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Chemistry: Post your doubts here!

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Can anyone give me the list of the bond angles in our syllabus ????????
2 bond pairs 0 lone pairs 180 degree shape is linear
3 bond pairs 0 lone pairs 120 degree shape is trigonal planor
4 bond pairs 0 lone pairs 109.5 degree shape is tetrahedral
3 bond pairs 1 lone pairs 107 shape is trigonal pyramedal
2 bond pairs 2 lone pairs 104 bent shaped
5 bond pairs 0 lone pairs 90,120 degree shape is trigonal bipyramedal
6 bond pairs 0 lone pairs 90 degree shape is octahedral


Hope that helps:)
 
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How to approach this question?Detailed method please..View attachment 7074View attachment 7074
How to approach this question?Detailed method please..
capture-png.7074
capture-png.7074
in glyceryl trieleosterate the first chain has 31 hydrogen 2 chain has 30 hydrogen and 3 chain has 31 hydrogen making a total of 31+30+31=92 hydrogen


in soft margarine the first chain has 31 hydrogen and the two similar chains have 33+33=66 hydrogen making a total of 31+66=97 hydrogen so for converting one mole of glyceryl into soft margarine u need 97-92 =5 hydrogen so the answer is B
 
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w11/13 http://www.xtremepapers.com/papers/... AS Level/Chemistry (9701)/9701_w11_qp_13.pdf
Plz with little explanations
21, 35, 36, 37, 39, 40

and also
w11/12 http://www.xtremepapers.com/papers/... AS Level/Chemistry (9701)/9701_w11_qp_12.pdf
Q4(i tried twice plz with sol)
Q6, Q8
Q9(how, when temp in so pressure must be decreases to act like a ideal gas)
Q10(Hat is enthalphy change when one mole of gaseous atom is formed from its element under stp,,, plz briefly describe whats this is going on)

I would be really thankful, May Allah grant us the best success in our papers Ameen..
 
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How to approach this question?Detailed method please..View attachment 7074View attachment 7074

Look at those side chains (in the original molecule) carefully, the three of them are equal and have 3 alkene bonds each. ( Overall 9 alkene bonds). Look at the side chains we are to substitute it with : 2 side chains containing one alkene bond and 1 side chain with two alkene bonds. ( Overall 4 alkenes left).

Therefore in the process of the reaction, 9-4= 5 alkene bonds are broken. So, we need 5 molecules of H2.
 
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i really need help with these questions.. 1,4,8,15,16,25,26,29,36.. in some of them i do have an explanation for my answer but apparently they:re wrong so if someone could please correct me!
in 16 isnt the major product of the electrolysis of brine, NaOH?
1n 25 (2.76 - 2.07)/2.76 x 100 gives us 75%. so why isnt B the answer?
in 29..why cant C be the right answer? im getting 3 Ch3 and 3 Ch2..
in 36..if we consider S then it doesnt form a basic hydride?
http://www.xtremepapers.com/papers/... AS Level/Chemistry (9701)/9701_w11_qp_11.pdf
i know these are quite a lot of questions but please do help me out! thanks!
 
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can somebody please explain how to get the ration of j/k in November 2011 paper 43 q5 part b iv?
 
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a friend is having a problem with this question.
Which of the following has dipole-dipole interactions between its molecules, but no
hydrogen bonding?
A Methane, CH4
B Methanol, CH3OH
C Ammonia, NH3
D Hydrogen iodide, HI
the answer is D

i personally think it's BS.
 

Nibz

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a friend is having a problem with this question.
Which of the following has dipole-dipole interactions between its molecules, but no
hydrogen bonding?
A Methane, CH4
B Methanol, CH3OH
C Ammonia, NH3
D Hydrogen iodide, HI
the answer is D

i personally think it's BS.

B has hydrogen bonding, so has C.
A doesn't have dipole-dipole interaction.
D has dipole-dipole interaction because Iodine is highly electronegative, with Hydrogen being partially positive and Iodine partially negative.
 
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yes i know this. but Iodine has a negative dipole so won't it form a hydrogen bond with the neighboring Hydrogen (+ve) in the neighboring HI?
 

Nibz

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No, it won't. You need to have a very strongly electronegative element present for Hydrogen bond to form. Iodine is not that strongly electronegative.
 
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Although one might expect hydrogen bonding to occur between Hl molecules. However, Iodine is too large, and thus the lone pairs are too diffuse ie. not concentrated enough, and hence hydrogen bond attractions between l's lone pairs and hydrogen atoms can not form.

answered it for myself hehe
 
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Thanks Nibz :D

Another question
At which stage is the oxidation state of Nitrogen different to the changes in the other steps

NO3- (aq) -----------(A)----> NO2- (aq) ----(B)--> NO(g) ----(C)--> N2O(g) -----(D)--> N2(g)


so in A its +5 to +3
in B +3 to +2
in C it's +2 to +1
in D it's +1 to 0

How is it B?
 

Nibz

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Thanks Nibz :D

Another question
At which stage is the oxidation state of Nitrogen different to the changes in the other steps

NO3- (aq) -----------(A)----> NO2- (aq) ----(B)--> NO(g) ----(C)--> N2O(g) -----(D)--> N2(g)


so in A its +5 to +3
in B +3 to +2
in C it's +2 to +1
in D it's +1 to 0

How is it B?

It's A. Open your eyes and check the answer again.

http://www.xtremepapers.com/CIE/International A And AS Level/9701 - Chemistry/9701_s11_er.pdf
Paper 11. MCQ number 2 - A
 
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WTH.....examiner report and ms are contradictory ....examiner report is ryt???or ms???
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can't we just find the no of moles of CaSO4 reacting with the acid rain...but if we do so then the loss in mass comes 0.61kg but the mark scheme answer is 0.748kg !!
 
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Chemistry Equilibrium question pwease help !
Chemistry Equilibrium question pwease help !

Can anyone pretty please help me with this question ? ^.^
coz seriously , its driving me crazy >-> :%) :%)

A mixture contains 0.5 mol of ethanoic acid ,0.5 mol of ethanol,0.1 mol of ethyl ethanoate and 0.1 mol of water was set up and allowed to come to an equilibrium at 298K.
CH3CO2H + C2H5OH = CH3CO2C2H5
Calculate the amount of ,in moles, of each of the substance present at the equilibrium, if Kc is 4.0.
Note-(the reaction is reversible)



which year queestion is this????[/quote]

the possible answer to this question is.... ester=0.21 mol and alcohol and carboxylic acid-0.29 and 0.29 mol
 
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