Chemistry: Post your doubts here!

[AbbbbY]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s13_qp_11.pdf
Q20

B

Cis-trans: Presence of double bond and the heavy-side light-side together rule.
Optical isomerism: Chiral center.


A- Chiral, but no cis-trans.
B- Chiral and cis trans
C- No Chiral but cis-trans
D- Chiral, but no cis-trans

 

[Metanoia]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w07_qp_1.pdf. Q 9 , 32 and 37 please

Thanks , what about q37 in the same paper do u know how to solve it ?

w07qp1

Q37
1. (Correct) CaO + ND4Cl --> CaCl2 + D2O + ND3
2. (Correct) CH3CN + NaOD + D2O --> CaCOONa + ND3
3. (wrong) NaOD + NDH3Cl --> NaCl + D2O + NH3 or NaCl + DHO + NDH2

 

[Abdel Moniem]

B

Cis-trans: Presence of double bond and the heavy-side light-side together rule.
Optical isomerism: Chiral center.


A- Chiral, but no cis-trans.
B- Chiral and cis trans
C- No Chiral but cis-trans
D- Chiral, but no cis-trans

How is B a cis trans

 

[AbbbbY]

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

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.

 

[AbbbbY]

How is B a cis trans

The Carbon on the left has Cl and I. They are both halides so have different masses. The Carbon on the right has CH(Cl)(Br) and H attached directly to it. 4 different groups across the double bond. Most definitely cis-trans. Don't even have to work the masses out!

 

[Metanoia]

someone pls explain the following!!


View attachment 43906
Answer: A
C

A quick way to do this question is to express it as mass fraction of TiO2 and multiply by the mass of the ore.

(Mr of TiO2/Mr of FeTiO3) x 19 tonnes

 

[Abdel Moniem]

Thank you

The Carbon on the left has Cl and I. They are both halides so have different masses. The Carbon on the right has CH(Cl)(Br) and H attached directly to it. 4 different groups across the double bond. Most definitely cis-trans. Don't even have to work the masses out!

 

[immie.rose]

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.

Thanks a tonne!

 

[immie.rose]

A quick way to do this question is to express it as mass fraction of TiO2 and multiply by the mass of the ore.

(Mr of TiO2/Mr of FeTiO3) x 19 tonnes

Thanks!

 

[IGCSE13]

http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_w08_qp_1.pdf q39

 

[Metanoia]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w08_qp_1.pdf q39

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

 

[IGCSE13]

Ya

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

B is the answer thanks

 

[Gehad Mohamed]

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

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!

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

 

[IGCSE13]

http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_s09_qp_1.pdf. Q9 ans A , q15 ans A , q17 ans D ,q19 ans D , q20 ans C q 35 ans B
Sorry they are a lot

 

[Awesome12]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s09_qp_1.pdf. Q9 ans A , q15 ans A , q17 ans D ,q19 ans D , q20 ans C q 35 ans B
Sorry they are a lot

I will answer 9
Electropostive is the act of the ion's attraction to the electrons. The more electropositive the metal, the greater the force of attraction, and thus the greater the reactivity. Conversely silver is less reactive and thus less electropositive

 

[Awesome12]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s09_qp_1.pdf. Q9 ans A , q15 ans A , q17 ans D ,q19 ans D , q20 ans C q 35 ans B
Sorry they are a lot

How is 35 B? I mean there are numbers only.

 

[Metanoia]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s09_qp_1.pdf. Q9 ans A , q15 ans A , q17 ans D ,q19 ans D , q20 ans C q 35 ans B
Sorry they are a lot

Q15. Check that the oxidation state of silver before and after the reaction is +1, it means that it is neither oxidized or reduced. This effectively eliminates options B, C and D. silver chloride is insoluble and the silver complex formed is soluble.

Q9. less electropositve means less easily oxidised. It is harder to oxidise Ag (into Ag+ ions) than Cu into Cu2+ ions, so Ag remains in solid state while Cu dissolves.

Q17.
A: SO3 is acidic and NH3 is basic
B: There is a dative bond between N and 1 of the H in NH4+
C: Ionic bond between ammonium ion and sulfate ion
D: No change in oxidation state of the atoms

Q19. Hard to describe by typing, most easily missed out by students are the 3 chiral carbons that each contain a H which are not reflected in the diagram (check the "Y junctions")

Q20. all 3 C=C bonds are capable of cis-trans isomerism, so max isomers = 2^3 = 8

Q35.
1. Refactory linings require substances to withstand high temperatures
2. The carbonate will react with acidic gases to form solids and prevent the acidic gases from being released into the atmosphere.

 

[sadiaali]

http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_w04_qp_1.pdf
Please anyone explain me Q40 , I cant understand a word

 

[Metanoia]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w04_qp_1.pdf
Please anyone explain me Q40 , I cant understand a word

In direction of oxidation,

Primary alcohol functional group --> aldehyde function group --> acid functional group

secondary alcohol functional group --> ketone functional group

Q40.
1. (Correct) 1st compound aldehyde functional group is oxidised to 2nd compound acid group
2. (Correct) 2nd compound ketone functional group is reduced to 3rd compound alcohol group
3. (Correct) 1st compound alcohol group is oxidised to 2nd compound ketone group

 

[Metanoia]

I'm new to this forum, suppose I want to attach a diagram to explain some of the questions, do I need to host the image elsewhere or can I upload directly from my hard drive?