Chemistry: Post your doubts here!

[twisty]

assalamualaikum can some one plsssss clear my doubttttttt plssss
it comes under solubility of product lesson in the equilibrium chapter a2 chem


solubility produst of agcl is 1.46(multiplied)10(to the power of -7) mol(squared)dm(to the power negative 6) at 25(degrees celcius)

0.1 moldm(to the powernegative3) NaCl was added to a saturated soluton of AgCl
calculate the new solubility of AgCl and comment on ure findings

 

[daredevil]

http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_s09_qp_4.pdf

plleasee someone explain question 1 cii

 

[snowbrood]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s09_qp_4.pdf

plleasee someone explain question 1 cii

 

[snowbrood]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s09_qp_4.pdf

plleasee someone explain question 1 cii

can u please help me sketch that titration curve q1b(iii)

 

[Yash Jain]

friends i have doubt in
9701_w07_qp_1 => Q n0. 13 and 14

 

[Namehere]

friends i have doubt in
9701_w07_qp_1 => Q n0. 13 and 14

For Q13: All of the ions in the question will have an electronic configuration similar to that of Ar. However, the only difference is the nuclear charge. The answer is A - P3- and the reason being it has the least nuclear charge which makes the electrons less attracted compared to the other species. It has also been added 3 extra electrons so the extra repulsion felt by those electrons makes them want to spread out more and so causing a larger radius.

For Q14: A is the only option which makes sense. D cannot be because if the first ionisation energy were to increases down the group the atomic radius would have to decrease down the group, which obviously isn´t true. C cannot be because of similar reason to D. B cannot be because what determines the melting point of a substance (in this case Group II metals) is the charge on the cation and the cloud/sea of delocalised electrons. The greater the charge on the cation, the greater the contribution to the delocalised sea of electrons and so the stronger the metallic bond is, making it have a higher melting point. So B cannot be because all have the same charge on the cation, since they are all in group II. A makes sense because if you look at the periodic table, across the period (atomic number increases by 1) and the mass number increases by more than 1 and so the graph has this kind of curve shape.

Hope it helps.

 

[daredevil]

can u please help me sketch that titration curve q1b(iii)

see u have a weak base at first so pH slightly lower than 7 and a strong base so pH will rise up to almost a full 14 at the end when all is in...

curve and tilt the corners of the graph but u get the idea
and i missed somehting.... the equilibrium point will be marked with an 'x' in the halfway of the vertical line if needed... here it isn't needed i think.
tell me if u dont get this

 

[snowbrood]

see u have a weak base at first so pH slightly lower than 7 and a strong base so pH will rise up to almost a full 14 at the end when all is in...

curve and tilt the corners of the graph but u get the idea
and i missed somehting.... the equilibrium point will be marked with an 'x' in the halfway of the vertical line if needed... here it isn't needed i think.
tell me if u dont get this

i know it should start at 1.94 and end at 14
there should be a vertical line at 10vol added but at what ph should that curve join the vertical line i dont get it and the vertical line should go how much units upwards

 

[daredevil]

i know it should start at 1.94 and end at 14
there should be a vertical line at 10vol added but at what ph should that curve join the vertical line i dont get it and the vertical line should go how much units upwards

it's no hard and fast rule... even the starting point is at 1.94 bcz u calculated it. and u have the equation so u know what ratio of volumes would do the complete reaction but u dont know the pH at which neutralisation is complete so u just have to work with the concept that as it was a weak acid and stronger base, the mid point of the vertical line should be slightly above pH 7 ... and r u sure about the 1.94?? bcz that does not seem like the pH a WEAK acid would have. 1.94 is QUITE acidic.
we only work with the numericals we have and other than that we need to show the right concept and it's done

 

[aaaamfa]

2) In the preparation of soft margarine, glyceryl trieleostearate
is suitably hydrogenated so that, on average, one of its side-chains is converted into the
CH3(CH2)4CH=CHCH2CH=CH(CH2)7CO2 residue and two side-chains are converted into the
CH3(CH2)7CH=CH(CH2)7CO2 residue.
How many moles of hydrogen are required to convert one mole of glyceryl trieleostearate into the
soft margarine?
A 4 B 5 C 6 D 9

need help with the explaination how do we get to the ans in such questions

 

[aaaamfa]

Section A question no. 6 i dont know what to do and how to do

 

[aaaamfa]

section A
question no. 9
question no. 10
question no. 11
Section B
question no. 33

 

[aaaamfa]

^^^^^
srry for posting many questions at a time
but i am preparing for as at home and need a little co-operation frm the ppl of this site

 

[daredevil]

^^^^^
srry for posting many questions at a time
but i am preparing for as at home and need a little co-operation frm the ppl of this site

ummm is D the answer??

 

[snowbrood]

it's no hard and fast rule... even the starting point is at 1.94 bcz u calculated it. and u have the equation so u know what ratio of volumes would do the complete reaction but u dont know the pH at which neutralisation is complete so u just have to work with the concept that as it was a weak acid and stronger base, the mid point of the vertical line should be slightly above pH 7 ... and r u sure about the 1.94?? bcz that does not seem like the pH a WEAK acid would have. 1.94 is QUITE acidic.
we only work with the numericals we have and other than that we need to show the right concept and it's done

i dont kw i think examiner did a blunder here i dont get the graph at all half equavelnce point equavalence point all are against the rules lets see maybe i have to ratafy it well u can check the ms

 

[crys123]

the answer is D. Why is C wrong? C and D both have one chiral centre and they both have -COOH group!

 

[crys123]

Ans is C.

 

[crys123]

the ans is D

 

[crys123]

this paper is absolutely killing me! ans is D

 

[hino]

this paper is absolutely killing me! ans is D

tertiary halogenoalkanes are the ones that only proceed through Sn1 mechanism and Dz the only option which consists of a tertiary halogenoalkane
A and B both are primary ones while C is a secondary one
Just in case if you are unsure about what a tertiary one is :In a tertiary halogenoalkane, the carbon atom holding the halogen is attached directly to three alkyl groups which in this case is (Ch3)3CI