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• Flying grenade

  17 Apr 16, 19:45

  Originally posted by UltimaOnline:

  
  Yes, mathematically, a solution can be considered a valid buffer as long as the ratio of molarities of conjugate acid-base pairs lies within 1:10 and 10:1, ie. pH = pKa (+1 or - 1)

  Woah damn cool add 5moldm-3 of acid/base to a buffer solution comprising of 0.5moldm-3 weakacid/base , ph still remains unchanged 

• UltimaOnline

  17 Apr 16, 19:55

  Originally posted by Flying grenade:

  Woah damn cool add 5moldm-3 of acid/base to a buffer solution comprising of 0.5moldm-3 weakacid/base , ph still remains unchanged 

  
  A buffer solution's pH changes *slightly* when a *small* amount of strong acid or base is added to it. Note that highlighted words. And Cambridge will require you to write equations to explain exactly how a buffer works. The exact equations will of course, depend on exactly what is the conjugate acid-base pair of the buffer solution.

  Qn : Why did the acid go to the gym?
  Ans : http://www.infinity.usanethosting.com/Tuition/#BedokFunlandJC_Chemistry_Qns

• Flying grenade

  17 Apr 16, 19:57

  Regarding " As a chemist, ...Edited by UltimaOnline" post , schools (or at least my current jc) still uses mercury thermometers ! Darn it

  Thanks for the toxicity of mercury tip and advice!

• Flying grenade

  17 Apr 16, 20:00

  Originally posted by UltimaOnline:

  
  A buffer solution's pH changes *slightly* when a *small* amount of strong acid or base is added to it. Note that highlighted words. And Cambridge will require you to write equations to explain exactly how a buffer works. The exact equations will of course, depend on exactly what is the conjugate acid-base pair of the buffer solution.

  Qn : Why did the acid go to the gym?
  Ans : http://www.infinity.usanethosting.com/Tuition/#BedokFunlandJC_Chemistry_Qns

  Thank you for elaborating!!!

  Yep sorry, i meant to write pH remains almost unchanged. I'm just marvelled by the fact that buffer solution can buffer up to 10x its concentration

• Flying grenade

  17 Apr 16, 20:40

  http://www.infinity.usanethosting.com/Tuition/H2Chem_What_is_this_Functional_Group.jpg

  For qn 1

  Either ester or tertiary amide is it?

• UltimaOnline

  17 Apr 16, 21:18

  Originally posted by Flying grenade:

  http://www.infinity.usanethosting.com/Tuition/H2Chem_What_is_this_Functional_Group.jpg

  For qn 1

  Either ester or tertiary amide is it?

  
  Wrong. Go ask your school teacher or private tutor.
• Flying grenade

  18 Apr 16, 17:51

  pH=pKa is at maximum buffer capacity.

• Flying grenade

  18 Apr 16, 17:54

  Can i ask, rj book page 274, qn part iii) ,

  What shld i notice about the ICE table? Why can skip ICE table? I understand why can skip icF table already 

• UltimaOnline

  18 Apr 16, 17:55

  Originally posted by Flying grenade:

  Hi ultima,

  May i ask, pH=pKa at equivalence point or maximum buffer capacity? 

  
  This is the kind of question that with a little googling, you can easily settle by yourself. Don't disappoint me.
• UltimaOnline

  18 Apr 16, 17:56

  Originally posted by Flying grenade:

  Can i ask, rj book page 274, qn part iii) ,

  What shld i notice about the ICE table? Why can skip ICE table? I understand why can skip icF table already 

  
  If you don't see why you can skip the table (either ICE or ICF), then you should not skip it. Note that ICF tables are actually more critical than ICE tables, in certain contexts or scenarios.
• Flying grenade

  18 Apr 16, 18:20

  Originally posted by UltimaOnline:

  
  This is the kind of question that with a little googling, you can easily settle by yourself. Don't disappoint me.

  Hey sorry, just now my brain keesiao. And yep, i know the answer alr. Thanks and sorry .

• Flying grenade

  18 Apr 16, 19:37

  Hi ultima, hope you could lend help for this qn,

  Qn 1 in here : https://www.dropbox.com/s/3o7vr03sfgpyhm6/20160418_193429.jpg?dl=0

  Qn and attempted answer is in the pic. Suggested soln is C, but i cant figure out why. I dun have solution for mcq, only final answer alphabet

   

  Thanks

• Flying grenade

  18 Apr 16, 20:16

  Hi ultima, hope you could help me here,

  For qn 2

  https://www.dropbox.com/s/ses61cnec6rtqpo/20160418_201120.jpg?dl=0

   

  Suggested answer is D. 

  update : sorry, for careless mistake. Ionic charge on BrF2^+ ion is one.

