Chemistry: Post your doubts here!

[aalmuhannadi]

Would someone mind explaining why the answer for this is C?

 

[aalmuhannadi]

 

[scouserlfc]

Would someone mind explaining why the answer for this is C?
View attachment 19836

Have u read the word decreased by 10*C if not then thats the problem because in a decrease in temp. the boltzman distribution graph gets more narrower and elongated so L will definitely increase but as it gets narrower M will decrease and definitiely N is to decrease to so its C otherwise when the temp. increases graph becomes more wider and less elongated and Activation energy also increases

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for this one is the answer A ?? tell me the answer then ill explain !

 

[aalmuhannadi]

Have u read the word decreased by 10*C if not then thats the problem because in a decrease in temp. the boltzman distribution graph gets more narrower and elongated so L will definitely increase but as it gets narrower M will decrease and definitiely N is to decrease to so its C otherwise when the temp. increases graph becomes more wider and less elongated and Activation energy also increases


for this one is the answer A ?? tell me the answer then ill explain !

Alright I understood the first one, the second one the answer is A but I'm confused as to why you can't also have hydrogen bonding with B. I know C and D are wrong.

 

[scouserlfc]

Alright I understood the first one, the second one the answer is A but I'm confused as to why you can't also have hydrogen bonding with B. I know C and D are wrong.

Thats because in B u have another aldehyde which has a functional group of O--C--H so in this u dont have any hydrogen bonded to an electronegative element just like in the methanal they said in the question so if this is used u wont have hydrogen bonds but dipole-dipole attractions for hydrogen bonds u need either F,O and N with H so if H is with these then u have Hydrogen bonds and in A we have H with O

 

[aalmuhannadi]

Thats because in B u have another aldehyde which has a functional group of O--C--H so in this u dont have any hydrogen bonded to an electronegative element just like in the methanal they said in the question so if this is used u wont have hydrogen bonds but dipole-dipole attractions for hydrogen bonds u need either F,O and N with H so if H is with these then u have Hydrogen bonds and in A we have H with O

Oh I thought that H was also bonded with O in B, but we didn't cover any organic chemistry in school yet so I guess that's why I got it wrong, many thanks!

 

[scouserlfc]

Oh I thought that H was also bonded with O in B, but we didn't cover any organic chemistry in school yet so I guess that's why I got it wrong, many thanks!

Yeah ! ! ! Organic is a bit difficult

 

[MKh@lid]

What are the industrial uses of NaOH and Chlorine(obtained by using diaphragm cell)??

 

[iKhaled]

can someone explain to me question 4(c) of this paper. it is so confusing!

http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_s07_qp_4.pdf

 

[guitar]

Hey guys !! ive got my exams coming so i was practicing past papers so i came across a qn which i culdnt figure out so i thought you guys could help

http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_w11_qp_21.pdf
its qn no 3 d

 

[aalmuhannadi]

Can someone please just check if my explanation behind this is correct or not?
The answer is D because conc. H2SO4 isn't a strong enough oxidising agent to oxidise chlorine ions, which is why the gas made is colourless HCl. However, Iodine is a weak reducing agent due to its position in the group and is oxidised by conc. H2SO4 to Iodine vapour.

 

[aalmuhannadi]

And in this question, the answer is somehow D.. I know that NaI reacts with hot H2SO4 to give HI, I, H2O and H2S, so shouldn't they also be the products for the reactions between, say NaAs with H2SO4 but with As instead of I?

 

[aalmuhannadi]

Uploaded 3 more questions. In the first question, how do you deduce whether or not the reaction is endothermic or exothermic? In Q11, how do you recognise that the ions act as catalysts? Q16: how are the volumes achieved? I know how to get the molar quantities of Cl2 and H2 from their ionic equations but what about NaOH?

 

[scouserlfc]

What are the industrial uses of NaOH and Chlorine(obtained by using diaphragm cell)??

Many uses, NaOH is used as a base in many acid base reactions for producing various salts, Chlorine can be combined with NaOH both cold and hot to produce NaClO (bleach) and NaClO3(weed killer) respectively . NaOH may be used as an electrolyte . Chlorine as a water disinfectant this is its most important use worldwide and also for producing white paper by bleaching it so its a bleaching agent . remember the saponofication from OLevels dont worry if u dont just remember NaOH is used for that as well !
Chlorine really has so many uses used in PVCs, CFCs and other polymers and organic reactions

 

[scouserlfc]

View attachment 19883
Can someone please just check if my explanation behind this is correct or not?
The answer is D because conc. H2SO4 isn't a strong enough oxidising agent to oxidise chlorine ions, which is why the gas made is colourless HCl. However, Iodine is a weak reducing agent due to its position in the group and is oxidised by conc. H2SO4 to Iodine vapour.

