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Chemistry: Post your doubts here!

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qhttp://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Chemistry%20(9701)/9701_s13_qp_13.pdf 8
 
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That'll be 3:2

From half equation,
Tl+ --> Tl3+ + 2e

So 3 moles of Ti+ loses 6 moles of electrons to 2 moles of VO3+

1 mole of VO3+ gains 3 moles of electrons (oxidation state will decrease by 3 units)

V in VO3+ has original oxidation number of +5.

New oxidation number = +5 - 3 = +2
 
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From half equation,
Tl+ --> Tl3+ + 2e

So 3 moles of Ti+ loses 6 moles of electrons to 2 moles of VO3+

1 mole of VO3+ gains 3 moles of electrons (oxidation state will decrease by 3 units)

V in VO3+ has original oxidation number of +5.

New oxidation number = +5 - 3 = +2
Got it! Thank you so much!
 
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according to boyle's law, p1v1 = p2v2
so when the 2 flask connect together, the total volume is now 1 + 2 = 3dm3
using the formula above , for flask X
- 1 x 2 = p x 3
- p = 2/3
for flask Y
- 2 x 1 = p x 3
- p = 2/3
therefore, the total pressure after both flask connect is 2/3 + 2/3 = 4/ 3
so answer is A.
i dont get it
if we calulate total volume to b 3 dm3 thn total pressure should b 2/3 :(
 
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Oct/ nov 2008
PLEASE HELP!!
Question 2 ans is C
4 answer is C
I don't understand 3 :$ answer is D
30 !! Answer is C

Please HELPPPP!!! :(

w08qo1

Q2
Picture 2.png

Q4. HCO3- total electrons= 1 from H + 6 from C + 3 x 8 from O + 1 from negative charge = 32

Q30.
moles of ethanol = 30/46 = 0.652
moles of ethanoic acid =30/60 = 0.5 (limiting)
Based on limiting reactant, we expect 0.5 moles of ester = 0.5 x 88 = 44g
Since we only obtained 44g, yield = 44/88 = 50%
 
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i dont get it
if we calulate total volume to b 3 dm3 thn total pressure should b 2/3 :(
What i am doing is finidnt the pressure of the gas in flask X in Joined
and the presuure of gas in flask Y when joined
they both apply pressure in the final volumne
 
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need help in this question
Q: flask X contain 1 dm3 of helium at 2kpa pressure and flask Y contain 2 dm3 of neon at 1kpa pressure.
if the flasks are connected at constant temperature, what is the final pressure?
A: 4/3 kpa
B:3/2 kpa
C; 5/3 kpa
D:2 kpa
how do we solve it and answer is A

Focus on individual gases first.

Helium exerted 2kPa of pressure when occupying 1dm^3 at start, what pressure did it exert when it occupies 3 dm^3 at the end?
Pressure (start) x volume (start) = pressure (end) x volume (end)
2 x 1 = pressure (end) x 3
pressure (end) = 2/3 kPa from helium


Neon exerted 1kPa of pressure when occupying 2dm^3 at start, what pressure did it exert when it occupies 3 dm^3 at the end?
Pressure (start) x volume (start) = pressure (end) x volume (end)
1 x 2 = pressure (end) x 3
pressure (end) = 2/3 kPa from Neon

Total pressure of Ne + He = 4/3 kPa
 
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Q14. In a way, this is more of a recall question (although of course students can study the reasons behind the trends).

For the solubility , if it helps, recall that Barium sulfate is insoluble.

So it gets less soluble for the sulfates down the group.

For hydroxides, the solubility is the opposite.

thank u :')

and thanks for the link _Ahmad
 
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Focus on individual gases first.

Helium exerted 2kPa of pressure when occupying 1dm^3 at start, what pressure did it exert when it occupies 3 dm^3 at the end?
Pressure (start) x volume (start) = pressure (end) x volume (end)
2 x 1 = pressure (end) x 3
pressure (end) = 2/3 kPa from helium


Neon exerted 1kPa of pressure when occupying 2dm^3 at start, what pressure did it exert when it occupies 3 dm^3 at the end?
Pressure (start) x volume (start) = pressure (end) x volume (end)
1 x 2 = pressure (end) x 3
pressure (end) = 2/3 kPa from Neon

Total pressure of Ne + He = 4/3 kPa
thanku vry much
u guys r awsome
 
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s11qp12

Q10. There are many maths approach, but i think least stressful is to do trial and error with the 4 options and tally with the Mr.

Q29. Perhaps you can show how many chiral carbons (and where) you are getting, then we can see whats missing?

Q33. Equation 2 is out. pV= nRT, not MRT
That makes only equation 1 to be true.

pV= nRT
p=nRT/V
p=mRT/MV

p= density x RT/M
 
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