P1 MCQ's preparation thread for chemistry ONLY!!!!

[Obsidian Fl1ght]

But all we need to know is that with the removal of one H, the C gains a +ve charge. Is that right?

Yes it is.
Thanks anyways.

And referring to my question its may/ june 2012

Thanks.

MJ/12/qp11 Q14 was daunting.

Let's form the equation first.

Al + Ba(NO3)2 ---> Al2O3 + BaO + N2
They said only metal oxides and Nitrogen is produced hence the above.

Balance it. (This balancing was slightly cumbersome)
Anyway:

I got:
10/3 Al + Ba(NO3)2 -------> N2 + BaO + 5/3 Al2O3
Multiply be 3:

10Al + 3Ba(NO3)2 ----------> 3N2 + 3BaO + 5 Al2O3
...............3(261) ......................3(28)
................783............................ 84
..............0.783........................... x

x = 0.783 * 84 / 783 = 0.-84 g N2
Convert it into moles:
0.84/28 = 0.003mol

Find volume.
0.003 * 24000 = 72 cm3

 

[1357911]

10Al + 3Ba(NO3)2 ----------> 3N2 + 3BaO + 5 Al2O3
3(261) 3(28)
783 84
0.783 x

I didnt get this part. Can u plz explain??

 

[HubbaBubba]

21 An alkene has the formula CH3CH=CRCH2CH3 and does not possess cis-trans isomers. What is R?
A) H
B) Cl
C) CH3
D) C2H5

Why can't it be B? Why is the answer D?

 

[Obsidian Fl1ght]

Refresh the page and look at my reply again. Hopefully you'll understand then.
If not, tell me.

And can u please tell me Q11/MJ11/12?

 

[Manobilly]

21 An alkene has the formula CH3CH=CRCH2CH3 and does not possess cis-trans isomers. What is R?
A) H
B) Cl
C) CH3
D) C2H5

Why can't it be B? Why is the answer D?

You have to see the position it will be attached to the same carbon not diff.

 

[Obsidian Fl1ght]

21 An alkene has the formula CH3CH=CRCH2CH3 and does not possess cis-trans isomers. What is R?
A) H
B) Cl
C) CH3
D) C2H5

Why can't it be B? Why is the answer D?

Yeah this is a good question.
I've realised there are TWO cases for cis trans.
1) Two same groups, two different groups.
OR
2) (ANd I found this one out by myself):
4 DIFFERENT groups.

Thus, it can't be B coz then alkene'll have 4 different grps and will exhibit the cis-trans isomerism/

 

[1357911]

Refresh the page and look at my reply again. Hopefully you'll understand then.
If not, tell me.

And can u please tell me Q11/MJ11/12?

Someone else posted this question too. Its really hard. Im still trying to figure it out

 

[Obsidian Fl1ght]

10Al + 3Ba(NO3)2 ----------> 3N2 + 3BaO + 5 Al2O3
3(261) 3(28)
783 84
0.783 x

I didnt get this part. Can u plz explain??

Refresh the page and look at my reply again. Hopefully you'll understand then.
If not, tell me. I've edited it.

And can u please tell me Q11/MJ11/12?

 

[Manobilly]

Someone else posted this question too. Its really hard. Im still trying to figure it out

It will be at the same carbon so how can it posses cis trans? Hope this helps. @hubba

 

[1357911]

Thanks.

MJ/12/qp11 Q14 was daunting.

Let's form the equation first.

Al + Ba(NO3)2 ---> Al2O3 + BaO + N2
They said only metal oxides and Nitrogen is produced hence the above.

Balance it. (This balancing was slightly cumbersome)
Anyway:

I got:
10/3 Al + Ba(NO3)2 -------> N2 + BaO + 5/3 Al2O3
Multiply be 3:

10Al + 3Ba(NO3)2 ----------> 3N2 + 3BaO + 5 Al2O3
...............3(261) ......................3(28)
................783............................ 84
..............0.783........................... x

x = 0.783 * 84 / 783 = 0.-84 g N2
Convert it into moles:
0.84/28 = 0.003mol

Find volume.
0.003 * 24000 = 72 cm3

Im really sorry to disturb u again and again. And Im still not getting it.From where did u get 261 and 28. Can u explain. Sorry again

 

[Pwincessajwa]

http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_w09_qp_12.pdf can soneone help me with q14,20,27,28,30,40 plz ​

 

[1357911]

It will be at the same carbon so how can it posses cis trans? Hope this helps. @hubba

Which question does this diagram refers too?