Yup its absolutely correct ! But make this deduction that when u say Cl ions then say I ions as well and dont say iodine because then u may contradict compare with like i think and u can say I ions are more stronger reducing agents ! (just remember Iodine is a weak reducing agent but iodide is a strong one )

And in this question, the answer is somehow D.. I know that NaI reacts with hot H2SO4 to give HI, I, H2O and H2S, so shouldn't they also be the products for the reactions between, say NaAs with H2SO4 but with As instead of I?
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this should be D as you say as again as you go down grp 7 the ions of halogens become more and more stronger reducing agents like u can see from the reaction of HCl,then HBr and HI the products formed increase and they end up oxidising themselves always with more vigour also here i think they want you to just tell the main thing formed and not all products so since As forms in almost all four out of the three reactions so its made into the answer ! this reaction will be very similar to that of Iodine forget similar its exactly what u said (CIE went nuts here )

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Uploaded 3 more questions. In the first question, how do you deduce whether or not the reaction is endothermic or exothermic? In Q11, how do you recognise that the ions act as catalysts? Q16: how are the volumes achieved? I know how to get the molar quantities of Cl2 and H2 from their ionic equations but what about NaOH?

1.Basically just remember this that any decomposition or disassociation reaction is always endothermic as you break down the strong bonds of the substance and moreever if my memory is right i read that PCl5 was an ionic solid at stp so its more endothermic ! then abt the shape thats simple u can do it yourself !

11.If u look at the reaction 2 and 3 closely and combine them u will see that Cu+ and Cu2+ can be cancelled and u will have the reaction 1 only in the forward direction so u can easily use the catalysts definition and state that Cu+ and Cu2+ ions are catalysts (if u dont get this please ask as this was complex for me too)

16.i dont know if you people know this equation or not but here it is !
2H2O + 2e ---> H2 + 2(OH)- so from here you see that 2 OH ions are produced so u will need exactly same no of Na+ ions so NaOH has a molar ratio of 2 and Cl will be 1 and H will also be 1

Remember how u got the equation: 2H+ ----> 2e + H2
you got it from the decomposition of: H2O ---> H+ + (OH)-
so just add 2e- here and multiply by 2 : 2H2O +2e---> H2 + 2(OH)-

 

[scouserlfc]

can someone explain to me question 4(c) of this paper. it is so confusing!

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s07_qp_4.pdf

this is A2 so i cant do this so wait for others !

Hey guys !! ive got my exams coming so i was practicing past papers so i came across a qn which i culdnt figure out so i thought you guys could help

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w11_qp_21.pdf
its qn no 3 d

ill do this later its such a big question but dont worry today or tomorrow it will be done or someone else will do it !

 

[avenger]

can someone explain to me question 4(c) of this paper. it is so confusing!

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s07_qp_4.pdf

(i) Solution C has almost zero absorbance (i.e. it reflects all) of wavelength 650, which has red colored photons, so C has red colored solution (remember, color is caused by the light that is being reflected!)
Solution D has minimum absorbance for 400-450 wavelength (violet/blue color) so it appears blue.

(ii) Solution C will have higher energy gap because it absorbs photons of low wavelength (high frequency) and E=hf (physics is applied here :S )

hope u get it

 

[aalmuhannadi]

Here in Q11, I don't get why the answer is A, what's the effect of acidifying on this in chemical terms? and in Q33, I know 1 is correct but for 2, when i drew the ion i had a double bond between H and S so S had only 2 lone pairs, what's incorrect here?

 

[avenger]

View attachment 19931View attachment 19932
Here in Q11, I don't get why the answer is A, what's the effect of acidifying on this in chemical terms? and in Q33, I know 1 is correct but for 2, when i drew the ion i had a double bond between H and S so S had only 2 lone pairs, what's incorrect here?

11. when u add acid, H+ ions react with OH-, reducing their conc. so the equilibrium shifts to right, producing more HOCl.

not sure about 33. have forgotten a lot of AS

 

[scouserlfc]

View attachment 19931View attachment 19932
Here in Q11, I don't get why the answer is A, what's the effect of acidifying on this in chemical terms? and in Q33, I know 1 is correct but for 2, when i drew the ion i had a double bond between H and S so S had only 2 lone pairs, what's incorrect here?

For 33 try contemplating on this that the SH can never be negative unless u have negative hydrogen ion attached to the sulfur atom so do this and you will have three lone pairs !

For 11 i need to revise this Cl thing tomorrow ill do it INSHAALLAH