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Chemistry: Post your doubts here!

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Any expected practical in 34? By your teacher?
I literally have no idea.
I think it'll have something about thermal decomposition, definitely titration and maybe maybe organic chemistry question in qualitative analysis, the last time they got organic was I guess 2016,
I hope it's organic cause it's much easier.
These are my assumptions okay (y)
 
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Could anyone please explain what the question asking exactly and how to do it?

View attachment 64754
Is it B?
If so, what I think is this,
we know the Mr of KNO3 is 101.1 , and they gave us the actual mass so we know we have 1 mole of KNO3. To define Mr, we say it's the mass of particles that have the same no of atoms as in C-12 isotope, basically its the mass of a particle when it's one mole. Since we know we have one mole of KNO3 this means that the actual mass of nitrogen is 14g
looking at the Mr of urea, the Mr is 60g, if we take 60g (Actual mass) then that means the actual mass of nitrogen is 28g since the number of moles is (60/60) and we want half the 28, so this means if I take 30g actual mass, my number of moles will be (30/60) so that multiply by urea equation means I have 14 g nitrogen since I'm halfing the equation.
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Chem 34 practical on 28th may.(COPIED DONT KNOW ABOUT AUTHENTICITY)
1. Redox titration
2. Enthalpy change through titration
3. Salt analysis (FeSO4 and CaCl2)
 
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Is it B?
If so, what I think is this,
we know the Mr of KNO3 is 101.1 , and they gave us the actual mass so we know we have 1 mole of KNO3. To define Mr, we say it's the mass of particles that have the same no of atoms as in C-12 isotope, basically its the mass of a particle when it's one mole. Since we know we have one mole of KNO3 this means that the actual mass of nitrogen is 14g
looking at the Mr of urea, the Mr is 60g, if we take 60g (Actual mass) then that means the actual mass of nitrogen is 28g since the number of moles is (60/60) and we want half the 28, so this means if I take 30g actual mass, my number of moles will be (30/60) so that multiply by urea equation means I have 14 g nitrogen since I'm halfing the equation.
Was this right ? helpful?
Heyy thank you so much....yes you're right, its B
 
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I need genuine help here,
I don't understand option C, I said A because I confused with C, the actual answer is B (btw this is section B) so one and 2 only are right
upload_2019-5-28_22-5-12.png
 
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I need genuine help here,
I don't understand option C, I said A because I confused with C, the actual answer is B (btw this is section B) so one and 2 only are right
View attachment 64756

X+ is boron-10 (5 protons, 5 neutrons, 4 electrons)
Y+ is boron -11 (5 protons, 6 neutrons, 4 electrons)
Z+ is aluminium (13 protons, 14 neutrons, 12 electrons)
 
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Hello everyone!
could anyone please explain to me how the answers are begin obtained in the following qns.
qn6.-A
qn9.-B (is there a shorter method except trial and error)
qn28.-C
upload_2019-6-1_9-57-14.png

upload_2019-6-1_9-57-41.png

upload_2019-6-1_9-58-9.png
 
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Hy
Could anyone explain how to get answers to the following
8-A
18-D
39-B

View attachment 64762

View attachment 64763

View attachment 64764
For Q8 C2O4 has an overall charge of 2- since Oxygen will give - 8 charge you will have to +6 to make the overall charge to 2-,therefore each atom of C will have +3 charge and its oxidation as the the oxidation state changes from +3 to +4.
For Q18 you'll have to make equation of decomposition for each nitrate and use the total mass of gas obtained of No2 and O2 which is 108g and solve by mass ratios.
 
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Is electrolysis of brine and extraction of Aluminium in the syllabus for chem p1??
 
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For Q8 C2O4 has an overall charge of 2- since Oxygen will give - 8 charge you will have to +6 to make the overall charge to 2-,therefore each atom of C will have +3 charge and its oxidation as the the oxidation state changes from +3 to +4.
For Q18 you'll have to make equation of decomposition for each nitrate and use the total mass of gas obtained of No2 and O2 which is 108g and solve by mass ratios.
Thank you

Could you plz explain about the mass ratios and how you are calculating themm
 
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