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[Obsidian Fl1ght]

s_11_qp12
Q35.
Nitrogen ions are reduced to nitrogen atoms.NaOCl is a reducing agent(bleach). And NaCl gives a white precipitate with Silver nitrate.(Cations and anions Tests)

So, if something's being merely reduced, it still undergoes redox?
.
Tch, Look what I've written... this is pathetic. Okay Can you just define redox for me so I can clarify A once and for all?
I'm still confused.

 

[salvatore]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s09_qp_1.pdf
Could someone please explain question no. 15, 19, 20 and 28?

I don't know how to go about the organic questions

I'm reposting this.. someone pleaseee help me!

 

[1357911]

So, if something's being merely reduced, it still undergoes redox?
.
Tch, Look what I've written... this is pathetic. Okay Can you just define redox for me so I can clarify A once and for all?
I'm still confused.

Redox means when an element is oxidised or reduced. yes, Andif something's being merely reduced, it still undergoes redox

 

[polniks]

can someone please help me with this question I would really appreciate it thanks in advance:
http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_w11_qp_12.pdf q 26

 

[1357911]

http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_w11_qp_12.pdf q 26
C2H5OH on gentle haeting and imediate distillation produces CH3CHO
Take out the RMM for both
RMM= 46:44
do Ratio and proportion method
46:44
2.3:x find x x=1.54

 

[Obsidian Fl1ght]

A'ight ppl... this is the third time I'm re-asking this question - and they claim third time's lucky - So let's see about that.
.
s_11_qp12 Q11



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[queen of the legend]

The equation below represents the combination of gaseous atoms of non-metal X and of
hydrogen to form gaseous X

2H6 molecules.
2X(g)

+ 6H(g) → X2H6(g) ΔH = –2775 kJ mol–1
The bond energy of an X–H bond is 395 kJ mol

–1.
What is the bond energy of an X–X bond? ans :+405 how ?

 

[queen of the legend]

A'ight ppl... this is the third time I'm re-asking this question - and they claim third time's lucky - So let's see about that.
.
s_11_qp12 Q11



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ok the way i have thought of it without any calculation ....how many electrons does carbon want = 4 ...how many will Al lose 3
so if we have 3 Al x 4 = 12 electrons and no. of carbons will be 4 x 3 = 12 i knw its kinda wierd but when u got thinkk out of the box smtimes nd if ur lucky enough it ill work so we get Al4 C3

 

[1357911]

9701_s12_qp_11.pdf
q-1 can anyone help?

 

[Obsidian Fl1ght]

9701_s12_qp_11.pdf
q-1 can anyone help?

Answer is A.
Consider BF3.
B is in Grp III.
F is in Grp VII
When B bonds covalently with 3 Fluorine atoms, it forms 3 single CVBs.

So three electrons offered by Boron while each Fluorine atom gives one electron each.
Total CVBs = 3 (Single).
Total electrons shared = 6.
Electronic configuration of Boron in BF3: 1s2. 2s2, 2p6.
Incomplete valence shell... hence A.

 

[1357911]

Answer is A.
Consider BF3.
B is in Grp III.
F is in Grp VII
When B bonds covalently with 3 Fluorine atoms, it forms 3 single CVBs.

So three electrons offered by Boron while each Fluorine atom gives one electron each.
Total CVBs = 3 (Single).
Total electrons shared = 6.
Electronic configuration of Boron in BF3: 1s2. 2s2, 2p6.
Incomplete valence shell... hence A.

What about the other 3 options???????

 

[1357911]

9701/11/M/J/12 q-14 Please help

 

[Obsidian Fl1ght]

What about the other 3 options???????

B
CH3-
First look at CH3.
Three single CVBs between C and H.
C has 4 + 3 = 7 electrons.
But now notice that lil' sign on top of CH3. : -
Hence C has an extra electron.
Valence now has: 7 +1 = 8 electrons.

C
F2O:
O has 6 electrons. Two partake in CVB formation with F. (So 4 remain)
Two CVBs: 4 shared electrons.
Now add shared electrons to remaining (aka the two lone pairs) of O... 4 + 4 = 8


D:
H3O+
Three CVBs.
6 shared electrons.(3 electrons provided by O for this - 6-6 = 3 left)
Add shared to remaining electrons.
6 + 3 = 9 electrons.
Now due to the +ve sign, take one electron away... 9 - 1 = 8

There.

 

[Obsidian Fl1ght]

M

9701/11/M/J/12 q-14 Please help

MJ12/variant 11?

Btw after this, kindly tell me about the different radicals that can be formed during free radical substitution reactions!
I've no idea how to do them!

eg:
CH3CH2(C)(CH3)3
What are the radicals possible?

 

[Manobilly]

O/n 2004 q10 can't attach the link

 

[1357911]

CH3CH2(C)(CH3)3
What are the radicals possible?

You mean when it reacts with a halogen atom?

 

[Obsidian Fl1ght]

CH3CH2(C)(CH3)3
What are the radicals possible?

You mean when it reacts with a halogen atom?

Yeah.

 

[1357911]

Im sorry, im not able to understand this.

 

[Obsidian Fl1ght]

Im sorry, im not able to understand this.

I've come across questions in which they ask the possible radical formations of the organic specie during Free-radical substitution.
I can't find those questions right now.
But all we need to know is that with the removal of one H, the C gains a +ve charge. Is that right?

And confirm the year. MJ12? Or MJ11? (Referring to yr question)

 

[1357911]

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