Chemistry: Post your doubts here!

[daredevil]

* A restriction for the bonds to rotate, thus becoming less flexible.

why is there a restriction for rotation??

 

[Snow Angel]

regarding chemistry as level practicals, how are we supposed to estimate the rough titre? and how to manage time with doing the experiment as well as doing the calculations? please answer as possible. thanks in advance.

 

[daredevil]

regarding chemistry as level practicals, how are we supposed to estimate the rough titre? and how to manage time with doing the experiment as well as doing the calculations? please answer as possible. thanks in advance.

for the time part be as deft as possible and should practice titration as much as u can in the laboratory along with the calculations becz there just is no other way to perfect it....

furthermore... where choosing the rough titration value u let the titre pour into the the titrant at a steady stream and keep looking at the titrant. as soon as it changes colour close the tap. the reading u get here is the rough titration value and is more often than not larger than the actual titration result. this reading gives u an idea about where ur actual value will lie. e.g. if the rough value = 27cm3 u can reduce the stream to drops when u reach let's say 20cm3 so that u can later close the tap exactly when the titrant changes colour. i hope it's not too confusing

 

[Namehere]

why is there a restriction for rotation??

Well, dont know if im right so if im not please do correct me!

Its because when you have double bonds like in ethene, the pi system (which has a pi lobe above and below the plane) doesn´t allow rotation because there is no space for it to rotate (since its occupied by the pi lobe)

 

[Namehere]

Well, dont know if im right so if im not please do correct me!

Its because when you have double bonds like in ethene, the pi system (which has a pi lobe above and below the plane) doesn´t allow rotation because there is no space for it to rotate (since its occupied by the pi lobe)

...and i guess that if you want that rotation to occur you will have to break the pi system, and that requires energy. I´m not sure about this, so if anyone knows the answer please do say it.

 

[sidbloom1995]

why is there a restriction for rotation??

Well, dont know if im right so if im not please do correct me!

Its because when you have double bonds like in ethene, the pi system (which has a pi lobe above and below the plane) doesn´t allow rotation because there is no space for it to rotate (since its occupied by the pi lobe)

yes the restriction is there because of pi-bond
think of it as a ball and it is inserted with a stick like in lollipop....but not stuck so if you spin it, it will spin in it's place but now if you attach two sticks with it, the ball wont spin any more because if it tries to rotate around one stick the other will stop it! so it's movement have been restricted!
that is the image i have of it in my mind

 

[daredevil]

yes the restriction is there because of pi-bond
think of it as a ball and it is inserted with a stick like in lollipop....but not stuck so if you spin it, it will spin in it's place but now if you attach two sticks with it, the ball wont spin any more because if it tries to rotate around one stick the other will stop it! so it's movement have been restricted!
that is the image i have of it in my mind

...and i guess that if you want that rotation to occur you will have to break the pi system, and that requires energy. I´m not sure about this, so if anyone knows the answer please do say it.

ooohh i got it thankks!! )

 

[mahabaloch]

http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_w07_qp_1.pdf
Plz help with Q1, 10, 19 and 37

 

[ZohaKhan]

http://papers.xtremepapers.com/CIE/...l O Level/Chemistry (5070)/5070_s11_qp_21.pdf
A2 part a.. how is it sulphur dioxide in MS..
anyone?????

 

[ZohaKhan]

http://papers.xtremepapers.com/CIE/...l O Level/Chemistry (5070)/5070_s11_qp_21.pdf
B6 e)ii) n e)iii ???

 

[ZohaKhan]

http://papers.xtremepapers.com/CIE/...l O Level/Chemistry (5070)/5070_s12_qp_21.pdf
A2 part a all parts
??

 

[Anzuman turin]

@sidbloom1995

 

[Snow Angel]

for the time part be as deft as possible and should practice titration as much as u can in the laboratory along with the calculations becz there just is no other way to perfect it....

furthermore... where choosing the rough titration value u let the titre pour into the the titrant at a steady stream and keep looking at the titrant. as soon as it changes colour close the tap. the reading u get here is the rough titration value and is more often than not larger than the actual titration result. this reading gives u an idea about where ur actual value will lie. e.g. if the rough value = 27cm3 u can reduce the stream to drops when u reach let's say 20cm3 so that u can later close the tap exactly when the titrant changes colour. i hope it's not too confusing

thank you very much, i shall try it next time when im practising

 