• UltimaOnline

  18 Apr 16, 21:08

  Originally posted by Flying grenade:

  Hi ultima, hope you could lend help for this qn,

  Qn 1 in here : https://www.dropbox.com/s/3o7vr03sfgpyhm6/20160418_193429.jpg?dl=0

  Qn and attempted answer is in the pic. Suggested soln is C, but i cant figure out why. I dun have solution for mcq, only final answer alphabet

   

  Thanks

  
  2015 ACJC P1 Q9

  Work out the Kc value for original conditions.

  Kc = (183x183)/90 = 372.1

  Apply the same Kc value for new conditions.

  Concordantly, equilibrium partial pressure of hydrogen gas = Square root of (150x372.1) = 236.25 kPa.

• UltimaOnline

  18 Apr 16, 21:16

  Originally posted by Flying grenade:

  Hi ultima, hope you could help me here,

  For qn 2

  https://www.dropbox.com/s/ses61cnec6rtqpo/20160418_201120.jpg?dl=0

   

  Suggested answer is D. 

  Qn: how to draw the structure for BrF2+? Is 2 bond pairs, 1.5 lone pairs with respect to(w.r.t.) Br atom correct? 

  Thanks 

  
  2015 DHS P1 Q4

  Since the electrical conductivity of BrF3 decreases (ie. position of equilibrium shifts to LHS) with increasing temperature, hence heat may be regarded as a product. Hence reaction is exothermic.

  The cation is the unipositive BrF2+, for which due to its larger atomic radius (to minimize steric strain caused by van der Waals repulsions), as well as the fact that a positive formal charge is more stable on a less electronegative atom (Br is less electronegative than F), means that the central atom must be the Br atom with a unipositive formal charge, which means that the Br atom must have (in order to have a unipositive formal charge yet have a stable octet configuration) must have 2 lone pairs and 2 bond pairs, which means the electron geometry is tetrahedral, and making lone pairs invisible, means the ionic geometry of BrF2+ is therefore V-shape or bent.

• Flying grenade

  18 Apr 16, 21:32

  Originally posted by UltimaOnline:

  
  2015 ACJC P1 Q9

  Work out the Kc value for original conditions.

  Kc = (183x183)/90 = 372.1

  Apply the same Kc value for new conditions.

  Concordantly, equilibrium partial pressure of hydrogen gas = Square root of (150x372.1) = 236.25 kPa.

  How would u know the values for CO (g) (as values for CO not given) ?? 

  By avogadro's law? But n= v/vm , vm same for all gaseous products, since same temp and pressure, all gaseous products is 1 mol from this eqn, wouldn't volume be different? 

   

  So next time i just use the same partial pressure value of a gaseous product, of same mol, for the partial pressure value of a unknown gaseous molecule of same mol ??

• UltimaOnline

  18 Apr 16, 21:36

  Originally posted by Flying grenade:

  How would u know the values for CO (g) ?? 

  By avogadro's law? But n= v/vm , vm same for all gaseous products, since same temp and pressure, all gaseous products is 1 mol from this eqn, wouldn't volume be different? 

   

  So next time i just use the same partial pressure value of a gaseous product, of same mol, for the partial pressure value of a unknown gaseous molecule of same mol ??

  
  I've already given you detailed solution liao. If you still can't figure out this little bit by yourself (even if it didn't occur to you before reading my solution, after reading my solution you should now be able to figure out why), you'll have to go ask your school teacher or private tutor. Try to refrain from repeatedly asking too much on a single question, especially after I've already given you the answer on the forum.
• Flying grenade

  18 Apr 16, 21:37

  Originally posted by UltimaOnline:

  
  2015 DHS P1 Q4

  Since the electrical conductivity of BrF3 decreases (ie. position of equilibrium shifts to LHS) with increasing temperature, hence heat may be regarded as a product. Hence reaction is exothermic.

  The cation is the unipositive BrF2+, for which due to its larger atomic radius (to minimize steric strain caused by van der Waals repulsions), as well as the fact that a positive formal charge is more stable on a less electronegative atom (Br is less electronegative than F), means that the central atom must be the Br atom with a unipositive formal charge, which means that the Br atom must have (in order to have a unipositive formal charge yet have a stable octet configuration) must have 2 lone pairs and 2 bond pairs, which means the electron geometry is tetrahedral, and making lone pairs invisible, means the ionic geometry of BrF2+ is therefore V-shape or bent.