 

[Manobilly]

Which question does this diagram refers too?

The one HubbaBubba posted ,The R one .

 

[lee mee..]

Someone please help with these questions... PLEEEEAAASSSSS.....
http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_w12_qp_11.pdf
Q 4 and 17.
http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_w11_qp_11.pdf
Q15, 16 (why D is incorrect? ), 18 and 25.
http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_s11_qp_11.pdf
Q2 and 8.
http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_w10_qp_12.pdf
Q7 and 13
http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_s07_qp_1.pdf
Q9, 12, 18 and 26.

 

[Obsidian Fl1ght]

Im really sorry to disturb u again and again. And Im still not getting it.From where did u get 261 and 28. Can u explain. Sorry again

No matter.
Okay.
The equation is:
10Al + 3Ba(NO3)2 ----------> 3N2 + 3BaO + 5 Al2O3

We have given mass of Ba(NO3)2 and we need to fid volume of N2.
For this volume, we first need moles of Nitrogen formed.
There are two ways to do this:

1) - The method I used.
We'll only consider Ba(NO3)2 and N2 right now. We couldn't care any less about the others.
Well, I don't at any rate.

Write down their total Mr.
Mr Ba(NO3)2 = 261, Mr N2 = 28

Now write the TOTAL Mr of the two, inclusive the amount used. This means you'll use the moles used of each as well.

10Al + 3Ba(NO3)2 ----------> 3N2 + 3BaO + 5 Al2O3
...............3(261) ......................3(28)
................783............................ 84
So we have 783 for Ba(NO3)2 and 84 for N2.
Since these are a measure of 'mass', we'll use good old ratio method to find MASS (In g) of N2 formed from given MASS (in g) of Ba(NO3)2


10Al + 3Ba(NO3)2 ----------> 3N2 + 3BaO + 5 Al2O3
...............3(261) ......................3(28)

................783............................ 84
..............0.783........................... x

x = 0.783 * 84 / 783
x = 0.084 g of N2 formed.

U know the rest.

Method 2:
U just use MOLAR ratio method to find moles of N2 formed.

For given mass of Ba(NO3)2, moles are:
0.783/261 = 0.003 moles used.

Molar ratio:
N2 : Ba(NO3)2
3..........3
x.........0.003

x = 0.003 moles (Seriously don't use a calculator for the ratio step.)
0.003 moles of N2 formed.

U know the rest.

I seriously hope u got it.

 

[1357911]

​

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w09_qp_12.pdf can soneone help me with q14,20,27,28,30,40 plz ​

Q-14
A and D cannot be because the bond length increases and the bond energy decreases down the group. As the bond length and the bond energy increases so the boiling point increases too. But as u go down the group the no pf electrons increases so eventually the vanderwaals forces will increases too. That is option B

 

[HubbaBubba]

Yeah this is a good question.
I've realised there are TWO cases for cis trans.
1) Two same groups, two different groups.
OR
2) (ANd I found this one out by myself):
4 DIFFERENT groups.

Thus, it can't be B coz then alkene'll have 4 different grps and will exhibit the cis-trans isomerism/

So for cis trans to occur, either two same groups, or four different groups? :S

It will be at the same carbon so how can it posses cis trans? Hope this helps. @hubba

But if you put Cl, there would be no same groups, so no cis-trans as well!

 

[Pwincessajwa]

Q-14
A and D cannot be because the bond length increases and the bond energy decreases down the group. As the bond length and the bond energy increases so the boiling point increases too. But as u go down the group the no pf electrons increases so eventually the vanderwaals forces will increases too. That is option B

ahan got it thanks what about other questions plx ?

 

[Obsidian Fl1ght]

So for cis trans to occur, either two same groups, or four different groups? :S



But if you put Cl, there would be no same groups, so no cis-trans as well!

Yeah. Something like that.

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[Obsidian Fl1ght]

s11_12_Q11