Chemistry: Post your doubts here!

[AbbbbY]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w10_qp_12.pdf


Q3
31
Question 36 ( why 2 is wrong ?)

please help !

Q3: C
A Charge on the cation is ruled out. Mg is 2+ and Na is 1+ yet they both have the same lattice.
B Ratio of the ionic charges is ruled out. The ratio is the same for all three lattices.
C NaCl = (102/181 = 0.56) MgO = ( 72/140 = 0.51) CsCl = (167/181=0.92). So definitely this.
D Ruled out. Sum of ionic charges is 0 in all cases.


Q31: D in a heartbeat.
2 is ruled out the second you see it. P4O10 will react with the ammonia produced. This alone eliminates A B C. 3 is wrong anyway and even an O Level kid will tell you that!

36: D again.

1- Increased surface area = more catalyst present = reduced NOx conc.
2- More exhaust gases same surface area. Catalyst being used completely. Some gas escapes without making contact with the catalyst. NOx goes up
3- Higher temperature = much more NOx produced.

 

[Haya Ahmed]

Q3: C
A Charge on the cation is ruled out. Mg is 2+ and Na is 1+ yet they both have the same lattice.
B Ratio of the ionic charges is ruled out. The ratio is the same for all three lattices.
C NaCl = (102/181 = 0.56) MgO = ( 72/140 = 0.51) CsCl = (167/181=0.92). So definitely this.
D Ruled out. Sum of ionic charges is 0 in all cases.


Q31: D in a heartbeat.
2 is ruled out the second you see it. P4O10 will react with the ammonia produced. This alone eliminates A B C. 3 is wrong anyway and even an O Level kid will tell you that!

36: D again.

1- Increased surface area = more catalyst present = reduced NOx conc.
2- More exhaust gases same surface area. Catalyst being used completely. Some gas escapes without making contact with the catalyst. NOx goes up
3- Higher temperature = much more NOx produced.

Loads of Thanks

 

[AbbbbY]

Plz help me with these questions
Octobernovember 2009 paper42 q5 part A.
Octobernovember 2009 paper41 q3 bi and bii
October November 2008 paper4 q2 part b.
THANKS!!

ON09 42 Q5a.
I've answered this before a day or two ago but I'll do it again anyway.
Look. This stuff isn't in the syllabus or taught. This is precisely why it has shown you the 4 figures above the actual question. The examiner wants you to analyze those figures, see whats going on, and apply THAT to the question.

The question shows that all carbons in Benzene can be made to lie on the same plane, but not in cyclohexane. The question also shows that butane Carbons can be made to lie on the same plane, but methylpropane, also an alkane, cannot.
This tells us that if a compound has a ring, OR branching, it cannot be made to lie on the same plane.

All of them are co-planer apart from B.

C is coplaner because you don't have to consider the Oxygen atoms. The others are unbranched and two individual chains lying on the same plane.
Get it?

ON 41

Q3bi
[Cu(H2O)6]2+ = Light blue
[Cu(NH3)4(H2O)2]2+ = Deep blue.

You don't even need to know how to analyze absorption spectrums to know this. Remember the Cu ion test in p3?

Q3biii
I'd make a graph between the two shown. Meaning, with a lower amplitude than (NH3)4 but higher than (H2O)6 and with the maximum point around 700nm.


ON 08
2b:
1 1 0
1 1 1
1 2 2

I won't lie. I got this question wrong too when I attempted it back in my mocks. Asked a teacher to explain it to me.
Basically, you have to consider ALL steps above the slowest step identified.

As in, if you're considering step 3 to be the slowest, you have to use the values from step 2 and 1 as well. If you're considering step 2 slowest, you've to use values from step 1 as well and so forth. I still can't come up with a logical answer as to why we do this. Maybe someone else can help clarify. Namehere

 

[sitooon]

Q3: C
A Charge on the cation is ruled out. Mg is 2+ and Na is 1+ yet they both have the same lattice.
B Ratio of the ionic charges is ruled out. The ratio is the same for all three lattices.
C NaCl = (102/181 = 0.56) MgO = ( 72/140 = 0.51) CsCl = (167/181=0.92). So definitely this.
D Ruled out. Sum of ionic charges is 0 in all cases.