  Sorry, found out my careless mistake. Unipositive charge on BrF2+ ion, not dipositive charge.

• UltimaOnline

  18 Apr 16, 21:50

  Originally posted by Flying grenade:

  Thank you for godly explanation.

   

  But i dun understand, br has 7 valence e-, bond with 2 F atoms, left 5 unpaired e-, lose 2 electrons as indicated by 2+ ionic charge of polyhalogen cation, shouldn't br be left with 3 unpaired e- , thats why im confused

  
  For every question's discussion, there will come a point where your own further doubts and misconceptions will have to be clarified face-to-face with a school teacher or private tutor, as it will be too inefficient to attempt further elaboration on the forum, especially after I've already described and explained the essential answer for everyone reading this (ie. the essential solution is already given, each person reading it will have unique further thoughts and doubts, for which they will have to figure out the nitty-gritty details for themselves, and/or consult their school teacher or private tutor).

  You've reached that point for this question. Have a nice day.

• Flying grenade

  19 Apr 16, 17:53

  From 2015 h2 chem A level thread,

  for the part that asks us to distinguish between hbr and hcl, is by adding h2so4?

  Unfortunately that doesn't work with "dilute aqueous solutions" as in this question. You'll require gaseous HBr and concentrated H2SO4 (which is a serious safety hazard and thus shouldn't be used for distinguishing tests). But you may still get 1 out of 2 marks, provided you correctly described the smell of SO2(g) evolved as a "strong, choking odor".

   

  Are we supposed to know hbr can react with h2so4 and hcl can't? Never thought of or saw h-x reaction with h2so4 before 

• Flying grenade

  19 Apr 16, 18:10

  Cs toh advanced study guide page 238 "group VII elements are typical non metals ( each element has 7 valence electrons) "

  I know by observation that non metals have large number of valence e-, but is large number of valence e- a *characteristic* of non metals?

   

  From wiki " ...chemically, they tend to have high ionization energy and electronegativity values, and gain or share electrons when they react with other elements or compounds. " 

• Flying grenade

  19 Apr 16, 18:31

  thinning of ozone layer *reduces* global warming, not increases it!

   

  Ideally ozone layer should be thin ,without holes, is it?

• UltimaOnline

  19 Apr 16, 18:37

  Originally posted by Flying grenade:

  From 2015 h2 chem A level thread,

  for the part that asks us to distinguish between hbr and hcl, is by adding h2so4?

  Unfortunately that doesn't work with "dilute aqueous solutions" as in this question. You'll require gaseous HBr and concentrated H2SO4 (which is a serious safety hazard and thus shouldn't be used for distinguishing tests). But you may still get 1 out of 2 marks, provided you correctly described the smell of SO2(g) evolved as a "strong, choking odor".

   

  Are we supposed to know hbr can react with h2so4 and hcl can't? Never thought of or saw h-x reaction with h2so4 before 

  
  Yes, you're supposed to know. Look at your syllabus (or CS Toh's A Level Study Guide) under Group VII (called Group 17 in new 2017 syllabus).
• UltimaOnline

  19 Apr 16, 18:41

  Originally posted by Flying grenade:

  Cs toh advanced study guide page 238 "group VII elements are typical non metals ( each element has 7 valence electrons) "

  I know by observation that non metals have large number of valence e-, but is large number or valence e- a *characteristic* of non metals?

   

  From wiki " ...chemically, they tend to have high ionization energy and electronegativity values, and gain or share electrons when they react with other elements or compounds. " 

  
  By definition, metals are electropositive, non-metals are electronegative. To be electronegative, you need more protons in your nucleus vis-à-vis the other elements in your same period. Condordantly, as a non-metal you need to be on the right side of the periodic table. Ergo, you'll have a large number of valence electrons. So it's not a definition of non-metals, it's just a characteristic. Transition metals also have a large number of valence electrons, because both the (n-1)d and ns electrons (eg. 3d and 4s), are both considered valence electrons.
• UltimaOnline

  19 Apr 16, 18:42

  Originally posted by Flying grenade:

  thinning of ozone layer *reduces* global warming, not increases it!

   

  Ideally ozone layer should be thin ,without holes, is it?

  
  But if the ozone layer is thin, it will have holes. You can't have your cake and eat it too. You can't lepak for A levels and still score distinction A grade (unless you're me, Mr BedokFunland JC).