• We need your support!

    We are currently struggling to cover the operational costs of Xtremepapers, as a result we might have to shut this website down. Please donate if we have helped you and help make a difference in other students' lives!
    Click here to Donate Now (View Announcement)

Chemistry: Post your doubts here!

AbbbbY

AbbbbY

Messages
675
Reaction score
862
Points
103
immie.rose said:
someone pls explain the following!!


View attachment 43906
Answer: B

View attachment 43907
Answer: B


View attachment 43908
Answer:C
Click to expand...

Are you sure you didn't read the answers wrong? I've gotten to A with the first one, I'm damn sure the second one is D. Could you please recheck?

For the last one,

One mol of alcohol reacts with one mol of the acid to form one mol of ester.

Ethanol: 46
Ethanoic Acid: 60
Ethyl Ethanoate: 88
Since ethanoic acid has the greater Mr and same mass is used, it's the limiting reagentt. 30/60 = 0.5 mol so 0.5 mol of the ester = 44g should be formed.

22/44 = 0.5 so, 50% yield.
 
immie.rose

immie.rose

Messages
124
Reaction score
343
Points
73
AbbbbY said:
Are you sure you didn't read the answers wrong? I've gotten to A with the first one, I'm damn sure the second one is D. Could you please recheck?

For the last one,

One mol of alcohol reacts with one mol of the acid to form one mol of ester.

Ethanol: 46
Ethanoic Acid: 60
Ethyl Ethanoate: 88
Since ethanoic acid has the greater Mr and same mass is used, it's the limiting reagentt. 30/60 = 0.5 mol so 0.5 mol of the ester = 44g should be formed.

22/44 = 0.5 so, 50% yield.
Click to expand...

OMG sorry!! Yes the first one is A, and next is D. M all mess atm, sry agn.
Thanks!!
 
AbbbbY

AbbbbY

Messages
675
Reaction score
862
Points
103
Gehad Mohamed said:
Anyone plz Help me in Q1 Q 20 Q27 Q28 Q29 Q35 http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w09_qp_12.pdf
Click to expand...

1:
D
__
CO2 = 0.8 mol
H2O = 0.8 mol

0.2 : 0.8 : 0.8
= 1 : 4 : 4
So, the Hydrocarbon is giving 4 mols of CO2 and H2O on combustion of 1 mol of the HC. The only options that can give 4 mols of CO2 are C and D. Now, I know by looking the an alkene cannot give me 1:1 CO2:H2O, but even if you don't know that, just combust the hydrocarbon and pick.
C4H4 + O2 -> 4CO2 + 2H2O
C4H8 + O2 -> 4CO2 + 4H2O

20:
A
__

Aah. I've answered this so many times here. Should save the answer in notepad or sth :p

Anyway.
You need to understand that everytime a ring is formed, or a double bond is formed the compound loses TWO hydrogens.

C20H28O.
So if it were a fully saturated HYDROCARBON, it'd have 42 Hydrogens. C20H42.
We have 28, so 14 are missing.
You have an Aldehyde, a ring, and a double bond in the ring. Thats 6 carbons hydrogens.
14-6 = 8. 8/2 = 4 further double bonds.

Since you're asked total number of double bonds, 4+1 = 5 double bonds.
I hope this makes sense :)

27:
C?

You basically need a central C with OH, CH2CO2CH3, H and R attached.
A is out. CO2CH3, not CH2CO2CH3
B is out. CH3CO2CH2CH, not CH2CO2CH3
D is out. While CH2CO2CH3 is correct, it can't have (CH3)2 as the Carbonyl is an aldehyde.

28:
B

I often find it much easier to solve these questions the other way around.

Since ethanolic NaCN is used, the starting material has to be a haloalkane giving you a nitrile. The nitrile undergoing acid hydrolysis will give you a carboxylic acid.

Haloalkane -> Nitrile -> Carboxylic acid.

Start with 2-ethyl-3-methylbutanoic acid. Remove the -OOH and replace in with N. Careful not to add an additional C here. A lot of people make this mistake. The addition C is already present since you're backsolving!
Now, just remove the entire CN terminal and replace it with Br. Re-arrange to get B.

BTW, this is a very common approach in SATII. Idk if you've given that, but if you have, use the SAT tricks in MCQs. They help a lot.
If this doesn't make sense, let me know and I'll draw it out for you.