[sidbloom1995]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w07_qp_1.pdf
Plz help with Q1, 10, 19 and 37

Q1. the mass that is given is the mass of Ag so use it to find out the moles of Ag
n of Ag = 0.216/108 = 2 x 10^-3
now multiply this with avagadro's constant to find the no. of atoms in this much mass (or moles) of Ag
no. of atoms of Ag present = [2 x 10^-3] X [6.02 x 10^23] = 1.204 x 10^21
now they asked us no.of atoms per unit area (on 1 cm2) and have already given us the area 150 cm2
divide the total no. of atoms by the given area and you will have atoms per unit area
no. of atoms per unit area = [1.204 x 10^21]/150 = 8.03 x 10^18, thus A is your answer

Q10. moles of O2 is given use this to find out the moles of NO and moles of NO2 is also given
use cross multiplication by seeing the ratio of moles from the equation provided
2 : 1
x : 0.8

x = 2 x 0.8 = 1.6
so we have moles of each as
O2 = 0.8
NO = 1.6
NO2 = 4 - 1.6 = 2.4 (because 1.6 moles of NO2 have been converted to NO)

Kc = ([NO]^2 x [O2])/[NO2]^2
Kc = ([1.6]^2 x [0.8])/ [2.4]^2
thus our answer is D

Q19. I was stuck at it so i opened the Examiner's Report
and this was the answer in it

In Question 19 candidates were asked to recognise any chiral centre in three compunds for which only the
molecular formula was given. Many candidates failed to notice that C3H6I2 could have the structure
CH3CHICH2I.

so our answer is B

Q37. no idea sorry

 

[sidbloom1995]

@sidbloom1995

can you please post the MS
I need to make sure of my answers before giving them
thank you

 

[sidbloom1995]

http://papers.xtremepapers.com/CIE/Cambridge International O Level/Chemistry (5070)/5070_s11_qp_21.pdf
A2 part a.. how is it sulphur dioxide in MS..
anyone?????

they said that the gas is evolved and to check the identity of the gas, they tested it with filter paper dipped into acidified potassium dichromate(VI), and the colour changed from orange to green.
this is particularly the test for SO2 gas and it's result tell us that the gas evolved it SO2
i know that you will be confused because normally speaking when we react Cu with H2SO4 we get CuSO4 and H2, but not all of the H2SO4 will undergo the same reaction and some will react differently, forming a different product
and one more thing, your question is from O level this is A level doubt post (i dont mean it in a rude way, just informing you )

 

[sidbloom1995]

http://papers.xtremepapers.com/CIE/Cambridge International O Level/Chemistry (5070)/5070_s11_qp_21.pdf
B6 e)ii) n e)iii ???

you have to look at the data provided in the table
if you notice, the amount of mass deposited on the cathode is 0.12 g for exp. 1
in reaction 2, the current is doubled so the mass deposited will also double thus out mass in exp. 2 is 0.24 g
in exp. 3, the time is doubles but the current is same, 2A so the mass on cathode is again doubled
in part -ii- they are asking the mass lost from anode (A), look at it in terms of loss and gain, the mass gained by cathode (C)is also the mass lost by anode! so our equation will be
Mass lost by A = 1.45 - 0.24 = 1.21 g

in part -iii- they increased the current to 8 A and decreased the time to 90s
one is increased and the other is decreased so the total result will remain the same!
due to that out mass that is gained by the cathode will remain same, 0.24 g
and if we add this to the mass of cathode we will get the mass gained by the cathode
mass gained by C = 1.51 + 0.24 = 1.75 g

 

[ZohaKhan]

they said that the gas is evolved and to check the identity of the gas, they tested it with filter paper dipped into acidified potassium dichromate(VI), and the colour changed from orange to green.
this is particularly the test for SO2 gas and it's result tell us that the gas evolved it SO2
i know that you will be confused because normally speaking when we react Cu with H2SO4 we get CuSO4 and H2, but not all of the H2SO4 will undergo the same reaction and some will react differently, forming a different product
and one more thing, your question is from O level this is A level doubt post (i dont mean it in a rude way, just informing you )

ooooooooooooooooooooooooooooooo i seeee

 

[sidbloom1995]

ooooooooooooooooooooooooooooooo i seeee

is that for the answer or for informing you that it is A level post? -_-

 

[ZohaKhan]

is that for the answer or for informing you that it is A level post? -_-

Answer