Q31: D in a heartbeat.
2 is ruled out the second you see it. P4O10 will react with the ammonia produced. This alone eliminates A B C. 3 is wrong anyway and even an O Level kid will tell you that!

36: D again.

1- Increased surface area = more catalyst present = reduced NOx conc.
2- More exhaust gases same surface area. Catalyst being used completely. Some gas escapes without making contact with the catalyst. NOx goes up
3- Higher temperature = much more NOx produced.

Abby can you answer the question of paper5 posted ab ahmed abdullah

 

[AbbbbY]

Abby can you answer the question of paper5 posted ab ahmed abdullah

Will do in a lil while. A bit stuck up + I loathe P5's

 

[sara kamal]

pH = pKa + lg([salt]/[acid]) (or pH = pKa - lg([acid]/[salt]) )
5.5 = -lg(1.79 × 10^-5) + lg([Sa]/[A])
5.5 - 4.75 = lg([Sa]/[A])
lg([Sa]/[A]) = 0.75
[Sa]/[A] = 10^0.75
[Sa]/[A]= 5.62
S = 5.62A
A+S = 100
100-A=5.62A
A = 15.1 = 15
So S = 85.

Acid = 15, Salt = 85. Is this the answer?


Edit: Somehow [ S ] wasn't being posted without the gaps so I just changed it to Sa.

Jazak Allah khair for yr help,May Allah grant you with good grades !

 

[NaNinG]

Reposting.....:/
Ny A2 lvl genius pls help!!!!!!!!(simple question)

http://www.sheir.org/a-level-chemistry-51-nov2010.pdf
Q 2 (c)
y axis-----> Freezing point
x axis------> Molality of G

Can some1 please suggest which sequence of range of number do we asign for Y axis (especially) n X axis.
really get stuck here ;/

 

[AbbbbY]

Posting this for more than once
Paper 5 , http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w12_qp_53.pdf

Question 2b(iii) ( moles )
Question 3 , how to Complete the table when two rows not given


& Whats is the max. Volume a gas syringe can hold which we are supposed to label here ?

Sorry for the delays. Got a bit busy + don't really like opening p5's, personally.
__
2biii:
This is a pretty simple question with a horribly messed up wording. See. Collecting 30dm3 and 24dm3 is not possible with normal lab equipment. Collecting 10cm3 or 20cm3 is.

We know that 1 mol = 123.5g and this produced 30000 cm3 in 2.1 and 24000 in 2.2

So, simply using a ratio, we can find the mass that'd produce x cm3, a measurable value, of gas.

30,000 : 123.5
10 : x

Now you'll simple use this value of x found as the mass of CuCO3 to be used. If it produces 10cm3, it'll be following 2.1. If not, it'll be following 2.2. This can be repeated for several values of gas released and for either of 2.1 or 2.2.
Get it?

3a)

Again. Ratio.

Volume of Sulfuric acid:

10g of H2SO4 : 5.476 cm3
20g = x
x = 10.952

Volume of water:
70g of water : 70.211cm3 of water
20g of water : x
x = 20.060

and so on.

Total volume of 100g of solution:
Simple! Add the volume of Sulfuric acid and Water found.

Density of the solution:

mass / vol
so 100/volume of 100g sol

Easiest question I've seen concerning calculations tbh. If you don't get it let me know and I'll explain it in a more organized manner.

 

[itallion stallion]

ON09 42 Q5a.
I've answered this before a day or two ago but I'll do it again anyway.
Look. This stuff isn't in the syllabus or taught. This is precisely why it has shown you the 4 figures above the actual question. The examiner wants you to analyze those figures, see whats going on, and apply THAT to the question.