29:
D

It's important to understand what's going on. It's taking an ester and using an acid to hydrolyse it. Your products will be an Alchol and an Acid. The Acid half always has the C=O linkage, so HCOOH + RCH(CH3)OH

35:
D

Beautifully made question I must say. Tricked me. Tricked most of my friends too. I got this wrong back when I did it. Class test or sth iirc.

Yes, the FINAL reaction will give you Br2 + SO2. However, it's important to note that 2 and 3 are incorrect regardless.

1- Correct. HBr is an intermediate which later changes to Br2.
2- WRONG! Sulfuric Acid is an oxidising agent so it is itself REDUCED. Besides, it's going from +6 to +4. That's clearly reduction, NOT oxidation.
3- Wrong again! Br- -> Br2. -1 -> 0. That's oxidation!
 
Last edited:
Metanoia

Metanoia

Messages
603
Reaction score
1,102
Points
153
Last edited:
AbbbbY

AbbbbY

Messages
675
Reaction score
862
Points
103
immie.rose said:
OMG sorry!! Yes the first one is A, and next is D. M all mess atm, sry agn.
Thanks!!
Click to expand...

Which is primarily why I ask people to link to the marking schemes and question papers instead!

Anyway.

The first one is A because:

FeTiO3 = 151.7g
TiO2 = 79.9g

You just need the TiO2 mass ratio out of it. (FeTiO3 -> TiO2 + FeO, not that it matters)

Since it's a ratio, grams, tonnes, doesn't matter as long as I keep it the same throughout.

151.7 tonnes will give me 79.9 tonnes

so:
151.7 : 79.9
19 : x
x = 10.007 tonnes

__

This one's really simple. At AS, iirc all they ask is combustion and formation. Not atomization ionization and electron affinity (basically not the born haber cycle questions) though I've seen concepts from Born-Haber cycles asked in the form of Hess' Law.

Anyway. Remember this. You can't use one formation value and one combustion value. They have to be both formation or both combustion. At times, though, formation values are also combustion values.
For example, C + O2 -> CO2. This is combustion, as well as formation. You've to be careful to spot 'wordplay' like this.

My suggestion? In questions like this, first see for two commons. 2 combustion values or 2 formation values. In this case, D. Formation values of both PbO and Pb3O4 will get you the enthalpy change (Products - Reactants. Fpr). But lets say in another question like this, such an option was not present, OR, not suitable. In that case, you'd look to see which of the formation/combustion values could be used as either.
 
immie.rose

immie.rose

Messages
124
Reaction score
343
Points
73
AbbbbY said:
Which is primarily why I ask people to link to the marking schemes and question papers instead!

Anyway.

The first one is A because:

FeTiO3 = 151.7g
TiO2 = 79.9g

You just need the TiO2 mass ratio out of it. (FeTiO3 -> TiO2 + FeO, not that it matters)

Since it's a ratio, grams, tonnes, doesn't matter as long as I keep it the same throughout.

151.7 tonnes will give me 79.9 tonnes

so:
151.7 : 79.9
19 : x
x = 10.007 tonnes

__

This one's really simple. At AS, iirc all they ask is combustion and formation. Not atomization ionization and electron affinity (basically not the born haber cycle questions) though I've seen concepts from Born-Haber cycles asked in the form of Hess' Law.

Anyway. Remember this. You can't use one formation value and one combustion value. They have to be both formation or both combustion. At times, though, formation values are also combustion values.
For example, C + O2 -> CO2. This is combustion, as well as formation. You've to be careful to spot 'wordplay' like this.

My suggestion? In questions like this, first see for two commons. 2 combustion values or 2 formation values. In this case, D. Formation values of both PbO and Pb3O4 will get you the enthalpy change (Products - Reactants. Fpr). But lets say in another question like this, such an option was not present, OR, not suitable. In that case, you'd look to see which of the formation/combustion values could be used as either.
Click to expand...

Thanks a tonne! :):)
 
Metanoia

Metanoia

Messages
603
Reaction score
1,102
Points
153
IGCSE13 said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w08_qp_1.pdf q39
Click to expand...
From a quick look , it seem that ans is B (1 and 2 is correct)? It might be faster if you can post the corresponding answers for your selected questions.

