splitting of double bond (-C=C-), so the ring breaks where the double bond occurs. Each carbon involved in the double bond becomes attached to an oxygen by a double bond (-C=O and O=C-), making it an aldehyde or ketone. If it's an aldehyde group, it is further oxidised to a carboxylic acid.
Apply this to each of the structures.
1. You get two compounds, both have a ketone group and an acid group.
2. Only one double bond, so you don't actually get two organic products, so B is not possible.
3. You get one compound which is a di-oic acid and another compound which is a di-ketone. So 3 doesn't work either.