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it wont playy
For moles practice as much as you can , I know there is less time but still. Once you get the hang of it , its damn easy
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Post the link of the other video
it wont playy
Can you tell me which paper is this? as in the year and the session.The answer is actually C even i thought it was C at first
May/June 2014 p11Can you tell me which paper is this? as in the year and the session.
It is Bromine that will be released into the anode. it is because bromine is lesser in the displacement series, lower than OH- .Can anyone tell me what will be produced at anode in electrolysis of aqueous solution of KBr? (it isn't specified if it is dilute or concentrated)
Isn't OH at the end?It is Bromine that will be released into the anode. it is because bromine is lesser in the displacement series, lower than OH- .
It is Bromine that will be released into the anode. it is because bromine is lesser in the displacement series, lower than OH- .
And why B is the correct answer in this one?View attachment 60447
Yeah I thought so too, but the marking scheme states bromine is discharged. So confusingI believe hydrogen will be produced. Since the solution is aqueous OH- will be selectively discharged. Generally in past paper questions unless the solution is stated to be concentrated it always means it is dilute. Also the displacement series is only taken into regard when deciding on displacement of cations.
Yeah I thought so too, but the marking scheme states bromine is discharged. So confusing
No its Bromine. This question is about a concentrated solution as it only says aqueous' so we assume in this case that it is concentrated unless they themselves have said that it is dilute. For concentrated solutions the rule is slightly different. The solute ie any compound you add (other than water) will always discharge provided that it is not TOO reactive like SO4. Bromine is not , there fore it will be discharged at the anodeThe mark scheme must be wrong then
No its Bromine. This question is about a concentrated solution as it only says aqueous' so we assume in this case that it is concentrated unless they themselves have said that it is dilute. For concentrated solutions the rule is slightly different. The solute ie any compound you add (other than water) will always discharge provided that it is not TOO reactive like SO4. Bromine is not , there fore it will be discharged at the anode
A, because plants make oxygen using photosynthesis which takes place when there is sunlight. During midnight (12 a.m) there would be no sunlight and no photosynthesis so no oxygen.
B, because in the structure of a plant stem or tree bark, phloem is located on the outer region and xylem in the inner region. When the ring of bark is removed , phloem is removed and since it is responsible for translocation of nutrients (amino acids , sucrose) from leaves to roots (and other storage organs) , the plant will die due to lack of nutrition.
Hope this helped, good luck for exams!
How much have you done bro?Thanks a lot and good luck to you too:')
Only a few days remaining and the stress just adds up..
I don't but if it isn't mentioned that something is dilute, we automatically assume that it is aqueous.Do you have a link to the question by any chance? I want to see it for myself. Is it p1 or p2?
If it is concentrated then obviously bromine will be discharged
Yeah set square one was confusing. As for last question, one way to do it was to find the circumference and then use the formula for circumference to find the diameter.what about physics atp.. it seemed to be very strange this time.. no question was from electronics.. wasnt the paper a bit difficult.. how did u use the set squares to find the rod was horizontal (i knew how to do it with a ruler).. and what about the very last experiment?? what did u do in that?
what do you think about the threshold this time, will it be higher or will it go down??
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