The question shows that all carbons in Benzene can be made to lie on the same plane, but not in cyclohexane. The question also shows that butane Carbons can be made to lie on the same plane, but methylpropane, also an alkane, cannot.
This tells us that if a compound has a ring, OR branching, it cannot be made to lie on the same plane.

All of them are co-planer apart from B.

C is coplaner because you don't have to consider the Oxygen atoms. The others are unbranched and two individual chains lying on the same plane.
Get it?

ON 41

Q3bi
[Cu(H2O)6]2+ = Light blue
[Cu(NH3)4(H2O)2]2+ = Deep blue.

You don't even need to know how to analyze absorption spectrums to know this. Remember the Cu ion test in p3?

Q3biii
I'd make a graph between the two shown. Meaning, with a lower amplitude than (NH3)4 but higher than (H2O)6 and with the maximum point around 700nm.


ON 08
2b:
1 1 0
1 1 1
1 2 2

I won't lie. I got this question wrong too when I attempted it back in my mocks. Asked a teacher to explain it to me.
Basically, you have to consider ALL steps above the slowest step identified.

As in, if you're considering step 3 to be the slowest, you have to use the values from step 2 and 1 as well. If you're considering step 2 slowest, you've to use values from step 1 as well and so forth. I still can't come up with a logical answer as to why we do this. Maybe someone else can help clarify. Namehere

Thanks a lot!it was really helpful.Still stuck up on the coplaner.u said that if a compound has ring or branching it can't be made coplaner.how come A is coplaner as it has ring.
Thanks again.

 

[AbbbbY]

Reposting.....:/
Ny A2 lvl genius pls help!!!!!!!!(simple question)

http://www.sheir.org/a-level-chemistry-51-nov2010.pdf
Q 2 (c)
y axis-----> Freezing point
x axis------> Molality of G

Can some1 please suggest which sequence of range of number do we asign for Y axis (especially) n X axis.
really get stuck here ;/

What are your max and min values for the x and y axis. Cba to calculate it all.

 

[AbbbbY]

Thanks a lot!it was really helpful.Still stuck up on the coplaner.u said that if a compound has ring or branching it can't be made coplaner.how come A is coplaner as it has ring.
Thanks again.

It has an arene ring/benzene ring. Not a cyclic ring. Look at the 4 diagrams shown in the question! It blatantly states that benzene is co-planer.

 

[NaNinG]

What are your max and min values for the x and y axis. Cba to calculate it all.

well for da Y-axis---------> Max (4.37) /*C Min(1.03) mol/kg

X-axis--------->Max (2.35) Min (0.556)

i jus messup when deciding which range of numbers to apply ,n how much cube/block u may call to space.
thanx in advance Abby

 

[AbbbbY]

& Whats is the max. Volume a gas syringe can hold which we are supposed to label here ?

Sorry just saw this.

There are several, available, but in P5, due to common-lab equipment restrictions, we're to use 100cm3 gas syringes.

 

[sitooon]

Order wrt ( H+ )

 

[AbbbbY]

well for da Y-axis---------> Max (4.37) /*C Min(1.03) mol/kg

X-axis--------->Max (2.35) Min (0.556)

i jus messup when deciding which range of numbers to apply ,n how much cube/block u may call to space.
thanx in advance Abby

That's low enough to take real-axes

x axis: I'd take 0 origin and 3 end.
y axis: I'd take 0 origin and 5 or 6 end.

9 big black lines on the x axis,
12 on the y axis.

X: 9/3 = 3 big boxes = 1.00
Y: 12/6 = 2 big boxes = 1.00

Not saying that everyone should use this, just showing how I do it.

Oh, and if I were you, I'd plot from 103 to 437 for increased accuracy. As in, every 2 big boxes = 100

 

[AbbbbY]

Order wrt ( H+ )

View attachment 40016

Order wrt H2O2 = Order 1
Order wrt H+ = Order 0


A 0.07 0.05 1.4
B 0.09 0.07 1.8

I need to convert A to 0.09 0.05 1.4 to compare the two (or convert B to 0.07 0.07 1.8. Same thing).