For propane, two possible radicals can be formed. The unpaired electron could be on the 1st ( CH2CH2CH3) or 2nd carbon (CH3CHCH3)

So we can have some combinations for the termination step.

Option 1, formed by CH2CH2CH3 + CH3CHCH3
Option 2, formed by CH3CHCH3 + CH3CHCH3
 
I

IGCSE13

Messages
73
Reaction score
133
Points
18
Ya
Metanoia said:
From a quick look , it seem that ans is B (1 and 2 is correct)? It might be faster if you can post the corresponding answers for your selected questions.

For propane, two possible radicals can be formed. The unpaired electron could be on the 1st ( CH2CH2CH3) or 2nd carbon (CH3CHCH3)

So we can have some combinations for the termination step.

Option 1, formed by CH2CH2CH3 + CH3CHCH3
Option 2, formed by CH3CHCH3 + CH3CHCH3
Click to expand...
B is the answer thanks
 
Gehad Mohamed

Gehad Mohamed

Messages
116
Reaction score
142
Points
53
AbbbbY said:
1:
D
__
CO2 = 0.8 mol
H2O = 0.8 mol

0.2 : 0.8 : 0.8
= 1 : 4 : 4
So, the Hydrocarbon is giving 4 mols of CO2 and H2O on combustion of 1 mol of the HC. The only options that can give 4 mols of CO2 are C and D. Now, I know by looking the an alkene cannot give me 1:1 CO2:H2O, but even if you don't know that, just combust the hydrocarbon and pick.
C4H4 + O2 -> 4CO2 + 2H2O
C4H8 + O2 -> 4CO2 + 4H2O

20:
A
__

Aah. I've answered this so many times here. Should save the answer in notepad or sth :p

Anyway.
You need to understand that everytime a ring is formed, or a double bond is formed the compound loses TWO hydrogens.

C20H28O.
So if it were a fully saturated HYDROCARBON, it'd have 42 Hydrogens. C20H42.
We have 28, so 14 are missing.
You have an Aldehyde, a ring, and a double bond in the ring. Thats 6 carbons hydrogens.
14-6 = 8. 8/2 = 4 further double bonds.

Since you're asked total number of double bonds, 4+1 = 5 double bonds.
I hope this makes sense :)

27:
C?

You basically need a central C with OH, CH2CO2CH3, H and R attached.
A is out. CO2CH3, not CH2CO2CH3
B is out. CH3CO2CH2CH, not CH2CO2CH3
D is out. While CH2CO2CH3 is correct, it can't have (CH3)2 as the Carbonyl is an aldehyde.

28:
B

I often find it much easier to solve these questions the other way around.

Since ethanolic NaCN is used, the starting material has to be a haloalkane giving you a nitrile. The nitrile undergoing acid hydrolysis will give you a carboxylic acid.

Haloalkane -> Nitrile -> Carboxylic acid.

Start with 2-ethyl-3-methylbutanoic acid. Remove the -OOH and replace in with N. Careful not to add an additional C here. A lot of people make this mistake. The addition C is already present since you're backsolving!
Now, just remove the entire CN terminal and replace it with Br. Re-arrange to get B.

BTW, this is a very common approach in SATII. Idk if you've given that, but if you have, use the SAT tricks in MCQs. They help a lot.
If this doesn't make sense, let me know and I'll draw it out for you.

29:
D

It's important to understand what's going on. It's taking an ester and using an acid to hydrolyse it. Your products will be an Alchol and an Acid. The Acid half always has the C=O linkage, so HCOOH + RCH(CH3)OH

35:
D

Beautifully made question I must say. Tricked me. Tricked most of my friends too. I got this wrong back when I did it. Class test or sth iirc.

Yes, the FINAL reaction will give you Br2 + SO2. However, it's important to note that 2 and 3 are incorrect regardless.

1- Correct. HBr is an intermediate which later changes to Br2.
2- WRONG! Sulfuric Acid is an oxidising agent so it is itself REDUCED. Besides, it's going from +6 to +4. That's clearly reduction, NOT oxidation.
3- Wrong again! Br- -> Br2. -1 -> 0. That's oxidation!
Click to expand...
Thanks sooooooo much all doubts are now cleared , never thought someone would answer all these ,most of them are sooo tricky and confusing , thnx again
 
You must log in or register to reply here.
Top