We know it's order 1. So, simply do this:

0.07 0.05 1.4
0.09 0.05 X

0.09/0.07 = X/1.4
X = 1.8

So order wrt H+ = 0.


I know it looks awfully long but I did all this in <30 seconds on my calculator. Understand the concept and practice it and you will too.

 

[NaNinG]

AbbbbY



how do u decide the origin n da ends...??

 

[AbbbbY]

AbbbbY



how do u decide the origin n da ends...??

Idk I just take somethign that fits on the divisions without awkward calculations, and something that has enough spread to cover 2/3rds of the graph. So, a slight overlap from the max value.

 

[sitooon]

Order wrt H2O2 = Order 1
Order wrt H+ = Order 0


A 0.07 0.05 1.4
B 0.09 0.07 1.8

I need to convert A to 0.09 0.05 1.4 to compare the two (or convert B to 0.07 0.07 1.8. Same thing).

We know it's order 1. So, simply do this:

0.07 0.05 1.4
0.09 0.05 X

0.09/0.07 = X/1.4
X = 1.8

So order wrt H+ = 0.


I know it looks awfully long but I did all this in <30 seconds on my calculator. Understand the concept and practice it and you will too.

1.8=1.8 , so its one ? why your answer and MS answer is zero

 

[Namehere]

ON 08
2b:
1 1 0
1 1 1
1 2 2

I won't lie. I got this question wrong too when I attempted it back in my mocks. Asked a teacher to explain it to me.
Basically, you have to consider ALL steps above the slowest step identified.

As in, if you're considering step 3 to be the slowest, you have to use the values from step 2 and 1 as well. If you're considering step 2 slowest, you've to use values from step 1 as well and so forth. I still can't come up with a logical answer as to why we do this. Maybe someone else can help clarify. Namehere

I´ll go in for the rescue!

Lets consider the first row, step 1 is the slowest: This means that the hydrogen peroxide and iodide are involved somehow in the rate determining step. Since they are only involved once (directly or indirectly, you´ll see what I mean later on) and they are in the rate equation, then a=1 and b=1, c=0 because there is no hydrogen ions reacting in step 1.

Lets consider the second row, step 2 is the slowest: This means that the IO- and hydrogen ion are involved somehow in the rate determining step, but the IO- ion doesn´t appear in the rate equation! First, it is clear that the hydrogen ion will give us c=1, since it is only involved once and appears on the rate equation. Secondly, in order for the IO- to be produced, the hydrogen peroxide and hydrogen ion will have to react, and since these are involved in the rate equation, it follows that a=1 and b=1 (since each specie was involved only once), though this time the hydrogen peroxide and hydrogen ion were involved indirectly.

Lets finally consider the third row, step 3 is the slowest: This means that HOI, hydrogen ion and iodide are involved somehow in the rate determining step. Lets analyze HOI. We can see that for HOI to be formed, the IO- and hydrogen ion had to react, but the IO- which was needed for reaction 2 was produced by reaction of hydrogen peroxide with iodide from reaction 1! By looking at the rate equation we see that hydrogen peroxide was involved indirectly once to produce HOI and so we can see a=1. Lets continue with the hydrogen ion. We can see that it is in the rate equation and that it is involved twice, one directly from reaction 3 and one indirectly from reaction 2, since hydrogen ions reacted with IO- to form HOI, one of the reactants somehow involved in the rate determining step in reaction 3. Since it is involved twice, it follows that c=2. Finally with iodide we can also see it is in the rate equation. Here the iodide ion has been involed twice, again once directly from reaction 3 and one indirectly from reaction 1 (since iodide was needed in reaction 1 to produce IO-, which was needed in reaction 2 to make HOI, which was needed in reaction 3 - the rate determining step. Since it is involved twice, it also follows that b=2.

Hope it helps. It is a bit difficult to explain these types of questions by simply writing it. Do ask if you are still unsure about something.